Second Quantization

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Appendix A Second Quantization

An exact description of an interacting many-body system requires the solving of the corresponding many-body Schrodinger¬ equations. The formalism of second quantization leads to a substantial simpliﬁcation in the description of such a many- body system, but it should be noted that it is only a reformulation of the original Schrodinger¬ equation but not yet a solution. The essential step in the second quantization is the introduction of so-called creation and annihilation operators. By doing this, we eliminate the need for the laborious construction, respectively, of the symmetrized or the anti-symmetrized N-particle wavefunctions from the single-particle wavefunctions. The entire statistics is contained in fundamental commutation relations of these operators. Forces and interactions are expressed in terms of these “creation” and “annihilation” operators. How does one handle an N-particle system? In case the particles are distinguishable, that is, if they are enumerable, then the method of description follows directly the general postulates of quantum mechanics: H(i) {|φ(i) } 1 : Hilbert space of the ith particle with the orthonormal basis α :

(i) (i) φα |φβ =δαβ (A.1)

HN : Hilbert space of the N-particle system H = H(1) ⊗ H(2) ⊗···⊗H(N) N 1 1 1 (A.2)

with the basis {|φN }: |φ =|φ(1)φ(2) ···φ(N) =|φ(1) |φ(2) ···|φ(N) (A.3) N α1 α2 αN α1 α2 αN

An arbitrary N-particle state |ψN , |ψ = c(α ···α )|φ(1)φ(2) ···φ(N) (A.4) N 1 N α1 α2 αN α1···αN

underlies the same statistical interpretation as in the case of a 1-particle system. The dynamics of the N-particle system results from the formally unchanged Schrodinger¬ equation:

W. Nolting, A. Ramakanth, Quantum Theory of Magnetism, 491 DOI 10.1007/978-3-540-85416-6 BM2, C Springer-Verlag Berlin Heidelberg 2009 492 A Second Quantization ∂|ψ i N = H |ψ (A.5) ∂t N

The handling of the many-body problem in quantum mechanics, in the case of distinguishable particles, confronts exactly the same difﬁculties as in the classical physics, simply because of the large number of degrees of freedom. There are no extra, typically quantum mechanical complications.

A.1 Identical Particles

What are “identical particles”? To avoid misunderstandings let us strictly separate “particle properties” from “measured quantities of particle observables”. “Parti- cle properties” as e.g. mass, spin, charge, magnetic moment, etc. are in principle unchangeable intrinsic characteristics of the particle. The “measured values of particle observables” as e.g. position, momentum, angular momentum, spin projection, etc., on the other hand, can always change with time. We define “identical particles” in the quantum mechanical sense as particles which agree in all their particle properties. Identical particles are therefore particles, which behave exactly in the same manner under the same physical conditions, i.e. no measurement can differentiate one from the other. Identical particles also exist in classical mechanics. However, if the initial conditions are known, the state of the system for all times is determined by the Hamilton’s equations of motion. Thus the particles are always identifiable! In quantum mechanics, on the contrary, there is the principle of indistinguishability, which says that, identical particles are intrinsically indistinguishable. In quantum mechanics, in some sense, identical particles lose their individuality. This originates from the fact that there do not exist sharp particle trajectories (only “spreading” wave packets!). The regions, where the probability of finding the particle is unequal to zero, overlap for different particles. Any question whose answer requires the observation of one single particle is physically meaningless. We face now the problem, that, essentially for calculational purposes, we cannot avoid, to number the particles. Then, however, the numbering must be so that all physically relevant statements, that are made, are absolutely invariant under a change of this particle labelling. How to manage this is the subject of the following considerations. We introduce the permutation operator P, which, in the N-particle state, inter- changes the particle indices:

P|φ(1)φ(2) ···φ(N) =|φ(i1)φ(i2) ···φ(iN ) (A.6) α1 α2 αN α1 α2 αN

Every permutation operator can be written as a product of transposition operators Pij. On application of Pij to a state of identical particles, results, according to the principle of indistinguishability, in a state, which, at the most, is different from the initial state by an unimportant phase factor λ = exp(iη): A.1 Identical Particles 493

P |···φ(i) ···φ( j) ··· = |···φ( j) ···φ(i) ··· ij αi α j αi α j

= λ |···φ(i) ···φ( j) ··· (A.7) αi α j

2 = λ =± Since Pij 1, it is necessary that 1. Therefore, a system of identical particles must be either symmetric or antisymmetric against interchange of particles! This deﬁnes two different Hilbert spaces: H(+) |ψ (+) N : the space of symmetric states N

(+) (+) Pij |ψN =|ψN (A.8)

H(−) |ψ (−) N : the space of antisymmetric states N

(−) (−) Pij |ψN =−|ψN (A.9)

In these spaces, Pij are Hermitian and unitary! What are the properties the observables must have for a system of identical particles? They must necessarily depend on the coordinates of all the particles

A = A(1, 2, ··· , N) (A.10) and must commute with all the transpositions (permutations) , = Pij A − 0 (A.11)

This is valid specially for the Hamiltonian H and therefore also for the time evolution operator i U t, t = exp − H t − t H = H t ( 0) ( 0) ;( ( )) (A.12)

, = Pij U − 0 (A.13)

That means the symmetry character of an N-particle state remains unchanged for all times! Which Hilbert space out of H(+) and H(−) is applicable for which type of particles is established in relativistic quantum ﬁeld theory. We will take over the spin-statistics theorem from there, without any proof. H(+) N : Space of symmetric states of N identical particles with integral spin. These particles are called Bosons. H(−) N : Space of antisymmetric states of N identical particles with half-integral spin. These particles are called Fermions. 494 A Second Quantization

A.2 Continuous Fock Representation

A.2.1 Symmetrized Many-Particle States

H(ε) Let N be the Hilbert space of a system of N identical particles. Here + : Bosons ε = (A.14) − : Fermions

Let φ be a 1-particle observable (or a set of 1-particle observables) with a continuous spectrum, φα being a particular eigenvalue corresponding to the 1-particle eigenstate |φα : φ |φα =φα |φα (A.15)

φα|φβ =δ(φα − φβ )(≡ δ(α − β)) (A.16)

dφα|φα φα|=1l i n H1 (A.17)

H(ε) A basis of N is constructed from the following (anti-) symmetrized N-particle states: ) * (ε) 1 p (1) (2) (N) |φα ···φα = ε P |φ |φ ···|φ (A.18) 1 N α1 α2 αN N! P p is the number of transpositions in P. The summation runs over all the possible permutations P. The sequence of the 1-particle states in the ket on the left-hand side of (A.18) is called the standard ordering. It is arbitrary, but has to be ﬁxed right at the beginning. Interchange of two 1-particle symbols on the left means only a constant factor ε on the right. One can easily prove the following relations for the above introduced N-particle states: Scalar product:

(ε) φ ···|φ ··· (ε) β1 α1 ) * 1 pα = ε Pα δ(φβ − φα ) ···δ(φβ − φα ) (A.19) N! 1 1 N N Pα

The index α shall indicate that Pα permutes only the φα’s. Completeness relation: ··· φ ··· φ |φ ··· (ε)(ε) φ ···|= d β1 d βN β1 β1 1l (A.20) A.2 Continuous Fock Representation 495

This leads to a formal representation of an observable of the N-particle system, which will be used later a few times: = ··· φ ··· φ φ ··· φ ∗ A d α1 d αN d β1 d βN ∗|φ ··· (ε)(ε) φ ···| |φ ··· (ε)(ε) φ ···| α1 α1 A β1 β1 (A.21)

A.2.2 Construction Operators

H(ε) | We want to build up the basis states of N , step by step, from the vacuum state 0 ( 0|0 =1) with the help of the operator † ≡ † cφα cα These operators are deﬁned by their action on the states: √ † (ε) (ε) c |0 = 1|φα ∈ H (A.22) α1 1 1 √ † (ε) (ε) (ε) c |φα = 2|φα φα ∈ H (A.23) α2 1 2 1 2 Or, in general √ † (ε) (ε) c |φα ···φα = N + 1 |φβ φα ···φα (A.24) β 1 N 1 N ∈H(ε) ∈H(ε) N N+1

† cβ is called the creation operator.Itcreates an extra particle in the N-particle state. The relation (A.24) is obviously reversible:

(ε) 1 † † † |φα φα φα ···φα = √ cα cα ···cα |0 (A.25) 1 2 3 N N! 1 2 N |φ ··· (ε) | The N-particle state αi can be built up from the vacuum state 0 by apply- ing a sequence of N creation operators, where the order of the operators is to be strictly obeyed. For a product of two creation operators, it follows from (A.24)

† † (ε) (ε) c c |φα ···φα = N(N − 1)|φα φα φα ···φα (A.26) α1 α2 3 N 1 2 3 N If the sequence of the operators is reversed, then we have

† † (ε) c c |φα ···φα α2 α1 3 N

= − |φ φ φ ···φ (ε) N(N 1) α2 α1 α3 αN

= ε − |φ φ φ ···φ (ε) N(N 1) α1 α2 α3 αN (A.27) 496 A Second Quantization

The last step follows because of (A.18). Since the states in (A.26) and (A.27) are basis states, by comparing, we can therefore read off the following operator identity: c† , c† = c† c† − ε c† c† = 0 (A.28) α1 α2 −ε α1 α2 α2 α1 The creation operators commute in the case of Bosons (ε =+1) and anticommute (ε =−1) in the case of Fermions. † We now consider the operator cα which is the adjoint of cα. Because of (A.24) and (A.25), we can write √ (ε) (ε) φα ···φα | cγ = N + 1 φγ φα ···φα | 1 N 1 N (A.29) (ε) 1 φα ···φα |= √ 0| cα ···cα cα 1 N N! N 2 1

The meaning of cγ is made clear by the following consideration:

(ε) (ε) φβ ···φβ |cγ | φα ···φα 2 N 1 N ∈H(ε) ∈H(ε) N−1 N √ = (ε) φ φ ···φ |φ ···φ (ε) N γ β2 βN α1 αN √ N ) = ε pα P δ φ − φ δ φ − φ ··· α ( γ α1 ) ( β2 α2 ) N! P α * ···δ φ − φ ( βN αN )

Here we have used in the ﬁrst step (A.29) and in the second step (A.19). We can further rewrite the right-hand side

(ε) (ε) 1 1 ···|cγ |··· = √ ∗ − ⎧ N (N 1)! ⎨ pα ∗ δ(φγ − φα ) ε Pα(δ(φβ − φα ) ···δ(φβ − φα )) ⎩ 1 2 2 N N P α + εδ φ − φ ε pα P δ φ − φ δ φ − φ ···δ φ − φ ( γ α2 ) α( ( β2 α1 ) ( β3 α3 ) ( βN αN )) Pα + ···+ + εN−1δ φ − φ ε pα P δ φ − φ δ φ − φ ( γ αN ) α( ( β2 α1 ) ( β3 α2 ) Pα * ···δ φ − φ ( βN αN−1 )) ) 1 (ε) (ε) = √ δ(φγ − φα ) φβ ···φβ |φα ···φα N 1 2 N 2 N + εδ φ − φ (ε) φ ···φ |φ φ ···φ (ε) ( γ α2 ) β2 βN α1 α3 αN + ···+ * + εN−1δ φ − φ (ε) φ ···φ |φ φ ···φ (ε) ( γ αN ) β2 βN α1 α2 αN−1 A.2 Continuous Fock Representation 497

(ε) φ ···φ | − With this, the action of cγ is clear, since β2 βN is an arbitrary (N 1)- particle bra basis-state:

|φ ···φ (ε) = cγ α1 αN

1 (ε) = √ {δ(φγ − φα )|φα ···φα + N 1 2 N + ···+ + εN−1δ φ − φ |φ ···φ (ε)} ( γ αN ) α1 αN−1 (A.30) cγ annihilates a particle in the state |φγ and that is why it is called the annihilation operator. From (A.28), it immediately follows that

† † † cα , cα =−ε c , c = 0 (A.31) 1 2 −ε α1 α2 −ε

The annihilation operators commute in the case of Bosons (ε =−1) and anticommute (ε =+1) in the case of Fermions. From (A.24) and (A.30) one can show that (Problem A.1)

† − ε † |φ ···φ (ε) = δ φ − φ |φ ···φ (ε) (cβ cγ cγ cβ ) α1 αN ( β γ ) α1 αN

From this it follows that , † = δ φ − φ cβ cγ −ε ( β γ ) (A.32)

(A.28), (A.31) and (A.32) are the three fundamental commutation rules of the con- † struction operators aγ and aγ .

A.2.3 Many-Body Operators

We start with the formal (“spectral”) representation of the N-particle observable A φ ≡ α as given in (A.21). Using (A.25) and (A.29), we can write (d αi d i ) 1 A = ··· dα ···dα dβ ···dβ ∗ N! 1 N 1 N

† † (ε) (ε) ∗ c ··· c |0 φα ···|A|φβ ··· 0|cβ ···cβ α1 αN 1 1 N 1 (A.33)

Normally, such an operator consists of 1-particle and 2-particle parts:

i= j N 1 A = A(1) + A(2) (A.34) i 2 ij i=1 i, j 498 A Second Quantization

Using this, we will calculate the matrix elements in (A.33). We start with the 1- particle part:

N (ε) (1) (ε) 1 pα +pβ φα ···| A |φβ ··· = ε ∗ 1 i 1 (N!)2 i=1 Pα Pβ ) * N < = (N) (1) † (1) (N) (N) ∗ φ |··· φ | P A Pβ |φ ···|φ αN α1 α i β1 βN i=1 (A.35)

One can easily see that every summand in the double sum Pα Pβ gives the same contribution, since every permuted ordering of {|φα } or {|φβ } can be restored to the standard ordering by renumbering the integration variables in (A.33). In order to bring back the creation and annihilation operators, which have been reindexed in the process, into the “correct” sequence, we require, according to (A.28) and (A.31), a factor ε pα +pβ , which along with the corresponding factor in the above equation (A.35) gives a factor +1. Thus for (A.33), we need (A.35) only in the following simpliﬁed form:

N (ε) φ ···| (1)|φ ··· (ε) ⇒ α1 Ai β1 i=1 ) * N < = φ(N)|··· φ(1)| A(1) |φ(1) ···|φ(N) (A.36) αN α1 i β1 βN i=1

Substituting this in (A.33), we get

N (1) 1 † † A = ··· dα1 ···dβN cα ···cα |0 ∗ i N! 1 N i=1 < ∗ φ(1)|A(1)|φ(1) φ(2)|φ(2) ··· φ(N)|φ(N) α1 1 β1 α2 β2 αN βN = + φ(1)|φ(1) φ(2)|A(1)|φ(2) ··· φ(N)|φ(N) +··· ∗ α1 β1 α2 2 β2 αN βN

∗ 0|cβ ···cβ N 1 1 † † = ··· dα1 ···dαN cα ···cα |0 ∗ N! 1 N (1) (1) (1) ∗ dβ φ |A |φ 0|cα ···cα cβ 1 α1 1 β1 N 2 1 (2) (1) (2) + dβ φ |A |φ 0|cα ···cα cβ cα +··· 2 α2 2 β2 N 3 2 1 A.2 Continuous Fock Representation 499 1 (1) (1) (1) † 1 = dα1dβ1 φα |A |φβ aα N 1 1 1 1 (N − 1)! † † ··· dα ···dα c ···c |0 0|cα ···cα cβ 2 N α2 αN N 2 1 1 (2) (1) (2) † 1 + dα2dβ2 φα |A |φβ cα N 2 2 2 2 (N − 1)! ··· dα dα ···dα c† c† ···c† |0 1 3 N α1 α3 αN = | ··· ε2 +··· 0 cαN cα3 cα1 cβ2 (A.37)

The factor ε2 in the last line stems from the (anti)commutation c† ↔ c† and α2 α1 † c† ↔ c . Due to the analogous rearrangements all the other terms get a fac- α1 β2 tor ε2m which in any case is equal to +1. The term in each of the curly brack- H(ε) ets is the identity for the Hilbert space N−1 as given in (A.20). Therefore what remains is

N N (1) 1 (i) (1) (i) † A = dαi dβi φα |A |φβ cα cβ (A.38) i N i i i i i i=1 i=1

The matrix elements are naturally the same for all the N identical particles. There- fore, the factor 1/N and the summation cancel out:

N (1) = α β φ |(1)|φ † Ai d d α A β cαcβ (A.39) i=1

The remaining matrix element is in general easy to calculate. On the right-hand side, we do not have the particle number any more. It is of course implicitly present † because, in between cα and cβ , there appears in principle an identity 1l, correspond- H(ε) ing to the Hilbert space N−1. For the two-particle part of the operator A, at ﬁrst, exactly the same considerations are valid which led us from (A.35) to (A.36). Therefore, we can use in (A.33)

i= j (ε) 1 (2) (ε) φα ···| A |φβ ··· ⇒ 1 2 ij 1 i, j ) * i= j < = (N) (1) 1 (2) (1) (N) φα |··· φα | A |φβ ···|φβ (A.40) N 1 2 ij 1 N i, j 500 A Second Quantization

Substituting this we are left with

i= j 1 (2) 1 1 † † A = ··· dα1 ···dβN cα ···cα |0 ∗ 2 ij 2 N! 1 N i, j < ∗ φ(2)| φ(1)|A(2)|φ(1) |φ(2) φ(3)|φ(3) ··· φ(N)|φ(N) α2 α1 12 β1 β2 α3 β3 αN βN = + ··· | ··· 0 cβN cβ1 1 1 † † † = ··· dα1 ···dαN dβ1dβ2cα cα ···cα |0 ∗ 2 N! 1 2 N (2) (1) (2) (1) (2) ∗ φ | φ |A |φ |φ 0|cα ···cα cβ cβ +··· α2 α1 12 β β N 3 2 1 1 2 1 = ··· dα dα dβ dβ ∗ 2N(N − 1) 1 2 1 2 ∗ φ(1)φ(2)|(2)|φ(1)φ(2) † † ∗ α α A12 β β cα cα 1 2 1 2 1 2 1 † † ∗ ··· dα3 ···dαN cα ···cα |0 (N − 2)! 3 N = | ··· +··· 0 cαN cα3 cβ2 cβ1

H(ε) The curly bracket now gives the identity in the Hilbert space N−2, so that we have

i= j i= j 1 1 A(2) = ··· dα dα dβ dβ × 2 ij 2N(N − 1) i j i j i, j i, j

(i) ( j) (2) (i) ( j) † † × φ φ |A |φ φ c c cβ cβ (A.41) αi α j ij βi β j αi α j j i

In a system of identical particles, naturally, all the summands on the right-hand side give identical contributions. Therefore, we have

i= j 1 1 A(2) = ··· dα dα dβ dβ ∗ 2 ij 2 1 2 1 2 i, j (1) (2) (2) (1) (2) † † ∗ φ φ | |φ φ β β α α A12 β β cα cα c 2 c 1 1 2 1 2 1 2 (A.42)

The remaining matrix element on the right-hand side can be built either with unsymmetrized two-particle states

φ φ |= φ | φ | |φ φ =|φ |φ α1 α2 α1 α2 ; β1 β2 β1 β2 (A.43) A.3 Discrete Fock Representation (Occupation NumberRepresentation) 501 or also with the symmetrized states < = (ε) 1 (1) (2) (2) (1) |φβ φβ = |φβ |φβ +ε|φβ |φβ (A.44) 1 2 2! 1 2 1 2

What have we achieved? We have, through (A.25) and (A.29), replaced the laborious building up of symmetrized products from the single-particle states, by the application of products of construction operators to the vacuum state |0 . The operation is quite simple, for example,

cα|0 =0 (A.45) and the whole statistics is taken care by the fundamental commutation relations (A.28), (A.31) and (A.32). The N-particle observables can also be expressed by the construction operators, (A.39) and (A.42) where the remaining matrix elements can be, usually, easily calculated with single-particle states. Note that the choice of the single-particle basis {|φα } is absolutely arbitrary, a great advantage with respect to practical purposes. We will demonstrate this in (A.4) with a few examples. The theory developed so far is valid for continuous as well as discrete single- particle spectra. Only δ-functions have to be replaced by Kronecker-δs and integrals by summations in the case of discrete spectra.

A.3 Discrete Fock Representation (Occupation Number Representation)

H(ε) Let N be again the Hilbert space of a system of N identical particles. Now, let φ be a single-particle observable with discrete spectrum. In principle, the same considerations are valid as in (A.2).

A.3.1 Symmetrized Many-Particle States

H(ε) We will use the following (anti-) symmetrized N-particle states as the basis of N : ) * (ε) p (1) (N) |φα ···φα = Cε ε P |φ ···|φ (A.46) 1 N α1 αN P

Up to the still to be determined normalization constant Cε, this deﬁnition agrees with the corresponding deﬁnition (A.18) of the continuous case. However, now for the single-particle states, we have φα|φβ =δαβ ; |φα φα|=1l i n H1 (A.47) α 502 A Second Quantization

One can see that (A.46) can be written for Fermions (ε =−1), as a determinant |φ(1) |φ(2) ··· |φ(N) α1 α1 α1 (1) (2) (N) |φα |φα ··· |φα ....2 2 2 (−) |φα ···φα = C− (A.48) 1 N .... .... |φ(1) |φ(2) ··· |φ(N) αN αN αN

This is known as the Slater determinant. In case two sets of quantum numbers are equal, say, αi = α j , then, two rows of the Slater determinant are identical, which means, the determinant is equal to zero. Consequently, the probability that such a situation exists for a system of identical Fermions is zero. This is exactly the statement of the Pauli’s principle! We deﬁne = |φ ni occupation number, i.e. the frequency with which the state αi appears in |φ ···φ (−) the N-particle state α1 αN . Naturally, we have ni = N (A.49) i where the values ni can be

n = 0, 1 forFermions i (A.50) ni = 0, 1, 2, ··· f or Bosons

First, we want to ﬁx the normalization constant, which we assume to be real:

! (ε) (ε) 1 = φα ···|φα ··· 1 1 ) * ) * = C2 ε p+p φ(N)|··· P† P |φ(1) ··· (A.51) ε αN α1 P P

In the case of Fermions (ε =−1), every state is occupied only once. Therefore, in the sum, only the terms with P = P are unequal to zero and due to (A.47), each term is exactly equal to 1. Therefore, we get

1 C− = √ (Fermions) (A.52) N!

In the case of Bosons (ε =+1), the summands are then unequal to zero, when P differs from P at the most by such transpositions for which only the groups of ni identical single-particle states |φi are interchanged among themselves. There are naturally ni ! such possibilities. With this argument, we get for Bosons A.3 Discrete Fock Representation (Occupation NumberRepresentation) 503

−1/2 C+ = (N! n1! n2! ···ni ! ···) (Bosons) (A.53)

Obviously, a symmetrized basis state can be completely speciﬁed by giving the occupation numbers. This permits the representation by the Fock states

(ε) (ε) |N; n n ···n ···n ··· ≡|φα ···φα 1 2 ⎧ i j 1 N ⎫ ⎨⎪ ⎬⎪ P (1) (2) (r) (r+1) = Cε ε P |φα |φα ··· ···|φα |φα ······ (A.54) ⎪ 1 1 i i ⎪ P ⎩ ⎭ n1 ni

One has to give all the occupation numbers, even those with ni = 0. Completeness and orthonormality follow from the corresponding relations for the symmetrized states: . (ε) ··· ···|˜ ··· ··· (ε) = δ δ N; ni N; n˜ i N N˜ ni n˜ i (A.55) i

(ε)(ε) ··· ···|N; ···ni ··· N; ···ni ···|=1l (A.56) n1 n2 ni = ( i ni N)

A.3.2 Construction Operators

Up to normalization constants, we deﬁne the construction operators exactly in the same manner as we have done for the case of continuous spectrum in (A.2.2):

† (ε) † (ε) c |N˜ ; ···n˜ ··· = c |φα ···φα αr r αr 1 N

= + |φ φ ···φ (ε) nr 1 αr α1 αN

ε = εNr + |φ ···φ ···φ ( ) nr 1 α1 αr αN

Nr (ε) = ε nr + 1|N + 1; ···nr + 1 ··· (A.57)

Here, Nr is the number of transpositions necessary to bring the single-particle state |φ αr to the “correct” place:

r−1 Nr = ni (A.58) i=1 504 A Second Quantization

Equation (A.57) does not, as yet, contain the Pauli’s principle in a correct way. The operation of the so-called creation operator is precisely deﬁned as follows.

Bosons : √ c† |N; ···n ··· (+) = n + 1 |N + 1; ···n + 1 ··· (+) αr r r r (A.59) Fermions : † (−) N (−) c |N; ···n ··· = (−1) r δ , |N + 1; ···n + 1 ··· αr r nr 0 r

Any N-particle state can be built up by repeated application of the creation operators on the vacuum state |0 :

= .n p N ε 1 † n p |N; n n ··· ( ) = c εN p |0 (A.60) 1 2 αp p n p!

The annihilation operator is again deﬁned as the adjoint of the creation operator: † † cα = c (A.61) r αr

The action of this operator becomes clear from the following:

(ε) (ε) N; ···nr ···|cα |Nø ; ···nør ··· √ r N (ε) (ε) = ε r nr + 1 N + 1; ···nr + 1 ···|Nø ; ···nø r ··· √ = εNr + δ δ ···δ ··· nr 1 N+1,Nø n ,nø nr +1,nør √ 1 1 ø = εNr δ δ ···δ ··· nø r N,Nø −1 n ,nø nr ,nør −1 √ 1 1 Nø (ε) (ε) = ε r nø r N; ···nr ···|Nø − 1; ···nø r − 1 ···

(ε) Nø r is deﬁned as in (A.58). Since N1 ···Nr ···|is an arbitrary bra-basis state, we have to conclude

ε √ ε | ··· ··· ( ) = εNr | − ··· − ··· ( ) cαr N; nr nr N 1; nr 1 (A.62) or, more speciﬁcally,

Bosons : √ | ··· ··· (+) = | − ··· − ··· (+) cαr N; nr nr N 1; nr 1 (A.63) Fermions : − − | ··· ··· ( ) = δ − Nr | − ··· − ··· ( ) cαr N; nr nr ,1( 1) N 1; nr 1

With (A.59) and (A.63), one can easily prove three fundamental commutation rules (Problem A.2): A.3 Discrete Fock Representation (Occupation NumberRepresentation) 505 † † cα , cα = c , c = 0 r s −ε αr αs −ε † (A.64) cα , c = δ r αs −ε rs

We further introduce two special operators, namely, the occupation number operator:

† n = c cα (A.65) r αr r and the particle number operator: N = nr (A.66) r The Fock states are the eigenstates of nr as well as of N. One can easily show with (A.59) and (A.63) that

(ε) (ε) nr |N; ···nr ··· = nr |N; ···nr ··· (A.67)

Thus nr refers to the number of particles occupying the rth single-particle state. The eigenvalue of N is the total number of particles N: (ε) (ε) N|N; ···nr ··· = nr |N; ···nr ··· r (ε) = N|N; ···nr ··· (A.68)

The following relations are valid for both Bosons and Fermions (Problem A.3): † † nr , cs = δrs cr ; [nr , cs ]− =−δrs cr − (A.69) † † N , cs = cs ; N , cs =−cs − −

In order to transform a general operator A (A.34) with a discrete spectrum into the formalism of second quantization, we have to follow the same procedure as was done in the case of the continuous spectrum. We only have to replace integrations by summations and delta functions by Kronecker deltas. Thus, we have expressions analogous to (A.39) and (A.42): (1) † A ≡ φα |A |φα c cα r r αr r r,r 1 (1) (2) (2) (1) (2) † † + φα φα |A |φα φα cα cα cα cα 2 r s r s r s s r r,r s,s (A.70) 506 A Second Quantization

In contrast to the case of continuous spectrum, here, the matrix elements must be calculated with non-symmetrized two-particle states. The reason for this is the different normalization used here.

A.4 Examples

In this section, we want to transform a few of the most often used operators from the ﬁrst to second quantized form.

A.4.1 Bloch Electrons

We consider electrons in a rigid ion-lattice. The electrons interact with the lattice potential but do not interact with each other:

Ne = + (0) = (i) H0 He,kin Hei h0 (A.71) i=1

He,kin is the operator of the kinetic energy

Ne 2 pi H , = (A.72) e kin 2m i=1

(0) Ne is the number of electrons which interact with the rigid ion-lattice via Hei :

Ne N (0) = v v = − Hei (ri ); (ri ) Vei (ri Rα) (A.73) i=1 α=1

N is the number of lattice atoms, whose equilibrium positions are given by Rα. v(ri ) has the same periodicity as the lattice:

v(ri ) = v(ri + Rα) (A.74)

H0 is obviously a single-particle operator. The eigenvalue equation for

p2 h = + v(r) (A.75) 0 2m deﬁnes the Bloch function ψk(r) and the Bloch energy ε(k):

h0 ψk(r) = ε(k) ψk(r) (A.76) A.4 Examples 507

For ψk(r), we make the usual ansatz

ik·r ψk(r) = uk(r) e (A.77) where the amplitude function uk(r) has the periodicity of the lattice. The Bloch functions build a complete orthonormal set: ∗ 3 ψ ψ = δ d r k (r) k (r) kk (A.78)

1st B.Z. ψ∗ ψ = δ − k (r) k(r ) (r r ) (A.79) k

Neither H0 nor h0 contains spin terms. Therefore the complete solutions are

|kσ ⇔ r|kσ =ψkσ (r) = ψk(r)χσ (A.80)

1 0 χ↑ = ; χ↓ = (A.81) 0 1

We deﬁne † ckσ (ckσ ) - creation (annihilation) operator of a Bloch electron Since H0 is a single-particle operator, according to (A.70), we can write † = σ| | σ H0 k h0 k ckσ ck σ (A.82) k,σ k,σ

The matrix element is given by 3 kσ|h0|k σ = d r kσ|r r|h0|k σ ∗ = ε 3 ψ ψ (k ) d r kσ (r) k σ (r)

= ε(k) δkk δσσ (A.83)

So that we can ﬁnally write = ε † H0 (k) ckσ ckσ (A.84) k,σ 508 A Second Quantization

The Bloch operators satisfy the fundamental commutation relations of Fermion operators: † † , = , = [ckσ ck σ ]+ ckσ ckσ 0 + (A.85) † ckσ , c σ = δkk δσσ k +

In the special case, where one can neglect the crystal structure, e.g., as in the Jellium model, the Bloch functions become plane waves

2 2 1 ik·r k ψk(r) ⇒ √ e ; ε(k) ⇒ :(v(r) ≡ const) (A.86) V 2m

A.4.2 Wannier Electrons

The representation using Wannier functions

. . 1 1st B Z −ik·Ri wσ (r − Ri ) = √ e ψkσ (r) (A.87) N k is a special, frequently used position representation. The typical property of the Wannier function is its strong concentration around the respective lattice site Ri : 3 ∗ d r wσ (r − Ri ) wσ (r − R j ) = δσσ δij (A.88)

We deﬁne † ciσ (ciσ ) - creation (annihilation) operator of an electron with spin σ in a Wannier state at the lattice site Ri These construction operators satisfy the following commutation relations: † † † ciσ , c jσ = c σ , c σ = 0; ciσ , c σ = δσσ δij (A.89) + i j + j +

In this basis, H0 looks as follows: † = H0 Tijciσ c jσ (A.90) i, j,σ

3 ∗ Tij = d r wσ (r − Ri ) h0 wσ (r − R j ) (A.91) A.4 Examples 509

Tij is known as the hopping integral. In this form, H0 describes, more transparently, the hopping of an electron of spin σ from the lattice site R j to the lattice site Ri . The connection with the Bloch representation (A.4.1) is easy to establish by using (A.78) and (A.87):

1st B.Z. 1 · − T = ε(k)eik (Ri R j ) (A.92) ij N k

. . 1 1st B Z ik·Ri ciσ = √ e ckσ (A.93) N k

A.4.3 Density Operator

The operator for the electron density

Ne ρ(r) = δ(r −ri ) (A.94) i=1 is another example for a single-particle operator. It should be noticed that the electron positionri is an operator but not the variable r: † ρ = σ|δ − | σ (r) k (r ri ) k ckσ ck σ (A.95) kσ, kσ

The matrix element is given by kσ |δ(r −r)|kσ = d3r kσ|δ(r −r)|r r|kσ 3 = δσσ d r δ(r − r ) kσ |r r |k σ ∗ = δ ψ ψ σσ kσ (r) k σ (r) (A.96)

If one restricts oneself to plane waves (v(r) = const.), then

1 iq·r kσ|δ(r −r )|k + qσ =δσσ e (A.97) V

Using this in (A.95) we get 1 † iq·r ρ(r) = c c + σ e (A.98) V kσ k q kqσ 510 A Second Quantization

From the above equation, we therefore ﬁnd the Fourier component of the density operator: ρ = † q ckσ ck+qσ (A.99) kσ

A.4.4 Coulomb Interaction

Now we have to deal with a two-particle operators:

= 2 i j 1 e 1 HC = (A.100) 2 4πε r −r 0 i, j i j

How does this operator look in the formalism of second quantization? We again choose the momentum representation:

e2 HC = ∗ 8πε0 1 ∗ (k σ )(1)(k σ )(2)| |(k σ )(1)(k σ )(2) ∗ 1 1 2 2 (1) −(2) 3 3 4 4 ··· r r k1 k4 σ1···σ4 † † ∗ c c c σ c σ (A.101) k1σ1 k2σ2 k4 4 k3 3

Since the operator itself is independent of spin, the matrix element is surely nonzero only for σ1 = σ3 and σ2 = σ4. Therefore, we are left with the following matrix element:

v ··· = e2 (1) (2)| 1 | (1) (2) (k1 k4) πε k1 k2 |(1)−(2)| k3 k4 4 0 r r = e2 3 3 (1) (2)| 1 | (1) (2) (1) (2)| (1) (2) 4πε d r1d r2 k1 k2 |(1)−(2)| r1 r2 r1 r2 k3 k4 0 r r (A.102) = e2 3 3 1 (1) (2)| (1) (2) (1) (2)| (1) (2) πε d r1d r2 |r −r | k1 k2 r1 r2 r1 r2 k3 k4 4 0 1 2 e2 3 3 1 ∗ ∗ = d r1d r2 ψ (r1)ψ (r2)ψk (r1)ψk (r2) 4πε0 |r1−r2| k1 k2 3 4

Translational symmetry demands that we must have

k1 + k2 = k3 + k4 (A.103)

Thus we have obtained the following expression for the Coulomb interaction: 1 † † H = v(k, p, q) c c c σ c σ (A.104) C 2 k+qσ p−qσ p k kpq σσ A.5 Problems 511 with e2 v , , = 3 3 ψ∗ ψ ∗ (k p q) d r1 d r2 k+q(r1) p−q(r2) 4πε0 1 ∗ ψk(r1) ψp(r2) (A.105) |r1 − r2|

A.5 Problems

|ϕ ···ϕ (ε) Problem A.1 Let α1 αN be an (anti) symmetrized N-particle basis state for the case of a continuous one-particle spectrum. Then show that † − ε † |ϕ ···ϕ (ε) = δ ϕ − ϕ |ϕ ···ϕ (ε) cβ cγ cγ cβ α1 αN ( β γ ) α1 αN

Problem A.2 Prove the fundamental commutation relations † † cα , cα = cα , cα = 0 r s ∓ s r ∓ † cα , c = δ , r αs ∓ r s for the creation and annihilation operators for the Bosons and Fermions in the discrete Fock space. ≡ † ≡ † Problem A.3 cϕα cα und cϕα cα are the annihilation and creation operators for one-particle states |ϕα of an observable Φ with discrete spectrum. With the help of the fundamental commutation relations for Bosons and Fermions, calculate the following commutators: † 1.) nα, cβ 2.) nα, cβ 3.) N, cα − − − † Here N = α nα = α cαcα is the particle number operator. Problem A.4 Under the same assumptions as in Problem A.3 for Fermions, prove the following relations: 2 † 2 1. (cα) = 0; cα = 0 2 2. (nα) = nα † 3. cα nα = cα ; cα nα = 0 † † 4. nα cα = 0;nα cα = cα

† Problem A.5 Let |0 be the normalized vacuum state. Let cα und cα be the creation and annihilation operators for a particle in one-particle state |ϕα .Usingthe fundamental commutation relations derive the relation 512 A Second Quantization

† † 0| cβ ···cβ cα ···cα |0 N 1 1 N pα = (±) Pα [δ(β1,α1)δ(β2,α2) ···δ(βN ,αN )]

Pα

Pα is the permutation operator that operates on the indices αi . Problem A.6 For the occupation number density operator calculate the commutators , † . , 1) nα cβ − ;2) nα cβ −

Is there a difference for Bosons and Fermions? 8 ϕ ···ϕ (±) H(±) Problem A.7 The anti-symmetrized basis states α1 αN of N are built from continuous one-particle basis states. They are the eigenstates of the particle number operator N. Then show that 8 8 † ϕ ···ϕ (±) ϕ ···ϕ (±) 1) cβ α1 αN 2) cβ α1 αN

are also eigenstates of N and calculate the corresponding eigenvalues.

Problem A.8 AsystemofN electrons in volume V = L3 interact among themselves via the Coulomb interaction

= i j 2 1 (i, j) (i, j) e 1 V2 = V ; V = 2 2 2 4πε r −r i, j 0 i j

ri undr j are, respectively, the position operators of the ith and jth electron. Formu- late the Hamiltonian of the system in second quantization. Use as the one-particle basis plane waves which have discrete wavevectors k as a consequence of the periodic boundary conditions on V = L3

Problem A.9 Show that the Hamiltonian calculated in Problem A.8 of the interacting N-electron system † 1 † † H = ε (k) c c σ + v (q) c c c σ c σ N 0 kσ k 2 0 k+qσ p−qσ p k kσ kpq σσ

commutes with the particle number operator = † N ckσ ckσ kσ

What is the physical meaning of this? A.5 Problems 513

Problem A.10 Two identical particles move in a one-dimensional inﬁnite potential well: / 0for0≤ x ≤ a, V (x) = (A.106) ∞ for x < 0 and x > a

Calculate the energy eigenfunctions and energy eigenvalues of the two-particle system if they are (a) Bosons and (b) Fermions. What is the ground state energy in the case N 1 for Bosons and for Fermions? Problem A.11 There is a system of non-interacting identical Bosons or Fermions described by

N = (i) H H1 i=1

(i) The one-particle operator H1 has a discrete non-degenerate spectrum:

(i)|ϕ(i) = |ϕ(i) ϕ(i)|ϕ(i) =δ H1 r r r ; r s rs

|ϕ(i) | , ,... () r are used to build the Fock states N; n1 n2 . The general state of the system is described by the un-normalized density matrix ρ, for which the grand canonical ensemble (variable particle number) holds

ρ = exp[−β(H − μNˆ )]

1. How does the Hamiltonian read in second quantization? 2. Show that the grand canonical partition is given by / {1 − exp[−β( − μ)]}−1 Bosons, Ξ(T, V,μ) = Spρ = i i { + −β − μ } i 1 exp[ ( i )] Fermions.

3. Calculate the expectation value of the particle number:

1 Nˆ = ρNˆ ΞTr( )

4. Calculate the internal energy:

1 U = H = ρ H ΞTr( )

5. Calculate the average occupation number of the ith one-particle state 514 A Second Quantization

1 † nˆ = Tr(ρa a ) i Ξ i i

and show that U = i nˆ i ; Nˆ = nˆ i i i

are valid. Appendix B The Method of Green’s Functions

B.1 Linear Response Theory

We want to introduce the Green’s functions using a concrete physical context: How does a physical system respond to an external perturbation? The answer is provided by the so-called response functions. Well-known examples are the magnetic or electric susceptibility, the electrical conductivity, the thermal conductivity, etc. These are completely expressed by a special type of Green’s function.

B.1.1 Kubo Formula

Let the Hamiltonian

H = H0 + Vt (B.1) consist of two parts. Vt is the perturbation which is the interaction of the system with a possibly time-dependent external field. H0 is the Hamiltonian of the field-free but certainly interacting particle system. Thus H0 is already in general not exactly solvable. Let the perturbation be operating through a scalar field Ft which couples to the observable B :

Vt = B · Ft (B.2)

The restriction made now to a scalar ﬁeld can be easily removed later. One should note that B is an operator, whereas Ft is a c-number. Now let A be a not explicitly time-dependent observable. The interesting question is the following:

How does the measured quantity A react to the perturbation Vt ?

Without the ﬁeld holds

A 0 = Tr(ρ0 A)(B.3)

515 516 B The Method of Green’s Functions

exp(−β H0) ρ0 = (B.4) Tr (exp(−β H0)) where ρ0 is the statistical operator of the ﬁeld-free system, at the moment in the canonical ensemble. β = 1/kB T is as usual the reciprocal temperature. On the other hand, with ﬁeld holds:

A t = Tr(ρt A)(B.5) exp(−β H) ρ = (B.6) t Tr (exp(−β H)) where ρt is now the statistical operator of the particle system in the presence of the external field Ft . Thus a measure of the response of the system to the perturbation could be the change in the value of A under the influence of the external field:

ΔAt = A t − A 0 (B.7)

To calculate this we need ρt . In Schrodinger¬ picture ρt satisﬁes the equation of motion

iρút = [H0 + Vt ,ρt ]− (B.8) with the boundary condition that the ﬁeld for t →−∞is switched off:

lim ρt = ρ0 (B.9) t→−∞

Because of H0, the Schrodinger¬ picture is not convenient for perturbational approaches. More advantageous would be the Dirac picture:

i − i ρ D = H0t ρ H0t t (t) e t e (B.10)

One should note the two different time dependences. The lower index t denotes the possible explicit time dependence caused by the ﬁeld. On the contrary, the argument of the statistical operator represents the dynamic time dependence. The equation of motion now is determined solely by the perturbation: ρ D = D ,ρD i út (t) Vt (t) t (t) − (B.11)

The boundary condition follows from (B.9) and (B.10) where ρ0 commutes with H0:

D lim ρ (t) = ρ0 (B.12) t→−∞ t

We can now formally integrate (B.11): B.1 Linear Response Theory 517 i t ρ D = ρ − D ,ρD t (t) 0 dt Vt (t ) t (t ) − (B.13) −∞

This integral equation can be solved iteratively up to arbitrary accuracy in the perturbation Vt . For small perturbations one can restrict oneself to the ﬁrst non-trivial step (linear response). That means after back-transformation to Schrodinger¬ picture t i − i i ρ ≈ ρ − H0t D ,ρ H0t t 0 dt e Vt (t ) 0 − e (B.14) −∞

We now calculate A t approximately by substituting ρt in (B.5): t i i i − H0t D H0t A = A − dt Tr e V t ,ρ e · A t 0 t ( ) 0 − −∞ i t = − D ρ D − ρ D D A 0 dt Tr Vt (t ) 0 A (t) 0Vt (t )A (t) −∞ i t = − ρ D , D A 0 dt Tr 0 A (t) Vt (t ) − −∞

In the reformulation, we many times used the cyclic invariance of trace. We thus obtain the reaction of the system to external perturbation: t B A i Δ =− D , D At dt Ft A (t) B (t ) − (B.15) −∞ 0

As a consequence of linear response, the reaction of the system is determined by an expectation value for the field-free system. One should note that the Dirac representation of the operators A and B here correspond to the Heisenberg representation without field (H0 → H). One defines: “double-time, retarded Green’s function” B A ret , = ret =−Θ − , G AB(t t ) A(t); B(t ) i (t t ) A(t) B(t ) − (B.16) where Θ is the Heaviside step function. The operators are in the field-free Heisen- berg representation. The properties of the retarded Green’s function will be discussed in more detail in the following. Here, we recognize that it describes the reaction of the system in terms of the observable A when the external perturbation couples to the observable B

+∞ 1 Δ = ret , At dt Ft G AB(t t ) (B.17) −∞

Later we will assume that the (ﬁeld-free) Hamiltonian H is not explicitly dependent on time. In that case one can show that the Green’s function does not depend 518 B The Method of Green’s Functions on two times but only on the time difference. Then it is called homogeneous in time:

ret , → ret − G AB(t t ) G AB(t t )

In such a situation one has the Fourier transformation:

ret = ret G AB(E) A; B E +∞ i − = − ret − E(t t ) d(t t ) G AB(t t ) e (B.18) −∞

+∞ 1 − i − ret − = ret E(t t ) G AB(t t ) dE GAB(E) e (B.19) 2π −∞

In the following, all the time/energy functions will be transformed in this way. In particular, for the delta function holds:

+∞ 1 − i − δ(E − E ) = dt e (E E )t (B.20) 2π −∞ +∞ 1 i − δ(t − t ) = dEe E(t t ) (B.21) 2π −∞

We write the perturbing ﬁeld as a Fourier integral:

+∞ 1 i + − (E+i0 )t Ft = dE F(E) e (B.22) 2π −∞

The additional i0+ takes care of fulﬁlling the boundary condition (B.9). Substituting this expression now in (B.17) we ﬁnally obtain the Kubo formula:

+∞ 1 + − i + + Δ = ret + (E i0 )t At dE F(E) G AB(E i0 ) e (B.23) 2π2 −∞

In the following sections we will present a few applications of this important formula.

B.1.2 Magnetic Susceptibility

We consider a magnetic induction Bt which is homogeneous in space and oscillating in time as perturbation which couples to the total magnetic moment m of a localized spin system (e.g. Heisenberg model) (Fig. B.1):

m = mi (B.24) i B.1 Linear Response Theory 519

Fig. B.1 Interacting localized moments

i indexes the lattice sites at which the localized moments (spins) are present. The perturbation term then reads

(x,y,z) +∞ 1 α α − i + + V =−m · B =− dEm B E e (E i0 )t t t π ( ) (B.25) 2 α −∞

Interesting here is the reaction of the magnetization of the moment system:

1 1 M = m = m (B.26) V V i i

It can be approximately calculated with the help of the Kubo formula (B.17):

+∞ β β β 1 α β α ΔM = M − M =− dt B m (t); m (t ) (B.27) t t 0 t V −∞ α

Here M0 is the ﬁeld-free magnetization which is unequal zero only for ferromagnets. One deﬁnes “magnetic susceptibility tensor”

αβ μ β χ (t, t) =− 0 m (t); mα(t) (B.28) ij V i j

This function, which is so important for magnetism, is thus a retarded Green’s function: +∞ Δ β = 1 χ αβ , α Mt dt ij (t t )Bt (B.29) μ −∞ 0 i, j α

χ αβ , ≡ χ αβ − If the susceptibility is homogeneous in time, ij (t t ) ij (t t ), which we want to assume, then energy representation is meaningful:

+∞ β 1 αβ + α − i + + Δ = χ + (E i0 )t Mt dE ij (E i0 )B (E)e (B.30) 2πμ −∞ 0 i, j α

Of particular interest are • “longitudinal” susceptibility: μ χ zz(E) =− 0 mz; mz (B.31) ij V i j E 520 B The Method of Green’s Functions

With this function it is possible to make statements about magnetic stability.For example, if one calculates 1 · − χ zz(E) = χ zz(E)eiq (Ri R j ) (B.32) q N ij i, j

for the paramagnetic moment system, then in the limit (q, E) → 0 the singularities ! " −1 χ zz =! lim q (E) 0 (q,E)→0

give the instabilities of the paramagnetic state against ferromagnetic ordering. At these points even an infinetesimal, symmetry breaking field produces a finite χ zz magnetization. The poles of the function q (E) therefore describe the phase transition para-ferromagnetism. • “transverse” susceptibility: μ χ +−(E) =− 0 m+; m− (B.33) ij V i j E

Here holds

± = x ± y m j m j imj

For this function also the poles are interesting. They represent resonances or eigenoscillations. One obtains from them the energies of spin waves or magnons: χ +− −1 =! ⇐⇒ = ω q (E) 0 E (q) (B.34)

Thus the linear response theory is not only an approximate procedure for weak perturbations but also provides valuable information about the unperturbed system.

B.1.3 Dielectric Function

We want to discuss another application of the linear response theory. Into a system of quasifree conduction electrons (metal), an external charge density is introduced. Because of Coulomb repulsion, the charge carriers react to the perturbation.This leads to changes in the density of metal electrons in such a way that it causes effec- tively a more or less strong screening of the perturbing charge. This is described by

ε(q, E):dielectric function B.1 Linear Response Theory 521

We ﬁrst want to handle the problem classically and undertake the necessary quantization later. For the external charge density we write

+∞ 1 i + iq·r − (E+i0 )t ρext(r, t) = dE ρext(q, E) e e (B.35) 2πV −∞ q

A corresponding expression should hold for the conduction electrons:

e − eρ(r) =− ρ eiq·r (B.36) V q q

The interaction of the conduction electrons with the external charge density then reads as −e ρ(r) ρ (r, t) = 3 3 ext Vt d rd r (B.37) 4πε0 |r − r |

One shows (Problem 4.3) i(q·r+q·r) π 3 3 e 4 V d rd r = δ ,− (B.38) |r − r| q2 q q

Using this it holds with the Coulomb potential

1 e2 v (q) = (B.39) 0 2 V ε0 q after simple reformulation:

+∞ 1 i + v (q) − (E+i0 )t 0 Vt = dEe ρ−q ρext(q, E) (B.40) 2π −∞ −e q

We assume that without the external perturbation the charge densities of the conduction electrons and the positively charged ions of the solid exactly compensate each other. Then we have for the total charge density

ρtot(r, t) = ρext(r, t) + ρind(r, t) (B.41) where ρind(r, t) is the charge density induced by the displacement of charges. With the Fourier transformed Maxwell equations

iq · D(q, E) = ρext(q, E) (B.42) 1 iq · F(q, E) = (ρext(q, E) + ρind(q, E)) (B.43) ε0 522 B The Method of Green’s Functions where F is the electric ﬁeld strength and D is the electric displacement, and the material equation

D(q, E) = ε0ε(q, E) F(q, E) (B.44) holds for the induced charge density: ! " 1 ρ (q, E) = − 1 ρ (q, E) (B.45) ind ε(q, E) ext

We now translate the formulas obtained classically so far into quantum mechanics. When we do this, the electron density becomes the density operator ρ = † ρ = ρ† q ckσ ck+qσ ; −q q (B.46) k,σ

Then the interaction energy also becomes an operator. According to (B.40) it holds = ρ† Vt q Ft (q) (B.47) q

The perturbing ﬁeld

+∞ v (q) i + 0 − (E+i0 )t − eFt (q) = dEρext(q, E) e (B.48) 2π −∞ however, remains naturally, a c-number. How does the induced charge density (operator!) react to the perturbing ﬁeld produced by the external charge? ) * ρ , =− ρ − ρ =− Δ ρ ind(q t) e q t q 0 e q t (B.49)

We now use the Kubo formula (B.17) +∞ 1 † Δ ρ = ρ ρ q t dt Ft (q ) q(t); q (t ) (B.50) −∞ q

We assume a system with translational symmetry so that we can use

† † ρ ρ → δ ρ ρ q(t); q (t ) q,q q(t); q(t )

Then we have

+∞ −e † ρ , = ρ ρ ind(q t) dt Ft (q) q(t); q(t ) (B.51) −∞ B.2 Spectroscopies and Spectral Densities 523 or after Fourier transformation

v0(q) † ρ q, E = ρ q, E ρ ρ + + ind( ) ext( ) q; q) E i0 (B.52)

We compare this with the classical result (B.45) and then we can represent the dielectric function by a retarded Green’s function

1 1 † = 1 + v (q) ρ ; ρ ) + + (B.53) ε(q, E) 0 q q E i0

The following two limiting cases are interesting:

• ε(q, E) 1: ⇒ ρind(q, E) ≈−ρext(q, E), almost complete screening of the perturbing charge. • ε(q, E) → 0: ⇒ arbitrarily small perturbations produce ﬁnite density oscillations ⇒ “resonances”, i.e. collective eigenoscillations of the electron system.

† The poles of the Green’s function ρq; ρq) E+i0+ are just the energies of the so- called “plasmons”.

B.2 Spectroscopies and Spectral Densities

An additional important motivation for the study of Green’s functions is their close connection to “elementary excitations” of the system, which are directly observable by appropriate spectroscopies. Thus certain Green’s functions provide a direct access to experiment. This is more directly valid for another fundamental function, namely the so-called “spectral density” which has a close relation to the Green’s functions. Figure B.2 shows in a schematic form, which elementary processes are involved in four well-known spectroscopies for the determination of electronic structure. The photoemission (PES) and the inverse photoemission (IPE) are the so-called one-particle spectroscopies since the system (solid) contains after the excitation process one particle more (less) as compared to before the process. In the photoemission the energy ω of the photon is absorbed by an electron in a (partially) occupied energy band. This gain in energy makes the electron to leave the solid. Analysis of the kinetic energy of the photoelectron leads to conclusions about the occupied states of the concerned energy band. Then the transition operator Z−1 = cα corresponds to the annihilation operator cα, if the particle occupies the one-particle state |α before the excitation process. In the inverse photoemission,a somewhat reverse process takes place. An electron is shot into the solid and it lands in an unoccupied state |β of a partially ﬁlled energy band. The energy released is emitted as a photon ω which is analysed. Now the system contains one electron † more than before the process. This corresponds to the transition operator Z+1 = cβ . PES and IPE are in some ways complementary spectroscopies. The former provides 524 B The Method of Green’s Functions

Fig. B.2 Elementary processes relevant for four different spectroscopies: 1. photoemission (PES), 2. inverse photoemission (IPE), 3. Auger electron spectroscopy (AES), and 4. appearance-potential spectroscopy (APS). Z j is the transition operator, where j means the change in the electron numbers due to the respective excitation information regarding the occupied and the latter about the unoccupied states of the energy band. Auger-electron spectroscopy (AES) and appearance potential spectroscopy (APS) are two-particle spectroscopies. The starting situation for AES is characterized by the existence of a hole in a deep lying core state. An electron of the partially ﬁlled band drops down into this core state and transfers the energy to another electron of the same band so that the latter can leave the solid. The analysis of the kinetic energy of the emitted electron provides information about the energy structure of the occupied states (two-particle density of states). The system (energy band) contains two particles less than before the process: Z−2 = cαcβ . Almost the reverse process B.2 Spectroscopies and Spectral Densities 525 is exploited in APS. An electron lands in an unoccupied state of an energy band. The energy released is used in exciting a core electron to come up and occupy another unoccupied state of the same energy band. The subsequent de-exciting process can be analyzed to gain information about the unoccupied states of the energy band. In this case then the system (energy band) contains two electrons more after the process † † as compared to before the process. Therefore the transition operator is Z+2 = cβ cα. AES and APS are obviously complementary two-particle spectroscopies. We now, using simple arguments, want to estimate the associated intensities for the individual processes. • The system under investigation is described by the Hamiltonian:

H = H − μN (B.54)

Here μ is the chemical potential and N is the particle number operator. Here we use H instead of H since in the following we want to perform averages in grand canonical ensemble. This is necessary because the above discussed transition operators change the particle number. However, H and N should commute which means they possess a common set of eigenstates:

H |En(N) =En(N) |En(N) ; N |En(N) =N |En(N)

Then H satisﬁes the eigenvalue equation

H |En(N) =(En(N) − μN) |En(N) →En |En (B.55)

In order to save writing effort, in the following, so long as it does not create any confusion, we will write the short forms i.e. for the eigenvalue instead of (En(N) − μN) write En and for the eigenstates instead of |En(N) simply write |En . However, the actual dependence of the states and eigenenergies on the particle number should always be kept in mind. • The system at temperature T ﬁnds itself in an eigenstate |En of the Hamiltonian H with the probability

1 −β E Ξ exp( n)

where Ξ is the grand canonical partition function

Ξ = Tr (exp(−βH) (B.56)

• The transition operator Zr induces transitions between the states |En and |Em with the probability

2 | Em |Zr |En | r =±1, ±2 526 B The Method of Green’s Functions

• The intensity of the measured elementary process corresponds to the total number of transitions with excitation energies between E and E + dE: 1 −β I E = e En | E |Z |E |2 δ E − E − E r ( ) Ξ m r n ( ( m n)) (B.57) m,n

If the excitation energies are sufﬁciently dense, as is anyway the case for a solid, then Ir (E) is a continuous function of the energy E. • At this point we neglect certain secondary effects, which of course are important for a quantitative analysis of the respective experiments, but are not decisive for the actually interesting processes. This is valid, for example, in PES and AES for the fact that the photoelectron leaving the solid can still couple to the residual system (sudden approximation). In addition, the matrix elements for the transition from band level into vacuum are not taken into account here. However, the bare line forms of the mentioned spectroscopies should be correctly described by (B.57). One should note that for the transition operator holds

= † Zr Z−r (B.58)

That means complementary spectroscopies are related to each other in some way. This is now investigated in more detail: 1 β −β I E = e E e Em | E |Z |E |2 δ E − E − E r ( ) Ξ m r n ( ( m n)) m,n 1 β −β = e E e En | E |Z |E |2 δ E − E − E Ξ n r m ( ( n m )) n,m β E e −β E 2 = e n | E |Z− |E | δ −E − E − E Ξ m r n (( ) ( m n)) n,m

In the second step only the summation indices n and m are interchanged; the last transition uses (B.58). We then have derived a “symmetry relation” for the complementary spectroscopies:

β E Ir (E) = e I−r (−E) (B.59)

We now deﬁne the spectral density which is important for the following: 1 (±) β E S E = I− E ∓ I −E = e ∓ I −E r ( ) r ( ) r ( ) 1 r ( ) (B.60)

The freedom in the sign will be explained later. From (B.59) and (B.60) one recognizes that the intensities of the complementary spectroscopies are determined by the B.2 Spectroscopies and Spectral Densities 527 same spectral density in a simple manner:

1 I (E) = S(±)(−E) (B.61) r e−β E ∓ 1 r

β E e (±) I− (E) = S (E) (B.62) r eβ E ∓ 1 r

Thus the just introduced spectral density is closely related to the intensities of spectroscopies. Therefore we want to further investigate this function by performing a Fourier transformation into the time domain:

+∞ 1 i − E(t−t ) dEe I−r (E) 2π −∞ 1 1 −β E − i (E −E )(t−t) † = e n e m n E |Z− |E E |Z− |E 2π Ξ m r n n r m m,n 1 1 −β E i Ht − i Ht = e n E |e Z− e |E ∗ 2π Ξ m r n m,n i i Ht − Ht ∗ En|e Zr e |Em 1 1 −β † = e En E |Z (t)|E E |Z (t )|E 2π Ξ n r m m r n m,n 1 1 −β † = e En E |Z (t)Z (t )|E 2π Ξ n r r n n 1 = Z (t) Z †(t) 2π r r

Completely analogously one ﬁnds

+∞ 1 − i − 1 † E(t t ) − = dEe Ir ( E) Zr (t ) Zr (t) 2π −∞ 2π

That means with (B.60) for the double-time spectral density

+∞ η 1 − i − η ( ) , = E(t t ) ( ) Sr (t t ) dEe Sr (E) 2π −∞ 1 = Z (t), Z †(t) (B.63) 2π r r −η

Here η =±is at the moment only an arbitrary sign factor. [··· , ···]−η is either a commutator or an anti-commutator: , † = † − η † Zr (t) Zr (t ) −η Zr (t) Zr (t ) Zr (t ) Zr (t) (B.64) 528 B The Method of Green’s Functions

We have shown that spectral density in (B.63) is of central importance for the intensities of spectroscopies. In addition it can also be shown that a generalization of spectral density to arbitrary operators A and B is closely related to the retarded Green’s functions introduced in (B.16). This is also true for the other types of Green’s functions to be deﬁned in the next section. The “spectral density”

η 1 S( ) (t, t) = A(t), B(t) (B.65) AB 2π −η has the same place of importance as the Green’s functions in the many-body theory.

B.3 Double-Time Green’s Functions

B.3.1 Deﬁnitions and Equations of Motion

In order to construct the full Green’s function formalism the retarded Green’s function introduced earlier is not sufﬁcient. One needs two other types: Retarded Green’s function ret , = ret =−Θ − , G AB(t t ) A(t); B(t ) i (t t ) A(t) B(t ) −η (B.66)

Advanced Green’s function ad , = ad =+Θ − , G AB(t t ) A(t); B(t ) i (t t) A(t) B(t ) −η (B.67)

Causal Green’s function

c , = c =− { } G AB(t t ) A(t); B(t ) i Tη A(t)B(t ) (B.68)

The operators here are in their time-dependent Heisenberg picture:

i H − i H X(t) = e t Xe t (B.69) where, just as in the derivation of the spectral density in the last section, the transformation should be carried out with the grand canonical Hamiltonian H = H − μN. In doing this we assume that H is not explicitly dependent on time. The averages are performed in the grand canonical ensemble:

1 XY = Tr e−βH XY Ξ (B.70)

Ξ is the grand canonical partition function (B.56). The choice of the sign η =±is arbitrary and appears based on the convenience in a given situation. If A and B are pure Fermi (Bose) operators, then η =−(η =+) is convenient but not necessarily B.3 Double-Time Green’s Functions 529 required. The brackets in (B.66) and (B.67) are anti-commutators in the former case and commutators in the latter case as was ﬁxed in (B.64). The deﬁnition (B.68) of the causal Green’s function contains the Wick’s time ordering operator:

Tη {A(t)B(t )}=Θ(t − t ) A(t)B(t ) + η Θ(t − t) B(t )A(t) (B.71)

The step function Θ 1fort > t Θ(t − t ) = (B.72) 0fort < t is not defined for equal times. This is reflected in the Green’s functions and so has to be considered later. Because of the averaging in (B.70), the Green’s functions are also temperature dependent. As a result, later, an unusual but convenient relationship between the temperature and time variables will be established. In addition to the Green’s functions (B.66), (B.67) and (B.68), the spectral density (B.65) is of equal significance. We now prove “retrospectively” that if H is not explicitly dependent on time, then the Green’s functions and the spectral density are homogeneous in time:

∂H , , , , = 0 Gret ad c(t, t) = Gret ad c(t − t) (B.73) ∂t AB AB SAB(t, t ) = SAB(t − t ) (B.74)

The proof is complete provided the homogeneity of the “correlation functions”

A(t) B(t) ; B(t) A(t) is proved. This is achieved by using the cyclic invariance of trace: −βH −βH i H − i H i H − i H Tr e A(t) B(t ) = Tr e e t Ae t e t Be t − i H −βH i H − i H i H = Tr e t e e t Ae t e t B −βH − i H i H − i H i H = Tr e e t e t Ae t e t B = Tr e−βH A(t − t) B

Thus the homogeneity is shown as

A(t) B(t) = A(t − t) B(0) (B.75)

Analogously one ﬁnds

B(t) A(t) = B(0) A(t − t) (B.76) 530 B The Method of Green’s Functions

For the approximate determination of a Green’s function, one can use the respective equation of motion which is directly derived from the equation of motion of the corresponding time-dependent Heisenberg operators. With

∂ ∂ Θ(t − t) = δ(t − t) =− Θ(t − t) (B.77) ∂t ∂t one gets formally the same equation of motion for all the three Green’s functions:

∂ ret,ad,c i G (t, t ) = δ(t − t ) [A, B]−η ∂t AB ret,ad,c + [A, H]− (t); B(t ) (B.78)

The boundary conditions are, however, different:

ret , = < G AB(t t ) 0fort t (B.79) av , = > G AB(t t ) 0fort t (B.80) −i A(t − t ) B(0) for t > t Gc (t, t ) = (B.81) AB −iη B(0) A(t − t) for t < t

The boundary conditions for the causal function are quite unmanageable. For that reason this function does not play any role in the equation of motion method. That is why it will not be considered here any more. On the right-hand side of the equation of motion (B.78) appears a higher Green’s function as [A, H]− (t) is itself a time-dependent operator. In some special cases [A, H]− (t) ∝ A(t) holds. Then an exact solution of the equation of motion is directly possible. However, in general, such a proportionality does not hold. Then the higher Green’s function satisﬁes its own equation of motion of the form

∂ ret,ad i [A, H]− (t); B(t ) = ∂t = δ(t − t ) [A, H]− , B + −η + , H , H ret,ad [A ]− − (t); B(t ) (B.82)

On the right-hand side appears a still higher Green’s function for which again another equation of motion can be written. This leads to an inﬁnite chain of equations of motion which, at some stage, has to be decoupled physically meaningfully. Going from the time domain into energy domain also does not change anything for this chain of equations. However, with (B.18) and (B.21), one gets a pure alge- braic equation which can possibly be of advantage:

ret,ad = , + , H ret,ad E A; B E [A B]−η [A ]− ; B E (B.83) B.3 Double-Time Green’s Functions 531

The boundary conditions (B.79) and (B.80) manifest themselves as different analytical behaviours of the Green’s functions in the complex E-plane. This shall be investigated in detail in the next section.

B.3.2 Spectral Representations

In order to learn more about the analytical properties of these functions, with the help of the eigenvalues and eigenstates (B.55) of the Hamiltonian H, we derive the spectral representations of the retarded and advanced Green’s functions. The eigenstates constitute a complete orthonormal system: En|Em =δnm ; |En En|=1l (B.84) n

− i − − ∗e (En Em )(t t )

In the last step we simply interchanged the summation indices n and m. Completely analogously one ﬁnds the other correlation function i −β En − (En −Em )(t−t ) Ξ · B(t ) A(t) = e En|B|Em Em |A|En e n,m

Comparing these expressions with the Fourier representation of the spectral density

1 S (t, t) = A(t), B(t) AB 2π −η +∞ 1 i − E(t−t ) = dEe SAB(E) (B.85) 2π −∞ 532 B The Method of Green’s Functions we get the important “spectral representation of the spectral density”

S E = E |B|E E |A|E ∗ AB( ) Ξ n m m n n,m

−β En β E ∗e (e − η)δ (E − (En − Em )) (B.86)

The arguments of the delta functions contain the possible excitation energies! Let us compare this with (B.57), the intensity formula for r-particle spectroscopies and their simple relationships (B.61) and (B.62) with the respective spectral densities. We now try to express the Green’s functions in terms of spectral density. This is possible with the following representation of the step function: +∞ −ix(t−t) Θ − = i e (t t ) dx + (B.87) 2π −∞ x + i0

This is proved using the residue theorem (Problem B.4). Using this we ﬁrst rewrite the retarded function +∞ i − ret = − E(t t ) − Θ − π − G AB(E) d(t t )e i (t t ) 2 SAB(t t ) −∞ +∞ +∞ −ix(t−t) i − 1 e = − E(t t ) ∗ d(t t )e dx + −∞ 2π −∞ x + i0

+∞ 1 i − E (t−t ) ∗ dE SAB(E )e −∞

+∞ +∞ = 1 ∗ dE SAB(E ) dx + −∞ −∞ x + i0

+∞ 1 i − − − ∗ d(t − t )e (E E x)(t t ) 2π −∞

+∞ +∞ = 1 δ − − dE SAB(E ) dx + (E E x) −∞ −∞ x + i0 +∞ 1 S (E) = dE AB 1 − + + −∞ (E E ) i0

This gives the “spectral representation of the retarded Green’s function”: +∞ S E ret = AB( ) G AB(E) dE + (B.88) −∞ E − E + i0 B.3 Double-Time Green’s Functions 533

The advanced Green’s function can be treated completely analogously: +∞ i − ad = − E(t t ) + Θ − π − G AB(E) d(t t )e i (t t) 2 SAB(t t ) −∞ +∞ +∞ −ix(t−t) i − −1 e = − E(t t ) ∗ d(t t )e dx + −∞ 2π −∞ x + i0

+∞ 1 i − E (t−t ) ∗ dE SAB(E )e −∞ +∞ +∞ − = 1 ∗ dE SAB(E ) dx + −∞ −∞ x + i0 +∞ 1 i − + − ∗ d(t − t )e (E E x)(t t ) 2π −∞ +∞ +∞ − = 1 δ − + dE SAB(E ) dx + (E E x) −∞ −∞ x + i0 +∞ 1 S (E) = dE AB 1 − − + −∞ (E E ) i0

This gives the “spectral representation of the advanced Green’s function”: +∞ S E ad = AB( ) G AB(E) dE + (B.89) −∞ E − E − i0

The different sign of i0+ in the denominator of the integrand is the only but important difference between the retarded and the advanced functions. The retarded and advanced functions have poles, respectively, in the lower and upper half-planes. This results in different analytical behaviours of the two functions.

ret av G AB(E) in the upper and G AB(E) in the lower complex half-plane can be analytically continued!

Finally, we can substitute (B.86) in (B.88) and (B.89):

ret Gad E = E |B|E E |A|E ∗ AB( ) Ξ n m m n n,m β(E −E ) −β e n m − η ∗ En e + (B.90) E − (En − Em ) ± i0

We see that we have meromorphic functions with simple poles at the exact (!) excitation energies of the interacting system. With a suitable choice of the operators A 534 B The Method of Green’s Functions and B special poles can be extracted, i.e. E = En − Em appears as a pole only if En|B|Em = 0 and Em |A|En = 0. Because of the identical predictive power and the same equation of motion the retarded and advanced functions are considered to be the two branches of a uniﬁed Green’s function in the complex plane: +∞ S (E) ret > = AB = G AB(E),ifImE 0 G AB(E) dE ad < (B.91) −∞ E − E G AB(E),ifImE 0

In the following, we want to call this the combined Green’s function. Its poles lie on the real axis. With the “Dirac identity”,

1 = P 1 ∓ πδ − + i (x x0) (B.92) x − x0 ± i0 x − x0 where P represents the Cauchy’s principal value, one ﬁnds the following relation between the spectral density and the Green’s functions:

i S (E) = Gret (E) − Gav (E) (B.93) AB 2π AB AB

† If SAB(E) is real, which it is in many of the important cases (e.g. B = A ), this relation becomes even simpler:

1 ret S E =∓ ImGav E AB( ) π AB( ) (B.94)

Equations (B.88) and (B.89) show that the Green’s functions are completely ﬁxed by the spectral density. On the other hand according to (B.94) only the imaginary part of the Green’s function determines the spectral density. This means there must be relations between the real and the imaginary parts of the Green’s function; they are not independent of each other. These relations are called the Kramers–Kronig relations which we will explicitly derive in Sect. B.3.5.

B.3.3 Spectral Theorem

The discussion of the last section has shown that the Green’s functions and the spectral density, in addition to their importance in the context of the response functions (Sect. B.1) and intensities of certain spectroscopies, also provide valuable microscopic information. Their singularities correspond to the exact excitation energies of the system. We now want to demonstrate that the whole macroscopic thermodynamics can be determined using suitably deﬁned Green’s functions and spectral densities. B.3 Double-Time Green’s Functions 535

The starting point is the spectral representation of the correlation functions introduced in Sect. B.3.2. 1 −β − i − − B t A t = e En E |B|E E |A|E e (En Em )(t t ) ( ) ( ) Ξ n m m n n,m

Comparing this expression with that for the spectral representation of the spectral density (B.86), we get the fundamental spectral theorem:

+∞ (η) 1 S (E) − i − 1 = AB E(t t ) + + η B(t ) A(t) dE β e (1 ) D (B.95) −∞ e E − η 2

Except for the second summand, this result follows directly from the comparison with (B.86). The second term comes into play only for the commutator (η =+1) spectral density. It disappears for the anti-commutator function (η =−1). The reason for this is clear from (B.86). For E = 0, i.e. En = Em , the commutator spectral densities do not contribute the corresponding term because (eβ E − 1) = 0, even 1 though they may give a contribution D unequal zero in the correlation function, where

= En Em −β D = e En E |B|E E |A|E Ξ n m m n (B.96) n,m

Experience, however, shows that in fact in many cases D = 0. But the fact that this is not always necessary is made easily clear from the following example: The operator pairs A, B and A = A − A 1l , B = B − B 1l form identical spectral densities:

+ + + + S( )(t − t) ≡ S( )(t − t) ⇐⇒ S( )(E) ≡ S( )(E) AB AB AB AB

On the other hand holds

B(t) A(t) = B(t) A(t) − B(t) A(t) which would lead to a contradiction without the second term in (B.95) in case A(t) = 0 and B(t) = 0. Thus without the extra term D, the spectral theorem would be incomplete for the commutator spectral density. How does one determine D? It is possible with the combined Green’s function (B.91) whose spectral representation

β(E −E ) η −β e n m − η G( ) (E) = E |B|E E |A|E e En (B.97) AB Ξ n m m n E − (E − E ) n,m n m 536 B The Method of Green’s Functions leads to the following relation:

· (η) = − η lim E G AB(E) (1 ) D (B.98) E→0

The limit must be taken in complex plane. One recognizes the following important consequences:

• Even though D is needed for the commutator spectral density, the determination succeeds only using the anti-commutator Green’s function. • (+) = The commutator Green’s function G AB(E) is always regular at E 0, i.e. it has no pole there. • (−) = The anti-commutator Green’s function G AB(E), for D 0, has a pole of ﬁrst order with the residue 2D.

B.3.4 Spectral Moments

Green’s functions and spectral densities for realistic problems are in general not exactly solvable. Therefore, one must tolerate approximations. These approximations can be checked for correctness by using exactly solvable limiting cases, symmetry relations, sum rules, etc. In this sense, the moments of spectral density are found to be extraordinarily useful. Let n and p be non-negative integers:

n = 0, 1, 2, ··· ;0≤ p ≤ n

Then for the time-dependent spectral density holds ∂ n−p ∂ p i −i 2π S (t − t) = ∂t ∂t AB n−p p +∞ ∂ ∂ 1 − i − = − E(t t ) i i dE SAB(E) e ∂t ∂t −∞ +∞ 1 i n − E(t−t ) = dE E SAB(E) e −∞ ∂ n−p ∂ p =! i −i A(t), B(t) ∂t ∂t −η

For t = t from the ﬁrst part of the system of equations one gets the spectral moments

+∞ (n) = 1 n MAB dE E SAB(E) (B.99) −∞ B.3 Double-Time Green’s Functions 537

An alternative representation results from the last part by using the equation of motion for the time-dependent Heisenberg operators

M(n) = !AB " = ··· , H , H ··· , H , H, ··· , H, ··· [ [[A ]− ]− ]− [ [ B]− ]− −η (n−p)−fold p−fold (B.100)

With this last relation it is possible in principle to calculate exactly all the moments independently of the respective spectral densities if the Hamiltonian is known. Then one has the possibility to control certain approximate procedures for the spectral density using (B.99). With the spectral moments, it is possible to formulate a very often useful “high- energy expansion”. For the “combined Green’s function” (B.91) holds

+∞ = SAB(E ) G AB(E) dE −∞ E − E

+∞ = 1 SAB(E ) dE E E −∞ − 1 E ∞ +∞ n = 1 E dE SAB(E ) E −∞ E n=0

A comparison with (B.99) gives

∞ M(n) G (E) = AB (B.101) AB En+1 n=0

For the extreme high-energy behaviour (E →∞) this means

(0) G (E) ≈ M = [A, B]−η (B.102) AB E AB E

The right-hand side is in general easy to calculate and therefore, e.g. the high-energy behaviour of the response functions in Sect. B.1.1 is known.

B.3.5 Kramer’s–Kronig Relations

ret,ad We have already seen that the Green’s function G AB (E) is completely determined by the spectral density SAB(E) (B.88) and (B.89). In case it is real, then it can be 538 B The Method of Green’s Functions determined by the imaginary part of the Green’s function alone (B.94). Therefore the real and imaginary parts of the Green’s function cannot be independent of each other. We will now derive the relationship between them. One calculates the following integral for real E: 6 Gret Eˆ = ˆ AB( ) IC (E) d E + (B.103) C E − Eˆ − i0

Integration is performed over the path C which follows the real axis and is closed in the upper half of the complex plane (Fig. B.3). The integrand has a pole at Eˆ = E − i0+, i.e. in the lower half-plane. The retarded Green’s function also has a pole only in the lower half-plane so that there is no pole in the region enclosed by C. Therefore it holds

IC (E) = 0 (B.104)

The semicircle is extended to inﬁnity. The high-energy expansion (B.101) shows that the integrand in (B.103) then goes to zero at least as 1/Eˆ 2. As a result, the semicircle does not contribute to the integral (B.103). It remains when the Dirac identity (B.92) is used:

+∞ Gret Eˆ = ˆ AB( ) 0 d E + −∞ E − Eˆ − i0 +∞ Gret Eˆ = P ˆ AB( ) + π ret d E i G AB(E) −∞ E − Eˆ

That means i +∞ Gret Eˆ ret = P ˆ AB( ) G AB(E) d E (B.105) π −∞ E − Eˆ

ad Completely analogous considerations hold for the advanced function G AB(E). Now the semicircle will be in the lower half-plane and in the denominator of the integrand (B.103) −i0+ is replaced by +i0+ . Then we have

Fig. B.3 Integration path C in the complex E-plane for the integral (B.103) B.3 Double-Time Green’s Functions 539 i +∞ Gad Eˆ ad =− P ˆ AB( ) G AB(E) d E (B.106) π −∞ E − Eˆ

Thus it is not at all necessary to know the full Green’s function. The real and imaginary parts account for each other (“Kramers–Kronig relations”)

ret +∞ ret i ImGad Eˆ ad =∓ P ˆ AB( ) ReGAB(E) d E (B.107) π −∞ E − Eˆ

ret +∞ ret i ReGad Eˆ ad =± P ˆ AB( ) ImGAB(E) d E (B.108) π −∞ E − Eˆ

In case the respective spectral density is real the additional relation (B.94) holds. That leads to

ret =− ad =−π ImGAB(E) ImGAB(E) SAB(E) (B.109)

+∞ S Eˆ ret = ad = P ˆ AB( ) ReGAB(E) ReGAB(E) d E (B.110) −∞ E − Eˆ

The retarded and advanced Green’s functions are therefore very closely related.

B.3.6 Simple Applications

We want to apply the above introduced Green’s function formalism to two simple systems.

B.3.6.1 Free Band Electrons As the ﬁrst example, we discuss a system of electrons in a solid which do not interact with each other, interacting only with a periodic lattice potential. This is described by the following one-particle Hamiltonian: H = − μ , = ε † 0 H0 N H0 (k) ckσ ckσ (B.111) kσ = † N ckσ ckσ (B.112) kσ

All the interesting properties of the electron system can be calculated from the so- called one-electron Green’s function: 540 B The Method of Green’s Functions

ret,ad = † ret,ad Gkσ (E) ckσ ; ckσ E (B.113)

Since we are dealing with a pure Fermi system, the choice of the anti-commutator Green’s function is natural, however, not mandatory. The ﬁrst step is setting up the equation of motion (B.83):

ret,ad = , † + , H † ret,ad EGkσ (E) [ckσ ckσ ]+ [ckσ 0]−; ckσ (B.114)

With the help of the fundamental commutation relations for Fermions, one easily gets

[ckσ , H0]− = (ε(k) − μ) ckσ (B.115)

On substituting this leads to a simple equation of motion

ret,ad = + ε − μ ret,ad EGkσ (E) ( (k) ) Gkσ (E)

Solving this and introducing ±i0+ in order to satisfy the boundary conditions one gets

, Gret ad(E) = (B.116) kσ E − (ε(k) − μ) ± i0+

The Green’s function is singular at the energy which is required to add an electron of wavevector k to the non-interacting electron system. That means the singularities of the Green’s function (B.113) correspond to the one-particle excitations of the system. The combined Green’s function is naturally directly given from (B.116) by removing the inﬁnitesimal imaginary term:

G σ (E) = (B.117) k E − (ε(k) − μ)

The energy E is thought to be complex here. It is interesting to conﬁrm the result (B.117) by an exact evaluation of high-energy expansion (B.101) (Problem B.7). Finally the one-electron spectral density is another important quantity, for which with the Dirac identity (B.92) along with (B.94) directly from (B.116) follows:

Skσ (E) = δ (E − ε(k) + μ) (B.118)

Using the spectral theorem (B.95) one can easily calculate the average occupation number of the (k,σ)-level:

+∞ = † = 1 Skσ (E) = 1 nkσ ckσ ckσ dE β β ε −μ (B.119) −∞ e E + 1 e ( (k) ) + 1 B.3 Double-Time Green’s Functions 541

This is of course a well-known result of quantum statistics. The average occupation β(E−μ) −1 number is given by the Fermi function f−(E) = (e + 1) at E = ε(k). From nkσ by summing over all the wavevectors k and the two spin projections σ we can ﬁx the total number of electrons Ne:

+∞ = = 1 Skσ (E) Ne nkσ dE β −∞ e E + 1 kσ kσ 1 +∞ = dE f−(E)Skσ (E − μ) −∞ kσ

We denote by ρσ (E) the density of states per spin, where self-evidently for the free Fermion system ρσ (E) ≡ ρ−σ (E) holds, so that we can write Ne as

+∞ Ne = N dEf−(E)ρσ (E) σ −∞

Here N is the number of lattice sites with which ρσ (E) is normalized to 1 since the number of energy band states per each spin direction should be equal to the number of lattice sites. The comparison of the last two equations leads to the important deﬁnition of the “quasiparticle density of states”:

1 ρσ (E) = S σ (E − μ) (B.120) N k k

In the case of the non-interacting electrons considered here with (B.118)

1 ρσ (E) = ρ−σ (E) ≡ ρ (E) = δ (E − ε(k)) (B.121) 0 N k follows. Without the lattice potential, for the one-particle energies we get the well- ε = 2k2 known parabolic√ dispersion (k) 2m . One can then easily show that the density of states has a E-dependence. The considerations for the electron number are correct not only for free electron system but also valid in general. That is why (B.120) will be accepted as the general (!) deﬁnition of the quasiparticle density of states for any interacting electron system. The “internal energy” U as the thermodynamic expectation value of the Hamil- tonian is ﬁxed in a simple manner by nkσ : 542 B The Method of Green’s Functions U = H0 = ε(k) nkσ kσ 1 +∞ = ε(k) dEf−(E)Skσ (E − μ) −∞ kσ 1 +∞ = dE(E + ε(k)) f−(E)Skσ (E − μ) (B.122) 2 −∞ kσ

The bit more complicated representation of the last line will turn out to be the deﬁni- tion of U which is valid in general for an interacting electron system (see B.4.5). Finally the time-dependent functions are interesting. For the spectral density (B.118), it is trivial to perform the Fourier transformation:

1 i S σ (t − t ) = exp − (ε(k) − μ)(t − t ) (B.123) k 2π

This represents an undamped oscillations with a frequency that corresponds to an exact excitation energy of the system (Fig. B.4). This is typical for non-interacting particle systems. Exactly similarly we ﬁnd the time-dependent Green’s functions:

i Gret t, t =−iΘ t − t − ε k − μ t − t kσ ( ) ( )exp ( ( ) )( ) (B.124) i Gav t, t =+iΘ t − t − ε k − μ t − t kσ ( ) ( )exp ( ( ) )( ) (B.125) σ (0) k S 0 πℜ 2

0 (t − t’) Fig. B.4 Time dependence of the real part of the single-particle spectral density of non-interacting Bloch electrons B.3 Double-Time Green’s Functions 543

B.3.6.2 Spin Waves As another example we consider the (linear) spin waves (magnons) of a ferromagnet. At low temperatures the Heisenberg Hamiltonian (7.1) can be simpliﬁed in the HolsteinÐPrimakoff approximation (7.1.1) as follows: = + ω † HSW E0 (q) aqaq (B.126) q

† aq(aq) are the creation (annihilation) operators for magnons. According to this, the ferromagnet is modelled as a gas of non-interacting magnons with one-particle energies

2 ω(q) = 2S (J0 − J(q)) + gJ μB B0 (B.127)

J(q) are the exchange integrals with J0 = J(q = 0). The second summand describes the influence of an external magnetic field on the one-particle energies. A precondition for the concept of spin waves is that at T = 0 the system is ferromagnetic (E0 : ground state energy (7.246)). A symmetry breaking field

+ B0 ≥ 0 (B.128) has to be present necessarily. Particle number conservation does not hold for magnons. At a given temperature T , it gives exactly the magnon number for which the free energy is minimum: ∂ F =! ∂ 0 (B.129) N T,V

The differential fraction on the left-hand side is just the deﬁnition of the chemical potential μ. So for magnons holds

μ = 0 (B.130)

This means that we can substitute H = H − μN = H in the equation of motion for the Green’s function. We need the commutator † aq, HSW = ω(q ) aq, a aq − q − q † = ω(q ) aq, a aq q − q

= ω(q) aq

Then the equation of motion becomes very simple:

ret,av = + ω ret,av EGq (E) (q) Gq (E) 544 B The Method of Green’s Functions

Solving it along with the boundary conditions then gives Gret,av(E) = (B.131) q E − ω(q) ± i0+

Here also the pole represents the excitation energy which indicates either a creation or annihilation of a q-magnon. That is exactly ω(q) in the absence of interaction. With (B.92) and (B.94), the important “one-magnon spectral density” follows:

Sq(E) = δ (E − ω(q)) (B.132)

One can quickly calculate the time-dependent function and see that it represents as in the case of free Bloch electrons, an undamped oscillation:

1 S (t − t) = exp −iω(q)(t − t) (B.133) q 2π The frequency of oscillation again corresponds to an exact excitation energy of the system. With the help of the spectral theorem (B.95) and the spectral density (B.132) we obtain the “magnon occupation density”:

1 m = a†a = + D (B.134) q q q exp (βω(q)) − 1 q

As we started with the commutator Green’s function, we must determine the constant Dq using the respective anti-commutator Green’s function. The fundamental commutation relation for Bosons give for the inhomogeneity in the equation of motion , † = + aq aq + 1 2mq

Except for this the anti-commutator Green’s function satisﬁes the same equation of motion as the commutator Green’s function. One obtains (1 + 2m ) G(−)(E) = q (B.135) q E − ω(q)

In the presence of at least an inﬁnitesimal symmetry breaking external ﬁeld (B0 ≥ 0), the magnon energies are always unequal zero and are positive. That according to (B.98) means

= (−) = 2 Dq lim EGq (E) 0 E→0

So that for the occupation density we get

1 m = (B.136) q exp (βω(q)) − 1 B.4 The Quasiparticle Concept 545

This is the BoseÐEinstein distribution function which is a well-known result of elementary quantum statistics for the ideal Bose gas.

B.4 The Quasiparticle Concept

The really interesting many-body problems unfortunately can not be solved exactly. Therefore one must tolerate approximations. For describing the interacting many- particle systems, the concept of “quasiparticles” has proved to be very successful and will be discussed in this section. The basis for this is the following idea:

complex interacting systems of “real” particles =⇒ non- (or weakly-) interacting system of quasiparticles

This replacement is valid only if one assigns certain special properties to the quasiparticles which will be discussed in the following: • energy renormalization; • damping, ﬁnite lifetimes; • effective masses; • spectral weights, etc.

B.4.1 Interacting Electrons

In order to be concrete, we here want to concentrate on a system of electrons in a non-degenerate energy band interacting via Coulomb interaction. For such a system holds in Bloch representation: H = − μ = † H N ; N ckσ ckσ (B.137) kσ † 1 † † H = ε(k)c c σ + v (q)c c c σ c σ (B.138) kσ k 2 kp k+qσ p−qσ p k kσ kpqσσ

Let the matrix elements be built with Bloch functions ψk(r): ! " 2 ε(k) = d3r ψ∗(r) − Δ + V (r) ψ (r) (B.139) k 2m k ψ∗ (r )ψ∗ (r )ψ (r )ψ (r ) 1 3 3 k+q 1 p−q 2 p 2 k 1 vkp(q) = d r1 d r2 4πε0 |r1 − r2| (B.140)

vkp(q) = vpk(−q) (B.141) 546 B The Method of Green’s Functions

Fig. B.5 Diagram of the Coulomb interaction between electrons of a non-degenerate energy band

All the wavevectors k, p, q stem from the ﬁrst Brillouin zone. Actually the detailed structure and interpretation of the matrix elements is not important for the following considerations. What is decisive is that the interaction (Fig. B.5) makes the problem deﬁnitely unsolvable. In spite of that, is it possible to make a few basic statements? We will see in the following that also for interacting electrons the one-electron Green’s function can provide a large part of the interesting information:

ret,ad = † ret,ad Gkσ (E) ckσ ; ckσ E (B.142)

Completely equivalent to this is the one-electron spectral density:

+∞ 1 − i − † = − E(t t ) , Skσ (E) d(t t ) e ckσ (t) ckσ (t ) (B.143) 2π −∞ +

We want to determine the spectral density using the equation of motion of the Green’s function, where from now onwards, we consider only the retarded Green’s function and for convenience, drop the index ret. Starting point is the following commutator whose derivation is suggested as an exercise (Problem B.6): † , H = ε − μ + v [ckσ ]− ( (k) ) ckσ pk+q(q)cp+qσ cpσ ck+qσ (B.144) pqσ

With the higher Green’s function

σ σ † † Γ = pk;q(E) cp+qσ cpσ ck+qσ ; ckσ E (B.145) the equation of motion reads as − ε + μ = + v Γσ σ (E (k) ) Gkσ (E) pk+q(q) pk;q(E) (B.146) pqσ

The function Γ prevents the exact solution. A formal solution, however, is possible with the following separation B.4 The Quasiparticle Concept 547

, H − H † ≡! Σ · [ckσ 0]− ; ckσ E kσ (E) Gkσ (E) (B.147)

This equation deﬁnes the fundamental “self-energy” which, in general, is a complex function:

Σkσ (E) = Rkσ (E) + i · Ikσ (E) (B.148)

With this we can now write the Green’s function as follows: Gkσ (E) = (B.149) E − (ε(k) − μ + Σkσ (E))

One should note that in case the self-energy is real, one has to include the term + +i0 in the denominator. Let it be mentioned as a side remark that because of av ∗ = ret the KramersÐKronig relations (B.109) and (B.110), Gkσ (E) Gkσ (E) follows. Then it must also hold: Σav ∗ = Σret kσ (E) kσ (E) (B.150)

We can therefore restrict the discussion to only the retarded self-energy. For simplicity we from now on drop the index ret. If we switch off the interaction, then we get the Green’s function for the free (0) system (B.116) which we want to denote by Gkσ (E): G(0)(E) = (B.151) kσ E − (ε(k) − μ) + i0+

So the total inﬂuence of the interaction is contained in the self-energy, whose knowl- edge solves the many-body problem. If (B.150) is substituted in (B.149) then we have −1 (0) = + Σ Gkσ (E) Gkσ (E) kσ (E)Gkσ (E)

This leads to the “Dyson equation”:

(0) (0) 1 G σ (E) = G (E) + G (E) Σ σ (E)G σ (E) (B.152) k kσ kσ k k

This equation can be solved by iteration up to any given accuracy by knowing (approximately) the self-energy: ∞ n (0) (0) 1 (0) G σ (E) = G (E) + G (E) Σ σ (E)G (E) (B.153) k kσ kσ k kσ n=1 548 B The Method of Green’s Functions

B.4.2 Electronic Self-energy

We want to try to get a picture of the general structure of the fundamental many- body terms such as self-energy, Green’s function and spectral density, without being able to calculate explicitly the inﬂuence of the interaction. In doing this we want to restrict to the retarded functions whose relationship with other types is simple because of the KramersÐKronig relations (see (B.150)). First we rewrite a bit the formal solution for the Green’s function using (B.149):

[E − ε(k) + μ − Rkσ (E)] + iIkσ (E) G σ (E) = k − ε + μ − 2 + 2 [E (k) Rkσ (E)] Ikσ (E)

Skσ (E) is real and non-negative as one can easily see from the spectral representa- = = † tion (B.86) for the case A ckσ und B ckσ . Then holds (B.94)

Ikσ (E) S σ (E) =− (B.154) k π − ε + μ − 2 + 2 [E (k) Rkσ (E)] Ikσ (E)

Obviously for the imaginary part of the self-energy must hold

Ikσ (E) ≤ 0 (B.155)

In the following we want to analyse spectral density for the argument E − μ and we expect prominent maxima at points of resonances given by

! Enσ (k) − ε(k) − Rkσ (Enσ (k) − μ) = 0 n = 1, 2, ··· (B.156)

We must distinguish two cases:

(a) In the immediate E-neighbourhood of a resonance let

Ikσ (E − μ) ≡ 0 (B.157)

+ Then with Ikσ →−0 and the well-known expression

1 x δ(E − E0) = lim (B.158) → 2 2 π x 0 (E − E0) + x

for the delta function, the following representation holds for the spectral density:

Skσ (E − μ) = δ (E − ε(k) − Rkσ (E − μ)) (B.159)

For the case where there lies more than one resonance in the region under consideration (B.157), we use another well-known property of the delta function: B.4 The Quasiparticle Concept 549

1 δ ( f (x))) = δ(x − x ); f (x ) = 0 (B.160) | f (x )| i i i i

xi are the zeros of the function f (x). With this holds

n0 Skσ (E − μ) = αnσ (k) δ (E − Enσ (k)) (B.161) n=1

where the coefﬁcients αnσ (k) are known as spectral weights. −1 ∂ αnσ (k) = 1 − Rkσ (E − μ) (B.162) ∂ E E=Enσ (k)

Summation is over all resonances that lie in the region (B.157). Thus the spectral density appears as a linear combination of positively weighted delta functions in whose arguments the resonance energies appear. (b) Now let it hold

Ikσ (E − μ) ≡ 0 (B.163)

It may still be assumed that, however, in the immediate neighbourhood of a resonance holds:

|Ikσ (E − μ)| |ε(k) + Rkσ (E − μ)| (B.164)

Then a more or less prominent maximum is to be expected at E = Enσ (k).We therefore expand the bracket in the denominator of (B.154):

− ε − − μ Enσ (k) (k) Rkσ (Enσ (k) ) ∂ = 0 + (E − Enσ (k)) 1 − Rkσ (E − μ) ∂ E = E Enσ (k) 2 + O (E − Enσ (k))

We substitute this in (B.154) and also further assume that the imaginary part of the self-energy is a well-behaved function of energy, so that we can further simplify in the immediate neighbourhood of the resonance:

Ikσ (E − μ) ≈ Ikσ (Enσ (k) − μ) ≡ Inkσ (B.165)

This gives in the surroundings of resonance a Lorentzian structure of the spectral density:

2 α I σ S(n)(E − μ) ≈− nσ nk (B.166) kσ π 2 2 (E − Enσ (k)) + (αnσ Inkσ ) 550 B The Method of Green’s Functions

The results (B.161) and (B.166) lead ﬁnally to the classical quasiparticle picture. Under the assumptions (B.164) and (B.165) the spectral density is made up of one or more Lorentz curves or delta functions, whose widths and positions are to a large extent determined by the imaginary and the real parts of the self-energy, respectively. However, the assumptions made are not veriﬁable directly but only after the complete solution of the many-body problem. We want to physically interpret the general structure of the spectral density as depicted in Fig. B.6. For this, a Fourier transformation into the time domain is very useful:

+∞ 1 i − (E−μ)(t−t ) Skσ (t − t ) = dEe Skσ (E − μ) (B.167) 2π −∞

This acquires a particularly simple structure for the case (A). With (B.161) follows directly n0 1 i S σ (t − t ) = α σ (k)exp − (E σ (k) − μ)(t − t ) (B.168) k 2π n n n=1

Thus it is a sum of undamped oscillations with frequencies which correspond to the resonance energies. This is similar to the result (B.123) depicted in Fig. B.4 for the non-interacting electron system. The transformation of the function of type B is a little more complicated. For simplicity, we will assume that the Lorentz structure (B.166) is valid over the entire (n) − μ energy range for the spectral density Skσ (E ):

+∞ (n) 1 − i (E−μ)(t−t) S (t − t ) ≈ dEe α σ (k) ∗ kσ π 2 n !4 i −∞ ∗ 1 − − α E (Enσ (k) i nσ (k)Inkσ ) " 1 − (B.169) E − (Enσ (k) + iαnσ (k)Inkσ )

Ijkσ = 0 σ k S

Eiσ E σ Ekσ 0 j E Fig. B.6 Classical quasiparticle picture of the single-electron spectral density B.4 The Quasiparticle Concept 551

The evaluation is done using the residue theorem noting that αnσ (k)Inkσ ≤ 0. Therefore the first summand has a pole in the upper and the second summand in the lower half-plane. For (t − t) > 0 the integral in (B.169) is replaced by a contour integral where the path of integration runs along the real axis and closes with an infinite semicircle in the lower half-plane. Only the second summand in (B.169) then has a pole in the region enclosed by the integration path. The semicircle has no contribution because of the exponential function. For (t − t) < 0, for the same reason, the semicircle must be in the upper half-plane. In this region only the first summand in (B.169) has a pole. Then finally the residue theorem gives

π i (n) 2 i − (E σ (k)−μ)(t−t ) S (t − t ) ≈ α σ (k)e n ∗ kσ 4π 2i n 1 α − ∗ Θ t − t e nσ (k)Inkσ (t t ) − 1 α − +Θ t − t e nσ (k)Inkσ (t t )

This can obviously be summarized to (n) 1 i S (t − t ) ≈ α σ (k)exp − (E σ (k) − μ)(t − t ) ∗ kσ π n n 2 1 ∗ − α σ k I σ t − t exp n ( ) nk ( ) (B.170)

It now represents a “damped” oscillation. The frequency corresponds again to a resonance energy. The amount of damping is determined by the imaginary part Inkσ of the self-energy. Inkσ → 0 reproduces the result (B.108). The time-dependent spectral density Skσ (t−t ) for the interacting electron system consists of an overlap of damped and undamped oscillations with frequencies which correspond to the resonances Enσ (k). The resulting total time dependence can be naturally quite complicated. In the next section we want to make it clear what these (un)damped oscillations have to do with quasiparticles.

B.4.3 Quasiparticles

What does one understand by quasiparticles in many-body theory? It certainly has something to do with the resonance peaks discussed in the last section and shall at least be qualitatively explained here. For that, for simplicity, we consider the special case:

T = 0; |k| > kF ; t > t (B.171)

That means, we assume that there is something like a Fermi edge. In the case of non-interacting electrons it is just the Fermi wavevector, which is the radius of the Fermi sphere. |k| > kF here means only that at T = 0 the one-particle state 552 B The Method of Green’s Functions with |k| is unoccupied. Let the system be in the normalized ground state |E0 . At time t a(k,σ)-electron is introduced into the N-particle system. The resulting state

The spectral density is therefore the probability amplitude that the state that resulted by introducing of (k,σ)-electron at time t still exists at time t > t. That means it describes the “propagation” of an “extra electron” in an N-electron system. Simi- larly, for |k| < kF , the spectral density would describe the propagation of a hole. As typical limiting cases one recognizes • ϕ | ϕ 2 = . ↔ 0(t) 0(t ) const “stationary state” • ϕ | ϕ 2 = ↔ 0(t) 0(t ) (t−t→∞) 0 “state with ﬁnite lifetime”

First let us consider once again the case of non-interacting (band-)electrons of Sect. B.3.6.1 with the Hamiltonian (B.111) for the special case (B.171). Using the † | H commutator (B.115) we show that ckσ E0 is an eigenstate of 0 H † | = † H | + H , † | 0 ckσ E0 ckσ 0 E0 0 ckσ E0 − = + ε − μ † | (E0 (k) ) ckσ E0

With this we calculate

i † i |ϕ (t) =exp( H t) c exp(− H t)|E 0 0 kσ 0 0 i i † = exp(− E t)exp( H t) c |E 0 0 kσ 0 i i † = exp(− E t)exp( (E + ε(k) − μ) t) c |E 0 0 kσ 0

Then it holds

i |ϕ t = ε k − μ t |ϕ t = 0( ) exp( ( ( ) ) ) 0( 0) (B.174) B.4 The Quasiparticle Concept 553

Let us consider further the normalization of the state:

ϕ = | ϕ = = | † | 0(t 0) 0(t 0) E0 ckσ ckσ E0 = | − | † | E0 E0 E0 ckσ ckσ E0 = E0| E0 =1

We ﬁnally get for the probability amplitude

i ϕ (t)| ϕ (t) =exp(− (ε(k) − μ)(t − t)) = 2π S(0)(t − t) (B.175) 0 0 kσ

This naturally agrees exactly with the result (B.123) which we have found in Sect. B.3.6.1 using the equation of motion method for the free band electrons. It gives the undamped harmonic oscillations shown in Fig. B.4. In particular we have 2 ϕ0(t)| ϕ0(t ) = 1 (B.176)

Thus, for the case of free electrons it is a stationary state. This is not surprising |ϕ = † | H because 0 ckσ E0 turns out to be an eigenstate of 0. It is, however, no more the case for interacting (band-) electrons This one recognizes from the spectral representation of the spectral density. If one carries out the average over the ground state |E0 as is required by deﬁnition for T = 0, and then goes through exactly the same steps as in deriving (B.86), then one gets the spectral density † 2 i 2π S σ (t − t ) = E | c |E exp(− (E − E )(t − t )) (B.177) k n kσ 0 n 0 n

† | H In the free system, ckσ E0 is an eigenstate of 0, so that due to the orthogonality of the eigenstates, only one term in the sum contributes. That does not hold any more † | for the interacting system. ckσ E0 is no more an eigenstate, but can be expanded in terms of the eigenstates:

|ϕ = † | = γ | 0 ckσ E0 n En n where an arbitrary number, but at least two coefﬁcients γn are unequal zero. Every summand in (B.177) represents a harmonic oscillation but with different frequencies. The overlap sees to it that the sum is maximum for t = t. For increasing − − i − − t t the phase factors exp( (En E0)(t t )) distribute themselves over the entire unit circle in the complex plane and see to it that, because of the destructive interference, possibly a very complicated and certainly no more harmonic time- dependence results as depicted in Fig. B.7. The state created at t is not stationary 554 B The Method of Green’s Functions σ k S π ℜ 2 0

0 (t − t’)

Fig. B.7 Schematic plot of the time-dependent spectral density of an interacting electron system but has to some extent a ﬁnite lifetime. One could, however, imagine that the complicated time dependence could be simulated by a few weighted damped oscillations with well-deﬁned frequencies: i π S σ t − t = α σ k − η σ k − μ t − t 2 k ( ) n ( )exp( ( n ( ) )( )) (B.178) n

This expression formally has the same structure as for the free system (B.175) with, however, in general complex one-particle energies:

ηnσ (k) = Re ηnσ (k) + iImηnσ (k) (B.179)

In order to realize damping, one must have

Imηnσ (k) ≤ 0(t − t > 0) (B.180)

The ansatz (B.178) gives the impression as if the extra (k,σ)-electron decays into one or more quasiparticles with the following properties:

• Quasiparticle energy ⇔ Re ηnσ (k) −1 • Quasiparticle lifetime ⇔ · |Imηnσ (k)| • Quasiparticle weight (“spectral”) ⇔ αnσ (k) The lifetime is deﬁned here as the time that is required for the respective summand to decrease from its initial value by a fraction e. Because of the particle conservation, for the spectral weights of the quasiparticles αnσ (k) = 1 (B.181) n must still hold. Formally this follows from † 2π Skσ (0) = αnσ (k) = ckσ , c σ =1 k + n B.4 The Quasiparticle Concept 555

We will carry these interpretations now on to the results (B.108) and (B.110) of the last section

• Quasiparticle energy ⇔ Re Enσ (k)

! Enσ (k) = ε(k) + Rkσ (Enσ (k) − μ) (B.182)

• Quasiparticle lifetime ⇔ τnσ (k) τnσ (k) = (B.183) |αnσ (k) Ikσ (Enσ (k) − μ)|

• Quasiparticle weight (“spectral”) ⇔ αnσ (k) −1 ∂ αnσ (k) = 1 − Rkσ (E − μ) (B.184) ∂ E E=Enσ

The Lorentz-type peaks in the spectral weight are to be assigned to the quasiparticles, whose energies are given by the positions and their lifetimes by the widths of the peaks. The quasiparticle energies are given only by the real part and the lifetimes are given mainly by the imaginary part of the self-energy. Because of αnσ (k), in a limited way, the lifetime is of course influenced by the real part too. Ikσ = 0 always means an infinite lifetime. Delta functions in the spectral weight indicate stable, i.e. infinitely long-living quasiparticles. To conclude, it should be mentioned that these considerations for the classical quasiparticle picture are based on the preconditions (B.164) and (B.165) whose validity can be verified only after the complete solution of the many-body problem.

B.4.4 Quasiparticle Density of States

While discussing the free electrons as an application of the abstract Green’s function formalism in 3.6., we have learned about the important concept of quasiparticle density of states. Now we want to introduce this quantity for interacting electron system and understand its relation to the one-particle Green’s function and one- particle spectral density. The starting point is the average occupation number nkσ of the (k,σ)-level, which with the help of the spectral theorem can be expressed by the one-particle spectral density:

+∞ = † = 1 − μ nkσ ckσ ckσ dE f−(E)Skσ (E ) (B.185) −∞

Here f−(E) is again the Fermi function. A summation over all the wavevectors and both the spin directions gives the total number of electrons Ne 556 B The Method of Green’s Functions Ne = nkσ (B.186) kσ

Alternatively, the electron number can be expressed in terms of a density of states. ρˆσ (E)dE is the number of σ-states in the energy interval [E, E + dE] that can be occupied. Then, f−(E)ˆρσ (E)dE is the density of the occupied states. Therefore

+∞ Ne = f−(E)ˆρσ (E)dE (B.187) σ −∞ must hold. A non-degenerate energy band (s-band) contains 2N states, where N is the number of lattice sites. The factor 2 comes because of the two spin directions. For the completely occupied band ( f−(E) ≡ 1) therefore

+∞ dE f−(E)ˆρσ (E) = N −∞ holds. However, the density of states is usually normalized to one: ρσ (E) ≡ 1 ρ N ˆσ (E). A comparison of the two expressions for Ne then gives the “Quasiparticle density of states”

1 ρσ (E) = S σ (E − μ) (B.188) N k k

All the properties of the spectral density transfer to the quasiparticle density of states. It is in general dependent on temperature and particle number and naturally also depends on lattice structure. We have seen in Sect. B.2 that the one-electron spectral density has a direct relation to angle resolved photoemission. In contrast, the quasiparticle density of states is seen directly in angle averaged photoemission. We want to investigate the quasiparticle density of states for an illustrative special case. For that we consider a real, k-independent self-energy:

Rkσ (E) ≡ Rσ (E); Ikσ (E) ≡ 0 (B.189)

This corresponds to the case A of Sect. B.4.2. Therefore (B.159) holds:

Skσ (E − μ) = δ (E − Rσ (E − μ) − ε(k)) (B.190)

For the quasiparticle density of states this means

1 ρσ (E) = δ (E − Rσ (E − μ) − ε(k)) (B.191) N k

Comparing with the Bloch density of states for the non-interacting band electrons (B.121), we get in this special case B.4 The Quasiparticle Concept 557

ρσ (E) = ρ0(E − Rσ (E − μ)) (B.192)

ρσ (E) is unequal zero for such energies for which E − Rσ (E − μ) lies between the lower and the upper edge of the free Bloch band. If Rσ (E) is only a slowly varying smooth function of E, then ρσ (E) will only be slightly deformed from ρ0(E) (Fig. B.8). The inﬂuence of the particle interaction can possibly be taken into account by a renormalization of certain parameters. On the contrary, new kind of phenomena appear if E − Rσ (E − μ) is strongly structured, if as in Fig. B.9 the self-energy, e.g. shows a singularity in the interesting region. The result can be a band splitting which cannot be understood from the one-electron picture. At an appropriate band ﬁlling, it can happen that in one-electron picture (Bloch picture) the system is metallic, whereas in reality electronic correlations can make it an insulator. Such a system is called a “Mott–Hubbard insulator”.

Fig. B.8 Quasiparticle density of states for a “smooth” real part of the self-energy

Fig. B.9 Quasiparticle density of states of a self-energy with a singularity 558 B The Method of Green’s Functions

B.4.5 Thermodynamics

To conclude we want to show that the one-electron Green’s function (B.142) or the corresponding spectral density (B.143) can provide complete information regarding the macroscopic thermodynamics of the interacting electron system. We start with the “internal energy” which is deﬁned as the expectation value of the Hamiltonian (B.138): B A = = ε † + U H (k) ckσ ckσ kσ B A 1 † † + v , (q) c c c σ c σ 2 k p k+qσ p−qσ p k k,p,q,σ,σ (B.193)

We substitute q →−q and then k → k + q and use the symmetry relation (B.141). Then U can also be written as follows: B A = = ε † + U H (k) ckσ ckσ kσ B A 1 † † + v , + (q) c c c σ c + σ 2 p k q kσ p+qσ p k q k,p,q,σ,σ (B.194)

With the help of higher Green’s function (B.145) and the spectral theorem (B.95) it follows that # 1 +∞ dE 1 U = ε(k) − ImGkσ (E) −∞ exp(β E) + 1 π kσ ⎤ 1 1 σ σ ⎦ + v , + (q) − ImΓ (E) (B.195) 2 p k q π pk;q k,p,q,σ,σ

From the equation of motion we read off

1 1 σ σ v , + q − ImΓ E = p k q( ) π pk;q( ) 2 , , ,σ,σ k p q 1 1 = − Im((E − ε(k) + μ)G σ (E) − ) 2 π k kσ 1 1 = (E − ε(k) + μ) − ImG σ (E)) 2 π k kσ B.5 Problems 559

This we substitute in (B.195) 1 +∞ dE 1 U = (E + μ + ε(k)) − ImGkσ (E) (B.196) 2 −∞ exp(β E) + 1 π kσ

Once again substituting E → E − μ, we ﬁnally obtain 1 +∞ U = dE f−(E) (E + ε(k)) Skσ (E − μ) (B.197) 2 −∞ kσ

This is a very remarkable result because the contribution of a two-particle Coulomb interaction could be expressed in terms of one-particle spectral density. The result (B.197) was already formally obtained for the case of non-interacting band electrons (B.122). From U the “free energy” follows from the generally valid relation T U(T ) − U(0) F(T ) = U(0) − T dT (B.198) 2 0 T which we prove as problem (B.10). Therefore the whole of macroscopic thermodynamics is determined by the one-particle spectral density itself. In this context, for the various energy integrals, particularly the prominent quasiparticle peaks (see Fig. B.6) are important.

B.5 Problems

Problem B.1 Let A(t) be an arbitrary operator in the Heisenberg picture and ρ be the statistical operator: −β ρ = exp( H) Tr(exp(−β H))

Prove the Kubo identity: i β [A(t),ρ]− = ρ dλ Aú (t − iλ) 0

Assume that the Hamiltonian H is not explicitly time dependent! Problem B.2 With the help of the Kubo identity (Problem B.1) show that the retarded commutator Green’s function can be written as follows: 77 88 β 7 8 ret A(t); B(t) =− Θ(t − t) dλ Bú (t − iλ)A(t) 0 560 B The Method of Green’s Functions

Problem B.3 For the time-dependent correlation functions, show that

B(0) A(t + iβ) = A(t) B(0) holds provided the Hamiltonian does not explicitly depend on time. Problem B.4 Prove the following representation of the step function: +∞ − − Θ − = i exp( ix(t t )) (t t ) dx + 2π −∞ x + i0

Problem B.5 Show that a complex function F(E) has an analytical continuation in the upper (lower) half-plane, if its Fourier transform f (t)fort < 0(t > 0) vanishes.

Problem B.6 For an interacting electron system † 1 † † H = ε(k)c c σ + v (q)c c c σ c σ kσ k 2 kp k+qσ p−qσ p k kσ k,p,q σ,σ derive the equation of motion for the retarded one-particle Green’s function. Problem B.7 For a system of non-interacting electrons = ε − μ † H ( (k) ) ckσ ckσ kσ

(n) calculate all the spectral moments Mkσ and from there the exact spectral density. Problem B.8 The BCS theory of superconductivity can be carried out with the sim- pliﬁed Hamiltonian

2 † † Δ H = (ε(k) − μ) c c σ − Δ b + b + kσ k k k V kσ k

Here

† = † † bk ck↑c−k↓ is the “Cooper pair creation operator” and Δ = Δ∗ = = † V bk V bk k k B.5 Problems 561

, † 1. Calculate the commutation relations of the operators bk bk. Are the Cooper pairs Bosons? 2. Using the one-electron Green’s function (B.142) calculate the excitation spectrum of the superconductor. Show that it has a “gap” Δ. 3. Determine the equation satisﬁed by Δ! Problem B.9 1. For the superconductivity model in Problem B.8, calculate all the spectral moments of the one-electron spectral density. 2. Choose a two-pole ansatz for the spectral density

2 Skσ (E) = αiσ (k) δ (E − Eiσ (k)) i=1

and determine the spectral weights αiσ (k) and the quasiparticle energies Eiσ (k)! By inspecting the spectral moments show that the above ansatz is exact. Problem B.10 Prove the following relation between the internal and the free energies: T U(T , V ) − U(0, V ) F(T, V ) = U(0, V ) − T dT 2 0 T

Problem B.11 Let |E0 be the ground state of the interaction free electron system (Fermi sphere). Calculate the time dependence of the state

† |ψ = | 0 ckσ ck σ E0

Is it a stationary state? Problem B.12 For the one-electron Green’s function of an interacting electron system Gret(E − μ) = (γ>0) kσ − ε + E2 + γ | | E 2 (k) ε(k) i E may hold.

1. Determine the electronic self-energy Σkσ (E). 2. Calculate the energies and lifetimes of the quasiparticles. 3. Under what conditions is the classical quasiparticle concept applicable? 4. Calculate the effective masses of the quasiparticles. Problem B.13 For an interacting electron system, let the self-energy

aσ (E + μ − bσ ) Σσ (E) = (aσ , bσ , cσ positive, real ; cσ > bσ ) E + μ − cσ be calculated. For the density of states of the interaction free system holds 562 B The Method of Green’s Functions 1 for 0 ≤ E ≤ W ρ (E) = W 0 0 otherwise

Calculate the quasiparticle density of states. Is there a band splitting? Appendix C Solutions to Problems

Problem 1.1 Let us take Ri = 0 With ∇×(ϕa) = ϕ ∇×a − a ×∇ϕ follows:

(i) =∇× = ∇× − ×∇ jm (mi f (r)) f mi mi f ∇× mi = 0, since mi : particle property

Substitute:

1 = 3 × (i) mi d r r jm 2 1 3 = d r [r × (∇ f × mi )] 2 1 3 = d r [∇ f (r · mi ) − mi (r ·∇f )] 2 1 3 1 3 = d r ∇ f (r · mi ) − d r mi (div( f r) − fdiv r) 2 2 1 3 1 3 = d r ∇ f (r · mi ) − mi d rdiv( f r) 2 2 3 3 + mi d rf 2 ≡1, cond.2. 6 1 3 1 = d3r ∇ f (r · m ) + m − m dS · ( f r) 2 i 2 i 2 i ∂T ≡0, cond.1.

Intermediate result:

563 564 C Solutions to Problems 1 3 3 mi = d r ∇ f (r · mi ) + mi 2 2 1 1 3 ⇔−mi = d r [∇( f (r · mi )) − f ∇(r · mi )] 2 2 1 3 1 3 = d r ∇( f (r · mi )) − mi d rf 2 2 ≡1, cond.2. 1 3 ⇔ 0 = d r ∇( f (r · mi )) 2 6 1 = dS f (r · m ) = 0 q.e.d. 2 i ∂T

Problem 1.2 From the deﬁnition of the canonical partition function Z = Tr e−β H

The average magnetic moment is given by 1 d m = Tr − He −β H Z dB 0 1 1 ∂ = Tr e−β H β Z ∂ B0 ∂ = 1 1 Z β Z ∂ B0

From this follows the susceptibility: μ0 ∂ χT = m V ∂ B0 T μ ∂ 2 ∂2 = 0 − 1 Z + 1 Z β 2 ∂ ∂ 2 V Z B0 Z B0

The ﬁrst term is clear: 1 ∂ Z 2 = β2 m 2 (1.1) 2 Z ∂ B0

The second term is somewhat more complicated: C Solutions to Problems 565 2 ∂ Z ∂ ∂ H 1 = 1 −β −β H 2 Tr e Z ∂ B Z ∂ B0 ∂ B0 0 2 2 −β ∂ H ∂ H = Tr e−β H − β e−β H ∂ 2 ∂ Z B0 B0 2 2 β ∂ H = Tr e−β H Z ∂ B0 = β2 m2

In the third step we have exploited the condition that we are dealing with a perma- nent magnetic moment. Therefore for the susceptibility we have μ μ 0 2 2 1 0 2 χT = β m − m = (m − m ) (1.2) V kB T V

Problem 1.3 In a magnetic ﬁeld B, the magnetic dipole m has the potential energy

V =−m · B

It will therefore try to orient itself parallel to the field. If the field is only a homogeneous external field ⎛ ⎞ B0 ⎝ ⎠ B0 = B0ex = 0 0 then m will be oriented parallel to x-axis. According to elementary electrodynamics, the current creates an additional azimuthal field of the form I B = μ eϕ I 0 2πρ

It is convenient to use cylindrical coordinates:

x = ρ cos ϕ, y = ρ sin ϕ, z = z

ρ = x2 + y2 ⎛ ⎞ ⎛ ⎞ − sin ϕ −y ⎝ ⎠ 1 ⎝ ⎠ eϕ = cos ϕ = x ρ 0 0

Total ﬁeld 566 C Solutions to Problems ⎛ ⎞ ⎛ ⎞ −y 1 I B = B + B = B ⎝ 0 ⎠ + μ ⎝ x ⎠ 0 I 0 0 π 2 + 2 0 2 (x y ) 0

Field at the position r0 = (x0, 0, 0):

μ0 I B(r0) = B0ex + ey 2π x0

Dipole is oriented parallel to B, i.e. it makes an angle α given by

I tan α = μ0 2π x0 B0 which for small angles gives

I α ≈ μ0 2π x0 B0 with the x-axis.

Problem 1.4 For a magnetic system, we use the version (1.80) of the ﬁrst law. Then the work done is given by

δW = VB0dM = μ0VHdM

M: magnetization; V : constant volume (not a real thermodynamic variable!). From Curie law follows: C (dM) = dH T T CV ⇔ (δW) = μ HdH T 0 T H2 μ CV VT ⇔ Δ = δ = 0 2 − 2 = μ 2 − 2 ( W)12 ( W)T H2 H1 0 M2 M1 H1 2T 2C

Problem 1.5 1. ∂U c = M ∂ T M

follows directly from the ﬁrst law (1.80). Further holds with U = U(T, M) and N = const.: C Solutions to Problems 567

TdS = dU − VB0dM ∂U ∂U = dT + dM − VB dM ∂ ∂ 0 T M M T ! " ∂U ∂ M c − c = − VB H M ∂ 0 ∂ M T T H

Curie law: ∂ M C M2 =− H =− ∂ 2 T H T CH

Substitute ∂U ∂U ∂ M ∂ M c = + − VB H ∂ ∂ ∂ 0 ∂ T M M T T H T H ∂U ∂U ∂ M V = + + μ M2 ∂ ∂ ∂ 0 T M M T T H C

Since U = U(T, M) further holds ∂U ∂U dU = dT + dM ∂ ∂ T M M T ∂ ∂ ∂ ∂ U = U + U M ∂ ∂ ∂ ∂ T H T M M T T H ∂U V c = + μ M2 H ∂ 0 T H C

2. It holds ∂ ∂ ∂ M = M T ∂ ∂ ∂ H S T S H S

The two factors are determined separately. (a)First law (1.80):

dU = TdS+ V μ0 HdM

adiabatic means dS = 0. Therefore ∂U ∂U dU = dT + dM = V μ HdM ∂ ∂ 0 T M M T 568 C Solutions to Problems ∂U ∂U dT = V μ H − dM ∂ 0 ∂ T M S M T S ∂ M = cM ∂T μ − ∂U S V 0 H ∂ M T

(b)Once again the ﬁrst law for adiabatic changes of state

0 = dU − V μ0 HdM

With U = U(T, H) and M = M(T, H) follows: ! " ∂U ∂ M = − V μ H dT + 0 ∂ 0 ∂ T H T H ! " ∂U ∂ M + − V μ H dH ∂ 0 ∂ H T H T

According to part 1 the ﬁrst parenthesis is equal to cH . Therefore what remains is ! " ∂ M ∂U c dT = V μ H − dH H 0 ∂ ∂ H T H T ! " ∂U = V μ M − dH 0 ∂ H T

That means ∂U ∂T V μ0 M − = ∂ H T ∂ H S cH

This equation is combined with the ﬁnal result of part 1: ∂U ∂ M c V μo M − = M · ∂ H T ∂ H c μ − ∂U S H V 0 H ∂ M T

That was the original proposition!

Problem 1.6

∂ S ∂ S S = S T, H ⇔ dS = dT + dH =! ( ) ∂ ∂ 0 T H H T C Solutions to Problems 569

This means ∂ ∂T S T ∂ S =− ∂ H T =− ∂ H ∂ S c ∂ H S ∂T H H T

Free enthalpy (N = const.):

dG =−SdT − MVμ0dH

The corresponding Maxwell’s equation gives ∂ S ∂ M C = V μ =−V μ H ∂ 0 ∂ 0 2 H T T H T

With this it ﬁnally follows: ∂T C = μ V H ∂ 0 H S cH T

Problem 2.1 • L > S

L+S (2J + 1) J=|L−S| L+S = (2J + 1) J=L−S 1 = (2S + 1) [2(L + S) + 1 + 2(L − S) + 1] 2 = (2S + 1)(2L + 1)

• L < S

L+S (2J + 1) J=|L−S| L+S = (2J + 1) J=S−L 1 = (2L + 1) [2(S + L) + 1 + 2(S − L) + 1] 2 = (2S + 1)(2L + 1) 570 C Solutions to Problems

Problem 2.2 1. We obviously have 01 01 10 σ 2 = · = = 1l x 10 10 01 2 0 −i 0 −i 10 σ 2 = · = = 1l y i 0 i 0 01 2 10 10 10 σ 2 = · = = 1l z 0 −1 0 −1 01 2

2. Similarly one can easily get σ ,σ x y + + − = 01 · 0 i − 0 i · 01 10 i 0 i 0 10 i 0 −i 0 = + = 0 0 −i 0 i

σ ,σ y z + − − = 0 i · 10 + 10 · 0 i i 0 0 −1 0 −1 i 0 0 i 0 −i = + = 0 i 0 −i 0

[σ ,σ] z x + = 10 · 01 + 01 · 10 0 −1 10 10 0 −1 01 0 −1 = + = 0 −10 10 3. σ ,σ x y − − − = 01 · 0 i − 0 i 01 10 i 0 i 0 10 − = i 0 − i 0 0 −i 0 i 10 = 2i = 2iσ 0 −1 z

The other two components are obtained analogously. C Solutions to Problems 571

4. − σ σ σ = 01 · 0 i · 10 x y z 10 i 0 0 −1 = 01 · 0 i = i 0 10 i 0 0 i

= i1l 2

Problem 2.3 One immediately recognizes (i, j = x, y, z) σi σ j 0 αi α j = 0 σi σ j

With the commutation relations for the Pauli spin matrices (see Problem 2.1) follows σ ,σ α , α = i j + 0 i j + σ ,σ 0 i j + 1l 2 0 = 2δij = 2δij1l 4 01l2

One can further calculate σ −σ 0 i 1l 2 0 0 i αi β = = σi 0 0 −1l 2 σi 0 σ σ 1l 2 0 0 i 0 i β αi = = 0 −1l 2 σi 0 −σi 0 ⇒ α , β = i + 0

Finally, it remains to be veriﬁed 2 1l 2 0 1l 2 0 1l 2 0 β = = = 1l 4 0 −1l 2 0 −1l 2 01l2

So that (2.23), (2.24) and (2.25) are proved.

Problem 2.4 In solving the problem one uses the commutation relations for the Pauli spin matrices proved in Problem 2.2: 572 C Solutions to Problems 2 σ ,σ , = i j − 0 si s j − σ ,σ 4 0 i j − 2 ε σ = 2i k ijk k 0 ε σ 4 02i k ijk k 2 σk 0 = i εijk 2 0 σk k = i εijksk k

Problem 2.5 (0) HD is deﬁned in (2.38). s commutes with the momentum operator p.Wenowhave to calculate σ 0 0 σ 0 σ σ s α = i j = i j i j 0 σ σ 0 σ σ 0 2 i j 2 i j 0 σ j σi 0 0 σ j σi α j si = = 2 σ j 0 0 σi 2 σ j σi 0

From the result of Problem 2.3 this means σ σ α = 0 i j − si j − σ σ 2 i j − 0 ε σ = 02i k ijk k ε σ 2 2i k ijk k 0 0 σk = i εijk = i εijkαk σk 0 k k

So that we have , α · = ε α = × [si p]− i ijk p j k i (p a)i jk

[s, α · p]− = i (p ×a)

To this one ﬁnds σ σ i 0 1l 2 0 i 0 si · β = = = β ·si 2 0 σi o −1l 2 2 0 −σi

So that C Solutions to Problems 573 s, H (0) = ic (p × α) D −

For the orbital angular momentum we use the well-known commutation relations with the linear momentum , = ε li p j − i ijk pk k So that it follows: , α · = ε α = α × [li p]− i ijk j pk i ( p)i k With [li , β]− = 0 it eventually follows: l, H (0) = ic (α × p) D −

Problem 2.6 First we use the commutation relations from Problem 2.2: σ ,σ = ε σ i j − 2i ijk k k σ ,σ = δ i j + 2 ij1l 2

From this we get the important relation: σi σ j = δij1l 2 + i εijkσk k Due to the commutability it holds (σ · a)(σ · b) = ai b j σi σ j = ai b j δij + i εijkσk , , i j i j k = ai b j + i εijkai b j σk i ijk

= (a · b)1l2 + ia · (b × σ)

Since σ commutes with a and b, we can cyclically permute the operators in second summand (triple product!). Problem 2.7

2 HD = cα · (p + eA) + βmec − eϕ 574 C Solutions to Problems

Heisenberg’s equations of motion:

d i r =[r, H ]− = c[r, p]− · α = iαc dt D rú(t) = cα

d i (p + eA) dt ∂ = [(p + eA, H ]− + i (p + eA) = D ∂t ∂A = ce[p,α · A]− − e[p,ϕ]− + ec[A,α · p]− + ie = ∂t ∂A = ec [−α ·∇A +∇(α · A)] − e ∇ϕ + ie = i i ∂t ∂A = ec (α × (∇×A)) − e ∇ϕ + ie i i ∂t

d ∂A (p + eA) =−ec(α × B) + e ∇ϕ + dt ∂t

With ∂A E =−∇ϕ − ∂t ﬁnally follows:

d (p + eA) =−e(rú × B + E). dt On the right-hand side is the Lorentz force. Problem 2.8 1.

3 3 , = λ , = λ , = [HSB Li ]− L j S j Li − L j Li − S j j=1 j=1 3 = λ ε jikLk S j i = iλ εkjiLk S j = j=1 k jk

= iλ(L × S)i

=⇒ [HSB, L]− = iλ(L × S) C Solutions to Problems 575

3 3 , = λ , = λ , = [HSB Si ]− L j S j Si − L j S j Si − j=1 j=1 3 = λ L j i ε jikSk = iλ εkjiSk L j = j=1 k jk

= iλ(S × L)i

=⇒ [HSB, S]− = iλ(S × L)

3 3 , 2 = λ , 2 = λ , 2 = HSB L − Li S j L − Li L − Si 0 i=1 i=1

3 , 2 = λ , 2 = HSB S − Li Si S − 0 i=1

5. From 1 and 2 follows:

[HSB, Ji ]− = 0fori = x, y, z =⇒ , 2 = , 2 = HSB J − HSB Ji − 0 i

Problem 2.9

1. According to (2.162) it must hold

+k ε (k) −1 ε = ε (k) (k) Rz( )Tq Rz ( ) (Rz( ))qq Tq q=−k

Here we have according to (2.145)

i R ε = − εJ z( ) 1 z

and according to (2.158) 576 C Solutions to Problems 7 8 (k) (R (ε)) = kq |R (ε)| kq z q q & z % i = kq − εJ kq (1 z) i = δ − εqδ qq qq

= (1 − iεq)δqq k ε (k) = − ε (k) (Rz( ))qq (1 i q)Tq q=−k

On the other hand:

i i R ε T (k) R−1 ε = − εJ T (k) + εJ z( ) q z ( ) (1 z) q (1 z) i = T (k) − ε J , T (k) + O(ε2) q z q −

Compare: , (k) = (k) Jz Tq − qTq

2. According to (2.160) holds

(k) 1 (R (ε)) = δ − iε k(k + 1) − q(q + 1)δ , + x q q q q 2 q q 1 1 − iε k(k + 1) − q(q − 1)δ , − 2 q q 1

So that we have

+k (k) (k) (k) 1 (k) (R (ε)) T = T − iε k(k + 1) − q(q + 1)T x q q q q 2 q+1 q=−k 1 − iε k(k + 1) − q(q − 1)T (k) 2 q−1

On the other hand we have as in 1.:

i R (ε)T (k) R−1(ε) = T (k) − ε J , T (k) + O(ε2) x q x q x q −

Then a comparison gives C Solutions to Problems 577

J , T (k) = k(k + 1) − q(q + 1)T (k) x q − 2 q+1 + k(k + 1) − q(q − 1)T (k) 2 q−1 Completely analogously one ﬁnds J , T (k) =−i k(k + 1) − q(q + 1)T (k) y q − 2 q+1 + i k(k + 1) − q(q − 1)T (k) 2 q−1

With J± = Jx ± iJy follows the ﬁnal result: , (k) = + − ± (k) J± Tq − k(k 1) q(q 1)Tq±1

Problem 2.10 We use (2.150)

[(n · J), (n · K)]− = i(n × n) · Kn, n : unit vectors

1. n = ez If further n = ez is valid, we have

× = , = = · · (1) n n 0 [Jz Kz]− 0 0 K0

For n = ex we have

n × n = ey [Jz, Kx ]− = iK y

For n = ey we have × =− , =− n n ex Jz K y − i Kx

(1) 1 K± =∓√ (Kx ± iKy) 1 2

(1) 1 Jz, K± =∓√ i(K y ∓ iKx ) 1 − 2 1 (1) = √ (−Kx ∓ iKy) = · (±1) · K± 2 1 578 C Solutions to Problems

Therefore altogether we have

, (1) = (1) Jz Kq − qKq

This is the ﬁrst part of (2.163). √ √ 2. k(k + 1) − q(q ± 1) = 2 − q(q ± 1)

(a) q = 0:

K (1) = K 0 z , (1) = , = × · =− Jx K0 [Jx Kz]− i (ex ez) K i K y − , (1) = × · = Jy K0 i (ey ez) K i Kx − (1) J±, K =−iK y ± i(iKx ) 0 − =∓(K ± iK ) √ x y = (1) 2K±1

For q = 0 we therefore have

, (1) = − ± (1) J± Kq − 2 q(q 1)Kq±1

(b) q =+1:

(1) 1 K+ =−√ K+ 1 2 , (1) =−√1 , + Jx K+ Jx Kx iKy − 1 − 2 i =−√ Jx , K y 2 =√ (ex × ey) · K = √ Kz 2 2

analogously C Solutions to Problems 579 , (1) =−√1 , + Jy K+ Jy Kx iKy − 1 − 2 =−√1 , Jy Kx − 2 i =−√ (ey × ex ) · K = i √ Kz 2 2 (1) J+, K+ = 0 1 − √ √ (1) (1) J−, K+ = 2Kz = 2K 1 − 0

For q =+1 then holds , (1) = − ± (1) J± Kq − 2 q(q 1)Kq±1

(1) 1 K− = √ K− 1 2 , (1) = √1 − , Jx K− ( i) Jx K y − 1 − 2 = √ (ex × ey) · K = √ Kz 2 2 , (1) = √1 , Jy K− Jy Kx − 1 − 2 = i √ (ey × ex ) · K =−i √ Kz 2 2 (1) J±, K− = √ (Kz ± Kz) 1 − 2

For q =−1 we can write the following: , (1) = − ± (1) J± Kq − 2 q(q 1)Kq±1

2a, 2b and 2c together give the second part of (2.163): , (1) = − ± (1) J± Kq − 2 q(q 1)Kq±1

Problem 2.11 , (2) =! (2) 1. Iz Tq − qTq One immediately recognizes (2) 2 2 Iz, T = Iz, I − 3I = 0 0 − z − 580 C Solutions to Problems

On the other hand with a little more effort we have

√ < = (2) 1 Iz, T± =± 6 Iz, Iz I± + Iz, I± Iz 1 − − − 2 < = 1√ =± 6 I I , I± + I , I± I 2 z z − z − z 1 √ = 6 {I I± + I± I } 2 z z =± (2) T±1

√ (2) 1 2 Iz, T± =− 6 Iz, (I±) 2 − 2 − √ < = =−1 , + , 6 I± Iz I± − Iz I± − I± 2 √ 2 =−(±) 6(I±) =± (2) 2 T±2

Thus the ﬁrst commutation relation is obviously fulﬁlled √ , (2) =! − ± (2) 2. I± Tq − 6 q(q 1) qTq±1

, (2) = , 2 − , 2 I± T0 I± I − 3 I± Iz − − = − , − , 0 3Iz I± Iz − 3 I± Iz − Iz =±3(I I± + I± I ) z z √ 1√ = 6 ± 6(I I± + I± I ) 2 z z √ = (2) 6T±1

√ (2) 1 I+, T+ = 6 I+, Iz I+ + I+, I+ Iz 1 − − − 2 1√ = 6 I+, I I+ + I+ I+, I 2 z − z − √ 1 2 = 6(−2(I+) ) 2 = (2) 2 T+2 C Solutions to Problems 581 √ (2) 1 I+, T− =− 6 I+, Iz I− + I+, I− Iz 1 − 2 − − √ 1 2 =− 6 I 2I − I+ I− − I− I+ + 2I 2 z z z 1√ =− 2 − 2 − 2 6 (4Iz 2Ix 2Iy ) 2√ =− 6(3I 2 − I 2) √ z = (2) 6T0

√ (2) 1 I−, T+ =− 6 I−, Iz I+ + I−, I+ Iz 1 − 2 − − 1√ = 6 (I (−2I ) + I− I+ + I+ I− + (−2I )I ) 2 z z z z 1√ = − 2 + 2 + 2 6 ( 4Iz 2Ix 2Iy ) 2√ = 6(I 2 − 3I 2) √ z = (2) 6T0

√ (2) 1 I−, T− =− 6 I−, Iz I− + I−, I− Iz 1 − 2 − − √ 1 2 =− 6((I−) + I−(+I−)) 2 = (2) 2 T−2

√ (2) 1 2 I+, T+ =− 6 I+, (I+) = 0 2 − 2 −

√ (2) 1 2 I+, T− =− 6 I+, (I−) 2 − 2 − 1√ =− 6(I−2I + 2I I−) 2 z z = (2) 2 T−1 582 C Solutions to Problems √ (2) 1 2 I−, T+ =− 6 I−, (I+) 2 − 2 − 1√ =− 6(−2I+ I − 2I I+) 2 z z = (2) 2 T+1

√ (2) 1 2 I−, T− =− 6 I−, (I−) = 0 2 − 2 −

Problem 2.12

+2 +2 † = (2) · (2) = A tq Tq Ai q=−2 i=−2 = 2 − 2 2 − 2 A0 J 3Jz I 3Iz

= 2 2 − 2 2 − 2 2 + 2 2 A0 J I 3Jz I 3Iz J 9Jz Iz

3 ± = ± + ± ∓ + ∓ A 1 2 (Jz J J Jz)(Iz I I Iz)

A+1 + A−1 3 = (J J+ I I− + J J+ I− I + J+ J I I− + J+ J I− I + 2 z z z z z z z z + Jz J− Iz I+ + Jz J− I+ Iz + J− Jz Iz I+ + J− Jz I+ Iz) 3 = (J I (J+ I− + J− I+) + (J+ I− + J− I+) J I ) + 2 z z z z 3 3 + J (J+ I− + J− I+) I + (J+ (J I ) I− + J− (J I ) I+) 2 z z 2 z z z z = 3 (Jz Iz (J+ I− + J− I+) + (J+ I− + J− I+) Jz Iz) + 3 3 + (J+ I− Iz − J+ Iz I−) + (−J− I+ Iz + J− Iz I+) 2 2 = 6 Jz Iz Jx Ix + Jy Iy + Jx Ix + Jy Iy Jz Iz + + 3 , + 3 , J+ I− Iz − J− Iz I+ − 2 2 = + + 2 − 2 2 − + 2 + 6 Jx Ix Jy Iy Jz Iz 6Jz Iz 6 Jx Ix Jy Iy

3 2 (J+ I− + J− I+) 2 C Solutions to Problems 583

2 2 2 3 2 A+ + A− = 6 (J · I) − 6J I − (J+ I− + J− I+) 1 1 z z 2 2 2 + 3 (J · I) − 3 Jz Iz

3 2 2 ± = A 2 2 J± I∓

3 2 2 2 2 A+ + A− = J+ I− + J− I+ 2 2 2

We now sum the terms

= 2 2 − 2 2 − 2 2 + 2 2 + · 2 A J I 3Jz I 3Iz J 9Jz Iz 6 (J I)

2 2 3 2 − 6J I − (J+ I− + J− I+) + z z 2 2 2 3 2 2 2 2 + 3 (J · I) − 3 J I + J+ I− + J− I+ z z 2 = 6 (J · I)2 + 32 (J · I) − 2J 2 I 2 + D

= 2 2 − 2 2 − 2 2 + 2 2− D 3J I 3Jz I 3Iz J 3Jz Iz

2 3 − 3 Jz Iz (J+ J− I− I+ + J− J+ I+ I−) − 2 = 2 + 2 2 − 2 + 2 2 − 3 Jx Jy I 3 Jx Jy Iz 3Re (J+ J− I− I+) − 32 J I z z = 2 + 2 2 + 2 − 3 Jx Jy Ix Iy − 2 + 2 + , 2 + 2 − , 3Re Jx Jy i Jy Jx − Ix Iy i Iy Ix − 2 − 3 Jz Iz 2 =−3Re ((−iJz)(−iIz)) − 3 Jz Iz = 0

Then we have

A = 6 (J · I)2 + 32 (J · I) − 2J 2 I 2

This corresponds to the quadruple term (2.245). 584 C Solutions to Problems

Problem 2.13

1. Build

div(gfj) = (gf)divj + j · grad(gf) = j · grad( fg) = f j ·∇g + gj ·∇f

That is exactly the integrand of D. With Gauss theorem we have D = d3r div(gfj) = df · (gfj) = 0

S→∞

since j vanishes at inﬁnity. 2. Set f ≡ 1, g = x, y, z: 3 3 0 = D = d r j · ex,y,z = d rjx,y,z d3r j(r) = 0

3. Set f = xi , g = x j with xi, j ∈ {x, y, z}: Then with 1. we have 3 0 = d r(xi j j + x j ji ) 3 3 d rxj ji =− d rxi j j

Now let a be an arbitrary vector: 3 3 a · d r r ji (r) = a j d rxj ji (r) j 1 = a d3r (x j (r) − x j (r)) 2 j j i i j j 1 = a ε d3r (r × j) 2 j jik k j 1 =− ε a d3r (r × j) 2 ijk j k j 1 =− a × d3r(r × j) 2 i C Solutions to Problems 585

This is valid for all the components i: 1 a · d3r rj(r) =− a × d3r(r × j) 2

Problem 3.1 Because of the circular motion, every electron has an angular momentum and therefore a magnetic orbital momentum m(i). In the absence of an external field, the orientations of the electron moments are statistically distributed and hence compensate each other. After the application of a field, the angular momenta precess about the direction of the field while the motion of the electrons in the orbital planes remains unchanged. Because of the precession there is an extra current:

−e −eω ΔI = = L τ 2π which according to classical electrodynamics induces an additional magnetic moment:

− ω 2 (i) e L 2 e 2 2 Δm = ΔI F = πr ⊥e =− x + y B e i 2π i z 4m i i 0 z

Fi is the vector area of the circle along which the i th electron moves. Finally the (average) magnetization of the diamagnet (N, the number of atoms; V ,thevolume, Ne, the number of electrons per atom) is given by

N Ne Ne2 Ne ΔM = M = Δm(i) =− B r 2 V 6mV 0 i i=1 i=1

In calculating this we have used

N N N N e e e 1 e x2 = y2 = z2 = r 2 i i i 3 i i=1 i=1 i=1 i=1

Then the diamagnetic susceptibility is given by ∂ M Ne2μ Ne χ dia = μ =− 0 r 2 0 ∂ B 6mV i 0 T i=1

This expression agrees with the quantum mechanically correct expression (3.21). Since the calculation is not strictly classical, there is no contradiction with BohrÐ van Leeuwen theorem. The assumption of stationary electron orbits is classically untenable! 586 C Solutions to Problems

Problem 3.2 T = 0: ⎛ ⎞ ≤ kkF 2 2 1 ⎝ ⎠ 2 1 k εø = 2 ε(k) = N N Δk 2m k Spin k≤kF

(2π)3 Δk = grid volume : Volume per k-state in k-space. V

kF V 4π 2 V 2 1 V 1 εø = 2 dk k4 = k5 = ε k3 N 8π 3 2m N 2mπ 2 5 F F N 5π 2 F 0

3 = π 2 N With kF 3 V follows: 3 εø = ε 5 F

Problem 3.3

ˆ = ε + = ε H (k)akσ akσ (k)nˆ kσ kσ kσ 1. Statistical operator of the grand canonical ensemble

ρ = exp(−β(Hˆ − μNˆ )) (unnormalized)

ˆ = + = N akσ akσ nˆ kσ kσ kσ ˆ , ˆ = H N − 0 combined eigenstates (Fock states)

So that = ∞ ( nkσ N) Tr ρ = exp(−β (ε(k) − μ)nkσ )

N=0 {nkσ } kσ = ∞ ( nkσ N) . = exp(−β(ε(k) − μ)nkσ ), (nkσ = 0, 1) = { } σ N0 nkσ k . = ··· exp(−β(ε(k) − μ)nkσ )

{ σ } { σ } σ nk1 1 nk2 2 k C Solutions to Problems 587 ⎡ ⎤ 2 = ⎣ −β ε − μ ⎦ × exp( ( (k1) )nk1σ1 )

nk σ =0,1 ⎡ 1 1 ⎤ 2 × ⎣ −β ε − μ ⎦ ··· exp( ( (k2) )nk2σ2 ) σ = , nk2 2 0 1

2 2 = 1 + exp(−β(ε(k1) − μ)) 1 + exp(−β(ε(k2) − μ)) ···

. 2 Ξμ(T, V ) = 1 + exp(−β(ε(kν ) − μ))

kν

2. Average occupation number:

1 nˆ kσ = Sp (ρnˆ kσ ) Ξμ 1 = nˆ kσ 2 σ 1 1 ∂ =− Ξμ T, V β ∂ε ln ( ) 2 (k) 2 ln[1+exp(−β(ε(kν )−μ))] kν −β ε − μ = exp( ( (k) )) 1 + exp(−β(ε(k) − μ)) 1 = = f−(ε(k)) exp(β(ε(k) − μ)) + 1

3. Entropy:

∂ S = k T Ξμ B ∂ ( ln ) T = kB ln 1 + exp(−β(ε(k) − μ)) + kσ 1 exp(−β(ε(k) − μ))(ε(k) − μ) + k T B k T 2 1 + exp(−β(ε(k) − μ)) B kσ

∂μ ≈ 0 ∂T 588 C Solutions to Problems 1 S = kB ln + kB β (ε(k) − μ) < nˆ kσ > 1− < nˆ σ > kσ k

− β (ε(k) − μ) = ln exp(−β (ε(k) − μ)) exp(−β (ε(k) − μ)) = ln (1 + exp(−β (ε(k) − μ))) 1 + exp(−β (ε(k) − μ)) 1 = ln < nˆ σ > − ln k 1 + exp(−β (ε(k) − μ))

= ln < nˆ kσ > − ln {1− < nˆ kσ >}

S =−kB {ln(1− < nˆ kσ >)+ < nˆ kσ > ln < nˆ kσ > − kσ

− < nˆ kσ > ln(1− < nˆ kσ >)}

⎧ ⎨ =− − < > − < > + S kB ⎩(1 nˆ kσ ) ln(1 nˆ kσ ) kσ holes ⎫ ⎬ < > < > nˆ kσ ln nˆkσ ⎭ electrons

S = 0 for ﬁlled band (< nˆ kσ >≡ 1) 3rd law: (Behaviour for T → 0) T →0 T →0 T →0 ε(k) >μ: < nˆ kσ > −→ 0: ln(1− < nˆ kσ >) −→ 0 S −→ 0 T →0 T →0 ε(k) <μ: < nˆ kσ > −→ 1: ln< nˆ kσ > −→ 0 T →0 S −→ 0 3rd law is satisﬁed! Problem 3.4 Periodic boundary conditions grid volume

2π (2π)2 Δk(1) = ; Δk(2) = L L x L y

1. d = 1 Density of states:

2 d ρ(1)(E) = ϕ (E) 0 Δk(1) dE 1 C Solutions to Problems 589

Phase volume: √ + 1 2mE 2 √ ϕ E = dk = dk = mE 1( ) 2 √ ε(k)≤E − 1 2mE ( d 1 1 2m ϕ1(E) = √ 2m = dE 2mE E2

/ √ d · √1 , if E > 0 2m (1) 1 E ρ (E) = ; d1 = L 0 0, otherwise π

2. d = 2 2 2 2mE ϕ (E) = d k = πk |ε ≤ = π 2 (k) E 2 ε(k)≤E d 2mπ ϕ (E) = dE 2 2

/ (2) d2 > 0, if E > 0 L x L y m ρ (E) = ; d2 = · 0 0, otherwise π 2

3. d arbitrary: 2 ρ (E) dE = dd k 0 Δ(d)k E≤ε(k)≤E+dE

ε = 2k2 Δ(d) = (2π)d (k) 2m ; cell volume: (k) Vd Phase volume:

1 2 1 = 2m 2 k0 k0 2 E d d Ω 2 d−1 k0 d 2m d ϕ (E) = Ω dk k = Ω = E 2 d d d d d 2 0

Ωd : Surface of the d-dimensional unit sphere

density of states 590 C Solutions to Problems

V d ρ (E) = 2 d ϕ (E)Θ(E) 0 (2π)d dE d d Ω 2 d 2m d −1 = V E 2 Θ(E) d (2π)d 2

The surface Ωd is still to be determined! Gauss integral in the d-dimensional space:

+∞ .d √ d −r 2 −x2 d d gd = d re = dxi e i = ( π) = π 2 i=1−∞

Spherical coordinates:

∞ d−1 −x2 gd = Ωd dx x e 0 ∞ 1 d −1 −y = Ω dy y 2 e d 2 0 = 1Ω Γ d 2 d 2

= 2 = = 1 √dy (y x dy 2xdy dx 2 y ) by comparing:

d π 2 Ω = 2 d Γ d ( 2 )

Density of states

d d (d) V π 2 m d − ρ = d · 1 2 2 2 1Θ 0 (E) 2 (2π)d Γ d 2 E (E) ( 2 )

Check: 1. d = 1: ( (1) 1 L 2m ρ (E) = d1 √ Θ(E); d1 = 0 E π 2

2. d = 2: L L m ρ(2)(E) = d Θ(E); d = x y 0 2 2 π 2 C Solutions to Problems 591

3. d = 3: √ ρ(3) = Θ 0 (E) d3 E (E) 3 3 3 V π 2 2 2m 2 V 2m 2 d = √ = 3 4π 3 π 2 2π 2 2 Problem 3.5

ε(k) = ε(k) = ck

1. Density of states: 2V d ρ(E) = ϕ(E) π 3 (2 ) dE π π 3 4 3 4 3 ϕ(E) = d k = k | = = E 3 E c k 3c33 ε(k)≤E V ρ(E) = αE2Θ(E) ; α = π 23c3

Fermi energy:

1 N 3 ε = ck = c 3π 2 F F V

where kF is derived from the particle number: 2V V 4π N = d3k = · k3 (2π)3 4π 3 3 F k≤kF 1 N 3 k = 3π 2 F V

2. Chemical potential: With the help of the particle number:

εF +∞

N = dEρ(E) = dE f−(E)ρ(E) −∞ −∞ N(T =0) N(T ) μ α π 2 ε3 = dEρ(E) + (k T )2ρ(μ) + ... 3 F 6 B −∞ (Sommerfeld expansion) 592 C Solutions to Problems

1 1 π 2 ε3 = μ3 + (k T )2μ + ... 3 F 3 3 B / 0 1 k T 2 ≈ μ3 1 + π 2 B 3 μ / 0 1 k T 2 ≈ μ3 1 + π 2 B 3 εF

⎧ ⎫− 1 ⎪ ⎪ 3 ⎪ ⎪ ⎨ 2⎬ 2 kB T μ ≈ εF 1 + π ⎪ ε ⎪ ⎩⎪ F ⎭⎪ ≈10−4 / 0 2 2 π kB T ≈ εF 1 − 3 εF

The same structure as in the non-relativistic case, only the numerical factor in π 2 π 2 front of the correction term is changed from 12 to 3 3. Internal energy and heat capacity:

εF α U(T = 0) = dE E · ρ(E) = ε4 4 F −∞ μ π 2 U(T ) = dE E · ρ(E) + (k T )2(3αμ2) + ... 6 B −∞ α π 2 = μ4 + (k T )2(3αμ2) + ... 4 6 B / 0 α μ 4 k T 2 μ 2 = ε4 + π 2 B + ... F 2 4 εF εF εF / 0 2 2 2.) 4 k T k T = U(0) 1 − π 2 B + 2π 2 B + ... 3 εF εF

2 π 2 U(T ) ≈ U(0) 1 + 2 kB T 3 εF

= α ε3 α = 3N = 3 ε N F ε3 U(0) N F 3 F 4 C Solutions to Problems 593

Heat capacity

4π 2 k2 k2 c = γˆ T ;ˆγ = U(0) B = Nπ 2 B V ε2 ε 3 F F

Non-relativistic:

1 k2 γ = Nπ 2 B 2 εF which means

1 γˆ εnr 2k2 2 k N 3 = 2 F = 2 F = · F = 3π 2 << 1 γ εr F 2mc kF m c mc V

Problem 3.6 1. Degree of degeneracy of a Landau level (3.117):

eL L 2g (B ) = 2 x y B y 0 2π 0

(0) Factor 2 because of the spin degeneracy. B0 is determined from = (0) Ne 2gy B0

Therefore π (0) = B0 Ne eLx L y

2. The degeneracy of the Landau level is independent of the quantum number n. The ﬁrst n0 levels are then exactly fully occupied and the levels n > n0 are − = (n0 1) completely empty if B0 B0 so that N − e = (n0 1) 2gy B0 n0 − 1 (n0 1) = (0) B0 B0 n 0

3. The degree of degeneracy at an arbitrary ﬁeld B0 = (0) · B0 = B0 2gy(B0) 2gy B0 (0) Ne (0) B0 B0 594 C Solutions to Problems

Now let

− (n0 1) ≥ ≥ (n0) B0 B0 B0

n0 levels are fully occupied and the (n0 + 1)th level is partially occupied. The number of electrons in the fully occupied levels:

∗ = · = B0 N n0 2gy(B0) n0 Ne (0) B0

The number of electrons in the uppermost level: = − ∗ = − B0 Nn0 Ne N Ne 1 n0 (0) B0

Energy contribution of the highest level: 1 E = ω∗ n + · N n0 c 0 2 n0 = ω∗ + 1 − + 1 B0 Ne c n0 n0 n0 2 2 B(0) 0 = μ∗ + − + B0 Ne B B0 2n0 1 n0(2n0 1) (0) B0

Energy contribution of the fully occupied levels:

− n0 1 ∗ = ω∗ + 1 B0 E c n Ne 2 B(0) n=0 0 B 1 1 1 = ω∗ N 0 n − + n c e (0) 0 2 2 2 0 B0 B = μ∗ 0 2 Ne B B0 (0) n0 B0

Total energy: = + ∗ = μ∗ + − + B0 E(B0) En0 E Ne B B0 2n0 1 n0(n0 1) (0) B0

This gives the curve of Fig. 3.5. = (n0) 4. B0 B0 With (2) we have C Solutions to Problems 595

B(n0) 0 = 1 (0) B n0 + 1 0 B(n0) (n0) = μ∗ (n0) + − + 0 E B0 Ne B B0 2n0 1 n0(n0 1) (0) B0 = μ∗ (0) Ne B B0

that is, independent of n0, that means for all critical ﬁelds it is the same!

Problem 3.7

1. Energy levels (3.112):

1 2k2 E (k ) = 2μ B n + + z n z B 0 2 2m

Degeneracy (3.117):

eL L g (B ) = x y B y 0 2π 0

Partition function:

+∞ ∞ 1 Z1 = dkz gy(B0)exp[−β En(kz)] 2π/Lz −∞ n=0 ⎡ ⎤ +∞ 2 ∞ eV B0 p −βμ −β μ = ⎣ dp exp −β z ⎦ e B B0 e 2 B B0n (2π)2 z 2m −∞ n=0 ' −βμ π B B0 = eV B0 2 m e − βμ (2π)2 β 1 − e 2 B B0 / m 3 2 βμ B e Z = V B 0 μ = 1 2 B 2π β sinh(βμB B0) 2m

2. Free energy:

dF =−SdT− mdB0

(to magnetization’s work see Sect. 1.5) 596 C Solutions to Problems

∂ F ∂ ∂ m =− = kB T ln Z N = NkB T lnZ1 = ∂ B0 ∂ B0 ∂ B0 ∂ βμ B = Nk T B 0 B ∂ ln βμ B0 sinh( B B0) d sinhx =−Nμ ln B dx x x=βμB B0

In the bracket is the classical Langevin function (see Problem 4.6):

1 L(x) = cothx − x μB B0 m =−NμB L kB T negative sign induced magnetic moment is oriented opposite to the ﬁeld Diamagnetism

Problem 3.8 For f (x) one can write 1 1 f (x) = δ x − + δ x + if − 1 ≤ x ≤+1 2 2 with f (x) = f (x + 2) f (x) is thus periodic with the period 2 and is also symmetric

f (−x) = f (x) Then an ansatz for the Fourier series is possible

∞ f (x) = f0 + [am cos(mπx) + bm sin(mπx)] = m 1 1 +1 f0 = f (x) dx = 1 2 − 1 +1 am = f (x) cos(mπx) dx −1 = + = 0form 2p 1 2(−1)p for m = 2p

bm ≡ 0 , since f (x) symmetric

Then it follows: C Solutions to Problems 597

+∞ f (x) = 1 + 2(−1)p cos(2pπx) p=1 +∞ = 1 + (−1)p ei2pπx + e−i2pπx p=1 +∞ = (−1)pei2pπx p=−∞

Problem 4.1 q = q 1 1 d3rei(q−q )·r = d3r = 1 V V V V q = q

1 d3rei(q−q )·r = V V ! 1 L x L y Lz n − n = dx dy dz exp 2πi x x x+ V 0 0 0 L x " ny − n n − n y y + z z z L y Lz

, , = , , = Here we must have (nx ny nz) (nx ny nz). For example, let nx nx . Then the integral over x gives

! " L x n − n dx exp 2πi x x x L 0 ! x " L n − n L x = x exp 2πi x x x π − 2 i nx nx L x 0 = − ∈ Z 0 since nx nx

So that the proposition is proved. Problem 4.2

1. The solid is a three-dimensional periodic array of primitive unit cells VUC (a1 · (a2 × a3)) with the total volume

V = N1 N2 N3 (a1 · (a2 × a3))

Periodic boundary conditions for Bloch functions: 598 C Solutions to Problems

! ! ! ψk(r) = ψk(r + N1a1) = ψk(r + N2a2) = ψk(r + N3a3)

The Bloch functions are

ik·r ψk(r) = e uk(r)

where the amplitude functions have the periodicity of the lattice:

n uk(r) = uk(r + R )

Periodic boundary conditions therefore demand

· ! · ! · ! eik (N1a1) = eik (N2a2) = eik (N3a3) = 1

This means

k · (Ni ai ) = 2πzi with zi ∈ Z

Then for the allowed wavevectors holds:

z1 z1 z1 k = b1 + b2 + b3 N1 N2 N3

Here bi are the primitive translation vectors of the reciprocal lattice, deﬁned by

a2 × a3 ai · b j = 2πδij ⇐⇒ b1 = 2π ··· a1 · (a2 × a3)

The ﬁrst Brillouin zone ≡ WignerÐSeitz cell of the reciprocal lattice. Therefore for the wavevectors of the ﬁrst Brillouin zone holds

1 1 − N < z ≤+ N 2 i i 2 i

2. The proposition is valid for k = k, since

1 1 = 1 N n

k = k: According to 1. holds

3 n − · n = π j − (k k ) R 2 (z j z j ) N j j=1 C Solutions to Problems 599

So that we calculate ⎛ ⎞ 3 n n i(k−k )·R = ⎝ π j − ⎠ = e exp 2 i (z j z j ) N j n n1n2n3 j=1 − .3 Nj 1 = π n j − exp 2 i (z j z j ) N j j=1 n j =0

= z j z j at least for one j. Then we have

− Nj 1 n j N z j − z 1 − a j exp 2πi j = = 0 N j 1 − a n j =0

This holds because

z j − z a = exp 2πi j = 1 N j

= − < − < + since z j z j and in addition N j z j z j N j . On the other hand N j = π − = a exp 2 i(z j z j ) 1

− since z j z j is an integer. Thus the proposition is proved.

3. The proposition is trivial for Rn = Rm. Therefore let Rn = Rm:

With

3 n m z j k · R − R = 2π (n j − m j ) N j j=1

now holds ⎛ ⎞ 1.BZ 3 ik·(Rn−Rm) ⎝ z j ⎠ e = exp 2πi (n j − m j ) = N j k z1z2z3 j=1 / .3 12N j n j − m j = exp 2πizj N j j=1 z j =−1/2N j +1

n j = m j at least for one j. Then we have 600 C Solutions to Problems

/ 12N j n j − m j exp 2πizj N j z j =−1/2N j +1 − Nj 1 1 n j − m j = exp 2πi(p j − N j + 1) 2 N j p j =0 N −1 j n − m p j ∝ exp 2πi j j N j p j =0 1 − bN j = = 0 1 − b

since n − m b = exp 2πi j j = 1 N j

because n j = m j and −N j + 1 ≤ n j − m j ≤ N j − 1. On the other hand it holds N j b = exp 2πi(n j − m j ) = 1

because n j − m j is an integer. The proposition is thus proved. Problem 4.3 For both the integrals, it is meaningful (because of V →∞) to introduce relative and centre of mass coordinates: 1 x = r − r ; R = (r + r) 2 1 1 r = x + R ; r =− x + R 2 2 With the help of the Jacobi determinant one can show

d3rd3r = d3 Rd3x

e−α|r−r | I = d3r d3r 1 |r − r| V V e−αx ∞ = d3 R d3x = V · 4π dx x e−α x x 0 d ∞ = V · π − dx e−α x 4 α d 0 d 1 = V · 4π − dα α C Solutions to Problems 601

So that we have

4πV I = 1 α2

2. From the result of Problem 4.1 it follows directly: exp(i(q · r + q · r)) = 3 3 I2 d r d r V V |r − r | 1 i = d3 Rd3x exp (q − q) · x ∗ x 2 ∗ exp i(q + q) · R = V δq,−q · I

In order to calculateI it is advisable to introduce a factor that ensures convergence: 1 I = lim d3x eiq·x e−αx α→0+ x ∞ +1 = lim 2π dx x d cos ϑ eiqx cos ϑ e−αx α→ + 0 0 −1 ∞ x = lim 2π dx e−αx eiqx − e−iqx α→ + 0 0 iqx 2π ∞ = lim dx eiqx−αx − e−iqx−αx α→ + 0 iq 0 2π 1 ∞ 1 ∞ = lim eiqx−αx − e−iqx−αx α→0+ iq iq − α 0 −iq − α 0 2π −1 1 = lim − α→0+ iq iq − α iq + α π − π = 2 2 = 4 iq iq q2

Thus we ﬁnd

4πV I = δ ,− 2 q2 q q 602 C Solutions to Problems

Problem 4.4 First we calculate 3 ik·r I1(r) ≡ d ke FK +1 kF = 2π dx dk k2eikrx − 1 0 2π kF = dk k 2i sin(kr) ir !0 " 4π 1 k kF = sin(kr) − cos(kr) 2 r r r 0 k r cos(kr) − sin(k r) =−4π F F r 3

Then we have the intermediate result I 2(r) I = d3r 1 r ∞ 1 = (4π)3 k4 dx (x cos x − sin x)2 F 5 0 x

To calculate the integral we use integration by parts a number of times. ∞ 1 I = dx (x cos x − sin x)2 2 5 0 x ∞ 1 2 =− (x cos x − sin x) + 4x4 0 1 ∞ dx + 2(x cos x − sin x)(−x sin x) 4 4 0 x 1 ∞ dx =− (x cos x − sin x)sinx 3 2 0 x ∞ 1 = (x cos x − sin x)sinx + 4x2 0 1 ∞ dx + (x(sin2 x − cos2 x) + sin x cos x) 2 4 0 x 1 ∞ dx 1 = (−x cos 2x + sin 2x) 2 4 0 x 2 1 ∞ dy = (sin y − y cos y) 2 4 0 y C Solutions to Problems 603 ∞ ∞ 1 1 dy =− (sin y − y cos y) + y sin y 4y 0 4 0 y ∞ 1 1 sin y =− sin y = lim → 4y 0 4 y 0 y = 1 4

Then we ﬁnally have the integral

= π 3 4 I 16 kF

Problem 4.5

• S integer To show

+ + S S 1 n2 = 2 n2 = S(S + 1)(2S + 1) 3 n=−S n=1

Proof by full induction: The statement is certainly true for S = 1

+1 2 n2 = 2 n=1

We assume that it is valid for S and substitute S S + 1

S+1 +S 2 n2 = 2 n2 + 2(S + 1)2 n=1 n=1 1 = S(S + 1)(2S + 1) + 2(S + 1)2 3 1 = (S + 1)(S(2S + 1) + 6(S + 1)) 3 1 = (S + 1)(2S2 + 7S + 6) 3 1 = (S + 1)(S + 2)(2S + 3)) 3

Thus the proposition is proved! • S half-integer 604 C Solutions to Problems

+ S+ 1 S 2 1 2 n2 = 2 n − 2 n=−S n=1 S+ 1 S+ 1 S+ 1 2 2 1 2 = 2 n2 − 2 n + 1 2 n=1 n=1 n=1 1 1 3 = S + S + (2S + 2) 3 2 2 1 1 3 1 1 −2 S + S + + S + 2 !2 2 2 2 " 1 1 3 = S + S + (2S + 2) − (S + 1) 2 3 !2 " 1 3 3 = (2S + 1) (S + 1) S + − 3 2 2 1 = S(S + 1)(2S + 1) 3

This is the proposition!

Problem 4.6

1. Energy of a magnetic dipole in a magnetic ﬁeld

E =−μ · B N ⇔ H1 =−μ B cos ϑi i=1

ϑi is the angle between the magnetic moment of the ith atom μi and the ﬁeld B. 2. Canonical partition function 1 Z (T, B) = d3N qd3N pe−β H(q,p) N h3N N! , +1 Z N (T 0) N = (2π) d cos ϑ1 ··· (4π)N − 1 +1 N βμB = cos ϑi ··· d cos ϑN e i 1 −1 +1 N 1 βμBx = Z N (T, 0) dx e 2 −1 C Solutions to Problems 605 eβμB − e−βμB N = Z (T, 0) N 2βμB sinh βμB N = Z (T, 0) N βμB

μi = μ (sin ϑi cos ϕi , sin ϑi sin ϕi , cos ϑi )

Average value: 3N 3N −β H(q,p) d qd p μi e m = i d3N qd3N pe−β H(q,p)

The ϕi -integrations give mx = m y = 0 and therefore

d m = m e m = ln Z (T, B) z d(β B) N

Partition function from 2. d m = N ln sinh μx − ln μx (x = β B) dx cosh μx 1 = Nμ − (x = β B) sinh μx μx

m = Nμ L(βμB)ez

1 L(x) = coth x − x

L(x) is the Langevin function. This is the classical Langevin paramagnetism. 4. Low temperatures, strong ﬁelds: βμB 1 That means

1 coth βμB → 1; → 0 βμB

The system is in saturation:

m ≈ Nμez 606 C Solutions to Problems

High temperatures, weak ﬁelds: βμB 1

1 x coth x = + + O(x3) x 3 x L(x) ≈ 3 This result is the Curie law: μB m ≈ Nμ ez 3kB T

Problem 4.7 1. Free energy (N = const.):

F = U − TS

dF = dU − TdS− SdT =−SdT + VB0dM

∂ F = μ VH ∂ 0 M T ∂2 F ∂ H μ V μ V = μ V = 0 = 0 ∂ M2 0 ∂ M ∂ M χ T T ∂ H T T

From the last equation follows: ∂ F M = μ V + f T ∂ 0 χ ( ) M T T

since χT is only a function of temperature. The comparison with the above expression requires f (T ) ≡ 0. Then we have

1 M2 F(T, M) = F(T, 0) + μ0V 2 χT

2. Entropy ∂ F 1 d S(T, M) =− = S(T, 0) − μ VM2 χ −1 ∂ 0 T T M 2 dT

Internal energy C Solutions to Problems 607

, = , + , U(T M) F(T M) TS(T M) 1 d = U(T, 0) + μ VM2 χ −1 − T χ −1 2 0 T dT T U(T, 0) = F(T, 0) + TS(T, 0)

We assume that the Curie law is valid: C T d T χ = ⇔ χ −1 = ⇔ T χ −1 = T T T C dT T C Then we have

U(T, M) = U(T, 0) 1 1 S(T, M) = S(T, 0) − μ VM2 2 0 C 1 T F(T, M) = F(T, 0) + μ M2 2 0 C 3. Third law: This requires that for T → 0 entropy vanishes, independent of the value of the second variable. We therefore must assume in particular

lim S(T, 0) = 0 T →0 If we take M(T ) = 0ifH = 0, then according to the entropy obtained in 2. it must be concluded that if the Curie law is valid, then the third law is violated. On the contrary, it must further hold

d dχ χ −1 = 1 T =! lim T (T ) lim 0 T →0 T →0 χ 2 dT T dT χ −1 → = + O 2 T (T 0) c (T )

with a constant c = 0. χT (T ) therefore should not diverge, specially for T → 0.

Problem 4.8 1. Internal energy U = U(T, M) and the equation of state in the form M = f (T, H) are given. First law (1.80)

dU = δQ + μ0VHdM ∂U c = M ∂ T M

So that it holds 608 C Solutions to Problems ! " ∂U δQ = c dT + − μ VH dM M ∂ 0 M T

! " δQ ∂U ∂ M c = = c + − μ VH H M ∂ 0 ∂ dT H M T T H

Therefore ! " ∂U ∂ M c − c = − μ VH H M ∂ 0 ∂ M T T H

2. Ideal paramagnet:

C ∂U M = H = C Curie constant ; ∂ 0 : T M T

∂ M C =− H ∂ 2 T H T μ V c − c = 0 M2 ≥ 0 H M C

3. (a)Maxwell relation obtained from the free energy is

=− + μ dF SdT 0VHdM ∂ S ∂ H =−μ V ∂ 0 ∂ M T T M

(b)Maxwell relation of the free enthalpy is

=− − μ dG SdT 0VMdH ∂ S ∂ M = μ V ∂ 0 ∂ H T T H

(c)According to 1. from the ﬁrst law follows: ! " ∂U δQ = TdS = c dT + − μ VH dM M ∂ 0 M T

! " ∂ S ∂U T = − μ VH ∂ ∂ 0 M T M T C Solutions to Problems 609

This is the proposition! 4. ! " 1.) ∂U ∂ M c − c = − μ VH H M ∂ 0 ∂ M T T H 3.c) ∂ S ∂ M = T ∂ ∂ M T T H 3.a) ∂ H ∂ M =−μ VT 0 ∂ ∂ T M T H

5. ∂ H M M ∂ M = c − c =−μ VT ∂ H M 0 ∂ T M C C T H

Equation of state:

M 1 dH = dT + (T − T )dM + 3bM2dM C C C

! " ∂ M 1 M T − T + bM2 =− ∂ ( C ) 3 T H C C

So that it follows:

TM2 c − c = μ V H M 0 2 2 C(T − TC ) + 3bC M

6. ∂c ∂ ∂ S M = T ∂ ∂ ∂ M T M T M T ∂ ∂ S = T ∂ ∂ T M T M 2 3a.) ∂ H = T (−μ V ) 0 ∂ 2 T M

= 0

7. Because of 6. it holds: 610 C Solutions to Problems ∂U = c T ∂ M ( ) T M

According to 3. it also holds ∂U ∂ S = T + μ VH ∂T ∂ M 0 M T ∂ H =−μ VT + μ VH 0 ∂ 0 T M M 1 =−μ VT + μ V (T − T )M 0 C 0 C C + μ 3 0VbM T = μ V bM3 − C M 0 C

Integration: 1 T U(T, M) = μ V bM4 − C M2 + f (T ) 0 4 2 C

Then it must be

cM (T ) = f (T )

So that we have 1 T U(T, M) = μ V bM4 − C M2 0 4 2 C T + cM (T )dT + U0

One gets entropy analogously: ∂ S 1 = c T ∂ M ( ) T M T

∂ S 3a.) ∂ H M =−μ V =−μ V ∂ 0 ∂ 0 M T T M C

M2 T c (T ) S(T, M) =−μ V + M dT + S 0 2 C T 0 C Solutions to Problems 611

That means for the free energy

F = U − TS 2 1 4 M = F0 + μ0VbM + μ0V (T − TC ) 4 2C T c (T ) + M dT T F0 = U0 − TS0

1 H = (T − T )M + bM3 C C

For H → 0 we get the trivial solution M = 0, but also ( 1 M =± (T − T ) S bC C

These solutions are real for T ≤ TC . For the free energy we take the result from 7.:

μ V 1 F(T, M) = f (T ) + 0 (T − T )M2 + μ VbM4 2C C 4 0

We substitute in this expression both the mathematical solutions:

F(T, M = 0) = f (T ) μ V 1 F(T ; M =±M ) = f (T ) + 0 (T − T ) (T − T ) S 2C C bC C 1 1 + μ Vb (T − T )2 4 0 b2C2 C 1 μ V = f (T ) − 0 (T − T )2 4 bC2 C

It obviously holds

F(T, M =±MS) ≤ F(T, M = 0)

The equality sign holds for T = TC . Therefore the ferromagnetic solution is stable. This exists as a real solution only for T ≤ TC . 9. Magnetic susceptibility 612 C Solutions to Problems ∂ M 1 5.) 1 χT = = = ∂ H ∂ H 1 − + 2 T ∂ M T C (T TC ) 3bM 1 C lim χT = = H→0 1 − + 2 2(T − T ) C (T TC ) 3bMS C

< χT diverges for T → TC , as required by the general theory of phase transitions. Finally we again use the result from 5.:

2 TMS lim (cH − cM ) = μ0V H→0 − + 2 2 C(T TC ) 3bC MS 1 − T bC (TC T ) = μ0V C(T − TC ) + 3C(TC − T ) μ V = 0 T bC2 For T → 0 the two heat capacities are therefore equal to zero as demanded by the third law.

Problem 4.9 1. According to Problem 4.7 we have

U(T, M) = U(T, 0) Since ∂U c = = γ T M ∂ T M=0 follows: 1 U(T, M) = U + γ T 2 0 2 Isothermal change of ﬁeld from 0 to H = 0:

ΔU = 0 That means H ΔQ =−ΔW =−V μ0 H dH 0 C dM = dH T 1 V μ C H μ VC ΔQ =− 0 H dH =− 0 H 2 < 0 T1 0 2T1 C Solutions to Problems 613

2. Adiabatic and reversible means: dS = 0, therefore

! S(T1, H) = S(T f , 0)

From Problem 4.7 we have

1 1 2 S(T, M) = S(T, 0) − μ0V M 2 C T , − = cM=0(T ) = γ S(T 0) S0 dT T 0 T 1 1 S(T, M) = S + γ T − μ V M2 0 2 0 C 1 H 2 S(T, H) = S + γ T − μ CV 0 2 0 T 2

Switching off the ﬁeld

2 , = + γ − 1μ H S(T1 H) S0 T1 0CV 2 2 T1 ! = S(T f , 0) = S0 + γ T f

ﬁnal temperature:

1 H 2 T = T − μ CV < T f 1 0 γ 2 1 2 T1

Cooling effect.

Problem 5.1 Commutation relations:

1. +, − = 2 + , + Si Si − ci↑ci↓ ci↓ci↑ − = 2 + − − + − ci↑(1 ni↓)ci↑ ci↓(1 ni↑)ci↓ 2 = n ↑(1 − n ↓) − n ↓(1 − n ↑) i i i i 2 = ni↑ − ni↓ = z 2 Si 614 C Solutions to Problems z + 1 2 + + S , S = ni↑, c ↑ci↓ − ni↓, c ↑ci↓ i i − i − i − 2 1 = 2 − + − + − (ni↑ ni↓)ci↑ci↓ ci↑ci↓(ni↑ ni↓) 2 ⎛ ⎞ ⎜ ⎟ 1 2 ⎜ + + + + ⎟ = ⎝ni↑c ↑ ci↓ − c ↑ ni↓ci↓ − c ↑ni↑ ci↓ + c ↑ ci↓ni↓⎠ 2 i i i i + ≡0 ≡0 ci↓ ci↑ = 2 + ci↑ci↓ = + Si

1 z, − = 2 − + − + − Si Si − (ni↑ ni↓)ci↓ci↑ ci↓ci↑(ni↑ ni↓) 2 ⎛ ⎞ ⎜ ⎟ 1 2 ⎜ + + ⎟ = ⎝− ni↓c ↓ ci↑ − c ↓ ci↑ni↑⎠ 2 i i + ci↑ ci↓ =−2 + ci↓ci↑ =− − Si

2 = x 2 + y2 + z2 Si Si Si Si = 1 + − + − + + z2 Si Si Si Si Si 2 ⎛ ⎞

2 ⎜ ⎟ 2 ⎜ + + + + ⎟ 2 2 = ⎝c ↑ci↓c ↓ ci↑ + c ↓ci↑c ↑ci↓⎠ + ( n ↑ + n ↓ −2ni↑ni↓) 2 i i i i 4 i i n ↑(1−n ↓) n ↓(1−n ↑) ni↑ ni↓ i i i i 2 3 3 = (n ↑ + n ↓) − n ↑n ↓ 4 i i 2 i i

3 2 = (n ↑ + n ↓ − 2n ↑n ↓) 4 i i i i 3 2 2 = (n ↑ − n ↓) 4 i i 1 =! 2 S(S + 1) (S = ) 2

only if there is exactly one electron per lattice site C Solutions to Problems 615

2. 1 + −iqRi +ikR j + cqσ , c σ = e ciσ , c σ k + N j + i, j δijδσσ 1 −i(q−k)R = δσσ e i N i

= δq,kδσσ

+ + , = , = cqσ ckσ + cqσ ckσ + 0

Problem 5.2 Free energy:

F = U − TS; dF =−SdT + μ0VHdM

Integrability condition: ∂ S ∂ H =−μ V ∂ 0 ∂ M T T M

CurieÐWeiss law:

C M = H (C : Curie constant) T − TC

Heat capacity: ∂ S c (T, M) = T M ∂T !M " ∂c ∂2 S ∂ ∂ S M = T = T ∂ M ∂ M∂T ∂T ∂ M T ! " ∂ ∂ H =−μ VT 0 ∂ ∂ T T M =0

cM (T, M) ≡ cM (T )

Internal energy: Equation (1.80): dU = TdS+ μ0VHdM 616 C Solutions to Problems ∂U ∂ S = T = c (T ) ∂T ∂T M M M ∂U ∂ S = T + μ VH ∂ M ∂ M 0 T T ∂ H s.o.=−μ VT + μ VH 0 ∂ 0 T M M T − T =−μ VT + μ V C M 0 C 0 C M =−μ V T 0 C C

This means

M2 U(T, M) =−μ VT + f (T ) 0 C 2C

Because of ∂U = f T = c T ∂ ( ) M ( ) T M altogether we have

T M2 U(T, M) = c (T )dT − μ VT + U M 0 C 2C 0 0

Entropy:

T c (T ) S(T, M) = M dT + σ(M) T 0 ∂ S ∂ H M = σ M =−μ V =−μ V ∂ ( ) 0 ∂ 0 M T T M C M2 σ(M) =−μ V + σ 0 2C 0 T c (T ) M2 S(T, M) = σ + M dT − μ V 0 T 0 2C 0

Free energy: C Solutions to Problems 617

F(T, M) =U(T, M) − TS(T, M) T T μ V =F (T ) + c (T ) 1 − dT + 0 M2(T − T ) 0 M T 2C C 0

F0(T ) = U0 − T σ0 Check: ∂ F(T, M) S T, M =− ( ) ∂ T M

Free enthalpy:

G(T, H)

=F − μ0VMH μ V =F − 0 (T − T )M2 C C T T μ V =F (T ) + c (T ) 1 − dT − 0 M2(T − T ) 0 M T 2C C 0 G(T, H) T T 1 1 = + − − μ 2 F0(T ) cM (T ) 1 dT 0VC H T 2 T − TC 0

Problem 5.3 1. ∂ S c T, M = = T M ( 0) ∂ T M=0 T c (T , M = 0) S(T, 0) = M dT = γ T T 0

With the result of Problem 5.2

M2 S(T, M) = γ T − μ V + σ 0 2C 0 Because 618 C Solutions to Problems

C M = H T − TC

directly follows:

1 H 2 S(T, H) = γ T − μ CV + σ 0 2 0 2 (T − Tc)

For the free energy F we use ∂ F =−S T, M ∂ ( ) T M 1 M2 F(T, M) = F (M) − γ T 2 + μ V T − σ T 0 2 0 2C 0 According to the considerations of Problem 5.2 we must have

1 μ V F(T, M) = F (T ) + γ T 2 − γ T 2 + 0 M2(T − T ) 0 2 2C C μ V F (M) = U − 0 M2T 0 0 2C C What remains is only

1 μ V F(T, M) = U − γ T 2 + 0 M2(T − T ) 0 2 2C C Internal energy again from Problem 5.2:

1 M2 U(T, M) = U + γ T 2 − μ VT 0 2 0 C 2C Check:

F(T, M) = U(T, M) − TS(T, M) 1 μ V = U − γ T 2 + 0 M2(T − T ) − σ T 0 2 2C C 0 σ0 = 0

2. Heat capacities ∂ S c (T, M) = T = γ T = c (T, M = 0) M ∂T M M ∂ S TH2 c (T, H) = T = γ T + μ CV H ∂ 0 − 3 T H (T TC ) C Solutions to Problems 619

Since T > TC ,wehavecH ≥ cM . From elementary thermodynamics

χ , = χ cM = C S(T H) T μ 2 − + 0CV H cH T TC 2 γ (T −TC )

Problem 5.4

1. Spontaneous sub-lattice magnetization (B0 = 0)

M1S(T ) =−M2S(T ) (i) = μ λ − ρ BA 0( )MiS(T ) = ∗ β μ μ λ − ρ MiS(T ) M0 BJ ( gJ J 0 B ( )MiS(T ))

T → Tc MiS will be very small

J + 1 M = (n∗g Jμ )βg Jμ μ (λ − ρ)M iS 3J J B J 0 B iS

μ μ2 Curie constant: C = n 0 B g2 J(J + 1); n∗ = N = 1 n 3kB J 2V 2

=∼ 1 λ − ρ MiS MiS 2 ( )C T

Condition for intersection at ﬁnite Mis:

M is

x M is =M is M 0

* M i σ=M 0 Bj (...)

M is M is

d (M0 B j (...)) ≥ 1 dM is Mis=0 620 C Solutions to Problems

1 C (λ − ρ) ≥ 1 2 T =⇒ = 1 λ − ρ TN 2 ( )C

2. High temperature behaviour: No spontaneous sub-lattice magnetization

βμB B0 << 1

J + 1 M (T, B ) =∼ (n∗g Jμ )βg Jμ (B + B(i)) i 0 3J J B J B 0 A = 1 C + (i) 1 (B0 BA ) 2 T μ0

Total magnetization:

M(T, B0) = M1(T, B0) + M2(T, B0) = μ C + 1 C (1) + (2) 1 0 B0 (BA BA ) T 2 T μ0 μ0(λ+ρ)(M1+M2)

1 C C M(T, B0) 1 − (λ + ρ) = B0 2 T μ0T C 1 M(T, B ) = B 0 − 1 λ + ρ μ 0 T 2 C( ) 0 Curie–Weiss law ∂ M C χ T = μ = ( ) 0 ∂ − Θ B0 T T

Θ = 1 λ + ρ 2 C( ) paramagnetic Curie temperature

3. necessary: ρ<0 Θ λ + ρ |ρ|−λ − = = TN ρ − λ |ρ|+λ

(α) λ>0: ferromagnetic coupling within the sub-lattice Θ − < 1 i.a.: Θ < 0(λ<|ρ|) TN C Solutions to Problems 621

Example: EuTe: Θ =−4.0K ; TN = 9.6K (β) λ<0: antiferromagnetic coupling in the sub-lattice in order to have TN > 0: |λ| < |ρ|

Θ |ρ|+|λ| − = always: Θ < 0 TN |ρ|−|λ| Θ − > 1 TN

Examples:

MnO NiO MnF2 − Θ 5.3 5.7 1.7 TN Typically: Θ < 0

χ−1

Θ TN T

Problem 5.5 According to Problem 4.6 the equation of state of Weiss ferromagnet reads μ0 H + λμ0 M M = M0 L m kB T

N 2 mλμ M M m λμ0 3k Cλ 0 = V = Mˆ B kB T M0 kB T kB T

Classical Curie constant (Problem 4.6)

N m2 C = μ0 V 3kB 622 C Solutions to Problems

Further using (5.13): TC = λC and then we have Mˆ Mˆ = L b + 3 ε + 1

2. Series expansion of the Langevin function

1 1 L(x) = x − x3 + O(x5) 3 45 B0 = μ0 H = 0 b = 0 < T → Tc Mˆ very small

Then we can write approximately

Mˆ 3 Mˆ 3 Mˆ ≈ − ε + 1 5 (ε + 1)3

For Mˆ = 0 that means

ε 3 Mˆ 2 5 ≈− Mˆ 2 ≈− ε(ε + 1)2 ε + 1 5 (ε + 1)3 3

2 Since (ε + 1) → 1forT → TC , it follows that ( 5 1 Mˆ ∼ (−ε) 2 3

The critical exponent β of the order parameter then has the classical value:

β = 1 2

3. Critical isotherm: T = Tc; B0 → 0

ε = 0 and Mˆ as well as b very small.

This means C Solutions to Problems 623

1 1 Mˆ ≈ b + Mˆ − (b + 3Mˆ )3 3 45 3 1 15b ≈ (b + 3Mˆ ) ↔ b + 3Mˆ ≈ (15b) 3 1 1 ↔ 3Mˆ ≈ (15b) 3 − b ≈ (15b) 3 , since b → 0 9 b ∼ Mˆ 3 5

From this we read off the critical exponent:

δ = 3

That is also well-known value for classical theories. 4. Susceptibility:

∂ μ ∂ ˆ χ = M = M0 0m M T ∂ ∂ H T kB T b T,b=0 N m2μ ∂ Mˆ = V 0 kB (ε + 1)Tc ∂b T,b=0 ∂ ˆ = 3 M λ(ε + 1) ∂b T,b=0

In the critical region Mˆ is very small. Therefore we can expand ∂ L ∂x 1 1 = − x2 ∂ ∂ b b=0 b 3 15 b=0 ∂ ∂ ˆ x = + 3 M ∂ 1 ε + ∂ b 1 b ∂ Mˆ 3 ∂ Mˆ 1 1 9Mˆ 2 = 1 + − ∂b ε + 1 ∂b 3 15 (ε + 1)2 b=0 / b=0 0 ∂ Mˆ 1 9 Mˆ 2 1 9 Mˆ 2 1 − + = 1 − ∂b ε + 1 5 (ε + 1)3 3 5 (ε + 1)2 b=0 − 9 Mˆ 2 ∂ Mˆ 1 1 ε+ 2 = 5 ( 1) ∂b 3 ε + 9 Mˆ 2 b=0 ε+1 5 (ε+1)3

T → Tc means Mˆ → 0: 624 C Solutions to Problems # $ −1 ∂ Mˆ 1 ε 9 Mˆ 2 9 Mˆ 2 ≈ + 1 + ∂b 3 ε + 1 5 (ε + 1)3 5 (ε + 1)2 = b 0 # $ −1 ε ˆ 2 ≈ 1 + 9 M 3 ε + 1 5 (ε + 1)2

> (a)T → Tc: T →T then Mˆ ≡ 0 and ε + 1 −→c 1: ∂ Mˆ 1 ≈ ε−1 ∂b 3 b=0 1 χ ∼ ε−1 T λ

critical exponent:

γ = 1

< (b)T → Tc: then according to part 2 we have ( 5 1 Mˆ ∼ (−ε) 2 3

This means ! "− ∂ Mˆ 1 ε −ε 1 ≈ + 3 ∂b 3 ε + 1 (ε + 1)2 b=0 T →T 1 −→c [−2ε]−1 3 1 χ ∼ (−ε)−1 ⇒ γ = 1 T 2λ Critical amplitudes:

1 1 C C ∼ ; C ∼ = 2 λ 2λ C γ γ c The results obtained for , and c are typical for the classical theories of phase transitions. The concluding remark is as follows. The sign “ ∼ in the above formulae need not necessarily mean proportionality but should be understood as, “for T → Tc behaves like”. C Solutions to Problems 625

Problem 5.6 1. σ (1) · σ (2) 2 = σ (1)σ (2) + σ (1)σ (2) + σ (1)σ (2) 2 = x x y y z z = σ (1) 2 σ (2) 2 + σ (1) 2 σ (2) 2 + σ (1) 2 σ (2) 2 + x x y y z z + σ (1) σ (2) σ (1) σ (2) + σ (1) σ (2) σ (1) σ (2) + x x y y x x z z + σ (1) σ (2) σ (1) σ (2) + σ (1) σ (2) σ (1) σ (2) + y y x x y y z z + σ (1) σ (2) σ (1) σ (2) + σ (1) σ (2) σ (1) σ (2) = z z x x z z y y = 2 − σ (1) σ (2) − σ (1) σ (2) − σ (1) σ (2) − 31l z z y y z z − σ (1) σ (2) − σ (1) σ (2) − σ (1) σ (2) = x x y y x x = 31l − 2σ (1) · σ (2)

Here the properties (5.81) and (5.82) of the Pauli spin matrices are used many times. 2. Representatively, we calculate the x-component

σ (1) Q12 x 1 = + σ (1) σ (2) + σ (1) σ (2) + σ (1) σ (2) σ (1) = 1l x x y y z z x 2 1 = σ (1) + σ (1) 2 σ (2) + σ (1)σ (1)σ (2) + σ (1)σ (1)σ (2) 2 x x x y x y z x z 1 = σ (1) + σ (2) − i σ (1)σ (2) + i σ (1)σ (2) 2 x x z y y z

σ (2) x Q12 1 = σ (2) + σ (1) σ (2) + σ (1) σ (2) + σ (1) σ (2) = x 1l x x y y z z 2 1 = σ (2) + σ (1) σ (2) 2 + σ (1)σ (2)σ (2) + σ (1)σ (2)σ (2) 2 x x x y x y z x z 1 = σ (2) + σ (1) + i σ (1)σ (2) − i σ (1)σ (2) 2 x x y z z y

That means

σ (1) = σ (2) , Q12 x x Q12

so that the proposition for the x-component is proved. Analogously one can prove for the other components. 3. We use 626 C Solutions to Problems σ 1 = 01 1 = 0 x 0 10 0 1 σ 0 = 01 0 = 1 x 1 10 1 0 1 0 −i 1 0 σ = = i y 0 i 0 0 1 0 0 −i 0 1 σ = =−i y 1 i 0 1 0 1 1 0 0 σ = ; σ =− z 0 0 z 1 1

So that one immediately recognizes

σ (1) · σ (2) |↑↑ = |↓↓ + i 2 |↓↓ + |↑↑ = |↑↑ σ (1) · σ (2) |↓↓ = |↑↑ + (−i)2 |↑↑ + |↓↓ = |↓↓

σ (1) · σ (2) |↑↓ + |↓↑ ( ) = |↓↑ + |↑↓ − i 2 |↓↑ − i 2 |↑↓ − |↑↓ − |↓↑ = (|↑↓ + |↓↑ )

σ (1) · σ (2) |↑↓ − |↓↑ ( ) = |↓↑ − |↑↓ − i 2 |↓↑ + i 2 |↑↓ − |↑↓ + |↓↑ = 3(|↑↓ − |↓↑ )

The spin states are therefore eigenstates of the operator σ (1) · σ (2) with the eigenvalues (1, 1, 1, −3). Therefore they are also the eigenstates of Q12 with the eigenvalues (1, 1, 1, −1).

Problem 5.7

∞ sin x 1 +∞ exp(ix) − exp(−ix) R = dx x = dx x 2 2 2 2 0 x − y 4i −∞ x − y

We use the residue theorem and for that we choose the integration path parallel to the real axis from −∞ + i0+ to +∞ + i0+, close it with a semicircle at inﬁnity in the upper half-plane for the ﬁrst exponential function and in the lower half-plane for the second exponential function. Then the exponentials see to it that the semicircles do not contribute to the integral. The poles at x =±y lie inside the (mathematically C Solutions to Problems 627 negatively running) integration path C only for the second term. Then from the residue theorem we get 1 1 1 R =− dx e−ix + 8i C x + y x − y 2πi = e+iy + e−iy 8i π = cos y 2

Problem 5.8 1. The diagonal terms μ = ν in H1 directly give HU : H1(μ = ν) = HU That is why from now on we will consider only the non-diagonal terms μ = ν. For these we have to calculate H1(μ = ν) μ=ν 1 † † = Uμν n μσ n νσ + Jμν c c c μσ c νσ 2 i i iμσ iνσ i i iσσ μν

The ﬁrst summand leads to μ=ν 1 H (μ = ν) = Uμν n μn ν 1a 2 i i iμν The second summand we somewhat reformulate H1b(μ = ν) μ=ν 1 † † = Jμν c c c μσ c νσ = 2 iμσ iνσ i i iσσ μν μ=ν = 1 † † + † † + Jμν ciμ↑ciν↑ciμ↑ciν↑ ciμ↑ciν↓ciμ↓ciν↑ 2 μν i + † † + † † ciμ↓ciν↑ciμ↑ciν↓ ciμ↓ciν↓ciμ↓ciν↓

μ=ν 1 + − − + = Jμν −σ σ − σ σ − n μ↑n ν↑ − n μ↓n ν↓ 2 iμ iν iμ iν i i i i iμν μ=ν 1 x x y y z z = Jμν −2σ μσ ν − 2σ σ − 2σ σ − 2 i i iμ iν iμ iν iμν 628 C Solutions to Problems " 1 − (n μ↑ + n μ↓)(n ν↑ + n ν↓) 2 i i i i

μ=ν μ=ν 1 4 =− Jμν n μn ν − Jμν s μ · s ν 4 i i 2 i i iμν iμν

We recognize H1(μ = ν) = H1a(μ = ν) + H1b(μ = ν) = Hd + Hex

Thus the proposition is proved. 2. As an example we consider EuO. This has 5d-conduction bands which are empty and seven half-ﬁlled 4f-bands (levels). For the empty (!) conduction bands HU does not play any role but for sufﬁciently large Uμμ, it splits each of the f-bands into two sub-bands out of which the lower one is occupied and the upper one is empty. This leads to the localized magnetic 4f-moment. Hd provides only an unimportant energy shift of the f-levels. Hex becomes important if μ is an index of the conduction band and ν is that of an f-band or the converse. Let us assume that the coupling between the conduction band and the f-level is same for all the conduction bands,

4 Jμ = Jμν ∀ν, and deﬁne as localized spin: Sif = siν ν

where the ν-summation runs exclusively over the f-levels, then, Hex becomes the interaction operator of the multi-band Kondo lattice model: Hdf =− Jμ Sif · σiμ iμ The μ-summation runs only over the conduction bands.

Problem 5.9 1. We use the matrix representation with the basis

| kσ α = ckσα | 0 α = A, B

Then the “free” matrix reads as (0) ≡ ε † Hkσ αβ (k)ckσαckσβ αβ C Solutions to Problems 629

with the elements

γδ (0) = | ε † † | Hkσ 0 ckσγ αβ (k)ckσαckσβckσδ 0 αβ = εαβ (k)δγαδβδ 0 | 0 =εγδ(k) αβ

Hamilton Matrix: ε(k) t(k) H (0) = kσ t∗(k) ε(k) Eigenenergies: − (0) =! det E0k Hkσ 0

2 2 ! (ε(k) − E0k) − |t(k)| = 0

(±) = ε ±| | E0k (k) t(k)

Eigenstates: ∓|t(k)| t(k) cA ∗ = 0 t (k) ∓|t(k)| cB

± ± t(k) c( ) =±γ c( ) ; γ = A B |t(k)| Normalization (±) 1 † † |E σ =√ γ c σ ± c σ |0 0k 2 k A k B

2. Schrodinger’s¬ ﬁrst-order perturbation theory:

± ± E( ) |H |E( ) 0kσ 1 0kσ = 1 −1 z γ ∗ ± † ∗ Jzσ Sα 0 ckσ A ckσ B ckσαckσα 2 2 α † † ∗ γ c σ ± c σ 0 k A kB 1 † † =− z |γ |2 | | − Jzσ S 0 ckσ Ackσ Ackσ Ackσ A 0 4 − | † † | 0 ckσ B ckσ B ckσ B ckσ B 0

1 z 2 =− Jzσ S (|γ | − 1) 0|0 4 630 C Solutions to Problems

The ﬁrst order therefore does not contribute

(±) ≡ E1kσ 0

The energy correction in second order: We need the following matrix element:

+ − E( ) |H |E( ) 0kσ 1 0kσ = 1 −1 z γ ∗ + † ∗ Jzσ Sα 0 ckσ A ckσ B ckσαckσα 2 2 α † † ∗ γ c σ − c σ 0 k A k B 1 † † =− z |γ |2 | | + Jzσ SA 0 ckσ Ackσ Ackσ Ackσ A 0 4 + z − | † † | SB ( 1) 0 ckσ B ckσ B ckσ B ckσ B 0

1 z 2 =− Jzσ S (|γ | + 1) 0|0 4 1 z =− Jzσ S 2 This gives as the energy correction in the second order: 2 (±) (∓) E |H |E 1 2 z 2 ± 0kσ 1 0kσ J S E( ) = = 4 2kσ (±) − (∓) ±2|t(k)| E0k E0k

Thus the Schrodinger’s¬ second-order perturbation theory leads to the following spin-independent expression:

z 2 ± 1 S E( ) = ε(k) ±|t(k)|± J 2 k 8 |t(k)|

3. BrillouinÐWigner perturbation theory: The energy correction in the ﬁrst order is identical to the Schrodinger’s¬ perturbation theory, i.e. it vanishes in this case also. In the second order, one has to calculate ± ∓ 2 E( ) |H |E( ) ± 0kσ 1 0kσ E( ) = 2kσ (±) − (∓) Ek E0k

(±) It should be noted that in the denominator the “full” eigenenergy Ek appears. We have already used the matrix element in part 3 We then have the following determining equation: C Solutions to Problems 631

1 2 z 2 ± J S E( ) = ε(k) ±|t(k)|+ 4 k (±) − ε ±| | Ek (k) t(k)

This gives a quadratic equation for the energy, ± 2 1 E( ) − ε(k) −|t(k)|2 = J 2 Sz 2 k 4 with the solution: ( ± 1 E( ) = ε(k) ± |t(k)|2 + J 2 Sz 2 k 4 4. The problem can be easily exactly solved: 1 z † H = εαβ (k) − Jzσ Sα δαβ c c σβ ≡ H σ 2 kσα k k kσαβ kσ

Matrix representation: ⎛ ⎞ 1 z ε(k) − Jzσ S t(k) ⎝ 2 ⎠ Hkσ = ∗ 1 z ε + σ t(k) (k) 2 Jz S

The secular determinant

! det (E − Hkσ ) = 0

is solved by

1 (E − ε(k))2 − J 2 Sz 2 =|t(k)|2 4 That means ( 1 E±(k) = ε(k) ± |t(k)|2 + J 2 Sz 2 4 The BrillouinÐWigner perturbation theory gives the exact result already in the second order. If the root is expanded for small J, then the coefﬁcient of the ﬁrst term in the expansion is the result of Schrodinger’¬ perturbation theory in the second order.

Problem 5.10 1. The total spin S = S1 +S2 has the quantum numbers S = 0, 1, 2, ··· , 2S and the z = z + z =−, − + , ··· , + z-component S S1 S2 the quantum numbers M S S 1 S. Additionally holds 632 C Solutions to Problems

+ 2 = 2 + 2 + · ≡ 2 + (S1 S2) S1 S2 2S1 S2 S(S 1)1l 1 S · S = 2 S(S + 1) − S(S + 1) 1l 1 2 2

Possible energy levels: 2 1 E =−2J S(S + 1) − S(S + 1) − Mgμ B SM 2 B

With the abbreviation,

b ≡ gμB B

we have for the canonical partition function: −β H Z S = Sp e 2S +S −β2 JS(S+1) β 1 2 JS(S+1) βbM = e e 2 e S=0 M=−S

M-summation as in (4.94):

+ S sinh(βb(S + 1 )) eβbM = 2 sinh( 1 βb) M=−S 2

Partition function:

−β2 JS(S+1) 2S e β 1 2 JS(S+1) 1 Z S = e 2 sinh(βb(S + )) sinh( 1 βb) 2 2 S=0

= = 1 2. Special case S1 S2 2 : − 3 β2 J e 4 1 β2 J 3 Z / = sinh( βb) + e sinh( βb) 1 2 1 β 2 2 sinh( 2 b)

With

sinh( 3 x) 2 = + 1 1 2 cosh(x) sinh( 2 x)

we ﬁnally get C Solutions to Problems 633 1 β2 J −β2 J Z1/2 = e 4 e + 1 + 2 cosh(βgμB B)

The magnetization satisﬁes the following implicit equation:

∂ M = nkB T ln Z1/2 ∂ B0 2sinh(βgμ B) = ngμ B B −β2 J e + 1 + 2 cosh(βgμB B)

= / = sinh x For J 0, and with tanh x 2 1+cosh x we get the result of the Weiss theory (5.8) for S = 1/2.

Problem 6.1 With the abbreviations

1 j = β J ; b = βμB B0 ; β = kB T holds ⎛ ⎞ e j+b − Ee− j T − E1l = ⎝ ⎠ e− j e j−b − E

The eigenvalues follow from 0 =! det(T − E1l ) = e j+b − E e j−b − E − e−2 j = E2 − E e j+b + e j−b + e2 j − e−2 j = E2 − Eej 2 cosh b + 2sinh2j 2 = E − e j cosh b − e2 j cosh2 b + 2sinh2j

This gives the eigenvalues 5 β J 2 −2β J E± = e cosh βμB B0 ± cosh βμB B0 − 2e sinh 2β J

Problem 6.2 Partition function (6.35) in thermodynamic limit: N N E− N Z N (T, B0) = E+ 1 + −→ E+ f”ur N →∞ E+ 634 C Solutions to Problems

E+ as in Problem 6.1. Free energy per spin:

1 f (T, B0) =−kB T lim ln TN (T, B0) =−kB T ln E+ = N→∞ N =−J − kB T ln (cosh βμB B0 ± sqrt) where 5 2 −2β J sqrt = cosh βμB B0 − 2e sinh 2β J

Magnetization: ∂ m =− f T, B ∂ ( 0) B0 T sinh b + cosh b sinh b = μ sqrt B cosh b + sqrt sqrt + cosh b = μ sinh b B (cosh b + sqrt) sqrt sinh b = μB cosh2 b − 2e−2β J sinh 2β J

That gives

sinh(βμB B0) m(T, B0) = μB 2 −2β J cosh βμB B0 − 2e sinh 2β J

Susceptibility: ∂m χ T, B = μ T ( 0) 0 ∂ B0 T / cosh b = βμ2 B 2 b − e−2β J β J cosh 2 sinh 2 ⎫ ⎬⎪ sinh2 b cosh b − 3 ⎭⎪ cosh2 b − 2e−2β J sinh 2β J C Solutions to Problems 635

cosh b = βμ2 ∗ B 2 − −2β J β cosh/ b 2e sinh 2 J 0 sinh2 b ∗ 1 − cosh2 b − 2e−2β J sinh 2β J cosh(βμ B ) 1 − 2e−2β J sinh 2β J = βμ2 B 0 B / 2 −2β J 3 2 cosh (βμB B0) − 2e sinh 2β J

In the limit B0 → 0 this expression simpliﬁes to

χ , → = βμ2 2β J T (T B0 0) B e

Problem 6.3 In the classical Ising Model the magnetic moment is given by m = μ Si i where μ is a positive constant. With the Hamiltonian function H =−J Si S j − mB0 ij the canonical partition function is given by ⎛ ⎛ ⎞⎞ ⎝ ⎝ ⎠⎠ Z(T, m) = exp −β −J Si S j − mB0

{Si } ij

Here the summation is over all conceivable spin conﬁgurations, where the individual spins can have the values ±1. Therefore the substitution Si →−Si ∀i cannot affect the partition function. Then in the exponents the ﬁrst term does not change the sign but for the second term however, m →−m. That means

Z(T, m) = Z(T, −m) so that

F(T, m) =−kB T ln Z(T, m) =−kB T ln Z(T, −m) = F(T, −m) 636 C Solutions to Problems

Problem 6.4

Z N (T, B0) = Tr (exp(−β H)) 1 1 = Tr(1l) − βTr(H) + β2Tr(H 2) − β3 Sp(H 3) +··· 2 3! ∞ 1 = (−β)l Tr(H l ) l! l=0 # $ ∞ (−β)l = Tr(1l) 1 + m l! l l=1

N Each spin has two possible orientations Si =±1. That gives a total 2 spin conﬁgurations. Therefore

Sp(1l) = 2N

2. ∂2 F (T, B ) c =−T N 0 B0 ∂T 2 B0 ∂2 =−T (−k T ln Z (T, B )) ∂T 2 B N 0 B0 ∂2 = k β2 ln Z (T, B ) B ∂β2 N 0 B0 ∂ ∂ Z N (T, B ) 2 ∂β 0 = kB β ∂β Z N (T, B0) B 0 1 ∂2 Z 1 ∂ Z 2 = k β2 N − N B 2 2 Z N ∂β Z ∂β # N ∞ − 2 Tr(1l) l(l 1) l−2 = kB β (−β) ml Z N l! l=1 ⎤ ∞ 2 Tr(1l) l l−1 ⎦ − (−β) ml Z N = l! l 1 2 = k β2 m − m2 + O(β) B 2! 2 1 1 = (m − m2) +··· 2 2 1 kB T C Solutions to Problems 637

In the last step we have restricted ourselves to the lowest term in 1/T .One should notice that the moments are temperature independent. This result for the high-temperature behaviour of the heat capacity is of course valid not only for the Ising spins but also for all magnetic systems (!).

Problem 6.5

1. According to Problem 1.2:

μ 1 0 2 χT = (m − m ) kB T V

This means

1 μ χ = 0 g2μ2 (S − S ) S − S T k T V B i i j j B ij

So that we directly get ⎛ ⎞ 1 μ χ = 0 ⎝g2μ2 S S − m 2⎠ T k T V B i j B ij

2. The spin chain shows no spontaneous magnetization. When an external ﬁeld is switched off, then, m ≡0. According to (6.19) for the spin correlation of a one-dimensional chain we have

|i− j| Si S j =v

One can easily see that, in the double summation, N terms with |i − j|=0give the contribution v0 = 1; 2(N − 1) terms with |i − j|=1 give the contribution v1;2(N − 2) terms with |i − j|=2 give the contribution v2; ....; and ﬁnally two terms with |i − j|=N − 1 give the contribution vN−1. Then all of them together give − 1 μ N 1 χ , = 0 2μ2 + − vk T (T B0) g B N 2 (N k) kB T V k=1

One calculates 638 C Solutions to Problems

− N 1 1 − vN 2 Nvk = 2N − 2N 1 − v k=1 − − N 1 N 1 d 1 − vN −2 kvk =−2 kvk = 2v dv 1 − v k=1 k=0 (1 − v)(−NvN−1) + (1 − vN ) = 2v (1 − v)2

− N 1 v 1 − vN 2 (N − k)vk = 2N − 2v 1 − v (1 − v)2 k=1

From this it follows: 1 μ 2v χ (T, B = 0) = 0 g2μ2 N 1 + T 0 k T V B 1 − v B 1 − vN −2v (1 − v)2

3. For N →∞the expression for susceptibility can be simpliﬁed:

1 χ (T, B = 0) N T 0 1 μ 1 + v = 0 2μ2 g B kB T V 1 − v 1 μ 1 + tanh(β J) = 0 2μ2 g B kB T V 1 − tanh(β J) 1 μ eβ J + e−β J + eβ J − e−β J = 0 g2μ2 B β J −β J β J −β J kB T V e + e − e + e 1 μ = 0 2μ2 2β J g B e kB T V

This agrees with the result (6.46) for the Ising ring in the thermodynamic limit.

Problem 6.6 1. The partition function N−1 Z N (T ) = ··· exp ji Si Si+1

S1 S2 SN i=1

where ji = β Ji is valid, we have already calculated with (6.14): C Solutions to Problems 639

N.−1 N Z N (T ) = 2 cosh ji i=1

Four-spin correlation function i = j:

S S + S S + i i 1 j j 1 − 1 N 1 = ··· Si Si+1 S j S j+1 exp ji Si Si+1 Z N S1 S2 SN i=1 ∂2 = 1 Z N Z N ∂ ji ∂ j j ··· ··· ··· = cosh j1 sinh ji sinh j j cosh jN−1 cosh j1 ···coshN−1

= tanh ji tanh j j

For i = j the four-spin correlation function is equal to 1. Let us now set ji = j ∀i, then we have

1 falls i = j S S + S S + = i i 1 j j 1 tanh2 j falls i = j

2. From Problem 6.4 we have

1 ∂2 Z 1 ∂ Z 2 c = k β2 N − N B0 B ∂β2 2 ∂β Z N Z N 2 2 2 = kB β H − H

With

2 2 H =J Si Si+1 S j S j+1 ij

follows: 640 C Solutions to Problems

N−1 = β2 2 cB0=0 kB J Si Si+1 S j S j+1 , = i j 1 − Si Si+1 S j S j+1 N−1 2 2 2 2 = kB β J δij + (1 − δij) tanh j − tanh j i, j=1 N−1 2 2 2 = kB β J δij(1 − tanh j) i, j=1

2 2 1 cB =0 = (N − 1)kB β J 0 cosh2 β J

Compare this result with (6.45).

Problem 7.1

1. x y z S , S = iS δij and cyclic i j − i

z, ± = z, x ± z, y Si S j Si S j − i Si S j − − = iSyδ ∓ −Sx δ i ij i ij =± x ± y δ Si iSi ij =± ±δ Si ij

2. S+, S− = Sx , Sx + Sy, Sy + i Sy, Sx − i Sx , Sy i j − i j − i j − i j − i j − = − z δ − z δ i( i Si ) ij i(i Si ) ij = zδ 2 Si ij

S± S∓ = (Sx ± iSy)(Sx ∓ iSy) i i i i i i = x 2 + y 2 ± y, x (Si ) (Si ) i Si Si − = 2 − z 2 ± z Si (Si ) Si 2 = 2 + (where Si S(S 1)1l in the space of the spin states) C Solutions to Problems 641

4. 1 + − + − + = 1 x x + y y + y x − x y+ Si S j Si S j Si S j Si S j iSi S j iSi S j 2 2 + x x + y y − y x + x y Si S j Si S j iSi S j iSi S j = x x + y y Si S j Si S j = · − z z Si S j Si S j 1 S · S = S+ S− + S− S+ + Sz Sz i j 2 i j i j i j 5. H =− JijSi · S j i, j 4.) 1 =− (S+ S− + S− S+)J − J Sz Sz 2 i j i j ij ij i j i, j i, j i= j =−1 + − − 1 + − − z z JijSi S j Jij S j Si JijSi S j 2 , 2 , , i j i j i j i↔ j (J =J ) ij ji =− + − + z z Jij Si S j Si S j i, j

Problem 7.2 − n, z = − n = , ,... 1. (Si ) Si − n (Si ) ; n 1 2 Complete induction: n = 1: −, z = − Si Si − Si n → n + 1: − n+1, z = − n, z − + − n −, z (Si ) Si − (Si ) Si − Si (Si ) Si Si − = − n − + − n − n (Si ) Si (Si ) Si = + − n+1 (n 1) (Si ) q.e.d. − n, z 2 = 22 − n + z − n = , ,... 2. (Si ) (Si ) − n (Si ) 2n Si (Si ) ; n 1 2 − n, z 2 = z − n, z + − n, z z (Si ) (Si ) − Si (Si ) Si − (Si ) Si − Si =1.) z − n + − n z Si n (Si ) n (Si ) Si =1.) z − n + 22 − n 2n Si (Si ) n (Si ) q.e.d. 642 C Solutions to Problems +, − n = z + 2 − − n−1 = , ,... 3. Si (Si ) − (2n Si n(n 1))(Si ) ; n 1 2 Complete induction: n = 1: +, − = z Si Si − 2 Si

n → n + 1: S+, (S−)n+1 i i − = +, − n − + − n +, − Si (Si ) − Si (Si ) Si Si − = z + 2 − − n−1 − + − n z (2n Si n(n 1))(Si ) Si (Si ) 2 Si =1.) z + 2 − − n + 2 − n + (2n Si n(n 1))(Si ) 2n (Si ) + z − n 2 Si (Si ) = + z + 2 + − n (2 (n 1)Si n(n 1))(Si ) q.e.d.

Problem 7.3 Starting point is the identity (7.485):

.+S z − = Si ms 0 m S =−S

= 1 S 2 2 0 = Sz + Sz − = Sz 2 − i 2 i 2 i 4 2 z 2 = Si 4 1 2 1 α = ,α = 0 0 2 4 1 2

S = 1 0 = Sz + Sz Sz − i i i z 3 =2 z Si Si 2 α0(1) = 0,α1(1) = ,α2(1) = 0

= 3 S 2 C Solutions to Problems 643 = z + 3 z + 1 z − 1 z − 3 0 Si Si Si Si 2 2 2 2 9 1 = Sz 2 − 2 Sz 2 − 2 i 4 i 4 5 9 z 4 = 2 z 2 − 4 Si Si 2 16 3 9 4 3 3 5 2 α0 =− ,α1 = 0,α2 = , 2 16 2 2 2 3 α = 0 3 2

S = 2 = z + z + z z − z − 0 Si 2 Si Si Si Si 2 = z 2 − 2 z 2 − 2 z Si 4 Si Si z 5 = 2 z 3 − 4 z Si 5 Si 4 Si 4 α0(2) = 0,α1(2) =−4 ,α2(2) = 0, 2 α3(2) = 5 ,α4(2) = 0

= 7 S 2 = z + 7 z + 5 z + 3 z + 1 · 0 Si Si Si Si 2 2 2 2 · z − 1 z − 3 z − 5 z − 7 Si Si Si Si 2 2 2 2 49 25 9 = z 2 − 2 z 2 − 2 z 2 − 2 ∗ Si Si Si 4 4 4 1 ∗ Sz 2 − 2 i 4 987 Sz 8 =212 Sz 6 − 4 Sz 4 + i i 8 i 3229 617 + 6 z 2 − 8 Si 16 8 7 617 8 7 7 3229 6 α0 =− ,α1 = 0,α2 = , 2 8 2 2 16 7 7 987 4 7 α3 = 0,α4 =− ,α5 = 0, 2 2 8 2 7 7 α = 212,α = 0 6 2 7 2 644 C Solutions to Problems

Problem 7.4 Eu2+ on f.c.c.-sites each atom of a (111)-plane has six nearest neighbours in the same (111)-plane and three each in the two neighbouring planes and six next nearest neighbours, three each in the two neighbouring (111)-planes. EuSe:

2.8K ≤ T ≤ 4.6K NNSS-Antiferromagnet

12 nearest neighbours, 9 in the same and 3 in the other sub-lattice.

⎧ ⎫ ⎨ ∈ ∈ ⎬ 2 1 2 k T = 2 S(S + 1) J − J B N 3 ⎩ 1 j 1 j ⎭ j j 2 = 2 S(S + 1) {9J + 3J − 3J − 3J } 3 1 2 1 2 2 2 = S(S + 1)(6J1) 3 ⎧ ⎫ ⎨ ∈ ∈ ⎬ 2 1 2 k Θ = 2 S(S + 1) J + J B 3 ⎩ 1 j 1 j ⎭ j j 2 = 2 S(S + 1) {9J + 3J + 3J + 3J } 3 1 2 1 2 2 = 2 S(S + 1) {12J + 6J } 3 1 2

= kB J1 42 S(S+1) TN

2 kB Θ − 2kB TN = 4 S(S + 1)J2

= kB Θ − J2 42 S(S+1) ( 2TN )

Problem 7.5

1. It is convenient to ﬁrst split the matrix as follows:

A = bA + a1l

Here the reduced matrix A is given by C Solutions to Problems 645 ⎛ ⎞ 01000··· 00 ⎜ ⎟ ⎜ 10100··· 00⎟ ⎜ ⎟ ⎜ 01010··· 00⎟ A = ⎜ ⎟ ⎜ ...... ⎟ ⎝ ...... ⎠ 00000··· 10

Eigenvalues of A:

λ = a + bλ ; λ : Eigenvalue of A

Now we have to solve

! 0 = det(A − λ 1l ) ≡ Dd (λ )

One recognizes

2 D1(λ ) =−λ ; D2(λ ) = λ − 1

In general one obtains by expanding after the ﬁrst row:

λ =−λ λ − ∗ Dd ( ) Dd−1( ) 1 11 000··· 00 0 −λ 100··· 00 01−λ 10··· 00 ∗ ...... ...... 00 000··· 1 −λ

=−λ Dd−1(λ ) − Dd−2(λ )

In the last step, the remaining determinant is expanded after the ﬁrst column. With the ansatz for solution

±idα Dd (λ ) = e

follows:

e±idα = λe±i(d−1)α − e±i(d−2)α e±iα =−λ − e∓iα λ = λ =−2 cos α α = arccos(− ) 2

General solution 646 C Solutions to Problems

Dd (λ ) = c1 cos(dα) + c2 sin(dα)

c1,c2 from the initial conditions:

D1(λ ) =−λ = c1 cos(α) + c2 sin(α) = = 2 cos(α) 2 D2(λ ) = λ − 1 = c1 cos(2α) + c2 sin(2α) = = 4 cos2(α) − 1

c1 = 2 − c2 tan α

2 2 4 cos α − 1 = c1(2 cos α − 1) + c2 2sinα cos α 2 = 4 cos α − 2 − 2c2 sin α cos α +

+ c2 tan α + 2c2 sin α cos α

c2 = cot α c1 = 1

Intermediate result: cos α D (λ) = cos(dα) + sin(dα) d sin α 1 = (sin α cos(dα) + cos α sin(dα)) sin α + α = sin ((d 1) ) sin α

Requirement: rπ D (λ) =! 0 α = ; r = 1,...,d d d + 1 rπ λ =−2 cos r d + 1

With this we have the Eigenvalues of the tridiagonal matrix A: rπ λ = a − 2b cos , r = 1,...,d r d + 1

2. Heisenberg model for films: Hamiltonian in molecular field approximation (7.124): =− z z HMFA 2 Jij S j Si i, j C Solutions to Problems 647 z / That means we have an effective paramagnet in the molecular field 2 ij Jij S j gJ μB . Due to the film structure, translational symmetry is applicable only in the plane of the film and not in three dimensions. Let BS be the Brillouin function in the following. Then according to (7.134) holds ⎛ ⎞ z = ⎝ β z ⎠ Si SBS 2 S Jij S j j

Exchange only between the nearest neighbours: z = z + z + z Jij S j qJ Sα pJ Sα+1 pJ Sα−1 j

z Sα : layer magnetization, where α numbers the layers; q(p): coordination number (the number of nearest neighbours) within the layer (between the layers). q + 2p = z: volume coordination number. Examples:

sc (100) : q = 4, p = 1 sc (110) : q = 2, p = 2 sc (111) : q = 0, p = 3

→ ≈ S+1 For T TC linearization of Brillouin function: BS(x) 3S x: z 2 2 z z z Sα = (S + 1)Sβ qJ Sα +pJ S +pJ S 3 α+1 α−1

This gives a system of homogeneous equations: z 0 = Mαγ Sγ γ

Here the matrix M is given by ⎛ ⎞ qJ − xpJ 000··· 0 ⎜ ⎟ ⎜ pJ qJ − xpJ 00··· 0 ⎟ ⎜ ⎟ M ≡ ⎜ 0 pJ qJ − xpJ 0 ··· 0 ⎟ ⎝ ...... ⎠ 0 ··· ··· pJ qJ − x

1 x = 2 2 + β 3 S(S 1) 648 C Solutions to Problems

> z We thus have a tridiagonal matrix. There exist non-trivial solutions ( Sγ → 0) only for

.d ! det M = 0 = λr r=1 λr is the rth eigenvalue of M. According to part 1, for the eigenvalues we have rπ λ = qJ − x − 2pJ cos r d + 1

At least one of these eigenvalues must be equal to zero. This gives the following equation for TC : 2 rπ k T (r) = 2 S(S + 1)J q − 2p cos B C 3 d + 1

The physical solution must satisfy the well-known limiting cases

2 k T (d = 1) = 2 S(S + 1)Jq B C 3 2 k T (d =∞) = 2 S(S + 1)J(q + 2p) B C 3

1 sc(100) sc(110) 0.8 sc(111)

0.6

0.4 [Tc(bulk)-Tc(d)]/Tc(bulk) 0.2

0 510d 15 20 Fig. C.1 Relative change of the Curie temperature of a ﬁlm with its thickness. d: number of mono- layers

That is possible only for r = d (Fig. C.1): 2 dπ k T = 2 S(S + 1)J q − 2p cos B C 3 d + 1 C Solutions to Problems 649

It is instructive to compare the critical temperature of the ﬁlm with that of the bulk material: T (∞) − T (d) z − z dπ C C = 2 S 1 + cos TC (∞) z d + 1

Here z = q + 2p is the coordination number of the bulk and zS = q + p that of the surface.

Problem 7.6 We use the Heisenberg Hamiltonian in the wavenumber representation (7.105):

1 ) * H =− J(k) S+(k)S−(−k) + Sz(k)Sz(−k) − N k 1 − g μ B Sz J B 0 (0)

For the spin operators (7.106) and (7.107) hold

z + S (k) |0 = NS|0 δk,0 ; S (k) |0 = 0

With this we calculate

1 − J(k)S+(k)S−(−k) |0 = N k 1 ) * =− J(k) 2Sz(0) + S−(−k)S+(k) |0 N k 1 = (22 NS|0 ) − J(k) N k 2 =−(2 NS|0 )Jii (7.12) = 0

1 z z z − J(k)S (k)S (−k) |0 =−S J(k)S (k)δ− , |0 N k 0 k k =−SJ(0)Sz(0) |0 2 2 =−N S J0 |0 (J(0) ≡ J0)

1 − g μ B Sz | =−Ng Sμ B | J B 0 (0) 0 J B 0 0 650 C Solutions to Problems

With this it is shown that |0 is the eigenstate with the eigenvalue

2 2 E0(B0) =−N S J0 − NgJ μB B0 S

Problem 7.7

Problem 7.8 HolsteinÐPrimakoff transformation: ( √ + = ϕ ϕ = − ni Si 2S (ni )ai ; (ni ) 1 √ 2S − = +ϕ Si 2Sai (ni ) z = − = + Si (S ni );ni ai ai

+ ai , ai : Bose operators + + + ai , a = δij; ai , a j = a , a = 0 j − − i j − + + ni , a j =−ai δij; ni , a = a δij − j − i C Solutions to Problems 651

1. Commutation relations: +, − = Si S j − = 2S2δ ϕ(n )a , a+ϕ(n ) ij i i i i − n = 2S2δ ϕ(n )(1 + n )ϕ(n ) − a+ 1 − i a ij i i i i 2S i n 1 = 2S2δ (1 + n ) 1 − i − n + a+n a ij i 2S i 2S i i i n n2 1 1 = 2S2δ 1 − i − i − n + n2 ij 2S 2S 2S i 2S i

= 2δij(S − ni ) = δ z 2 ijSi

√ z, + = − , ϕ Si S j (S ni ) 2S (n j )a j − √ − =−2 2S [n ,ϕ(n )a ] δ √ i i i − ij =−2 ϕ , δ 2S (ni ) [ni ai ]− ij − √ ai 2 = 2Sϕ(ni )ai δij =+δ + ijSi

√ z, − =−2 , +ϕ δ Si S j 2S ni ai (ni ) − ij − √ =−2 2S n , a+ ϕ(n )δ i i − i ij a+ √ i =− +ϕ δ 2Sai (ni ) ij =−δ − ijSi

2 2. Si :

1 S2 = S+ S− + S− S+ + (Sz)2 i 2 i i i i i 1 = 2 ϕ +ϕ + 2S (ni )ai ai (ni ) 2 n +a+ 1 − i a + 2(S − n )2 i 2S i i 652 C Solutions to Problems ⎛ ⎞ n n 1 = 2 S ⎝ 1 − i + n 1 − i + n − a+ n a ⎠ + 2S i 2S i 2S i i i −ai +ai ni +2 2 − + 2 (S 2Sni ni ) = 2 + − 2 + 2 − + 2 (S 2ni S ni S 2Sni ni ) = 2 S(S + 1)

Problem 7.9 DysonÐMaleev« transformation: √ S+ = 2Sα i i √ n S− = 2Sα+ 1 − i i i 2S z = − = α+α Si (S ni );ni i i α α+ i , i : Bose operators

1. Commutation relations: n +, − = 2 α ,α+ − j Si S j 2S i j 1 − 2S − = 2δ − 2δ α ,α+ 2S ij ij i i ni − = 2δ − 2δ − 2δ α+ α , 2S ij ijni ij i [ i ni ]−

αi 2 = 2 δij(S − ni ) = δ z 2 ijSi

√ z, + = 2 − ,α Si S j 2S S ni j − − √ =−2 δ ,α 2S ij [ni i ]− −α √ i 2 = 2Sαi δij = δ + ijSi

√ n z, − = 2 − ,α+ − j Si S j 2S S ni j 1 − 2S − √ n =−2 δ ,α+ − i 2S ij ni i 1 2S − √ n =−2 2Sδ n ,α+ 1 − i ij i i − 2S α+ i C Solutions to Problems 653 √ n =−2 2Sδ α+ 1 − i ij i 2S =−δ − ijSi

2 2. Si :

1 2 = + − + − + + z 2 Si Si Si Si Si (Si ) 2 n n = S2 α α+ 1 − i + α+ 1 − i α + i i 2S i 2S i + 2(S2 − 2Sn + n2) i i 1 = 2 + − + 2 + − S(1 ni ) (ni ni ) Sni 2 1 − α+ α + 2 − + 2 i ni i S 2Sni ni 2 1 1 1 1 = 2 S(S + 1) − n + n2 + n − n2 2 i 2 i 2 i 2 i = 2 S(S + 1)

Problem 7.10

√ √ + ≈ − ≈ + z = − Si 2Sai ; Si 2Sai ; Si (S ni )

Fourier transformation: + − · + = ik Ri S (k) e Si √i −ik·Ri ≈ 2S e ai i √ √ Nak = 2SNak

− − · − = ik Ri S (k) e Si √i − · + ≈ ik Ri 2S e ai i √ √ Na−k = + 2SNa−k 654 C Solutions to Problems − · + z = ik Ri − S (k) e ( S ai ai ) i − · 1 + − = δ − ik Ri i(k q)Ri SN k,0 e aq ak e N , i k q + = δ − δ SN k,0 aq ak k ,k+q , k q = δ − + SN k,0 aq ak+q q

Veriﬁcation of commutation relations:

1. + , − = 2 , + S (k) S (q) − 2SN ak a−q − 2 = 2SN δk+q ≈ 2Sz(k + q)

2. √ z + + S (k), S (q) =− a ak+q , aq 2SN − q − q √ 2 + =− 2SN a , aq ak+q q − q −δ √ qq 2 = 2SNak+q = S+(k + q)

3. √ z − 2 + + S (k), S (q) =− 2SN a ak+q , a− − q q − q √ 2 + + =− 2SN a a + , a q k q −q − q −δ √ q ,−q−k =−2 + 2SNa−q−k =−S−(k + q) C Solutions to Problems 655

Problem 7.11

1 −iq·Ri αq = √ e αi N i 1 iq·Ri αi = √ e αq N q 1 − J(q) = J eiq(Ri R j ) N ij i, j

DysonÐMaleev:« √ S+ = 2Sα i i √ nˆ S− = 2Sα+ 1 − i i i 2S z = − Si (S nˆ i )

Heisenberg model:

H = E0 + H2 + H4 = 2 − 2 α+α H2 2S J0 i nˆ i 2S i, j Jij i j

1 − − + = i(q q )Ri α α nˆ i e q q N , , i q q i + = α α δ q q qq , q q = α+α q q q

α+α = Jij i j i, j 1 + iq(Ri −R j ) −iq Ri iq R j = J(q)α α e e e 2 q q N , , , i j q q q + = α α δ δ J(q) q q q,q q,q , , qq q = α+α J(q) q q q 656 C Solutions to Problems

2 ω(q) = 2S (J0 − J(q))

= ω α+α H2 (q) q q q =−2 + 2 α+ α H4 i, j Jijnˆ i nˆ j i, j Jij i nˆ i j Jijnˆ i nˆ j = i, j 1 + + = α αq α αq J(q)∗ N 3 q1 3 q2 4 i, j q1···q4,q − − − + − ∗ e q(Ri R j )ei(q1Ri q3Ri q2R j q4R j ) 1 + + = J(q)α αq α αq δq,q −q δq,q −q N q1 3 q2 4 1 3 4 2 q1···q4,q 1 + + = J(q4 − q2)δq +q ,q +q α αq α αq N 1 2 3 4 q1 3 q2 4 q1···q4 1 + = J(q4 − q2)δq +q ,q +q α αq + N 1 2 2 4 q1 4 q1q2q4 1 + + + J(q4 − q2)δq +q ,q +q α α αq αq N 1 2 3 4 q1 q2 3 4 q1...q4 1 + = J(q1 − q2)α αq + N q1 1 q1,q2 1 + + + J(q4 − q2)δq +q ,q +q α α αq αq N 1 2 3 4 q1 q2 3 4 q1...q4

α+ α = Jij i nˆ i j i, j 1 + + −iq(Ri −R j ) = J(q)α α αq αq e ∗ N 3 q1 q2 3 4 i, j q,q1...q4 + − − ∗ ei(q1Ri q2Ri q3Ri q4R j ) 1 + + = J(q)α α αq αq δq,q δq,q +q −q N q1 q2 3 4 4 1 2 3 q1...q4,q 1 + + = J(q4)δq +q ,q +q α α αq αq N 1 2 3 4 q1 q2 3 4 q1...q4 C Solutions to Problems 657

In the ﬁrst term q1 and q2 commute:

2 H = (J(q ) − J(q − q )) ∗ 4 N 4 4 1 q1...q4 + + ∗δ + , + α α α α q1 q2 q3 q4 q1 q2 q3 q4

Here we exploit

1 1 − − J(q − q ) = J ei(q1 q2)(Ri R j ) N 1 2 N 2 ij q2 q2 i, j 1 − = J eiq1(Ri R j )δ N ij ij i, j 1 = J = 0 N ii i

hω

Problem 7.12

E0 = E (B = 0) =−NJ 2 S2 0 0 0 0 =< >= 0 + ω < > U HSW E0 (q) nˆ q q 2 2 ω(q) = 2S (J0 − J(q)) ≈ Dq for small |q| Low temperatures → only a few magnons are excited: 658 C Solutions to Problems

q2 U ≈ E0 + D 0 β Dq2 − q e 1 ∞ V 2 2 = E0 + D d3qq2e−β Dq e−nβ Dq 0 (2π)3 BZ n=0 ∞ ∞ DV 2 ≈ E0 + dq q4e−nβ Dq 0 π 2 2 = n 1 0

Substitution √ nβ D t = nβ Dq2 dt = 2nβ Dq dq = 2 t √ dq nβ D 1 dt dq = √ √ 2 nβ D t

≈ 0+ U E0 ∞ ∞ DV − 5 3 −t + (nβ D) 2 dt t 2 e 4π 2 n=1 0 √ Γ 5 = 3 Γ 3 = 3 Γ 1 = 3 π ( 2 ) 2 ( 2 ) 4 ( 2 ) 4 5 3DV 5 kB 2 5 ≈ 0 + ζ 2 = U E0 3 T 16π 2 2 D 5 = 0 + η 2 E0 T

∂U 5 3 C = = = ηT 2 B0 0 ∂T 2 B0=0 is experimentally uniquely conﬁrmed. → η → → = Measurement of CB0=0 D J0 z1 J1. Problem 7.13 1. Proof by complete induction: p = 0; trivial. p = 1 , † = † , † + †, † nˆ q aq − aq aq aq − aq aq − aq = † aq

p p + 1 C Solutions to Problems 659 † p+1 nˆ q, (a ) = q − = † p , † + , † p † (aq) nˆ q aq − nˆ q (aq) − aq = † p † + † p † (aq) aq p(aq) aq = + † p+1 (p 1)(aq) q.e.d.

= nk |ψ

|ψ is therefore an eigenstate of nˆ k with the eigenvalue nk. Thus |ψ is also eigenstate of H: H |ψ = E0(B0) + ω(k)nk |ψ k

Problem 7.14 ) * =< >= ˆ + <α+α > + <β+β > U HSW Ea Eα(q) q q Eβ (q) q q q

B0 = 0; BA = 0:

Eα(q) = Eβ (q) ≈ D · q = ε(q) q U = Eˆ + 2D a eβε(q) − 1 q 660 C Solutions to Problems

∞ ∞ DV U ≈ Eˆ + dq q3e−β Dqn a π 2 = n 1 0 ∞ ∞ DV 1 1 = Eˆ + dy y3e−y a π 2 (β D)4 n4 n=1 0 Γ(4)=3!=6

Abbreviation: 6DV C = ζ (4) 4 2 4 π (kB D)

4 U ≈ Eˆ a + C4 · T

Heat capacity: ∂ U 3 C = = = 4C · T B0 0 ∂T 4 B0=0

Problem 7.15 Sub-lattice magnetization of an antiferromagnet (Eq. (7.304)): / 2 η 1 N 2 cosh q MA(T ) = gJ μB S − sinh ηq − + V 2 eβ Eα (q) − 1 q q 0 2 η + sinh q eβ Eβ (q) − 1

B0 = 0; BA = 0:

J(q) (7.356) tanh 2η ≈− =−γ (T independent) q J(0) q

< = = 1 μ N − 2 MA(0) V gJ B 2 q sinh q

2 Eα(q) ≡ Eβ (q) = 2S (J0 + J(q))(J0 − J(q)) J −J(q)≈dJ q2 √ 0 ≈ 0 2| | · ≡ ε (2 S J0 2d) q (q) D C Solutions to Problems 661

1 1 + 2sinh2 η M (T ) − M (0) =− g μ q A A V J B eβε(q) − 1 q

T → 0; β →∞:

1 = eβε(q) − 1 q V e−βε(q) = d3q (without correction term) (2π)3 1 − e−βε(q) BZ ∞ V = d3q e−nβε(q) π 3 (2 ) = BZ n 1 ∞ ∞ V − β ≈ dq q2e n Dq (justiﬁcation as for ferromagnets!) π 2 2 = n 1 0 ∞ ∞ V 1 1 = dy y2e−y 2π 2 (β D)3 n3 n=1 0 Γ = = (3) 2! 2 V 1 = ζ (3) T 3 π 2 −1 3 (kB D)

μ = gJ B ζ Abbreviation: C3 π 2 −1 3 (3) (kB D)

3 MA(T ) − M(0) = C3 · T if without correction term

With correction term:

2 1 1 1 + 2sinh ηq = 5 = 5 2 − γ 2 1 − tanh 2ηq 1 q

Consider 662 C Solutions to Problems

2 γq ≈ 1 − dq 5 1 = 1 = − γ 2 2dq2 − d2q4 1 q

= √1 · 1 · 5 1 2d q − 1 2 1 2 dq − γ 2 ≈ 2 1 q 2dq (only small q play a role in the spin wave approximation)

+ 2 η 1 2sinh q ≈ √1 1 1 eβε(q) − 1 q eβε(q) − 1 q 2d q ∞ ∞ 1 V 1 1 = √ dy ye−y 2π 2 (β D)2 n2 2d n=1 0 =1

μ = √ gJ B ζ Abbreviation: C2 2 2 (2) 2d2π (kB D)

2 MA(T ) − M(0) = C2 · T

Problem 7.16 ) * = + + + + + + + + H Ea bA aq aq bB bq bq c(q) aqbq aq bq q q q

in the new operators: = + η · α+ + η · β ∗ H Ea bA cosh q sinh q q + ∗ cosh η · αq + sinh η · β + q + η · α + η · β+ ∗ bB sinh q cosh q q ∗ sinh η · α+ + cosh η · βq + ) q + η · α + η · β+ ∗ c(q) (cosh q sinh q ) q ∗ η · α+ + η · β + (sinh q cosh q ) C Solutions to Problems 663 + cosh η · α+ + sinh η · β ∗ q q * + ∗ sinh η · αq + cosh η · β q = + 2 η · α+α + 2 η · β β+ Ea bA cosh q q sinh q q q + sinh η · cosh η α+β+ + β α q q q q . + b sinh2 η · α α+ + cosh2 η · β+β + B q q q q + sinh η · cosh η(α β + β+α+) + q q q q + 2 η α β + α+β+ + c(q) cosh ( q q q q ) + sinh2 η(β+α+ + β α )+ q q q q + cosh η · sinh η α α+ + β+β + q q q q +α+α + β β+ q q q q

Use 1 sinh η · cosh η = sinh 2η 2 1 cosh2 η = (cosh 2η + 1) 2 1 sinh2 η = (cosh 2η − 1) 2

1 H =E + b (α+α − β β+) + b (−α α+ + β+β )+ a 2 A q q q q B q q q q q + α β + α+β+ − β+α+ − β α c(q) q q q q q q q q 1 + sinh(2η) b (α+β+ + β α ) + b (α β + β+α+)+ 2 A q q q q B q q q q q + α α+ + β+β + α+α + β β+ c(q) q q q q q q q q 1 + cosh(2η) b (α+α + β β+) + b (α α+ + β+β )+ 2 A q q q q B q q q q q + α β + α+β+ c(q) 2 q q 2 q q 1 =E + (b − b )α+α + (b − b )β+β − (b + b ) a 2 A B q q B A q q A B q 1 + sinh(2η) (b + b )(α β + α+β+)+ 2 a B q q q q q + α+α + β+β + c(q) 2 q q 2 q q 2 664 C Solutions to Problems 1 + cosh(2η) (b + b )(α+α + β+β ) + (b + b )+ 2 a B q q q q A B q α β + α+β+ 2c(q) q q q q

−(bA+bB ) tanh(2η) N 1 =E − (b + b ) + ((b + b ) cosh 2η+ a 4 A B 2 A B q + 2c(q)sinh2η)+ 1 + α+α (b − b ) + c(q)sinh2η+ q q 2 A B q 1 + (bA + bB ) cosh 2η + 2 1 + β+β (b − b ) + c(q)sinh2η+ q q 2 B A q 1 + (b + b ) cosh 2η 2 A B

1 c(q)sinh2η + (b + b ) cosh 2η = 2 A B tanh 2η 1 1 =c(q) + (bA + bB ) 1 − tanh2 2η 2 1 − tanh2 2η −2c(q) bA+bB 1 1 =c(q)5 + (bA + bB )5 − 4c2 2 − 4c2 1 2 1 2 (bA+bB ) (bA+bB ) − 2 + 2 = 4c (q) + 1 (bA bB ) 2 2 2 2 2 (bA + bB ) − 4c 2 (bA + bB ) − 4c 1 = (b + b )2 − 4c2 2 A B 1 =Eα(q) − (b − b ) 2 A B 1 =Eβ (q) + (b − b ) 2 A B

ˆ = − N + + 1 + 2 − 2 Ea Ea 4 (bA bB ) 2 q (bA bB ) 4c (q)

= ˆ + α+α + β+β H Ea q Eα(q) q q Eβ (q) q q C Solutions to Problems 665

Problem 7.17 1 iq·RΔ γq = e 1 z1 Δ1 z1: number of nearest neighbours

RΔ1 : lattice vector from the origin to a nearest neighbouring site γ = q−q1 nˆ q1

q1 1 − · Δ 1 † · − = i(q q1) R 1 iq1 (Ri R j ) e ai a j e z1 N q1 Δ1 i, j 1 † · Δ = iq R 1 δ ai a j e i− j,Δ1 z1 i, j Δ1 1 · Δ † = iq R 1 e ai ai−Δ1 z1 Δ1 i

Due to translational symmetry, the expectation value is the same for all nearest neighbours. That is, independent of any particular Δ1:

γ q−q1 nˆ q1 q1 1 † 1 · Δ = iq R 1 ai ai−Δ1 e z1 z1 i,Δ1 Δ1 = γ 1 † q ai ai−Δ1 z1 i,Δ1 1 1 −iq ·R iq (R −RΔ ) † = γ e 1 i e 2 i 1 a a q q1 q2 z1 N , i,Δ1 q1 q2 1 −iq ·RΔ † = γ e 2 1 δ , a a q q1 q2 q1 q2 z1 Δ , 1 q1 q2 1 −iq ·RΔ † = γ e 1 1 a a q q1 q1 z1 Δ q1 1 = γ γ q q1 nˆ q1

q1 666 C Solutions to Problems

Problem 7.18

1 iq·RΔ γq = e 1 z1 Δ1 ∈{ ± , , , , ± , , , , ± } RΔ1 a( 1 0 0) a(0 1 0) a(0 0 1)

= 1 − γ c1 N q(1 q):

1 1 = 1 N q 1 1 1 iq·RΔ γq = e 1 N N z1 q q Δ1 = 1 δ = Δ1,0 0 z1 Δ1

since R0 = (0, 0, 0) is not a nearest neighbour.

c1 = 1

= 1 − γ 2 c2 N q(1 q) :

1 c = 1 − 2γ + γ 2 2 N q q q · + 1 2 1 1 iq RΔ RΔ γ = e 1 1 N q N 2 q q z1 Δ Δ 1 1 1 = δΔ ,−Δ 2 1 1 z1 Δ Δ 1 1 1 = 1 z2 1 Δ1 = 1 z1 1 c2 = 1 + z1

= 1 − γ 3 c3 N q(1 q) : C Solutions to Problems 667

1 c = 1 − 3γ + 3γ 2 − γ 3 3 N q q q q · + + 1 3 1 1 iq RΔ RΔ RΔ γ = e 1 1 1 N q N 3 q q z1 Δ Δ Δ 1 1 1 1 = δΔ,−Δ −Δ 3 1 1 1 z1 Δ Δ Δ 1 1 1 = 0 since (−RΔ − RΔ ) for s.c. lattice cannot be 1 1 a nearest neighbour.

Problem 7.19 Calculation of the anisotropy contribution of the dipole interaction Ha =−3 Dij Si · eij S j · eij i, j in spin wave approximation: √ S+ = 2Sa i √ i † S− = 2Sa i i z = − † Si S ai ai

This gives Ha =−3 Dij Si · eij S j · eij i, j ) =− x + y + z ∗ 3 Dij Si xij Si yij Si zij , i j = ∗ x + y + z S j xij S j yij S j zij < =− 2 x x + 2 y y + 2 z z+ 3 Dij xijSi S j yijSi S j zijSi S j , i j + x y + y x + x z + z x + xijyij Si S j Si S j xijzij Si S j Si S j = + y z + z y yijzij Si S j Si S j < 2S † † =−32 D x2 a + a a + a + ij ij 4 i i j j i, j 668 C Solutions to Problems −2S † † + 2 − − + yij ai ai a j a j 4 + 2 2 − + + + zij S S ni n j ni n j + 2S + † − † + xijyij ai ai a j a j 4i + − † + † + ai ai a j a j √ + 2S + † − + xijzij ai ai S n j 2 + − + † + (S ni ) a j a j √ + 2S − † − + yijzij ai ai S n j 2i = + − − † (S ni ) a j a j < S † † † † =−32 D x2 a a + a a + a a + a a − ij 2 ij i j i j i j i j i, j S † † † † − 2 + − − + yij ai a j ai a j ai a j ai a j 2 + 2 − − + zijS S ni n j + S − † + † − † †+ xijyij ai a j ai a j ai a j ai a j 2i + + † − † − † † + ai a j ai a j ai a j ai a j √ 2S † † + S x z a + a + a + a + 2 ij ij i i j j √ = + 2S − † + − † S yijzij ai ai a j a j 2i =− 2 2 − − 3 S Dijzij (S 2ni ) i, j † † 1 1 − 2 2 − 2 + + 3 S Dij ai a j xij yij ixijyij , 2 2 i j 1 1 + a a x2 − y2 − ix y + i j 2 ij 2 ij ij ij † † 1 1 + a a + a a x2 + y2 + i j i j 2 ij 2 ij √ √ = + † − − † S 2Sxijzij ai ai iS 2Syijzij ai ai

The mixed terms vanish. In order to see that, hold zij ﬁxed. All Ri and R j with = , =− zij const deﬁne an x y-plane. In this for each R j there is a R j with xij xij C Solutions to Problems 669 =− | |=| | = = and yij yij and Rij Rij , i.e. Dij Dij. That means j Dijxijzij = 0 in the plane and therefore in the entire space. Analogously: i Dijyijzij 0. Therefore what remains is (Fig. C.2) =− 2 2 − − Ha 3 S Dijzij (S 2ni ) i, j < † † 1 1 − 32 S D a a x + iy 2 + a a x − iy 2+ ij i j 2 ij ij i j 2 ij ij i, j = + † 2 + 2 ai a j xij yij

Fig. C.2 Graphical illustration for the evaluation of the “mixed” terms in Ha (see text)

Problem 7.20 First derivative

dω eax(1 + ϕ)ea = xω − da (1 + ϕ)ea − ϕ 2 = ω(x − α) with

(1 + ϕ)ea ϕ α = = 1 + (1 + ϕ)ea − ϕ (1 + ϕ)ea − ϕ

Second derivative: 670 C Solutions to Problems

d2ω ϕ(1 + ϕ)ea = 2ω(x − α)2 + ω da2 (1 + ϕ)ea − ϕ 2 ϕ = 2ω(x − α)2 + 2ωα + ϕ a − ϕ (1 )e = 2ω (x − α)2 + α(α − 1)

Coefﬁcient of the second term in the differential equation:

(1 + ϕ) + ϕe−a ϕ = α + = α + α − 1 (1 + ϕ) − ϕe−a (1 + ϕ)ea − ϕ

Thus what remains is

2ω (x − a)2 + α(α − 1) + + 2ω(x − α)(2α − 1) − 2 S(S + 1)ω = 0 x2 − 2ax + α2 + α2 − α + 2xα− − x − 2α2 + α − S(S + 1) = 0 x2 − x − S(S + 1) = 0

x1 =−S; x2 = S + 1 general solution:

Ω(a) = c1ω(−S, a) + c2ω(S + 1, a)

Problem 8.1 For the Hamiltonian

= μν + H0 Tij ciμσ c jνσ ijσ μν holds after substituting the Fourier integrals: C Solutions to Problems 671 1 μν · − H = T eik (Ri R j ) ∗ 0 N k ijσ μ,ν i k ∗ 1 −iq·R −p·R + mμ mν ∗ e i j c c σ U U N qmσ pm qσ pσ i q,p,m,m ∗ μν μ ν + = m m δ δ Tk Uqσ Upσ k,q k,pcqmσ cpm σ k,q,p m,m σ,ν,μ ∗ μν ν mμ + = m Tk Ukσ Ukσ ckmσ ckm σ k m,m σ,ν,μ ∗ mμ mμ + = ε m (k) Ukσ Ukσ ckmσ ckm σ , σ,μ km m + = ε δ m (k) m ,m ckmσ ckm σ , , ,σ km m = ε + m (k)ckmσ ckmσ k,m,σ

Problem 8.2 Wannier representation:

† 1 H = T c c σ + U n σ n −σ ij iσ j 2 i i ijσ i,σ

Hopping integrals:

1 · − T = ε(k)eik (Ri R j ) ij N k

Construction operators:

1 ik·Ri ciσ = √ ckσ e N k

One-particle part: 672 C Solutions to Problems = † H0 Tijciσ c jσ ijσ 1 † ik·(R −R ) −ip·R iq·R = ε(k)c c σ e i j e i e j 2 pσ q N , , ,σ , k p q i j = ε † δ δ (k)cpσ cqσ k,p k,q , , ,σ k p q = ε † (k)ckσ ckσ kσ

Interaction part:

1 H = U n σ n −σ 1 2 i i iσ 1 1 † † = U c c σ c c −σ ∗ 2 k1σ k2 k3−σ k4 2 N ... ,σ k1 k4 − + − + ∗ ei( k1 k2)Ri ei( k3 k4)Ri i 1 † † = U δk +k ,k +k c σ ck σ c −σ ck −σ 2 1 3 2 4 k1 2 k3 4 k1...k4,σ k1 → k + q, k3 → p − q, k4 → p, k2 → k 1 † † ⇔ H = U c c − −σ c −σ c σ 1 2 k+qσ p q p k k,p,q,σ

Total: † 1 † † H = ε(k)c c σ + U c c − σ c −σ c σ kσ k 2 k+qσ p q p k kσ k,p,q,σ

Problem 8.3 Band limit: U→0 G σ (E) = k E − ε(k) + μ

Zero bandwidth limit (W = 0): − W=0 (1 n−σ ) n−σ Gσ = + E − T0 + μ E − T0 − U + μ

Stoner approximation: C Solutions to Problems 673

(Stoner) = Gkσ (E) E − ε(k) − Un−σ + μ obviously satisﬁes the band limit but not the limit of inﬁnitely narrow band. Problem 8.4 = εαβ † ,α,β∈ { , } H σ (k)ckασ ckβσ A B kσ,α,β

AA 1 1 εσ (k) = ε(k) + Un − zσ Um − μ 2 2 BB 1 1 εσ (k) = ε(k) + Un + zσ Um − μ 2 2 AB BA∗ εσ (k) = t(k) = εσ (k)

Here we have used

AA BB < nσ >=−< nσ >= nσ

m = n↑ − n↓; m A =−m B = m

Green’s functions:

α,β =<< † >> Gkσ (E) ckασ ; ckβσ E

1. Quasiparticle energies: Aβ [ckAσ , H]− = εσ (k)ckβσ β AA AB = εσ (k)ckAσ + εσ (k)ckBσ

Equation of motion:

− ε AA AA = + ε AB BA (E σ (k))Gkσ (E) σ (k)Gkσ (E)

With

BA BB [ckBσ , H]− = εσ (k)ckAσ + εσ (k)ckBσ

also follows: 674 C Solutions to Problems

− εBB BA = εBA AA (E σ (k))Gkσ (E) σ (k)Gkσ (E) εBA(k) G BA E = σ G AA E kσ ( ) BB kσ ( ) E − εσ (k)

AA Substituting in the equation of motion for Gkσ (E):

|ε AB(k)|2 E − ε AA k G AA E = + σ G AA E ( σ ( )) kσ ( ) BB kσ ( ) E − εσ (k) E − εBB(k) G AA(E) = σ kσ AA BB AB 2 (E − εσ (k))(E − εσ (k)) −|εσ (k)|

Poles:

AA BB AB 2 ! (E± − εσ (k))(E − εσ (k)) −|εσ (k)| = 0

1 AA BB E±(k) = (εσ (k) + εσ (k)) (2 1 AA BB 2 AB 2 ± εσ (k) − εσ (k) +|εσ (k)| 4

That means the spin-independent quasiparticle energies: ( 1 1 E±(k) = ε(k) + Un ± U 2m2 +|t(k)|2 − μ 2 4

2. Spectral weights:

E − εBB k AA = σ ( ) Gkσ (E) (E − E+(k))(E − E−(k)) α(±) = AA − σ (k) lim Gkσ (E)(E E±(k)) E→E±(k) BB E±(k) − ε (k) α(±) k = σ σ ( ) − E±(k) E∓(k)

(±) 1 Um ⇔ ασ (k) = 1 ∓ zσ 2 U 2m2 + 4|t(k)|2

Spectral weights are obviously spin dependent Spectral density:

(±) = α(+) δ − + α(−) δ − Skσ (E) σ (k) (E E+(k)) σ (k) (E E−(k)) C Solutions to Problems 675

Quasiparticle density of states:

) (±) 1 1 (+) ρ (E) = α (k)δ(E − μ − E+(k)) σ N σ k * (−) +ασ (k)δ(E − μ − E−(k))

k: wavevector of the ﬁrst Brillouin zone of the sub-lattice.

Problem 8.5 Let x = 0:

1 β lim = lim βe−β|x| = 0 2 β→∞ 1 + cosh(βx) β→∞

The expression diverges for x = 0. In addition it holds

+∞ ∞ 1 β β dx lim = lim dx , 2 β→∞ 1 + cosh(βx) β→∞ 1 + cosh(βx) −∞ 0

∞ ∞ ∞ β 1 1 dx = dy = dy 1 + cosh(βx) 1 + cosh(y) 2 y 2 cosh 2 0 0 0 ∞ 1 ∞ = dz = tanh z cosh2 z 0 0 = 1 − 0 = 1.

Thus we have the deﬁning properties of the δ-function satisﬁed!

Problem 8.6 According to (8.97) the Hubbard Hamiltonian can be written as follows:

+ 2U 1 z H = ε(k)c c σ − σ(k) · σ(−k) + U Nˆ − 2μ B σ (0) kσ k 3N 2 B 0 kσ k

1. We calculate the commutator termwise: 676 C Solutions to Problems # $ σ + , + (k) Tmncmσ cnσ m,n,σ − −ik·Ri + + = e Tmn c ↑ci↓, c σ cnσ i m − m,n,σ i − · + + = ik Ri δ δ − δ δ e Tmn im σ ↓ci↑cnσ in σ ↑cmσ ci↓ i m,n,σ + − · + − · = ik Rm − ik Rn Tmn cm↑cn↓e cm↑cn↓e m,n − · − · + = ik Rm − ik Rn Tmn e e cm↑cn↓ m,n

We further calculate

# $ σ +(k), σ(p)σ(−p) ! p − 1 = σ +(k),σz(p)σ z(−p + σ +(p)σ −(−p) 2 p " 1 + σ −(p)σ +(−p) 2 − < = σ z σ + ,σz − + σ + ,σz ∗ (p) (k) ( p) − (k) (p) − p ∗σ z − + 1σ + σ + ,σ− − ( p) (p) (k) ( p) − 2 1 + σ +(k),σ−(p) σ +(−p) 2 − ) = −σ z(p)σ +(k − p) − σ +(k + p)σ z(−p)+ p * +σ +(p)σ z(k − p) + σ z(k + p)σ +(−p) = 0

One recognizes this when one replaces p by p + k in the term before the last and p by p − k in the last term: C Solutions to Problems 677 + −ik·Ri + + σ (k), Nˆ = e c ↑ci↓, c σ cmσ − i m − i,m,σ < = − · + + = ik Ri δ δ − δ δ e σ ↓ imci↑cmσ im σ ↑cmσ ci↓ i,m,σ < = − · + + = ik Ri − e ci↑ci↓ ci↑ci↓ i = 0 σ + ,σz =−σ + (k) (0) − (k)

Therefore what remains is σ + , = μ σ + (k) H − 2 B B0 (k) − · − · + + ik Rm − ik Rn Tmn e e cm↑cn↓ m,n

This corresponds to (8.106).

2. For the double commutator we need the results of 1.: −ik·Rm −ik·Rn ik·Ri + + Tmn e − e e c ↑cn↓, c ↓ci↑ m i − m,n i −ik·Rm −ik·Rn = Tmn e − e ∗ m,n < = · + + ∗ ik Ri δ − δ e ni cm↑ci↑ mici↓cn↓ )i − + = ik(Rn Rm ) − − Tmn e 1 cm↑cn↑ , m n * − + − − ik(Rm Rn ) δ 1 e micm↓cn↓

So that it follows: σ +(k), H ,σ−(−k) − − · − + = μ σ z + izσ k (Rn Rm ) − 4 B B0 (0) Tmn e 1 cmσ cnσ mnσ

This is exactly (8.107).

Problem 8.7 We use (8.137) and (8.138) and then have the following determining equation for the chemical potential μ: 678 C Solutions to Problems

2 f−(T0) ! n = nσ + n−σ = = 1 1 + f−(T0) − f−(T0 + U)

That is the same thing as

f−(T0) = 1 − f−(T0 + U) β(T0+U−μ) 1 = − 1 = e β −μ 1 β + −μ β + −μ e (T0 ) + 1 e (T0 U ) + 1 e (T0 U ) + 1 β −μ −β + −μ e (T0 ) + 1 = 1 + e (T0 U )

T0 − μ =−(T0 + U − μ)

2T0 + U = 2μ U μ = T + 0 2

Problem 8.8 Substituting the spectral density (8.139) for

1 nσ = n−σ = 2 in (8.77), it directly follows:

1 Z = =− f− (T ) + f− (T + U) W 0 2 0 0

On the other hand substituting in (8.78) gives

NW=0 =−f−(T0) + f−(T0 + U)

For the susceptibility holds (8.79)

ZW=0 χø W=0 = 2μB 1 + NW=0

The chemical potential for half-ﬁlling is known from Problem 8.7:

U μ(n = 1) = T + 0 2

So that we calculate C Solutions to Problems 679 + − = 1 − 1 ( f−(T0 U) f−(T0)) β + −μ β −μ (T0 U ) + (T0 ) + e 1 e 1 = 1 − 1 β U −β U e 2 + 1 e 2 + 1 −β U β U 4 − 4 = e e β U −β U e 4 + e 4 W N = =−tanh(β ) W 0 4

For ZW=0 we need the derivative of the Fermi function:

β(E−μ) e f− (E) =−β (eβ(E−μ) + 1)2 1 =−β 2 1 β(E−μ) − 1 β(E−μ) e 2 + e 2

Then it follows: ⎛ 1 β ⎜ 1 − − + − + = ⎝ + f (T0) f (T0 U) 2 2 2 −β U +β U e 4 + e 4 ⎞ 1 ⎟ + ⎠ 2 +β U −β U e 4 + e 4 β 1 1 = · 2 · · 2 4 2 β U cosh ( 4 ) 1 1 Z = = β W 0 4 2 β U cosh ( 4 )

Susceptibility:

1 1 cosh2(β U ) χø = βμ 4 2 B − β U 1 tanh( 4 ) 1 1 + tanh(β U ) = βμ 4 2 B 2 β U − 2 β U cosh ( 4 )(1 tanh ( 4 )) 1 1 + tanh(β U ) = βμ 4 2 B 2 β U − 2 β U cosh ( 4 ) sinh ( 4 )

Then we ﬁnally have 680 C Solutions to Problems 1 βU χø = = βμ 1 + tanh W 0 2 B 4

Problem 8.9 Complete basis: A (1) = + | E1 c1σ 0 A (1) = + | E2 c2σ 0

Hamiltonian matrix: 0 t H (1) = t 0

Diagonalization: Det H (1) − E =! 0 −Et Det = E2 − t2 t −E ⇔ (1) =− (1) =+ E1 t ; E2 t

(1) For E1 : tt α ! 1 = 0 tt α2

⇔ t (α1 + α2) = 0 ⇔ α2 =−α1

Normalization: A (1) 1 + + E = √ c σ |0 − c σ |0 1 2 1 2 (1) For E2 :

−tt β ! 1 = 0 t −t β2

⇔ t (−β1 + β2) = 0 ⇔ β1 = β2

Normalization: A (1) 1 + + E = √ c σ |0 + c σ |0 2 2 1 2 C Solutions to Problems 681

Problem 8.10 1. We use + ciσ , c σ = δijδσσ j + + + ciσ , c jσ = c σ , c σ = 0 + i j + + c σ |0 = 0; 0| c = 0 Bi A jσ ε(2) | ε(2) = | + + | 1 1 0 c2−σ c1σ c1σ c2−σ 0 = | + | = | = 0 c2−σ c2−σ 0 0 0 1 analogously: B A ε(2) ε(2) = = , ··· , i i 1 i 2 4 B A ε(2) ε(2) = | † † | 1 2 0 c2−σ c1σ c2σ c1−σ 0 B A = † † 0 c2−σ c2σ c1−σ c1σ 0 = 0 B A ε(2) ε(2) = | † † | = 1 3 0 c2−σ c1σ c1σ c1−σ 0 0 B A ε(2) ε(2) = | † † | = 1 4 0 c2−σ c1σ c2σ c2−σ 0 0 B A ε(2) ε(2) = | † † | = 2 3 0 c1−σ c2σ c1σ c1−σ 0 0 B A ε(2) ε(2) = | † † | = 2 4 0 c1−σ c2σ c2σ c2−σ 0 0 B A ε(2) ε(2) = | † † | = 3 4 0 c1−σ c1σ c2σ c2−σ 0 0

Therefore it holds B A ε(2) ε(2) = δ , = ... i j ij; i j 1 4 682 C Solutions to Problems

A (2) † † † † ε = + | + H 2 t c1σ c2σ c2σ c1σ c2σ c1−σ 0 σ 1 † † + U n σ n −σ c c |0 2 i i 2σ 1−σ iσ † † † † = tc c |0 − tc c |0 1σ 1−Aσ A2−σ 2σ = ε(2) + ε(2) t 3 4

A (2) † † † † ε = + | + H 3 t c1σ c2σ c2σ c1σ c1σ c1−σ 0 σ 1 † † + | U n1σ n1−σ c1σ c1−σ 0 2 σ † † † † † = | + | t c1σ c2σ c1σ c1−σ 0 Un1σ n1−σ c1σ c1−σ 0 σ = † † − + † | + † † | tc2σ c1−σ tc2−σ c1σ 0 Uc1σ c1−σ 0 A A A = ε(2) + ε(2) + ε(2) t 1 2 U 3 C Solutions to Problems 683 A (2) † † † † ε = + | + H 4 t c1σ c2σ c2σ c1σ c2σ c2−σ 0 σ 1 † † + | U niσ ni−σ c2σ c2−σ 0 2 iσ = † † | − † † | + t c1σ c2−σ 0 c1−σ c2σ 0 † † + Un σ n −σ c c |0 2 2 2σ 2−σ = † † | + † † | + † † | t c1σ c2−σ 0 c2σ c1−σ 0 Uc2σ c2−σ 0 A A A = ε(2) + ε(2) + ε(2) t 1 2 U 4

So that we get from the matrix elements B A ε(2) ε(2) , = ... i H j ; i j 1 4 the Hamiltonian matrix: ⎛ ⎞ 00 tt ⎜ 00 tt⎟ H (2) = ⎜ ⎟ ⎝ ttU0 ⎠ tt0 U

3. The eigenvalues are determined from the secular determinant: det H (2) − E1l =! 0

⎛ ⎞ −E 0 tt ⎜ 0 −Et t⎟ det ⎜ ⎟ ⎝ ttU− E 0 ⎠ tt 0 U − E ⎛ ⎞ −Et t =−Edet ⎝ tU− E 0 ⎠ + t 0 U − E ⎛ ⎞ ⎛ ⎞ 0 −Et 0 −Et + tdet ⎝ tt 0 ⎠ − tdet ⎝ ttU− E ⎠ ttU− E tt 0 = (−E) (−E)(U − E)2 − 2t2(U − E) + + t(t3 − t3 + E(U − E)t) − t(−E(U − E)t + t3 − t3) = E(U − E)(E(U − E) + 4t2) 684 C Solutions to Problems

It must therefore hold

0 = E(U − E)(E(U − E) + 4t2)

Eigenvalues

(2) = E1 E− (2) = E2 0 (2) = E3 U (2) = E4 E+ with

( 1 1 E± = U ± U 2 + 4t2 2 4

Eigenstates: (2) = E1 E−

⎛ ⎞ ⎛ ⎞ −E− 0 tt x1 ⎜ ⎟ ⎜ ⎟ ⎜ 0 −E− tt⎟ ⎜ x2 ⎟ ! ⎝ ⎠ ⎝ ⎠ = 0 ttU− E− 0 x3 tt 0 U − E− x4

−E−x1 + t(x3 + x4) = 0

−E−x2 + t(x3 + x4) = 0

t(x1 + x2) + (U − E−)x3 = 0

t(x1 + x2) + (U − E−)x4 = 0 E− x = x ; x = x ; x = x ; 1 2 3 4 3 2t 2

We deﬁne (8.168)

E± γ± = 2t

Normalization: C Solutions to Problems 685

2 + 2 + 2 + 2 = x1 x2 x3 x4 1 2 + 2 = 2x1 2x3 1 2 + γ 2 2 = 2x1 2 −x 1 1 2 = x1 2 2(1 + γ−) 1 x1 = 5 = x2 2 2(1 + γ−) γ− x3 = 5 = x4 2 2(1 + γ−)

A A A 1 (2) = 5 ε(2) + ε(2) E1 1 2 2 2(1 + γ−) A A + γ ε(2) + ε(2) − 3 4 ; (8.164)

(2) = E2 0

⎛ ⎞ ⎛ ⎞ 00 tt y1 ⎜ ⎟ ⎜ ⎟ ⎜ 00 tt⎟ ⎜ y2 ⎟ ! ⎝ ⎠ ⎝ ⎠ = 0 ttU0 y3 tt0 U y4

t(y3 + y4) = 0

t(y1 + y2) + Uy3 = 0

t(y1 + y2) + Uy4 = 0

y3 =−y4; y3 =+y4; y3 = y4 = 0; y1 =−y2;

Normalization:

1 y1 = √ =−y2 2

A A A 1 E(2) = √ ε(2) − ε(2) ; (8.165) 2 2 1 2

(2) = E3 U 686 C Solutions to Problems ⎛ ⎞ ⎛ ⎞ −U 0 tt z1 ⎜ ⎟ ⎜ ⎟ ⎜ 0 −Utt⎟ ⎜ z2 ⎟ ! ⎝ ⎠ ⎝ ⎠ = 0 tt00 z3 tt00 z4

−Uz1 + t(z3 + z4) = 0

−Uz2 + t(z3 + z4) = 0

t(z1 + z2) = 0

z1 = z2; z1 =−z2; z1 = z2 = 0; z3 =−z4;

Normalization:

1 z3 = √ =−z4 2

A A A 1 E(2) = √ ε(2) − ε(2) ; (8.166) 3 2 3 4

(2) = E4 E+

⎛ ⎞ ⎛ ⎞ −E+ 0 tt w1 ⎜ ⎟ ⎜ ⎟ ⎜ 0 −E+ tt⎟ ⎜ w2 ⎟ ! ⎝ ⎠ ⎝ ⎠ = 0 ttU− E+ 0 w3 tt 0 U − E+ w4

(2) = γ γ Formally as for E1 E−, only − is replaced by +:

A A A 1 (2) = 5 ε(2) + ε(2) + E4 1 2 2 2(1 + γ+) A A + γ ε(2) + ε(2) + 3 4 ; (8.167)

4. (i) C Solutions to Problems 687 B A B A (2) † (1) 1 (2) † † † E c σ E = √ E c σ (c −σ − c −σ ) 0 1 1 1 2 1 1 1 2 B A B A 1 1 = √ E(2) ε(2) − √ E(2) ε(2) 2 1 3 2 1 1 1 1 = 5 (γ− − 1)√ 2 2(1 + γ−) 2

1 γ− − 1 = 5 ; 2 2 1 + γ− (2) − (1) = + E1 E1 E− t

(ii)

B A B A B A (2) † (1) 1 (2) (2) (2) (2) E c σ E = √ E ε − E ε 2 1 1 2 2 3 2 1 1 = ; 2 (2) − (1) =+ E2 E1 t

(iii)

B A B A B A (2) † (1) 1 (2) (2) (2) (2) E c σ E = √ E ε − E ε 3 1 1 2 3 3 3 1 1 = ; 2 (2) − (1) = + E3 E1 U t

(iv)

B A B A B A (2) † (1) 1 (2) (2) (2) (2) E c σ E = √ E ε − E ε 4 1 1 2 4 3 4 1 1 1 = 5 (γ+ − 1) ; 2 2 1 + γ+ (2) − (1) = + E4 E1 E+ t

(v) 688 C Solutions to Problems B A B A B A (2) † (1) 1 (2) (2) (2) (2) E c σ E = √ E ε + E ε 1 1 2 2 1 3 1 1 1 1 = 5 (1 + γ−); 2 2 1 + γ− (2) − (1) = + E1 E2 E− t

(vi) B A B A B A (2) † (1) 1 (2) (2) (2) (2) E c σ E = √ E ε + E ε 2 1 2 2 2 3 2 1 1 = ; 2 (2) − (1) =− E2 E2 t

(vii) B A B A B A (2) † (1) 1 (2) (2) (2) (2) E c σ E = √ E ε + E ε 3 1 2 2 3 3 3 1 1 = ; 2 (2) − (1) = − E3 E2 U t

(viii) B A B A B A (2) † (1) 1 (2) (2) (2) (2) E c σ E = √ E ε + E ε 4 1 2 2 4 3 4 1 1 1 = 5 (γ+ + 1) ; 2 2 1 + γ+ (2) − (1) = − E4 E2 E+ t

For the partition function of the one-particle system holds

Z = e+βt + e−βt ; and therefore

−β E(1) βt e 1 = e = 1 Z eβt + e−βt 1 + e−2βt −β E(1) −βt −2βt e 2 = e = e Z eβt + e−βt 1 + e−2βt C Solutions to Problems 689

Then with (8.147) the density of states is given by

ρ(−σ ) = ρ(−σ ) + ρ(−σ ) σ (E) tσ (E) Uσ (E)

1 ρ σ (E) = ∗ t + e−2βt 4(1 ) 2 (1 − γ−) ∗ δ(E − (E− + t)) + δ(E − t)+ 1 + γ 2 − + γ 2 (1 −) −2βt −2βt + δ − − + + δ + 2 e (E (E t)) e (E t) 1 + γ−

1 ρ σ (E) = ∗ U + e−2βt 4(1 ) 2 (1 − γ+) ∗ δ − + + + δ − + + 2 (E (E t)) (E (U t)) 1 + γ+ + γ 2 (1 +) −2βt + e δ(E − (E+ − t)) 1 + γ 2 + = +e−2βt δ(E − (U − t))

Problem 8.11 A (3) 1. We demonstrate the correctness of the eigenstates by substituting of E1,2 in the corresponding eigenvalue equation: A (3) 1 † † = √ + ∗ H0 E1,2 t c1σ c2σ c2σ c1σ 2 σ ∗ † ∓ † † † | c1−σ c2−σ c1σ c2σ 0 t † † = √ + ∗ c1−σ c2−σ c2−σ c1−σ 2 ∗ † ∓ † † † | c1−σ c2−σ c1σ c2σ 0 t † † = √ ∓ − + c1−σ (1 c2−σ c2−σ ) 2 + † − † † † | c2−σ (1 c1−σ c1−σ ) c1σ c2σ 0 t † † † † = √ ∓ | c2−σ c1−σ c1σ c2σ 0 2 A =∓ (3) t E1,2 690 C Solutions to Problems A (3) 1 1 = √ · + ∗ H1 E1,2 U (n1σ n1−σ n2σ n2−σ ) 2 2 σ ∗ † ∓ † † † | c1−σ c2−σ c1σ c2σ 0 U = √ (n1σ n1−σ + n2σ n2−σ ) ∗ 2 ∗ † ∓ † † † | c1−σ c2−σ c1σ c2σ 0 U † † † † = √ n1σ n1−σ c −σ ∓ n2σ n2−σ c −σ c σ c σ |0 2 1 2 1 2 U † † † † = √ n1σ c −σ ∓ n2σ c −σ c σ c σ |0 2 1 2 1 2 U † † † † † † = √ −c σ c −σ c σ ∓ c σ c −σ c σ |0 2 1 1 2 2 2 1 U † † † † = √ c −σ ∓ c −σ c σ c σ |0 2 1 2 1 2 A = (3) U E1,2

Therefore they are indeed eigenstates with eigenenergies:

(3) = − (3) = + E1 U t; E2 U t

2. Density of states

B A 2 ρ(σ,−σ ) = (3) † (2) δ − (3) − + σ (E) En c1σ E1 (E (En E−)) n B A 2 + (1) | | (2) δ − − (1) Em c1σ E1 (E (E− Em )) m

A = ρ(σ,−σ ) (2) At T 0, for the averaging in σ (E), the ground state E1 of the two- electron system can be used (see Problem 8.10, Eq. (8.164)) with the ground (2) = state energy E1 E− (8.163). Matrix elements: C Solutions to Problems 691 B A (3) † (2) E1,2 c1σ E1 B 1 † † † = 5 (3) + E1,2 c1σ c1σ c2−σ 2 2(1 + γ−)

2 A + † † + γ † † c2σ c1−σ − ciσ ci−σ 0 i=1 B 1 † † † = 5 (3) + E1,2 (c1σ c2σ c1−σ 2 2(1 + γ−) A + γ † † † −c1σ c2σ c2−σ ) 0 B A 1 1 † † † = 5 0 c2σ c1σ c1−σ c1−σ c1σ c2σ ) 0 2 2 (1 + γ−) B A 1 1 † † † ∓ 5 γ − 0 c2σ c1σ c2−σ c1σ c2σ c2−σ ) 0 2 2 (1 + γ−) 1 1 1 1 = 5 0 | 0 ∓ 5 γ− 0 | 0 2 2 2 2 (1 + γ−) (1 + γ−) ∓ γ = 1 51 − 2 2 (1 + γ−)

With (8.160) and (8.161) we calculate

B A B A (1) (2) 1 (2) E , c1σ E = √ 0 |(c1−σ ∓ c2−σ )c1σ | E 1 2 1 2 1

Density of states: 692 C Solutions to Problems

− γ 2 (σ,−σ ) (1 −) ρ = {δ − − − − + σ (E) 2 (E (U t E )) 4(1 + γ−) +δ(E − (E− + t))}+ 2 (1 + γ−) {δ − + − − + 2 (E (U t E )) 4(1 + γ−) +δ(E − (E− − t))}

Problem 8.12 The expression

i= j B A 1 † T c c −σ (2n σ − 1) N ij i−σ j i i, j should be expressed in terms of the one-electron spectral density!

(a) Spectral theorem:

i= j B A 1 † T c c −σ N ij i−σ j i, j B A B A 1 † † = T c c −σ − T c c −σ N ij i−σ j 0 i−σ i i, j +∞ 1 1 = T dE − ImG −σ (E − μ) f−(E) − N ij π ji i, j −∞ +∞ 1 − T dE − ImG −σ E − μ f− E 0 π ii ( ( ) −∞ +∞ 1 1 ik (R −R ) = e 1 i j ε(k ) dE f−(E) · N N 1 i, j k1 −∞ 1 ik (R −R ) 1 · e 2 j i − ImG −σ (E − μ − N π k2 k2 C Solutions to Problems 693

+∞ 1 1 − T dE f−(E) − ImG −σ (E − μ) 0 N π k −∞ k +∞ 1 = dE f−(E)δ , δ , ε(k )S −σ (E − μ) − N k1 k2 k1 k2 1 k2 k1,k2−∞ +∞ 1 − T dE f−(E)S −σ (E − μ) 0 N k k −∞ +∞ 1 = (ε(k − T ) dE f−(E)S −σ (E − μ) N 0 k k −∞

(b) Real expectation value:

B A B A B A † =! † = † ci−σ c j−σ niσ niσ c j−σ ci−σ c j−σ niσ ci−σ

i= j B A 1 † ⇔ T c c −σ n σ N ij i−σ j i i, j B A B A 1 † † = T c n σ c −σ − T c n σ c −σ N ij j−σ i i 0 i−σ i i i, j +∞ 1 1 = T − dE f−(E)ImΓ −σ (E − μ) − N ij π iii; j i, j −∞ +∞ 1 − T dE f− E − ImΓ −σ E − μ 0 ( ) π iii;i ( ) −∞

The “higher” Green’s function is deﬁned in (8.124). From (8.125) one reads off

1 ImΓ −σ (E − μ) = (Eδ − T )ImG −σ (E − μ) iii; j U im im mi m

Therewith follows: 694 C Solutions to Problems

i= j 1 † T < c c −σ n σ > N ij i−σ j i i, j +∞ 1 = (T − T δ ) dE f−(E) ∗ N ij 0 ij i, j −∞ 1 ∗ (Eδ − T )S −σ (E − μ) U mi im mj m 1 1 − = eik1(Ri R j )(ε(k − T ) ∗ N N 3 1 0 i, j k1,k2,k3 +∞ 1 ik (R −R ) ik (R −R ) ∗ dE f−(E) e 2 i m e 3 m j ∗ −∞ U m ∗ − ε − μ (E (k2))Sk3−σ (E ) +∞ 1 1 = (ε(k ) − T ) dE f−(E) (E − ε(k )) ∗ N 1 0 U 2 k1,k2,k3 −∞ ∗ − μ δ δ δ Sk3−σ (E ) k1,−k2 k2,k3 k1,−k3 +∞ 1 1 = (ε(k) − T ) dE f−(E) (E − ε(k)) ∗ N 0 U k −∞

∗Sk−σ (E − μ)

Together with the result of (a) we have found

n−σ (1 − n−σ )B−σ i= j B A 1 † = T c c −σ (2n σ − 1) N ij i−σ j i i, j +∞ 1 2 = (ε(k) − T ) dE f−(E) (E − ε(k)) − 1 ∗ N 0 U k −∞

∗Sk−σ (E − μ)

This is exactly (8.217)

Problem 8.13 0th spectral moment:

B A (0) † (0) M σ = ciσ , c σ = δij ⇔ M σ = 1(a) ij j − k C Solutions to Problems 695

1st spectral moment:

B A (1) † M σ = [ciσ , H]− , c σ ij j + † [ciσ , H]− = (Tmn − μδmn) ciσ , c σ cnσ + m − m,n,σ 1 + , U ciσ nmσ nm−σ − 2 mσ = (Tmn − μδmn)δimδσσ cnσ + m,n,σ 1 + U δim(δσσ cmσ nm−σ + δσ −σ nmσ cm−σ ) 2 mσ = (Tin − μδin)cnσ + Uni−σ ciσ (b) n † [ciσ , H]− , c σ = (Tij − μδij) + Uδijni−σ (c) j + (1) = − μδ + δ Mijσ (Tij ij) Un−σ ij (d) (1) = ε − μ + Mkσ (k) Un−σ (e)

2nd spectral moment:

B A (2) † M σ = [ciσ , H]− , H , c σ (f) ij − j +

abbreviation : tij = Tij − μδij

Then with (b) holds

, , = , + , [ciσ H]− H − tin [cnσ H]− U ni−σ ciσ H − n 696 C Solutions to Problems , ni−σ ciσ H − † = tmn ni−σ ciσ , c σ cnσ + m − m,n,σ 1 + U n −σ c σ , n σ n −σ 2 i i m m − m,σ † = δ δ + δ δ − tmn( im σσ ni−σ cnσ im σ −σ ci−σ cnσ ciσ m,n,σ † 1 − δ δσ −σ c c −σ c σ ) + U (δ δσσ n −σ c σ n −σ + in mσ i i 2 im i m m mσ

+ δimδσ −σ ni−σ nmσ cm−σ ) = + † − tinni−σ cnσ tinci−σ cn−σ ciσ n n † 1 − t c −σ c −σ c σ + U(n −σ c σ n −σ + n −σ n −σ c σ ) mi m i i 2 i i i i i i m = + † − † + tim(ni−σ cmσ ci−σ cm−σ ciσ cm−σ ci−σ ciσ ) m

+ Uni−σ ciσ (g)

2 = Here we have used ni−σ ni−σ which is an identity valid for Fermions. Then what remains is [c σ , H]− , H i − = tin [cnσ , H]− + n + + † − † + U tim(ni−σ cmσ ci−σ cm−σ ciσ cm−σ ci−σ ciσ ) m 2 + U ni−σ ciσ (h)

With this follows: B A (2) = (1) + + δ † − Mijσ tinMnjσ Utijn−σ ijU tim ci−σ cm−σ n B A m − δ † + 2δ ijU tim cm−σ ci−σ U ijn−σ m C Solutions to Problems 697

Because of (8.268), the third and the fourth terms cancel each other. Then it remains with (d):

(2) = + + 2 δ Mijσ tintnj 2Utijn−σ U n−σ ij (i) n

After Fourier transformation follows:

(2) = ε − μ 2 + ε − μ + 2 Mkσ ( (k) ) 2( (k) )Un−σ U n−σ

3rd spectral moment:

& % (3) = , , , , † M σ [ciσ H]− H − H c σ ij − j +

For the triple commutator holds with (h):

[ciσ , H]− , H , H − − = , , + tin [cnσ H]− H − n † † + U tim (ni−σ cmσ + c −σ cm−σ ciσ − c −σ ci−σ ciσ ), H + i m − m + 2 , U ni−σ ciσ H − = , , + , , − tin [cnσ H]− H − U [ciσ H]− H − n < − , + , + U tin [cnσ H]− U tim ni−σ cmσ H − n m = † † + c −σ cm−σ ciσ , H − c −σ ci−σ ciσ , H (j) i − m −

Three commutators remain to be calculated: 698 C Solutions to Problems = , (I ) U tim ni−σ cmσ H − m † = U tim tst ni−σ cmσ , c σ ctσ + s − m s,t,σ 1 + 2 , U tim ni−σ cmσ nsσ ns−σ − 2 ,σ m s = U timtst {δmsδσσ ni−σ ctσ + , , ,σ m s t = † † +δ δ − δ δ + is σ −σ ci−σ ctσ cmσ ti σ −σ csσ ci−σ cmσ 1 2 + U t {δ δσσ n −σ c σ n −σ + 2 im ms i s s m,s,σ +δ δσ −σ n −σ n σ c −σ } ms i s s = + † − U timtmtni−σ ctσ U timtitci−σ ct−σ cmσ m,t m,t † 1 2 − U t t c −σ c −σ c σ + U t n −σ c σ n −σ + im si s i m 2 im i m m m,s m 1 2 + U t n −σ n −σ c σ 2 im i m m m = + † − U timtmtni−σ ctσ U timtit(ci−σ ct−σ cmσ , , m t m t − † ++ 2 ct−σ ci−σ cmσ ) U timni−σ nm−σ cmσ m

 † (I ), c σ = U timtmjni−σ + j + m + † − † + U tittij(ci−σ ct−σ ct−σ ci−σ ) t 2 + U tijni−σ n j−σ

We again use (8.268):

B A 7 8 † 2 (I ), c σ = Un−σ timtmj + U tij ni−σ n j−σ (k) j + m

In the same manner we calculate C Solutions to Problems 699 † (II) = U tim c −σ cm−σ ciσ , H i − m † † = U tim tst c −σ cm−σ ciσ , c σ ctσ + i s − m s,t,σ 1 2 † + U tim c −σ cm−σ ciσ , nsσ ns−σ 2 i − m s,σ < † = δ δ + U timtst σσ isci−σ cm−σ ctσ , , ,σ m s t = † † +δ δ − δ δ + σ −σ msci−σ ctσ ciσ σ −σ itcsσ cm−σ ciσ < 1 2 † + U t δ δσσ c c −σ c σ n −σ + 2 im is i−σ m s s m,s,σ † +δ δ + is σ −σ ci−σ cm−σ nsσ cs−σ † +δ δ + ms σ −σ ci−σ csσ ns−σ ciσ † +δ δ − ms σσ ci−σ nsσ cs−σ ciσ † −δ δσ −σ c n −σ c −σ c σ − is sσ s m i = † − δ δ is σσ nsσ cs−σ cm−σ ciσ = † + † − U timtitci−σ cm−σ ctσ U timtmtci−σ ct−σ ciσ , , m t m t − † + U timtsics−σ cm−σ ciσ m,s < 1 2 † † + U t c c −σ c σ n −σ + c c −σ n −σ c σ + 2 im i−σ m i i i−σ m i i m † † +c c −σ n σ c σ + c n σ c −σ c σ − i−σ m m i i−σ m m i = − † − † ci−σ niσ cm−σ ciσ niσ ci−σ cm−σ ciσ † † (II), c σ = U timtijc −σ cm−σ + j + i m + δ † − † + U ij tim(tmtci−σ ct−σ titct−σ cm−σ ) m,t < + 2 δ † + U tim ijci−σ cm−σ ni−σ m † † † + δ c c −σ n σ − δ c c −σ c c σ − ij i−σ m m mj i−σ m mσ i = −δ † − δ † † ijci−σ niσ cm−σ ijci−σ ciσ cm−σ ciσ = † + Utij timci−σ cm−σ m 700 C Solutions to Problems

+ δ † − † − U ij tim(tmtci−σ ct−σ titct−σ cm−σ ) m,t 2 † † − U tijc −σ c σ ciσ c j−σ + i j < = + 2δ † + † U ij tim ci−σ cm−σ ni−σ ci−σ cm−σ nmσ m † (III) =−U tim c −σ ci−σ ciσ , H m − m † † =−U tim tst c −σ ci−σ ciσ , c σ ctσ − m s − m s,t,σ 1 2 † − U tim c −σ ci−σ ciσ , nsσ ns−σ 2 m − m s,σ < † =− δ δ + U tim tst is σσ cm−σ ci−σ ctσ m s,t,σ † +δ δσ −σ c c σ c σ − is m−σ t i = † −δ δ − mt σ −σ csσ ci−σ ciσ < 1 2 † − U t δ δσσ c −σ c −σ c σ n −σ + 2 im is m i s s m,s,σ † +δ δ + is σ −σ cm−σ ci−σ nsσ cs−σ † +δ δ + is σ −σ cm−σ csσ ns−σ ciσ † +δ δ − is σσ cm−σ nsσ cs−σ ciσ † −δ δσ −σ c n −σ c −σ c σ − ms sσ s i i = † −δ δ ms σσ nsσ cs−σ ci−σ ciσ =− † − U timtitcm−σ ci−σ ctσ , m t − † + U timtitcm−σ ct−σ ciσ , m t + † − U timtsmcs−σ ci−σ ciσ m,s < 1 2 † − U t c −σ c −σ c σ n −σ + 2 im m i i i m + † − cm−σ ci−σ ni−σ ciσ − † − cm−σ nmσ ci−σ ciσ C Solutions to Problems 701 = − † nmσ cm−σ ci−σ ciσ =− † + † + U timtit(cm−σ ci−σ ctσ cm−σ ct−σ ciσ ) , m t + † − U timtsmcs−σ ci−σ ciσ m,s < = − 2 † − † U tim cm−σ ci−σ ni−σ ciσ cm−σ ci−σ nmσ ciσ m

† † (III), c σ =−U timtijc −σ ci−σ + j + m m + δ † − U ij timtitcm−σ ct−σ , m t − δ † − U ij timttmct−σ ci−σ , mt − 2δ † + U ij timcm−σ ci−σ ni−σ m + 2δ † − U ij timcm−σ ci−σ nmσ m − 2 † † U tijc j−σ ci−σ c jσ ciσ

With this follows: B A † (II) + (III), c σ j + B A B A = † − † + Utij tim ci−σ cm−σ cm−σ ci−σ m B A B A + δ † − † − U ij timtmt ci−σ ct−σ ct−σ ci−σ m,t B A B A − δ † − † + U ij timtit ct−σ cm−σ cm−σ ct−σ m,t B A B A + 2δ † − † U ij tim ci−σ cm−σ ni−σ cm−σ ci−σ ni−σ m B A B A + 2δ † + † − U ij tim ci−σ cm−σ nmσ cm−σ ci−σ nmσ B m A B A − 2 † † + 2 † † U tij ci−σ c jσ ciσ c j−σ U tij c j−σ c jσ ci−σ ciσ

Translational symmetry: 702 C Solutions to Problems B A B A † − † tim ci−σ cm−σ cm−σ ci−σ m B A B A 1 † † = t c c −σ − c −σ c −σ N im i−σ m m i i,m B A 1 † = (t − t ) c c −σ = 0 N im mi i−σ m i,m B A B A † − † timtmt ci−σ ct−σ ct−σ ci−σ m,t B A 1 † = (t t − t t ) c c −σ = 0 N im mt tm mi i−σ t m,t,i B A B A † − † timtit ct−σ cm−σ cm−σ ct−σ m,t B A = − † = (timtit tittim) ct−σ cm−σ 0 m,t B A B A † − † tim ci−σ cm−σ ni−σ cm−σ ci−σ ni−σ m B A = δ − † tim imn−σ cm−σ ci−σ m B A = − μ − † (T0 )n−σ tim cm−σ ci−σ m

Real expectation values: B A B A † =! † ci−σ cm−σ nmσ nmσ cm−σ ci−σ

Then we ﬁnally have B A + + , † (I ) (II) (III) c jσ + B A = 2δ − μ − † + U ij (T0 )n−σ tim cm−σ ci−σ m + Un−σ timtmj + m B A + 2δ † + U ij tim 2 cm−σ ci−σ nmσ m <7 8 B A + 2 + † † + U tij ni−σ n j−σ c j−σ c jσ ci−σ ciσ B A= † † c jσ ci−σ ciσ c j−σ C Solutions to Problems 703

Fourier transformation:

<7 8 B A 1 − − † † ik(Ri R j ) + + e tij ni−σ n j−σ c j−σ c jσ ci−σ ciσ N , i j B A= + † † = c jσ ci−σ ciσ c j−σ i= j 1 −ik(R −R ) = t {n−σ − 2 n −σ n σ } + e i j t {···} 0 i i N ij i, j 2 = (T0 − μ) (n−σ − 2 niσ ni−σ ) + n−σ (ε(k) − T0) +

+n−σ (1 − n−σ )Fk−σ

The second summand is exactly the bandwidth correction of (8.213):

B A 1 − − † ik(Ri R j )δ e ij tim 2 cm−σ ci−σ nmσ N , i j m B A − † = cm−σ ci−σ 1 = t (···) N im i,m i=m 1 = t (···) + t (2 n −σ n σ − n−σ )) N im 0 i i i,m

= n−σ (1 − n−σ )B−σ + (T0 − μ) (2 ni−σ niσ − n−σ )

We have used here the deﬁnition (8.212) of spin-dependent band shift. Then ﬁnally what remains is

B A 1 † −ik(Ri −R j ) e (I ) + (II) + (III), c σ = N j + i, j 2 2 = Un−σ (ε(k) − μ) + U n−σ (T0 − μ)+ 2 2 2 + U n−σ (ε(k) − T0) + U n−σ (1 − n−σ )Bk−σ (l)

We substitute (l)in(j) and then have 704 C Solutions to Problems

(3) = ε − μ (2) + (2) − ε − μ (1) + Mkσ ( (k) )Mkσ UMkσ U( (k) )Mkσ 2 + Un−σ (ε(k) − μ) + 2 2 2 + U n−σ (T0 − μ) + U n−σ (ε(k) − T0) + 2 + U n−σ (1 − n−σ )Bk−σ 3 2 = (ε(k) − μ) + 2(ε(k) − μ) Un−σ + 2 3 + 2U (ε(k) − μ)n−σ + U n−σ + 2 2 + Un−σ (ε(k) − μ) + U n−σ (T0 − μ) + 2 2 2 + U n−σ (ε(k) − T0) + U n−σ (1 − n−σ )Bk−σ 3 2 = (ε(k) − μ) + 3Un−σ (ε(k) − μ) + 2 3 + U n−σ (ε(k) − μ)(2 + n−σ ) + U n−σ + 2 + U n−σ (1 − n−σ )(Bk−σ + T0 − μ)

This is the 3rd spectral moment (8.224).

Problem 8.14 Hamiltonian of the Stoner model (8.34): = ε + − μ † HS ( (k) Un−σ )ckσ ckσ kσ

That means

, = ε + − μ [ckσ H]− ( (k) Un−σ )ckσ n ···[c σ , H]− ··· , H = (ε(k) + Un−σ − μ) c σ k − k n− fold

Spectral moments:

(n) = ε + − μ n Mkσ ( (k) Un−σ )

Then it holds

Δ(0) ≡ M(0) = 1 kσ kσ (0) (1) Δ(1) = Mkσ Mkσ kσ (1) (2) Mkσ Mkσ = (0) (2) − (1) 2 Mkσ Mkσ (Mkσ ) = 0

Therefore the spectral density is a one-pole function! C Solutions to Problems 705

Problem 8.15

1 1. H = T n σ + U n σ n −σ 0 i 2 i i i,σ i,σ

It holds

[ciσ , niσ ]− = δσσ ciσ

With this it directly follows:

, H = − μ + [ciσ ]− (T0 )ciσ Uciσ ni−σ , H = , H ciσ ni−σ − [ciσ ]− ni−σ

= (T0 − μ + U)ciσ ni−σ

2 = Here we have used once more ni−σ ni−σ

(0) = Miiσ 1 (1) M = T − μ + Un−σ iiσ ( 0 ) 1 1 1 = (T0 − μ) + (T0 + U − μ) − (T0 − μ) n−σ

Complete induction: Let the proposition be true for n. That means ··· , H , H ··· , H = [ciσ ]− − − − n fold n n n = (T0 − μ) ciσ + (T0 + U − μ) − (T0 − μ) ciσ ni−σ ··· , H , H ··· , H = [ciσ ]− − − (n+1)− fold = − μ n , H − (T0 ) [ciσ ]− − + − μ n − − μ n , H (T0 U ) (T0 ) ciσ ni−σ − = − μ n − μ + + (T0 ) ((T0 )ci−σ Un i−σ ciσ ) n n + (T0 − μ + U) − (T0 − μ) (T0 − μ + U)ciσ ni−σ n+1 n = (T0 − μ) ciσ + U(T0 − μ) ni−σ ciσ + n+1 + (T0 − μ + U) ciσ ni−σ − n − (T0 − μ) (T0 − μ + U)ciσ ni−σ = − μ n+1 + (T0 ) ciσ n+1 n+1 + (T0 − μ + U) − (T0 − μ) ciσ ni−σ 706 C Solutions to Problems

So that it holds ··· , H , H ··· , H = [ciσ ]− − − n n n = (T0 − μ) ciσ + (T0 − μ + U) − (T0 − μ) ciσ ni−σ

Then the spectral moments are (n) = − μ n + − μ + n − − μ n Miiσ (T0 ) (T0 U) (T0 ) n−σ

2. Lonke theorem [24] (0) (1) Δ(1) = Miiσ Miiσ iiσ (1) (2) Miiσ Miiσ = M(0) M(2) − M(0) 2 iiσ iiσ ( iiσ ) 2 2 2 = (T0 − μ) + (T0 − μ + U) − (T0 − μ) n−σ − 2 − (T0 − μ + Un−σ ) 2 2 =−2Un−σ (T0 − μ) − U n−σ + 2U(T0 − μ)n−σ + 2 + U n−σ 2 = U n−σ (1 − n−σ ) = 0 , If n−σ = 0, 1

For empty bands (n = 0), fully occupied bands (n = 2) and fully polarized and half-ﬁlled bands (nσ = 1, n−σ = 0) the spectral density consists of only one (!) δ-function. In all other cases

Δ(1) > iiσ 0

We now calculate ⎛ ⎞ (0) (1) (2) Miiσ Miiσ Miiσ Δ(2) = ⎝ (1) (2) (3) ⎠ iiσ Miiσ Miiσ Miiσ (2) (3) (4) Miiσ Miiσ Miiσ = (0) (2) (4) + (1) (3) (2) − Miiσ Miiσ Miiσ 2Miiσ Miiσ Miiσ − (M(2) )3 − M(0) (M(3) )2 − (M(1) )2 M(4) < iiσ iiσ iiσ = iiσ iiσ = (0) (2) − (1) 2 (4) + Miiσ Miiσ (Miiσ ) Miiσ < = + (1) (3) − (2) 2 (2) + Miiσ Miiσ (Miiσ ) Miiσ < = + (1) (2) − (0) (3) (3) Miiσ Miiσ Miiσ Miiσ Miiσ C Solutions to Problems 707

We calculate the individual terms with the abbreviation:

t0 = T0 − μ

< = (0) (2) − (1) 2 = Miiσ Miiσ (Miiσ ) 2 2 2 2 = t + t + U − t n−σ − t + t + U − t n−σ 0 (( 0 )) 0 ) ( 0 (*0 ) ) = 2 − 2 + + 2 − − 2 2 t0 t0 n−σ 2t0U U 2t0U n−σ U = 2 − U< n−σ (1 n−σ ) = (1) (3) − (2) 2 = Miiσ Miiσ (Miiσ ) = + 3 + + 3 − 3 − (t0 Un−σ )(t0 ((t0 U) t0 )n−σ − 2 + + 2 − 2 2 (t0 ((t0 U) t0 )n−σ ) = + 3 + 2 + 2 + 3 − (t0 Un−σ )(t0 n−σ (3t0 U 3t0U U )) − 2 + + 2 2 (t0 n−σ (2t0U U )) = 4 + 3 + 2 2 + 3 + 3 + t0 n−σ (3t0 U 3t0 U t0U ) Un−σ t0 + 2 2 + 2 + 3 − 4 − Un−σ (3t0 U 3t0U U ) t0 2 2 2 2 2 − n (2t U + U ) − 2n−σ t (2t U + U ) −σ 0 0 0 3 2 2 3 3 3 = n−σ 3t U + 3t U + t U + Ut − 4t U− 0 0 0 0 0 − 2t2U 2 + n2 3t2U 2 + 3t U 3 + U 4 − 4t2U 2− 0 −σ 0 0 0 4 3 − U − 4t0U = 2 2 + 3 + 2 − 2 2 − 3 n−σ (t0 U t0U ) n−σ ( t0 U t0U ) = 2 2 + − 2 2 2 + U n−σ (t0 t0U) U n−σ (t0 t0U) 2 = U n−σ (1 − n−σ )t0(t0 + U) < = (1) (2) − (0) (3) Miiσ Miiσ Miiσ Miiσ 2 2 2 = (t + Un−σ )(t + n−σ (t + U) − t ) − 0 0 0 0 3 3 3 − t + n−σ (t + U) − t 0 0 0 3 2 2 2 = t + n−σ 2t U + t U + Un−σ t + 0 0 0 0 2 + n−σ (2t0U + U ) − 3 2 2 3 − t + n−σ t U + t U + U ( 0 (3 0 3 0 )) 2 2 2 = n−σ 2t U + t U + Ut − 0 0 0 − 2 − 2 − 3 + 2 2 + 3 3t0 U 3t0U U n−σ 2t0U U 2 3 2 2 = n−σ (−2t0U − U ) + n−σ U (2t0 + U) 2 =−U n−σ (1 − n−σ )(2t0 + U) 708 C Solutions to Problems

Intermediate result:

Δ(2) iiσ = (4) + + (2) − Miiσ t0(t0 U)Miiσ U 2n−σ (1 − n−σ ) − + (3) (2t0 U)Miiσ

We now ﬁnally calculate the right-hand side:

Δ(2) iiσ = U 2n−σ (1 − n−σ ) 4 4 4 = t + t + U − t n−σ + 0 (( 0 ) 0 ) 2 2 2 + t (t + U) t + (t + U) − t n−σ − 0 0 0 0 0 3 3 3 − (2t + U) t + (t + U) − t n−σ 0 0 0 0 = t4 + t3(t + U) − 2t4 − Ut3 + 0 )0 0 0 0 2 2 + n−σ (t + U) − t (t (t + U)+ 0 0 0 0 * + (t + U)2 + t2 − (2t + U)((t + U)3 − t3) 0 ) 0 0 0 0 2 2 2 = n−σ (2t U + U )(3t + t U + 2t U + U )− 0 0 0 * 0 − (2t + U)(3t2U + 3t U 2 + U 3) 0 ) 0 0 2 2 2 2 2 2 2 2 = n−σ 6t U + 3U t − 6t U − 3t U + 0 0 0 * 0 3 3 3 3 + 2t0U + 3t0U − 2t0U − 3t0U = 0

With this it is proved that the one-electron spectral density in the limit of inﬁnitely narrow band is a two-pole function:

Sσ (E) = α1σ δ(E − E1σ ) + α2σ δ(E − E2σ )

3. Spectral moments:

(n) = − μ n − + + − μ n Miiσ (E) (T0 ) (1 n−σ ) (T0 U ) n−σ

On the other hand it follows from part 2

(n) = α n + α n Miiσ (E) 1σ E1σ 2σ E2σ

Compare:

E1σ = T0 − μ; α1σ = 1 − n−σ

E2σ = T0 + U − μ ; α2σ = n−σ C Solutions to Problems 709

This agrees with (8.129), (8.130) and (8.131)! Problem 8.16 One can easily calculate

, = ε − μ + [cdσ H]− ( d )cdσ # $ 1 + V c σ + c σ , U n σ n −σ kd k d 2 d d k σ −

With (8.334) follows then the equation of motion: BB AA + μ − ε − Σ = + † (E d dσ (E))Gdσ (E) Vkd ckσ ; cdσ E k

We calculate the “mixed” Green’s function

, = ε − μ + [ckσ H]− ( (BBk) )ckAAσ Vkd cdσBB AA + μ − ε † = † (E (k)) ckσ ; cdσ Vkd cdσ ; cdσ E

So that it follows: BB AA † = Vkd ckσ ; cdσ Gdσ (E) E E + μ − ε(k)

With the deﬁnition of (8.335) of the “hybridization function”, what remains is:

(E + μ − εd − Σdσ (E))Gdσ (E) = + Δ(E)Gdσ (E)

This proves the proposition: Gdσ (E) = E + μ − εd − Σdσ (E) − Δ(E)

Problem 8.17

B A − − ε = † − ndσ (1 ndσ )(Bdσ d ) Vkd ckσ cdσ (2nd−σ 1) k

We begin with B A † = Vkd ckσ cdσ k +∞ BB AA =− 1 † Im dE f−(E) Vkd cdσ ; ckσ (1) π E−μ −∞ k 710 C Solutions to Problems

We have calculated the mixed Green’s function in Problem 8.16: BB AA BB AA † Vkd † c σ ; c = G σ (E) = c σ ; c (2) k dσ E + μ − ε(k) d d kσ

Because of the assumption that Vkd is real, the last step can be easily proved: B A † = Vkd ckσ cdσ k +∞ 1 =− Im dE f− E Δ E − μ G σ E − μ π ( ) ( ) d ( )(3) −∞

Δ: “hybridization function” (8.335):

V 2 Δ(E) = kd E + μ − ε(k) k

It holds [cdσ , H]− = (εd − μ)cdσ + Ucdσ nd−σ + Vpd cpσ p

So that B A B A † 1 † c c σ n −σ =− (ε − μ) c c σ − kσ d d U d kσ d B A B A 1 † 1 † − V c c σ + c [c σ , H]− U pd kσ p U kσ d p

Now for the band shift we still have to calculate B A εd − μ † n σ (1 − n σ )(B σ − ε ) = −2 − 1 V c c σ − d d d d U kd kσ d k B A 2 † − V V c c σ + U kd pd kσ p k,p B A 2 † + V c [c σ , H]− (4) U kd kσ d k

The ﬁrst summand is known from (3). For the second summand we need < † > ckσ cpσ : C Solutions to Problems 711 BB AA BB AA + μ − ε † = δ + † (E (p)) cpσ ; ckσ pk Vpd cdσ ; ckσ E (2) Vkd Vpd = δ + G σ (E) pk E + μ − ε(k) d

That means

B A † = Vkd Vpd ckσ cpσ k,p +∞ BB AA =− 1 † Im dE f−(E) Vkd Vpd cpσ ; ckσ π E−μ −∞ k,p +∞ / 2 1 Vkd =− Im dE f−(E) + π E − ε(k) k −∞ ⎫ 2 2 ⎬ Vpd Vkd + G σ (E − μ) (E − ε(k))(E − ε(p)) d ⎭ k,p +∞ 1 =− Im dE f− E Δ E − μ ∗ π ( ) ( ) −∞

∗ { + Δ(E − μ)Gdσ (E − μ)} (5)

Finally it still holds

BB AA BB AA B A , H † = † − , † [cdσ ]− ; ckσ E cdσ ; ckσ cdσ ckσ E E B + A † (2) Vkd = E Gdσ (E) − cdσ , c σ E+μ−ε(k) k +

Spectral theorem:

B A † , H = Vkd ckσ [cdσ ]− k +∞ 1 =− Im dE f− E E − μ Δ E − μ G σ E − μ π ( )( ) ( ) d ( )(6) −∞

In (4) we need (3), (5) and (6): 712 C Solutions to Problems

εd − μ I = (−2 − 1)Δ(E − μ)G σ (E − μ) − U d 2 − Δ(E − μ) { + Δ(E − μ)G σ (E − μ)} + U d 2 + Δ(E − μ)(E − μ)G σ (E − μ) U d 2 =−Δ(E − μ) + Δ(E − μ)Gdσ (E − μ) ∗ U 2 ∗ −1 + (−ε + μ − Δ(E − μ) + E − μ) U d

Equation of motion:

(E + μ − εd − Σdσ (E) − Δ(E))Gdσ (E) = 2 ⇔ I =−Δ(E − μ) − Δ(E − μ)G σ (E − μ) + U d 2 + Δ(E − μ)( + Σdσ (E − μ)Gdσ (E − μ)) U 2 = Δ(E − μ)G σ (E − μ) Σ σ (E − μ) − 1 d U d

Then it ﬁnally follows:

ndσ (1 − ndσ )(Bdσ − εd ) = +∞ 1 =− Im dE f− E Δ E − μ G σ E − μ ∗ π ( ) ( ) d ( ) −∞ 2 ∗ Σ σ (E − μ) − 1 U d

This is exactly the proposition (8.352).

On the other hand it is also valid that C Solutions to Problems 713 † |ϕ ···ϕ (ε) = δ ϕ − ϕ |ϕ ϕ ···ϕ (ε) cγ cβ α1 αN ( β α1 ) γ α2 αN + ···+ + εN−1δ ϕ − ϕ |ϕ ϕ ···ϕ (ε) ( β αN ) γ α1 αN−1

One multiplies the last equation by ε and then subtracts one equation from the other to get † − ε † |ϕ ···ϕ (ε) = δ ϕ − ϕ |ϕ ···ϕ (ε) cβ cγ cγ cβ α1 αN ( β γ ) α1 αN

Problem A.2 8 ··· ··· ··· (+) Bosons : nαr nαs : arbitrary Fock state. r = s :

8 † † (+) cα cα ···nα ···nα ··· r s r s 8 (+) = nα + 1 nα + 1 ···nα + 1 ···nα + 1 ··· r s 8 r s † † (+) = cα cα ···nα ···nα ··· s r r s =⇒ c† , c† = 0. αr αs −

For r = s this relation is trivially valid. Since

† † † cα , cα = c , c r s − αs αr − directly follows: , = . cαr cαs − 0 r = s :

8 † (+) cα cα ···nα ···nα ··· r s r s 8 √ (+) = nα nα + 1 ···nα − 1 ···nα + 1 ··· r s r 8 s † (+) = c cα ···nα ···nα ··· . αs r r s r = s : 714 C Solutions to Problems 8 8 † (+) (+) cα cα ···nα ··· = cα nα + 1 ···nα + 1 ··· r r r r r r 8 (+) = (nα + 1) ···nα ··· , 8 r r 8 † (+) √ † (+) cα cα ···nα ··· = nα cα ···nα − 1 ··· r r r r r 8r (+) = nα ···nα ··· r r † =⇒ cα , c = δ , . r αs − r s

Fermions: 8 † 2 (−) (c ) ···nα ··· = 0 (Pauli principle: Problem A.4). αr r r < s : 8 † † (−) cα cα ···nα ···nα ··· r s r s 8 † (−) = − Ns δ ··· ··· + ··· cα ( 1) nα ,0 nα nα 1 r s r s 8 (−) = − Nr − Ns δ δ ··· + ··· + ··· , ( 1) ( 1) nα ,0 nα ,0 nα 1 nα 1 s 8 r r s † † (−) cα cα ···nα ···nα ··· s r r s 8 † (−) = − Nr δ ··· + ··· ··· ( 1) nα ,0cα nα 1 nα r s r s 8 (−) = − Nr − Ns δ δ ··· + ··· + ··· , ( 1) ( 1) nαr ,0 nα s ,0 nαr 1 nαs 1

 N = Ns + 1 s 8 † † † † (−) =⇒ (cα cα + cα cα ) ···nα ···nα ··· = 0 r s s r r s =⇒ c† , c† = 0. αr αs +

Since † † † cα , cα = ( c , c ) r s + αs αr + again the second anti-commutator relation follows directly: , = cαr cαs + 0 r = s : 8 8 † (−) (−) ··· ··· = − Nr δ ··· + ··· = cα cα nα cα ( 1) nα ,0 nα 1 r r r r r r 8 (−) = − 2Nr δ ··· ··· = ( 1) nα ,0 nα r 8 r = δ ··· ··· (−) , nα ,0 nα 8 r r 8 † (−) (−) c cα ···nα ··· = δ , ···nα ··· . αr r r nαr 1 r C Solutions to Problems 715

= Since in every case nαr 0or1,wehave 8 8 † † (−) (−) (cα c + c cα ) ···nα ··· = ···nα ··· . r αr αr r r r r < s : 8 † (−) cα cα ···nα ···nα ··· r s r s 8 (−) = − Ns δ ··· ··· + ··· cα ( 1) nα ,0 nα nα 1 r s r s 8 + (−) = − Nr Ns δ δ ··· − ··· + ··· , ( 1) nα ,1 nα ,0 nα 1 nα 1 r s 8 r s † (−) cα cα ···nα ···nα ··· s r r s 8 † (−) = − Nr δ ··· − ··· ··· cα ( 1) nα ,1 nα 1 nα s r r s 8 + (−) = − Nr Ns δ δ ··· − ··· + ··· , ( 1) nαr ,1 nα s ,0 nαr 1 nαs 1

 N = Ns − 1 s 8 † † (−) =⇒ (cα c + c cα ) ···nα ···nα ··· = 0. r αs αs r r s

So altogether we have † cα , c = δ , . r αs + r s

Problem A.3

1. Bosons:

† † † nα cβ = cαcαcβ † † † = cαcβ cα + δαβ cα † † † = cβ cαcα + δαβ cα † † = cβ nα + δαβ cα

So that we have † † nα, cβ = δαβ cα −

Fermions: 716 C Solutions to Problems

† † † nα cβ = cαcαcβ † † † =−cαcβ cα + δαβ cα † † † = cβ cαcα + δαβ cα † † = cβ nα + δαβ cα

So that just as in the case of Bosons we get † † nα, cβ = δαβ cα −

2. Bosons :

† nα cβ = cαcαcβ † † = cαcβ cα = cβ cαcα − δαβ cα = cβ nα − δαβ cα

With this follows: , =−δ nα cβ − αβ cα

Fermions:

† nα cβ = cαcαcβ † † =−cαcβ cα = cβ cαcα − δαβ cα = cβ nα − δαβ cα

With this, as in the case of Bosons we get , =−δ nα cβ − αβ cα

3. For Bosons as well as for Fermions, with part 1 , † = , † = δ † = † N cα − nγ cα − αγ cα cα γ γ

is valid. 4. For Bosons as well as Fermions with part 2 , = , = −δ =− N cα − nγ cα − ( αγ cα) cα γ γ

is valid. C Solutions to Problems 717

Problem A.4 1.

, = , = 2 = 2 = cα cβ + 0 [cα cα]+ 2cα 0 cα 0 † † † † † 2 cα, cβ = 0 cα, cα = 2 cα = 0 + + † 2 cα = 0 (Pauli principle)

2 † † † † nα = cαcαcαcα = cα 1 − cαcα cα † † 2 2 1.) = cαcα − cα (cα) = nα (Pauli principle)

† † 1.) cαnα = cαcαcα = 1 − cαcα cα = cα † † † 1.) cαnα = cαcαcα = 0

† 1.) nα cα = cαcαcα = 0 † † † † † 1.) † nα cα = cαcαcα = cα 1 − cαcα = cα

Problem A.5 Proof by complete induction N = 1:

† † 0| cβ c |0 = 0| δ(β ,α ) ± c cβ |0 1 α1 1 1 α1 1 † = δ(β ,α ) 0|0 ± 0| c cβ |0 = δ(β ,α ) 1 1 α1 1 1 1 > = . because cβ1 10 0

N − 1 −→ N : 718 C Solutions to Problems

† † 0|cβ ···cβ c ···c |0 N 1 α1 αN β $taking c 1 to the right † † =δ(β ,α ) 0| cβ ···cβ c ···c |0 + 1 1 N 2 α2 αN 1 † † † + (±) δ(β ,α ) 0| cβ ···cβ c c ···c |0 + 1 2 N 2 α1 α3 αN +···+ N−1 † † † + (±) δ(β ,α ) 0| cβ ···cβ c c ···c |0 = 1 N N 2 α1 α2 αN−1 $condition for induction pα =δ(β1,α1) (±) Pα [δ(β2,α2) ···δ(βN ,αN )] + P α 1 pα (±) δ(β1,α2) (±) Pα [δ(β2,α1)δ(β3,α3) ···δ(βN ,αN )] +

Pα +···+ N−1 pα + (±) δ(β1,αN ) (±) Pα [δ(β2,α1)δ(β3,α2) ··· P α ···δ(β ,α − ) N N 1 pα = (±) Pα [δ(β1,α1)δ(β2,α2) ···δ(βN ,αN )] q.e.d.

Pα

Problem A.6 1. , † = † † − † † nα cβ − cαcαcβ cβ cαcα † † † † † = δ(α − β) cα ± cαcβ cα − cβ cαcα † † † † † = δ(α − β) cα + cβ cαcα − cβ cαcα † = δ(α − β) cα

2. , = † − † nα cβ − cαcαcβ cβ cαcα † † = cαcαcβ − δ(α − β) cα ∓ cαcβ cα † † † = cαcαcβ − δ(α − β) cα − cαcαcβ =−δ(α − β) cα

These relations are equally valid for both Bosons and Fermions. Problem A.7

Nˆ = dα nˆ α. C Solutions to Problems 719

We ﬁrst calculate the following commutators:

 † † Problem A.6 † † Nˆ , cβ = dα nˆ α, cβ = dα cαδ(α − β) = cβ , − − ˆ , = α , Problem= A.6 α −δ α − β =− . N cβ − d nˆ α cβ − d [ ( )cα] cα

We therefore have

† † Ncˆ β = cβ (Nˆ + 1l ) ;

Ncˆ β = cβ (Nˆ − 1l ) .

8 8 ˆ † ϕ ··· (±) = † ˆ + ϕ ··· (±) N cβ α1 cβ (N 1l ) α1 8 = + † ϕ ··· (±) (N 1) cβ α1

† As proposed, it is an eigenstate. The eigenvalue is N + 1. The name creator for cβ is therefore appropriate. 2.

8 8 ˆ ϕ ··· (±) = ˆ − ϕ ··· (±) N cβ α1 cβ (N 1l ) α1 8 = − ϕ ··· (±) (N 1) cβ α1

8 ϕ ··· (±) ˆ cβ α1 is also an eigenstate of the particle number operator N with the eigenvalue N − 1. The name annihilator for cβ is therefore appropriate.

Problem A.8 Plane waves:

1 ik·r ϕk(r) = √ e = r|k V

Kinetic energy: One-particle basis: |kσ =|k |σ 720 C Solutions to Problems & % 2 2 2 7 8 2 2 p k k kσ k σ = kσ | k σ = δkk δσσ 2m 2m 2m & % N 2 2 pi p † = kσ k σ c σ ckσ 2m 2m k i=1 kkσσ 2 2 k † = c c σ 2m kσ k kσ

Interaction:

? @ 1 k1σ1, k2σ2 k3σ3, k4σ4 r (1) −r (2) ? @ 1 = δσ σ δσ σ k1 k2 k3 k4 1 3 2 4 r (1) −r (2)

The interaction is spin independent. Therefore the spin parts of the eigenstates can be evaluated directly and they yield the two Kronecker deltas. The two-particle states used are not symmetrized:

? @ 1 k1 k2 k3 k4 r (1) −r (2) ? @ 3 3 1 = d r1d r2 k1 k2 r1 r2 r1 r2 | k3 k4 r (1) −r (2) B AB A 1 = d3r d3r k(1) | r(1) k(2) | r(2) ∗ 1 2 | − | 1 1 2 2 r1 r2 B AB A ∗ (1) | (1) (2) | (2) r1 k3 r2 k4 1 1 − · − · = d3r d3r ei(k3 k1) r1 ei(k4 k2) r2 2 1 2 V |r1 − r2| 1 3 1 i(k −k )·r = δ + , + d r e 3 1 k1 k2 k3 k4 V r

The last step is obtained by introducing the centre of mass and relative coordinates. We get the Kronecker delta when the centre of mass part is integrated out. The remaining integral is calculated using a convergence ensuring factor α: C Solutions to Problems 721 1 +1 ∞ lim d3r eiq·r eαr = lim 2π dx dr r eiqrx eαr α→ α→ 0 r 0 −1 0 2π ∞ = lim dr eiqr − e−iqr e−αr α→ 0 iq 0 4π = lim α→0 q2 + α2 π = 4 q2

Then the interaction matrix element reads as ? @ 1 k1σ1, k2σ2 k3σ3, k4σ4 r (1) −r (2) 4π = δσ σ δσ σ δ + , + 1 3 2 4 k1 k2 k3 k4 2 V |k3 − k1|

Interaction operator in second quantization:

i= j 1 1 = 2 ri −r j ij ? @ 1 1 = k1σ1, k2σ2 k3σ3, k4σ4 ∗ 2 r (1) −r (2) k1σ1,k2σ2,k3σ3,k4σ4 † † ∗c c c σ c σ k1σ1 k2σ2 k4 4 k3 3 1 4π † † = c c c + − σ c σ 2 k1σ1 k2σ2 k1 k2 k3 2 k3 1 2 V |k3 − k1| k1σ1,k2σ2,k3

We further set

k1 → k + q ; k2 → p − q ; k3 → k ; σ1 → σ ; σ2 → σ and have the Hamiltonian of the N-electron system in second quantization: † 1 † † H = ε (k)c c σ + v (q)c c c σ c σ N 0 kσ k 2 0 k+qσ p−qσ p k kσ kpqσσ

2k2 e2 ε (k) = ; v (q) = 0 0 2 2m ε0Vq 722 C Solutions to Problems

Problem A.9

† = =⇒ , = nˆ kσ ckσ ckσ [nˆ kσ nˆ k σ ]− 0

Therefore the kinetic energy commutes in any case with Nˆ . Therefore we only have to calculate the commutator with the interaction: 1 † † † v0(q) c + σ c − σ cpσ ckσ , c ck σ k q p q k σ − 2 kpq σσ kσ < 1 † † = v δ δ − 0(q) kk σσ ck+qσ cp−qσ cpσ ck σ 2 k,p,q k,σ,σ ,σ † † − δ δ + p,k σ σ ck+qσ cp−qσ ckσ ck σ † † + δ − δσ σ c c c σ c σ − p qk k σ k+qσ p =k † † δ δσσ σ σ k+qk ck σ cp−qσ cp ck < 1 † † † † = v − + 0(q) ck+qσ cp−qσ cpσ ckσ ck+qσ cp−qσ ckσ cpσ 2 k,p,q σ,σ = † † † † + − = cp−qσ ck+qσ cpσ ckσ ck+qσ cp−qσ cpσ ckσ 0 =⇒ , ˆ = HN N − 0

HN and Nˆ have common eigenstates. The particle number is a conserved quantity. Problem A.10 1. Hamiltonian of the two-particle system:

2 H = H + H =− (Δ + Δ ) + V (x ) + V (x ) 1 2 2m 1 2 1 2 Unsymmetrized eigenstate:

(1) (2) |ϕα ϕα =|ϕ |ϕ 1 2 α1 α2

Position space representation:

(1) (2) |ϕ ϕ =ϕ ϕ χ χ x1x2 α1 α2 n(x1) m(x2) S(m S ) S(m S )

χS: spin function (identical particles have the same spin S)

α1 = (n, m S); α2 = (m, m S ) C Solutions to Problems 723

2. Solution of the one-particle problem: 2 − Δ + V (x) ϕ(x) = Eϕ(x) 2m

We ﬁrst have

ϕ(x) ≡ 0forx < 0 and x > a

For 0 ≤ x ≤ a we have to solve

2 − Δϕ(x) = Eϕ(x) 2m

Ansatz for solution:

ϕ(x) = c sin(γ1x + γ2)

Boundary conditions:

ϕ(0) = 0 =⇒ γ2 = 0, π ϕ(a) = 0 =⇒ γ = n ; n = 1, 2, 3,... 1 a

Energy eigenvalues:

2 2π 2 E = γ 2 =⇒ E = n2; n = 1, 2,... 2m 1 n 2ma2

Eigen functions: π ϕ (x) = c sin n x , n a ( a π 2 1 =! c2 sin2 n x dx =⇒ c = , 0 a a /5 2 π ≤ ≤ , a sin n a x for 0 x a ϕn(x) = 0 otherwise

3. Two-particle problem: ) (±) 1 (1) (2) |ϕ ϕ −→ √ ϕ ϕ χ χ ± α1 α2 n(x1) m(x2) S(m S ) S(m S ) 2 * ± ϕ ϕ χ (2) χ (1) n(x2) m (x1) S(m S ) S(m S ) 724 C Solutions to Problems

(+): Bosons, (−): Fermions: (n, m S) = (m, m S ) because of Pauli’s principle. 4. Ground state energy of the N-particle system: Bosons: All particles in the n = 1-state:

2π 2 E = N . 0 2ma2 Fermions:

N 2 2π 2 2π 2 N 3 E = 2 n2 ≈ 0 2ma2 2ma2 24 n=1

with

N 2 N 3 3 N1 2 1 N N n2 ≈ n2dn = − 1 ≈ 3 8 24 n=1 1

Problem A.11 1. Non-interacting, identical Bosons or Fermions:

N = (i) H H1 i=1

Eigenvalue equation:

(i)|ϕ(i) = |ϕ(i) , ϕ(i)|ϕ(i) =δ H1 r r r r s rs

One-particle operator in second quantization: = ϕ | |ϕ † = δ † H r H1 s ar as s rsar as , , r s r s =⇒ = † = . H r ar ar r nˆr r r

2. Unnormalized density matrix of the grand canonical ensemble:

ρ = exp[−β(H − μNˆ )], Nˆ = nˆ r r

The normalized Fock states C Solutions to Problems 725

() |N; n1n2 ...ni ...

are the eigenstates of nˆ r and therefore also of Nˆ und H: () () H|N; n1 ... = r nr |N; n1 ... , r () () Nˆ |N; n1 ... = N|N; n1 ...

That is why it is convenient to build the trace with these Fock states: 7 8() () N; n n ...exp[−β(H − μNˆ )] N; n n ... 1 #2 $ 1 2 = exp −β (r − μ)nr with nr = N r r

From this follows: # $ ∞ Trρ = exp −β (r − μ)nr N=0 {nr } r ( nr =N) ∞ . −β −μ = e ( r )nr N=0 {nr } r ( nr =N) . −β −μ = ...... e ( r )nr n1 n2 nr r −β −μ −β −μ = e n1( 1 ) e n2( 2 ) ...

n1 n2

Grand canonical partition function: . −β −μ Ξ(T, V,μ) = Trρ = e nr ( r )

r nr

Bosons (nr = 0, 1, 2,...): . Ξ , ,μ = 1 B (T V ) −β −μ 1 − e ( r ) r

Fermions (nr = 0, 1): . −β(r −μ) ΞF (T, V,μ) = 1 + e r 726 C Solutions to Problems

3. Expectation value of the particle number:

1 Nˆ = ρNˆ ΞSp( )

To build the trace Fock states are preferred because they are the eigenstates of Nˆ : / # $0 ∞ 1 Nˆ = N −β − μ n Ξ exp ( r ) r N=0 {nr } r ( nr =N) ∂ = 1 Ξ β ∂μ ln

With part 2 / 0 ∂ ∂ −β −μ Ξ = − − e ( r ) ∂μ ln B ∂μ ln 1 r −β −μ −β ( r ) =− e −β −μ 1 − e ( r ) r = β 1 , β −μ e ( r ) − 1 r/ 0 ∂ ∂ −β −μ Ξ = + e ( r ) ∂μ ln F ∂μ ln 1 r

−β(r −μ) = β e −β −μ 1 + e ( r ) r = β 1 β −μ e ( r ) + 1 r

This means / 1 β −μ Bosons ˆ = r e ( r )−1 N 1 r eβ(r −μ)+1 Fermions

4. Internal energy:

1 U = H = ρ H ΞTr( )

Fock states are the eigenstates of H and therefore appropriate for building the trace required here: C Solutions to Problems 727 # $ ∞ 1 −β ( −μ)n U = n e r r r Ξ i i N=0 {nr } i ( nr =N) ∂ =− Ξ + μ Nˆ ∂β ln −β −μ ∂ − μ ( r ) − Ξ = ( r )e ln B −β −μ ∂β 1 − e ( r ) r =−μ ˆ + r , N β −μ e ( r ) − 1 r −β −μ ∂ − − μ ( r ) − Ξ =− ( r )e ln F −β −μ ∂β 1 + e ( r ) r =−μ ˆ + r N β −μ e ( r ) + 1 r

We ﬁnally get

/ r r eβ(r −μ−1 Bosons U = r . r eβ(r −μ)+1 Fermions

5. Fock states are also eigenstates of the occupation number operator:

1 n = ρn ˆ i ΞTr( ˆ i ) ∞ 1 −β ( −μ)n = n e r r r Ξ i N=0 {nr } ( nr =N) 1 ∂ =− ln Ξ, β ∂i −β −μ ∂ +β ( r ) ∂ − 1 Ξ =+1 e r ln B −β −μ β ∂ β 1 − e ( r ) ∂ i r i = 1 β −μ (Bose function), e ( i ) − 1 −β −μ ∂ −β ( r ) ∂ − 1 Ξ =−1 e r ln F −β −μ β ∂ β 1 + e ( r ) ∂ i r i = 1 β −μ (Fermi function). e ( i ) + 1

It follows: 728 C Solutions to Problems / { β − μ − }−1 , = exp[ ( i ) 1 Bosons nˆ i −1 {exp[β(i − μ)] + 1} Fermions.

One immediately recognizes by comparison with the earlier problems: Nˆ = nˆ r ; U = r nˆ r r r

Problem B.1 β β i d ρ dλAú (t − iλ) = ρ dλ A(t − iλ) = dλ 0 0 i = ρ A t − iβ − A t = [ ( ) ( )] i i − β H − i − β H = ρ e ( i ) A t e ( i ) − A t = ( ) ( ) i = ρ eβH A t e−βH − A t = ( ) ( ) ! −βH βH −βH " = i e e A(t)e − ρ = −βH A(t) Sp(e ) − i i = A t ρ − ρ A t = A t ,ρ ( ( ) ( )) [ ( ) ]− q.e.d.

Problem B.2 B A < = , = ρ , A(t) B(t ) − Sp A(t) B(t ) − ) * = Sp ρ A(t)B(t) − ρ B(t)A(t) ) * = Sp B(t)ρ A(t) − ρ B(t)A(t) < = = ,ρ Sp B(t ) − A(t) (cyclic invariance of trace).

Substitute Kubo identity: B A ret =−Θ − , = A(t); B(t ) i (t t ) A(t) B(t ) − β ) * =−Θ(t − t) dλSp ρ Bú (t − iλ)A(t) =

0 β 7 8 =−Θ(t − t) dλ Bú (t − iλ)A(t) q.e.d.

0 C Solutions to Problems 729

Problem B.3 < = 1 −βH i H + β − i H + β B A t + iβ = e Be (t i ) Ae (t i ) (0) ( ) ΞSp < = 1 βH −βH i H −βH − i H = e e Be t e Ae t ΞSp < = 1 −βH i H − i H = e e t Ae t B ΞSp = A(t)B(0)

Here the cyclic invariance of trace has been used several times. Problem B.4 1. t − t > 0: + The integrand has a pole at x = x0 =−i0 . Residue:

−ix(t−t) e −ix(t−t) c− = lim (x − x ) = lim e = 1 1 → 0 + → x x0 x + i0 x x0

Since t − t > 0, the semicircle closes in the lower half-plane; then the exponential function sees to it that the contribution from the semicircle vanishes. The contour runs mathematically negatively. Therefore it follows that

i Θ(t − t) = (−2πi)1 = 1 2π

2. t − t < 0: In order that no contribution from the semicircle appears, now it closes in the upper half-plane. Then it follows that

Θ(t − t) = 0

as there is no pole in the region of integration. Problem B.5

+∞ f (ω) = dt øf (t)eiωt −∞ 730 C Solutions to Problems

Let the integral exist for real ω. Set

ω = ω1 + iω2 +∞ ω −ω =⇒ f (ω) = dt øf (t)ei 1t e 2t −∞

1. øf (t) = 0fort < 0:

∞ ω −ω =⇒ f (ω) = dt øf (t)ei 1t e 2t 0

Converges for all ω2 > 0, therefore it is possible to analytically continue in the upper half-plane. 2. øf (t) = 0fort > 0:

0 ω −ω =⇒ f (ω) = dt øf (t)ei 1t e 2t −∞

Converges for all ω2 < 0, therefore it is possible to analytically continue in the lower half-plane.

Problem B.6 With

H = ε − μ † 0 ( (k) )akσ akσ kσ we ﬁrst calculate

 † [akσ , H0]− = (ε(k ) − μ) akσ , a ak σ = k σ − k σ = (ε(k ) − μ)δkk δσσ akσ = (ε(k) − μ)akσ . kσ 

The interaction term requires more effort: C Solutions to Problems 731

[akσ , H − H0]− 1 † † = vk p(q) akσ , a a − σ apσ ak σ 2 k +qσ p q − kpq σ σ 1 † = v (q) δσσ δ , + a a σ a σ 2 k p k k q p−qσ p k kpq σ σ † −δσσ δ − σ kp qak +qσ ap ak σ 1 † = v − (q)a a σ a − σ 2 k qp p−qσ p k q pqσ 1 † − v + (q)a a + σ a σ 2 k k q k +qσ k q k kqσ

In the ﬁrst summand:

q →−q; vk+q,p(−q) = vp,k+q(q)

In the second summand:

k → p; σ → σ 

Then the two summands can be combined: † , H − H = v [akσ 0]− p,k+q(q)ap+qσ apσ ak+qσ pqσ 

Equation of motion:

(E − ε(k) + μ)Gret (E) kσ † † = + v ret p,k+q(q) ap+qσ apσ ak+qσ ; akσ E pqσ 

Problem B.7

H = † − μ ˆ = − μ † (k)akσ akσ N ( (k) )akσ akσ kσ kσ

One can easily calculate 732 C Solutions to Problems † , H = − μ , [akσ ]− ( (k ) )[akσ akσ ak σ ]− k σ = ((k ) − μ)δkk δσσ akσ kσ 

= ((k) − μ)akσ

From this it further follows that

2 [[akσ , H]−, H]− = ((k) − μ)[akσ , H]− = ((k) − μ) akσ

For the spectral moments this means

(0) = , † Mkσ [akσ akσ ]+ = 1 B A (1) = , H , † Mkσ [[akσ ]− akσ ]+ = − μ , † ( (k) ) [akσ akσ ]+ = ((k) − μ) B A (2) † M σ = [[akσ , H]−, H]−, a σ k k + = − μ 2 , † ( (k) ) [akσ akσ ]+ = ((k) − μ)2 . .

Then by complete induction one gets immediately

(n) = − μ n = , , ,... Mkσ ( (k) ) ; n 0 1 2

The relationship (B.99) with the spectral density,

+∞ (n) = 1 n Mkσ dE E Skσ (E) −∞ then leads to the solution:

Skσ (E) = δ(E − (k) + μ)

Problem B.8 1. Creation and annihilation operators for Cooper pairs:

† = † † = bk ak↑ a−k↓ ; bk a−k↓ ak↑ C Solutions to Problems 733

Fundamental commutation relations (a) † † [bk, bk ]− = b , b = 0 k k −

because the creation and annihilation operators of fermions anticommute among themselves. Therefore products of even number of Fermion construction operators then commute. (b) † † † bk, b = a−k↓ak↑, a ↑a− ↓ k − k k − † † = δ − δ kk a−k↓a−k↓ −k−k ak↑ak↑ = δkk (1 − nˆ −k↓ − nˆ k↑)

Therefore the Cooper pairs inspite of their total spin being zero are not real Bosons because only two of the three basic commutation relations are satisﬁed. (c) Since

 [bk, bk ]+ = 2bkbk = 0fork = k

they are naturally also not real Fermions, either, even though 2 † = 2 = bk bk 0

is valid for them. 2. Equation of motion: ∗ † akσ , H = t(p) akσ , a σ apσ − p − pσ † † − Δ akσ , a−p↓ap↑ + a ↑a− ↓ p p − p = t(p)δσσ δkpapσ pσ − Δ δ δ † − δ δ † kp σ ↑a−p↓ k−p σ ↓ap↑ p † = t(k)a σ − Δ(δσ ↑ − δσ ↓)a , / k −k−σ +1, for σ =↑, zσ = −1, for σ =↓

Then the equation of motion reads as 734 C Solutions to Problems

− = − Δ † † . (E t(k))Gkσ (E) zσ a−k−σ ; akσ

The Green’s function on the right-hand side prevents a direct solution. Therefore we set up the corresponding equation of motion for this: † ∗ a− −σ , H k − † † =−t(−k)a− −σ − Δ a− −σ , a−p↓ap↑ k k − p =− † − Δ δ δ † − δ δ t(k)a−k−σ kp −σ ↓ap↑ −kp −σ ↑a−p↓ p =− † − Δ t(k)a−k−σ zσ akσ

This gives us the following equation of motion:

+ † † = −Δ (E t(k)) a−k−σ ; akσ zσ Gkσ (E) † † zσ Δ a ; a = − G σ (E) −k−σ kσ E + t(k) k

ret This is substituted in the equation of motion for Gkσ (E): Δ2 E − t(k) − G σ (E) = E + t(k) k

Excitation energies:

E(k) =+ t2(k) + Δ2 −→ Δ (Energy gap). t→0

Green’s function:

(E + t(k)) G σ (E) = k E2 − E2(k)

Imposing the boundary conditions: ! " + − ret t(k) E(k) t(k) E(k) G σ (E) = − k 2E(k) E − E(k) + i0+ E + E(k) + i0+

3. For Δ we need the expectation value:

† † ak↑a−k↓

Its determination is via spectral theorem and the Green’s function used in part 2.: C Solutions to Problems 735

† † −Δ −Δ a ; a = G ↑(E) = −k↓ k↑ E + t(k) k E2 − E2(k)

Taking into account the boundary conditions we obtain for the corresponding retarded function:

† † a ; a ret −k↓ k↑! E " Δ = 1 − 1 2E(k) E + E(k) + i0+ E − E(k) + i0+

The spectral density corresponding to this Δ S− ↓ ↑(E) = [δ(E + E(k)) − δ(E − E(k))] k ;k 2E(k)

Spectral theorem:

† † ak↑a−k↓ +∞ 1 S− ↓ ↑(E) = dE k ;k exp(β E) + 1 −∞ Δ = 1 − 1 −β + β + 2E(k) exp( E(k)) 1 exp( E(k)) 1 Δ 1 = tanh β E(k) 2E(k) 2

Then we ﬁnally get 1 β 2 + Δ2 1 tanh 2 t (k) Δ = ΔV 2 2 + Δ2 k t (k)

Δ = Δ(T ) ⇒ Energy gap is T dependent. Special case: 1 T → 0 ⇒ tanh β t2(k) + Δ2 → 1 2 736 C Solutions to Problems

Problem B.9 1. We ﬁrst show that ··· , ∗ , ∗ ··· , ∗ [akσ H ]− H − H − 2 + Δ2 n , = = (t (k) ) if p 2n 2 + Δ2 n − Δ † , = + (t (k) ) (t(k) akσ zσ a−k−σ ) if p 2n 1

is valid. Here n = 0, 1, 2, ···. We prove this by complete induction. Induction’s start p = 1, 2: , ∗ = − Δ † akσ H − t(k)akσ zσ a−k−σ (see Problem B.8)

 , ∗ , ∗ = − Δ † akσ H − H t(k) t(k)akσ zσ a−k−σ − − Δ − † − Δ zσ t(k)a−k−σ zσ akσ 2 2 = t (k) + Δ akσ .

Induction’s end p −→ p + 1:

(a) p even: ... , ∗ , ∗ ,..., ∗ akσ H − H H − − (p+1)-fold commutator 2 2 p ∗ = (t + Δ ) 2 a σ , H k − p † = 2 + Δ2 2 − Δ (t ) takσ zσ a−k−σ

(b) p odd: ... , ∗ , ∗ ,..., ∗ akσ H − H H − − + (p 1)-fold commutator 1 − † ∗ = 2 + Δ2 2 (p 1) − Δ , (t ) takσ zσ a−k−σ H − 1 − † = 2 + Δ2 2 (p 1) − Δ (t ) t(takσ zσ a−k−σ ) − Δ − † − Δ zσ ( ta−k−σ zσ akσ ) 2 2 1 (p+1) = (t + Δ ) 2 akσ q.e.d. C Solutions to Problems 737

For the spectral moments of the one-electron spectral density we directly get n = 0, 1, 2,...

n M(2n) = t2(k) + Δ2 , kσ (2n+1) = 2 + Δ2 n . Mkσ t (k) t(k)

2. We use

+∞ (n) 1 n M = dEE S σ (E) kσ k −∞

Determining equations from the ﬁrst four spectral moments:

α1σ + α2σ = ,

α1σ E1σ + α2σ E2σ = t, α 2 + α 2 = 2 + Δ2 , 1σ E1σ 2σ E2σ (t ) α 3 + α 3 = 2 + Δ2 1σ E1σ 2σ E2σ (t )t

Reformulating them:

α σ (E σ − E σ ) = (t − E σ ), 2 2 1 1 2 2 α σ E σ (E σ − E σ ) = t + Δ − tE σ , 2 2 2 1 1 α 2 − = 2 + Δ2 − 2σ E2σ (E2σ E1σ ) (t )(t E1σ )

After division follows:

2 = 2 + Δ2 =⇒ =+ 2 + Δ2 ≡ E2σ t E2σ (k) t (k) E(k)

This has the further consequence:

2 2 2 t + Δ − tE σ Δ E(k) = 1 = t + t − E1σ t − E1σ −1 2 =⇒ (E(k) − t(k)) Δ = t(k) − E1σ (k) Δ2 E(k)t(k) − E2(k) =⇒ E σ (k) = t(k) − = 1 E(k) − t(k) E(k) − t(k)

=⇒ E1σ (k) =−E(k) =−E2σ (k)

Spectral weights: 738 C Solutions to Problems

α2σ (k)2E(k) = (t(k) + E(k)) t(k) + E(k) =⇒ α σ (k) = , 2 2E(k) E(k) − t(k) α1σ (k) = − α2σ (k) = , ! 2E(k) E(k) − t(k) =⇒ Skσ (E) = δ(E + E(k)) 2E(k) " E(k) + t(k) + δ(E − E(k)) 2E(k)

Problem B.10 Free energy: ∂ F F T, V = U T, V − TS T, V = U T, V + T ( ) ( ) ( ) ( ) ∂ T V

So that for the internal energy we get ∂ 1 U T, V =−T 2 F T, V ( ) ∂ ( ) T T V

F(0, V ) ≡ U(0, V )

∂ U(T, V ) − U(T, 0) =−T 2 (F(T, V ) − F(0, V )) ∂T

T U(T , V ) − U(0, V ) − dT = 2 0 T 1 1 = (F(T, V ) − F(0, V )) − lim (F(T, V ) − F(0, V )) T T →0 T

Third law: 1 ∂ F lim (F(T, V ) − F(0, V )) = (T = 0) → ∂ T 0 T T V =−S(T = 0, V ) =! 0

Then it follows that C Solutions to Problems 739 T U(T , V ) − U(0, V ) F(T, V ) = F(0, V ) − T dT 2 0 T

Problem B.11

H = − μ † 0 ( (k) )akσ akσ kσ

Then one can easily calculate

[akσ , H0]− = ((k) − μ)akσ , † , H =− − μ † , [akσ 0]− ( (k) )akσ † † † , H = , H + , H [akσ ak σ 0]− [akσ 0]−ak σ akσ [ak σ 0]− † † =− − μ + − μ ( (k) )akσ ak σ ( (k ) )akσ ak σ † = − ( (k ) (k))akσ ak σ

|ψ0 is eigenstate of H0, because

† † H |ψ = H | − , H | 0 0 akσ ak σ 0 E0 [akσ ak σ 0]− E0 = (E0 − (k ) + (k))|ψ0

Time dependence:

† |ψ = | 0(t) akσ (t)ak σ (t) E0 i H † − i H = 0t 0t | e akσ ak σ e E0 i i − E0t H0t = e e |ψ0 i i − E0t (E0+(k )−(k))t = e e |ψ0 i − ((k )−(k))t =⇒ | ψ0(t) =e |ψ0

Further with E0|E0 =1 follows:

† † ψ |ψ = | | 0 0 E0 akσ akσ akσ ak σ E0 † = | − | E0 akσ (1 nkσ )ak σ E0 † = | | > E0 akσ ak σ E0 (k kF ) † = | − | E0 (1 ak σ akσ ) E0 = E0|E0 (k < kF ) = 1 740 C Solutions to Problems

Therewith we ﬁnally have i ψ t |ψ t = − k − k t − t 0( ) 0( ) exp ( ( ) ( ))( ) 2 =⇒ | ψ0(t)|ψ0(t ) | = 1 : stationary state

Problem B.12

ret = − + μ − Σ , −1 Gkσ (E (k) σ (k E))

general representation

1. It must hold

2 ! E E − (k) + μ − Σσ (k, E) = E − 2(k) + + iγ |E| (k)

=⇒ Σσ (k, E) = Rσ (k, E) + iIσ (k, E) E2 = (k) + μ − − iγ |E| (k) E2 =⇒ Rσ (k, E) = (k) + μ − , Iσ (k, E) =−γ |E|. (k)

2 ! Eiσ (k) E σ = (k) − μ + Rσ (k, E σ (k)) = 2(k) − i i (k) 2 2 =⇒ E σ (k) + (k)Eiσ (k) = 2 (k), i 2 1 9 2 E σ (k) + (k) = (k) i 2 4

Then we get two quasiparticle energies:

E1σ (k) =−2(k); E2σ (k) = (k).

Spectral weights (B.162); C Solutions to Problems 741 ∂ −1 −1 Eiσ (k) αiσ (k) = 1 − Rσ (k, E) = 1 + 2 ∂ E (k) E=Eiσ 1 =⇒ α σ (k) = α σ (k) = 1 2 3

Lifetimes:

Iσ (k, E1σ (k)) =−2γ |(k)|=I1σ (k),

Iσ (k, E2σ (k)) =−γ |(k)|=I2σ (k) 3 3 =⇒ τ σ (k) = ; τ σ (k) = . 1 2γ |(k)| 2 γ |(k)

3. Quasi particle concept is applicable provided

|Iσ (k, E)||(k) − μ + Rσ (k, E)|

⇐⇒ | Iσ (k, Eiσ )||Eiσ (k)|

⇐⇒ γ |Eiσ (k)||Eiσ (k)| ⇐⇒ γ 1

4. ∂ , Rσ (k E) =−2E ∂ E (k) (k) 2 ∂ Rσ (k, E) E = 1 + ∂ 2 (k) E (k) ∗ 1 − 4 1 =⇒ m σ (k) = m =− m, 1 1 + 5 2 ∗ 1 + 2 m σ (k) = m = m 2 1 + 2

Problem B.13 The self-energy is real and k independent. Then with (B.192), E − bσ ρσ (E) = ρ0(E − Σσ (E − μ)) = ρ0 E − aσ E − cσ

Lower band edges: 742 C Solutions to Problems

! E − bσ 0 = E − aσ E − cσ 2 ⇐⇒ 0 = E − (aσ + cσ )E + aσ bσ 2 1 1 2 = E − (aσ + cσ ) + aσ bσ − (aσ + cσ ) 2 4 (l) 1 =⇒ E = aσ + cσ ∓ (aσ + cσ )2 − 4aσ bσ 1,2σ 2 Upper band edges:

! E − bσ W = E − aσ E − cσ 2 ⇐⇒ − cσ W = E − (aσ + cσ + W)E + aσ bσ 1 2 0 = E − (aσ + cσ + W) + (aσ bσ + cσ W) 2

1 2 − (aσ + cσ + W) 4 =⇒

(u) 1 E = aσ + cσ + W ∓ (aσ + cσ + W)2 − 4(aσ bσ + cσ W) 1,2σ 2 Quasi particle density of states: ⎧ ⎪ 1 , (u) ≤ ≤ (o) ⎨ W falls E1σ E E1σ 1 (u) (o) ρσ (E) = , falls E ≤ E ≤ E ⎩⎪ W 2σ 2σ 0, otherwise

Band splitting into two quasi particle sub-bands. Index

A shift, 177, 435, 436, 460, 461, 464, 465, ABAB-structure, 306 466, 468, 469, 475, 480, 483, 485, 488, Actinides, 16, 137 489, 703, 710 Adiabatic demagnetization, 174 structure, 185, 391 Advanced Green’s function, 528, 531, 533, 539 Bandwidth correction, 436, 437, 465, 468, 703 Alkali metals, 120, 121, 387 BCS theory, 560 Alloy analogy of the Hubbard model, 471 Bernoulli number, 355 Angle averaged photoemission, 556 Binding energy, 388 Angle resolved photoemission, 556 Bloch density of states, 398, 399, 402, 403, Anisotropy, 280, 281Ð283, 291, 315, 316, 337, 437, 445, 446, 455, 458, 462, 471, 473, 340, 341, 346, 348, 350, 351, 667 478, 556 Anisotropy ﬁeld, 283, 337, 340, 341 Bloch energy, 506 Annihilation operator, 177, 201, 202, 226, 275, Bloch function, 124, 171, 506, 507, 508, 545, 278, 322, 391, 477, 491, 497, 498, 504, 597 507, 508, 511, 523, 543, 732, 733 Bloch representation, 486, 509, 545 Anomalous Zeeman effect, 61, 80, 164, 280 Bloch’s T3/2 law, 299, 326, 334, 356 Antiferromagnet, 227, 286, 297, 303Ð316, 317, Bogoliubov inequality, 286, 287, 289, 411, 412 337, 340, 341, 343, 345, 346, 350, 382, Bogoliubov transformation, 341, 348, 384 384, 431, 486, 644, 660 Bohr magneton, 11, 16, 106 Antiferromagnetism, 18, 87, 175, 431 Bohr radius, 89, 151 Antisymmetric N-particle states, 197 BohrÐSommerfeld condition, 132, 133 Antisymmetric singlet state, 187 BoseÐEinstein distribution function, 324, 343, Appearance potential spectroscopy, 524 545 Atomic-limit self-energy, 418 Bose operator, 276, 277, 279, 323, 384, 528, Auger-electron spectroscopy (AES), 524, 525, 650, 652 526 Boson, 294, 343, 350, 493, 494, 496, 497, 502, Average occupation number, 134, 324, 458, 503, 504, 505, 511, 512, 513, 544, 561, 468, 513, 540, 541, 555, 587 713, 715, 716, 718, 724, 725, 726, 727, 728, 733 B Bound current density, 3 Band Bravais lattice, 171 correction, 435, 436, 439, 441, 465, 468, Brillouin function, 159, 160, 161, 181, 227, 472 298, 299, 300, 308, 319, 484, 647 ferromagnetic solid, 389 magnetism, 12, 176, 387, 389, 395 magnets, 12, 176, 184, 194, 233, 387, 388, C 389, 391, 393, 395 Callen decoupling, 371 occupation, 394, 451, 457, 460, 461, 464, Callen method, 371Ð381, 385 466, 467, 468, 473, 481, 483, 485 Callen theory, 380

743 744 Index

Canonical partition function, 20, 21, 95, 104, Construction operator, 278, 347, 391, 433, 434, 109, 134, 135, 136, 156, 172, 236, 237, 485, 495, 497, 501, 503Ð506, 508, 671, 241, 254, 255, 257, 261, 264, 265, 266, 733 270, 323, 421, 525, 528, 564, 604, 632, Contact hyperfine interaction, 71, 73, 201 635, 725 Continuity equation, 3, 5 Cauchy’s principal value, 534 Continuous Fock representation, 494Ð501 Causal Green’s function, 528, 529 Conventional alloy analogy, 471, 473, 483 Central field approximation, 74, 75, 80 Cooper pair creation operator, 560 Centre of gravity, 163, 434, 437, 460, 468 Coordination number, 468, 647, 649 Centre of gravity of the energy spectrum, 437 Core electrons, 388, 525 Centre of mass coordinates, 149, 600 Correlated electron hopping, 435, 473 Chain of equations of motion, 372, 530 Correlations, 153, 154, 155, 236, 238, 239, Charge density, 3, 6, 62, 520, 521, 522 271, 301, 376, 394, 405, 407, 427, 435, Chemical potential, 19, 95, 97, 100Ð101, 109, 436, 439, 441, 446, 451, 453, 455Ð457, 117, 119, 135, 324, 399, 409, 417, 474, 459, 461, 463, 465, 466, 467, 468, 469, 475, 476, 478, 481, 487, 525, 543, 591, 471, 473, 474, 475, 477, 479, 481, 483, 677, 678 488, 529, 531, 535, 560, 637, 639 Classical Langevin paramagnetism, 173, 605 energy, 153, 154 Classical limit, 133 function, 236, 238, 239, 271, 301, 436, Classical quasiparticle picture, 550, 555 439, 441, 465, 481, 529, 531, 535, 560, 639 Classical theories, 405, 623, 624 Correspondence principle, 9 Closed orbit, 125, 127, 132 Coulomb gauge, 9, 10, 68, 72 Closed paths, 262, 263 Coulomb integral, 146, 191 Closed polygons, 247 Coulomb interaction, 24, 74Ð75, 120, 142, 145, Cluster configuration, 212 147, 148, 151, 153, 154, 175, 176, 177, Cluster model, 210, 212, 213, 216, 217, 219 184, 186, 190, 195, 197, 198, 199, 200, CMR system, 217 213, 229, 389, 392, 402, 420, 427, 451, c-number, 31, 32, 43, 234, 297, 301, 304, 313, 456, 462, 473, 485, 510Ð511, 545, 546, 347, 353, 396, 515, 522 559 Coherent potential approximation (CPA), 471, Covalent bonding, 188 472, 473, 471 CPA equation, 472, 473 Collective eigenoscillations, 523 Creation operator, 177, 226, 277, 278, 323, Collective magnetism, 15, 17, 18, 85, 89, 142, 391, 495, 496, 504, 511, 560 175, 176, 184, 226, 229, 280, 281, 387, Criterion for ferromagnetism, 182, 402, 410, 471, 482 457Ð461, 463 Collective phenomena, 87, 175 Critical exponent, 228, 269, 270, 299, 405, Colossal magneto-resistance, 217 622, 623, 624 Combined Green’s function, 534, 535, 537, Critical phenomena, 234 540 Critical region, 299, 368, 381, 623 Commutation relation, 49, 50, 52, 53, 58, 66, Curie constant, 24, 161, 165, 173, 181, 300, 82, 83, 202, 203, 226, 274, 275, 276, 306, 311, 319, 485, 608, 615, 619, 621 277, 282, 287, 293, 339, 383, 410, 429, Curie law, 24, 161, 165, 170, 173, 245, 420, 491, 501, 508, 511, 540, 544, 561, 571, 566, 567, 606, 607 573, 580, 613, 651, 652, 654, 733 Curie temperature, 17, 18, 175, 179, 183, 227, Commutation relations for the spin operators, 253, 254, 267Ð268, 281, 299, 300, 301, 203 303, 308, 309, 317, 319Ð320, 321, 336, Compensation temperature, 317, 318 355, 369, 372, 381, 382, 402, 403, 405, Completeness relations, 206, 284, 429, 494 461, 468, 482Ð485, 620, 648 Conduction electrons, 12, 16, 90, 91, 104, 109, CurieÐWeiss law, 183, 184, 227, 300, 308, 321, 116, 117Ð121, 137, 138, 142, 154, 155, 371, 372, 381, 484, 615, 620 178, 194, 201, 202, 203, 210, 230, 387, Current density of polarization charges, 3 388, 520, 521 Cyclic invariance of trace, 517, 529, 728, 729 Index 745

Cyclotron frequency, 105, 127, 128 Dynamic susceptibility, 14 Cyclotron mass, 126Ð127 Dyson equation, 439, 441, 442, 448, 449, 454, Cyclotron orbit, 126 477, 480, 547 DysonÐMaleev« transformation, 329 D Darwin term, 44, 73, 76 E d-band degeneracy, 393 Easy axis, 310, 311, 312, 313, 316, 341 Degeneracy of the Landau levels, 106, 593 Easy direction, 310, 315, 316, 337 Degenerate electron gas, 118 Effective Hamiltonian, 187, 195, 200, 207, Degree of degeneracy, 107, 109, 110, 128, 129, 222, 428, 429 593 Effective Heisenberg Hamiltonian, 216 de Haas-von Alphen effect, 113, 121Ð134 Effective magneton number, 165 Delta function, 505, 518, 532, 548, 549, 550, Effective mass approximation, 207, 209 555 Effective masses, 104, 113, 120, 121, 127, 207, Density 209, 545, 561 correlation, 436 Eigenenergies of the transfer matrix, 243 parameter, 151, 152 Eigenspace, 75, 196, 197, 200 of states, 93Ð95, 98, 100, 109, 110, 130, Electric displacement, 522 135, 138, 139, 142, 143, 177, 185, 186, Electron hopping, 217, 422, 423, 427, 431, 387, 398, 399, 400, 402, 403, 418, 420, 435, 465, 473 421, 422, 423, 424, 425, 426, 431, 437, ElectronÐphonon interaction, 389 445, 446, 455, 458, 460, 461, 462, 466, Electron polarization, 468 467, 468, 469, 471, 473, 475, 476, 478, Electron spin, 28, 34, 35, 37, 39, 41, 45, 71, 76, 481, 482, 483, 486, 487, 488, 524, 541, 91, 106, 123, 139, 187, 200, 201, 202, 555, 556, 557, 561, 562, 588, 589, 590, 211, 212, 218, 427, 431, 484, 486, 487 591, 675, 689, 690, 691, 742 Elementary excitations, 277, 291, 337, 353, Diagram technique, 443 354, 523 Diamagnetic susceptibility, 88, 134, 585 Energy gap, 340, 427, 734, 735 Diamagnetism, 15, 85Ð136, 175, 596 Energy renormalization, 545 Dielectric function, 520Ð523 Entropy, 19, 24, 98, 134, 137, 173, 174, 239, Dipolar hyperfine interaction, 69, 70, 73 240, 243, 244, 269, 587, 606, 607, 610, DipoleÐdipole interaction, 179 616 Dipole interaction, 179, 180, 280Ð281, 345, Entropy per spin, 243, 244 346, 347, 349, 350, 351, 385, 667 Equal-time correlations, 406, 407 Dipole moment, 3, 23, 63 Equation of motion, 124, 352, 353, 360, 362, Dirac equation, 28Ð34, 38, 45 372, 373, 374, 375, 416, 417, 451, 452, foranelectroninanelectromagneticfield, 477, 516, 530, 534, 537, 540, 543, 544, 30 546, 553, 558, 560, 673, 674, 709, 712, for a free particle, 30 731, 733, 734 Dirac identity, 354, 455, 534, 538, 540 Equation of motion method, 416, 451, 530, 553 Dirac picture, 516 EuO, 17, 176, 179, 230, 273, 335, 336, 628 Dirac spin operator, 33 EuSe, 176, 307, 309, 382, 644 Dirac’s vector model, 195Ð200 EuTe, 176, 273, 307, 309, 310, 621 Direct Coulomb interaction, 148 Exactly half-filled energy band, 427 Direct terms, 229, 393 Exchange corrected susceptibility, 154 Discrete Fock representation, 501Ð506 Exchange corrections, 142Ð154 Distinguishable particles, 149, 492 Exchange field, 178Ð180, 184, 227, 230, 231, Double exchange, 217Ð226 297, 304, 305, 318, 395, 397 Double exchange Hamiltonian, 223, 225 Exchange integral, 176, 191, 193, 210, 280, DoubleÐhopping correlation, 465 287, 304, 309, 326, 335, 382, 384, 431, Dressed skeleton diagrams, 442 543 d-states, 392 Exchange interaction, 17, 142, 149, 154, Dynamical mean-field theory (DMFT), 450, 175Ð231, 279, 280, 293, 336, 337, 346, 476, 478, 479, 480 357, 358, 359, 387 746 Index

Exchange operator, 200, 216, 281 Fourier transformation, 265, 411, 445, 448, Exchange parameter, 182, 194, 298, 305, 312, 452, 485, 518, 523, 527, 542, 550, 653, 319 697, 703 Exchange splitting, 397, 399, 435, 467, 468, Four-spin correlation function, 271, 639 474, 485 Free current density, 3, 7 Exchange terms, 229, 230, 281, 393 Free energy, 19, 20, 21, 98, 137, 142, 173, 174, Excitation energy, 294, 322, 542, 544 178, 182, 226, 227, 239, 240, 253, 323, 361, 401, 431, 543, 559, 595, 606, 608, F 611, 615, 616, 618, 634, 738 Families of loops, 258, 259 Free energy of the two-dimensional Ising Fermi model, 254Ð270 edge, 97, 111, 130, 139, 468, 481, 482, 551 Free energy per spin, 242, 245, 255, 266Ð267, energy, 92, 97, 124, 134, 152, 186, 398, 634 399, 402, 403, 459, 461, 462, 591 Free enthalpy, 227, 569, 608, 617 function, 97, 98, 99, 110, 111, 115, 130, Frustration, 201 140, 141, 143, 399, 419, 437, 444, 468, 4f-systems, 177, 201 δ 541, 555, 679, 727 -function, 113, 114, 119, 130, 141, 486 layer, 97, 99, 120, 140 Fundamental commutation operator, 226, 276 relations, 50, 82, 202, 277, 383, 491, 501, sea, 128, 130 508, 511, 540, 544, 733 sphere, 92, 93, 109, 129, 172, 203, 205, rules, 497, 504 551, 561 surface, 130, 131 G temperature, 93 Garnet, 317 wavevector, 92, 551 GdCl3, 303, 304 Fermion, 96, 135, 175, 186, 195, 197, 202, Generalized susceptibility, 13 394, 396, 411, 417, 441, 493, 494, 496, GibbsÐDuhem relation, 21 497, 502, 504, 505, 508, 511, 512, 513, Grand canonical ensemble, 21, 22, 513, 525, 540, 541, 696, 714, 715, 716, 718, 724, 528, 586, 724 725, 726, 727, 728, 733 Grand canonical partition functions, 95, 104, Ferrimagnet, 317Ð321, 336 109, 134, 323, 525, 528, 725 Ferrimagnetism, 17Ð18, 87, 175 Grand canonical potential, 21, 104, 109Ð117, Ferrite, 317 119, 121, 129, 130, 324, 350 Ferroelectrics, 235 Grid volume, 128, 586, 588 Ferromagnet with dipolar interaction, 345Ð351 Ground state energy, 75, 151, 152, 153, 191, Ferromagnetic saturation, 292, 323, 382, 383, 192, 215, 216, 222, 322, 329, 339, 342, 468, 474 349, 360, 384, 427, 513, 543, 690, 724 Ferromagnetism, 17, 175, 178, 179, 180, 182, Ground state energy of the Jellium model, 152, 184, 185, 186, 194, 217, 226, 233, 234, 153 245, 389, 398, 402, 404, 405, 409, 410, Ground state of the Jellium model, 151, 153 415, 419, 432, 435, 436, 439, 441, 443, Group velocity, 124 451, 453, 455Ð464, 468, 469, 471, 473, G-type, 286, 306 474, 475, 479, 481, 482, 483, 484, 485, 488, 520 H Fine structure, 11, 45, 57, 72, 79, 156, 162, Half-ﬁlled band, 403, 419, 427Ð431, 435, 462, 163, 164, 166, 281 473, 480, 483, 706 Finite lifetimes, 470, 471, 484, 485, 545, 552, Hamilton function, 270 554 Hamiltonian matrix, 220, 487, 680, 683 First law of thermodynamics, 19, 302 Hard direction, 281, 315 Fluctuation-dissipation theorem, 23, 271 Harmonic approximation, 322, 345Ð351 Fock states, 503, 505, 513, 586, 713, 724, 725, Harmonic approximation for antiferromagnets, 726, 727 336Ð345 Four-component theory, 41, 42 Harmonic oscillator, 106, 276, 279, 323 Index 747

HartreeÐFock approximation, 442, 443, 444, Imaginary part of the self-energy, 470, 548, 451, 456, 457 549, 555 Hartree-potential, 74 Impurity diagrams, 477 Heaviside step function, 517 Impurity level, 477 Heisenberg antiferromagnet, 305 Impurity scattering, 450, 476 Heisenberg Hamiltonian, 176, 195, 200, 216, Impurity self-energy diagrams, 449 274, 275, 276, 277, 279, 292, 297, 322, Impurity spectral density, 479 329, 382, 383, 430, 543, 649 Indirect exchange, 194, 200Ð226, 273, 387 Heisenberg model, 56, 176, 188, 192, 194, Indistinguishability, 153, 212, 492 233, 234, 273Ð386, 387, 395, 410, 427, Indistinguishable fermions, 195 431, 518, 646, 655 Indistinguishable particles, 149 Heisenberg operator, 360, 437, 530, 537 Infinite lattice dimensions, 438, 441, 444 Heisenberg representation, 517 Infinitely narrow band, 415Ð420, 425, 427, HeitlerÐLondon method, 188Ð194 431, 436, 438, 451, 453, 454, 455, 471, Hermite polynomials, 106 486, 489, 673, 708 Hierarchy of equation of motion, 353 Internal energy, 19, 20, 98, 101, 137, 143, 173, Higher Green’s function, 353, 372, 373, 374, 226, 301, 302, 328, 350, 359, 361, 384, 416, 451, 530, 546, 558, 693 513, 541, 558, 592, 606, 607, 615, 618, Higher spectral density, 466 726, 738 High-temperature behaviour of the Interpolation method, 418, 453, 454Ð455 magnetization, 161, 357 Inverse paramagnetic susceptibility, 301, 468 High-temperature expansion, 255Ð256, 270, Inverse photoemission, 523, 524 381 Ion, 3, 4, 5, 6, 7, 8, 9, 12, 13, 25, 85, 86, 87, 88, High-temperature region, 369Ð371, 372, 381 89, 90, 145, 149, 155, 156, 194, 195, Hilbert space, 163, 491, 493, 494, 499, 500, 200, 201, 209, 210, 211, 213, 217, 218, 501 219, 282, 291, 307, 317, 387, 388, 389, H2-molecule, 188, 193 390, 451, 506, 521 HolsteinÐPrimakoff transformation, 276Ð279, Irreducible space, 51, 52 322, 329, 337, 346, 383, 650 Irreducible tensor operator of rank, 52, 53, 64, Hopping integral, 177, 390, 411, 425, 444, 66, 83 509, 671 Ising model, 233Ð271, 273, 283, 291, 635 Hubbard bands, 436, 437, 438, 458, 466, 468, Iterative perturbation theory, 480 481 Hubbard Hamiltonian, 395, 416, 486, 675 J Hubbard-I approximation, 451Ð453 Jellium model, 145Ð154, 387, 508 Hubbard-I solution, 418, 453, 455, 456, 457, jj-coupling, 81 458, 464, 465, 468, 483 Hubbard-model, 177, 229, 291, 387Ð489 K Hubbard splitting, 438 KleinÐGordon equation, 29, 30 Hund’s rules, 25Ð28, 81, 155, 165, 211, 216, Knot, 247, 248, 257, 258, 259 218 Kondo behaviour, 201 Hybridization function, 477, 478, 489, 709, 710 Kondo-lattice model, 177, 280, 291 Hydrogen atom, 57, 90, 189, 192, 193 Kondo resonance, 481, 482, 484 Hypercubic lattice, 445, 446, 481, 482 Kondo-type resonance, 482 Hyperfine interaction, 68, 69, 70, 71, 73, 201, KramersÐKronig relations, 534, 539, 547, 548 202 Kubo formula, 515Ð518, 519, 522

I L Ideal Fermi gas, 91, 109, 409, 444 Landau cylinder, 109, 127Ð128, 129, 130, 131, Ideal paramagnet, 24, 173, 174, 297, 358, 608 132, 134 Identical Fermions, 186, 502 Landau diamagnetism, 87, 104Ð121 Identical particles, 175, 492Ð493, 499, 500, Landau levels, 104Ð109, 128, 129, 133, 134, 501, 513, 722 135, 139, 593 748 Index

Landau quantum number, 128, 130 curve, 182, 183, 184, 335, 336, 409, 468, Lande-factor, 56, 59 484 Lande’s g-factor, 11, 40, 233 per spin, 242, 243 Lande’s interval rule, 79, 170 Magnon Langevin function, 160, 181, 228, 596, 605, interaction, 332 622 occupation density, 544 Langevin paramagnetism, 16, 164, 165, 170, vacuum state, 323 173, 605 Many-magnon states, 295 Larmor diamagnetism, 87Ð90, 134 Material equations, 7, 522 Lattice dynamics, 388, 389 Matsubara propagator, 441 Lattice gas model, 235 MaxwellÐBoltzmann distribution, 97 Lattice potential, 91, 104, 145, 177, 389, 390, Maxwell’s equations, 1, 2, 569 506, 539, 541 Mean field decoupling, 451 Lattice structure, 307, 317, 389, 393, 469, 485, MerminÐWagner theorem, 283Ð291, 298, 405, 556 410Ð415 Linear response theory, 515Ð523 MetalÐinsulator transitions, 395 Linear spin wave approximation, 322, 343, 383 Metallic bonding, 153 Local diagrams, 450 Method of double-time Green’s function, 351 Local exchange, 218 Microscopic dipole density, 3 Localized magnetism, 12, 177, 194, 201, 209 Microscopic Maxwell’s equations, 1 Local-moment magnetism, 177 MnO structure, 307 Local propagator, 448, 449 Modified alloy analogy, 472, 473, 483 Logarithmic divergence, 269 Modified perturbation theory, 479Ð482 Longitudinal susceptibility, 519 Molecular field approximation, 296Ð321, 326, Loop, 257Ð261, 262, 263, 264 341, 351, 353, 369, 372, 381, 382, 395, Lorentz force, 7, 82, 574 404, 405, 408, 410, 451, 646 Lower sub-band, 425, 426, 462 Molecular Heisenberg model, 188 Low-temperature region, 334, 367Ð368, 380 Moment method, 464, 465, 470 LS-coupling, 77, 155 Monopole moment, 63 Luttinger sum rule, 481, 482 MottÐHubbard insulator, 557 Luttinger theorem, 481 Mott transitions, 395 Multiple processes, 432 M Multipole expansion, 62, 63 Macroscopic Maxwell’s equations, 2 Magnetic anisotropy, 281Ð283, 316 N Magnetic insulators, 194, 210, 233, 234, 273, Narrow band, 387, 390, 415Ð420, 422, 425, 387 427, 431, 436, 438, 451, 453, 454, 455, Magnetic moment, 1, 5, 6, 7Ð13, 16, 20, 22, 23, 471, 486, 489, 673, 708 28, 85, 86, 87, 90, 134, 136, 137, 155, Narrow energy band, 387, 389Ð393, 395 156, 161, 166, 172, 173, 175, 176, 178, Neel state, 337, 338, 342, 343, 344 179, 180, 186, 197, 233, 270, 290, 294, Neel temperature, 18, 175, 306Ð308, 311, 317 307, 317, 346, 351, 381, 387, 484, 492, NNSS structure, 307 518, 520, 564, 565, 585, 596, 604, 635 Non-linear spin wave theory, 329 Magnetic moment operator, 9 Non-local diagrams, 450 Magnetic moment of spin, 29, 41 Non-locality, 450, 468 Magnetic part of the specific heat, 328 Non-orthogonality catastrophe, 192 Magnetic quantum number, 26, 51, 90, 158, Non-relativistic limit, 38, 41 203 Non-relativistic limit of the Dirac’s theory, 37, Magnetic stability, 468, 470, 483, 485, 520 39 Magnetic susceptibility tensor, 519 Normal ordering, 278 Magnetite, 317 Normal Zeeman effect, 61, 81, 158, 162, 164 Magnetization N-particle state, 196, 197, 491, 492, 493, 494, current density, 5, 7 495, 501, 502, 504 Index 749

NSNS structure, 307 Pauli paramagnetism, 16, 113, 165 Nuclear current density, 67 Pauli spin matrices, 31, 81, 82, 198, 205, 212, Nuclear magneton, 71 571, 625 Number operator, 21, 95, 96, 177, 229, 275, Pauli’s principle, 16, 27, 140, 147, 153, 154, 323, 343, 393, 396, 416, 422, 477, 505, 184Ð188, 197, 201, 431, 502, 504, 724 511, 512, 525, 719, 727 Pauli’s theory, 38, 39 Pauli susceptibility, 91, 140, 145, 154, 408 O Peierls argument, 245, 254 Occupation number Periodic table, 81, 155, 402 operator, 96, 177, 229, 343, 393, 396, 422, Periodic boundary conditions, 91, 92, 106, 477, 505, 727 145, 171, 240, 241, 264, 265, 512, 588, representation, 501Ð506 597, 598 Occupation probability, 95Ð97 Permutation, 147, 195, 196, 197, 198, 204, One-dimensional Ising model, 236, 237, 239, 492, 493, 494, 512 240, 242, 243, 244, 245, 270, 271, 291 Permutation operator, 197, 204, 492, 512 One-electron Green’s function, 416, 439, 441, Perturbation 451, 453, 454, 456, 465, 539, 546, 558, matrix, 196, 428 561 theory, 80, 145, 151, 152, 153, 195, 196, One-electron spectral density, 406, 418, 434, 200, 204, 218, 230, 427, 428, 431, 438, 437, 441, 488, 489, 540, 546, 556, 561, 443, 444, 479Ð482, 629, 630, 631 692, 708, 737 theory around HartreeÐFock, 444, 480 One-magnon state, 291Ð296 Phase transition of second, 269, 303 Onsager argument, 131Ð134 Phase volume, 94, 589 Open orbit, 125 Phonons, 120, 134, 292, 329, 388, 389 Orbital angular momentum, 10, 25, 34, 35, 37, Photoemission, 523, 524, 556 48, 57, 71, 82, 85, 134, 156, 160, 573 Plasmons, 523 Orbital angular momentum operator, 48, 82 Polarization Orbital hyperﬁne interaction, 68, 73 cloud, 201 Orbital quantization, 106, 139 density, 5 Orbital quantum number, 90 Position representation, 508 Order parameter, 228, 270, 299, 405, 622 Primitive translation vectors, 171, 598 Order of the vertex, 256 Principal quantum number, 45, 57, 90 Overlap integral, 190, 194, 200, 411 Principle of indistinguishability, 492 P Probability, 47, 62, 69, 73, 95, 96, 97, 158, Paramagnet, 24, 137, 157, 159, 160, 169, 170, 213, 246, 247, 249, 251, 387, 389, 390, 173, 174, 175, 180, 184, 245, 297, 358, 418, 431, 432, 451, 467, 473, 492, 502, 420, 608, 647 525, 552, 553 Paramagnetic Curie temperature, 18, 227, 300, Projection operator, 429 301, 308Ð310, 317, 321, 382, 620 Paramagnetism, 15Ð16, 18, 89, 90, 113, 120, Q 137Ð174, 178, 180, 182, 226, 233, 234, Quadrupole moment, 63, 64, 66 245, 298, 396, 401, 459, 463, 605 Quantum Monte Carlo calculation, 482 Partial density of states, 458, 467 Quasiparticle Partial spectral density, 434 concept, 354, 545Ð559 Particle density, 396, 398, 399, 400, 418, 420, damping, 470Ð475, 484 421, 423, 424, 425, 426, 444, 455, 460, density of states, 398, 400, 418, 419, 420, 462, 466, 467, 468, 469, 473, 475, 476, 421, 423, 424, 425, 426, 455, 460, 462, 486, 487, 524, 541, 555Ð557, 675, 742 466, 467, 468, 469, 475, 476, 486, 487, Particle-hole excitations, 205 541, 555Ð557 Particle number operator, 21, 95, 416, 505, energies, 361, 362, 376, 378, 380, 408, 511, 512, 525, 719 417, 456, 458, 464, 466, 470, 486, 489, Path, 115, 261, 262, 263, 264, 538, 551, 626 554, 555 Pauli operator, 275Ð276 levels, 432, 451 750 Index

lifetime, 554, 555 Self-energy, 418, 439, 440, 441, 442, 443, 444, magnon, 294, 329 447, 448, 449, 450, 451, 454, 455, 456, splitting, 467, 468, 473, 474 465, 470, 471, 473, 475, 476, 477, 478, sub-bands, 425, 432, 436, 438, 451, 456 479, 547, 548Ð551, 555, 556, 557, 561, weight, 554, 555 741 Quenching, 201 Self-energy part, 442 Self-intersection, 258, 259, 261, 263, 264 R Semiclassical vector model, 221 Random phase approximation, 330, 353, 361 sÐf interaction, 204 Rare earth, 16, 28, 137, 155, 209, 317 s-f model, 177 Recursion formula, 237, 366 Short-range ordering, 301 Reduced matrix element, 54, 67 Single-impurity Anderson model, 476, 479, Relativistic electrons, 29, 82 480 Renormalized spin wave energy, 331 Single-ion anisotropy, 282 Renormalized spin waves, 329Ð336 Residue theorem, 114, 115, 532, 551, 626, 627 Single step, 261, 262, 265 Resonance energies, 549, 550, 551 Skeleton diagram, 442, 443, 479, 480 Resonances, 71, 282, 481, 482, 484, 520, 523, Slater determinant, 502 548, 549, 550, 551 Solvability condition, 59, 349 Response functions, 13, 123, 515, 534, 537 Sommerfeld expansion, 99Ð103, 141, 591 Retarded Green’s function, 352, 442, 451, 517, Sommerfeld model, 90Ð103, 104, 123, 129, 519, 523, 528, 532, 538, 546 130, 131, 134, 135, 138, 139, 142, 143, Riemann’s ζ-function, 99 151, 152, 154, 172, 201, 387 RKKY-coupling constant, 209 Sommerfeld’s fine structure constant, 11 RKKY interaction, 194, 200Ð209, 217 Space of antisymmetric states, 493 Rotation Space of symmetric states, 493 matrix, 46, 51 Specific heat, 98, 102, 244, 268Ð269, 297, operator, 47Ð48, 51 301Ð303, 304, 313Ð315, 328, 340, 341 RPA decoupling, 353, 362, 363, 368, 371, 372, Specific heat of the antiferromagnet, 314 376 Spectral density ansatz, 441, 464 S approach, 469 Satellite peaks, 432, 434 Spectral moments, 434, 437Ð439, 443, 453, Saturation magnetization, 165, 228, 281, 295, 464, 465, 466, 472, 473, 480, 488, 489, 298, 311, 356 536Ð537, 560, 561, 694, 695, 697, 704, Scalar operator, 47, 53 706, 708, 732, 737 Scalar product, 10, 42, 54, 55, 56, 58, 84, 188, Spectral representation of the advanced 199, 207, 215, 217, 233, 274, 283, 284, Green’s function, 533 285, 494 Spectral representation of the retarded Green’s Schrodinger¬ picture, 516, 517 function, 532 Schwarz inequality, 283, 285, 286 Screening, 75, 90, 154, 395, 402, 520, 523 Spectral representation of the spectral density, Screening effects, 154, 395 532, 533, 553 SDA self-energy, 465, 470, 475 Spectral theorem, 354, 360, 362, 372, 376, 406, Second order perturbation theory, 200, 428, 419, 443, 447, 458, 466, 473, 534Ð536, 431, 438, 444, 479, 630 540, 544, 555, 558, 692, 711, 734, 735 Second order phase transitions, 484 Spectral weights, 417, 432, 434, 438, 451, 457, Second order self-consistent perturbation 458, 464, 466, 473, 489, 545, 549, 554, theory, 443 555, 561, 674, 737, 740 Second quantization, 275, 390, 491Ð514, 721, Spectroscopies, 523Ð528, 532, 534 724 Spherical harmonics, 53, 63, 64 Secular equation, 192, 212, 428 Spherical multipole moments, 63 s-electrons, 62, 69, 70, 71, 73, 75, 121, 155, Spin 1/2 particles, 29, 229 202, 203, 420 Spin correlation function, 236, 238, 271, 639 Index 751

Spin-dependent band shift, 177, 436, 460, 461, Stoner approximation, 396, 486, 672 464Ð470, 475, 480, 483, 485, 488, 489, Stoner criterion, 186, 402, 404, 409, 415, 458, 703 460, 461 Spin deviation, 275, 277, 292, 294, 295, 322, Stoner electron, 398 354 Stoner model, 395Ð409, 410, 443, 444, 451, Spin disorder, 343 453, 456, 457, 461, 486, 489, 704 Spin-flip correlation, 436 Strong-coupling limit, 432, 458 Spinflip operator, 274 Strong-coupling regime, 427, 431Ð437, 438, Spin-flip terms, 296 441, 464, 468 Spin-flop field, 315Ð316, 341 Strong ferromagnetism, 398 Spin glass behaviour, 201 Sub-lattice magnetization, 286, 304, 305, 306, Spin operator, 33, 35, 37, 65, 82, 83, 176, 178, 310, 311, 314, 317, 318Ð319, 340, 341, 198, 202, 203, 205, 211, 226, 228, 230, 343, 344, 384, 619, 620, 660 234, 273Ð279, 282, 287, 292, 322, 337, Sum rule, 425, 464, 481, 482, 536 346, 351, 361, 362, 381, 382, 383, 410, Superconductor, 15, 291, 561 429, 430, 487, 649 Superexchange, 194, 195, 209Ð216, 218, 221, SpinÐorbit coupling, 9, 12, 25, 29, 40Ð45, 57, 427 61, 75Ð76, 79, 156, 158, 162, 164Ð165, Susceptibility, 1, 13Ð15, 16, 18, 23, 88, 89, 90, 281 91, 104, 117Ð124, 129, 134, 137, 138, SpinÐorbit interaction, 25, 41, 43, 44, 57, 72, 140, 142, 145, 154, 155, 161, 164, 165, 74, 77, 78, 80, 83, 157, 158Ð164, 203, 166, 170, 173, 174, 183, 184, 227, 244, 280, 281 245, 270, 271, 291, 292, 300, 301, 308, Spin-statistics theorem, 493 309, 310Ð312, 315, 317, 320Ð321, 340, Spin wave approximation, 278, 322, 336, 341, 345, 369, 370, 371, 381, 405Ð409, 419, 342, 343, 351, 375, 383, 384, 385, 662, 420, 461Ð464, 468, 470, 481, 482Ð485, 667 487, 515, 518Ð520, 564, 565, 585, 611, Spin wave branches, 340 623, 634, 638, 678, 679 Spin waves, 276, 278, 292, 294, 295, 296, 299, Susceptibility of the conduction electrons, 104, 322Ð336, 337, 339, 340, 341, 342, 343, 117Ð121, 138 345, 347, 348, 349, 351, 354, 355, 356, Symmetric decoupling, 373, 374 357, 360, 368, 372, 375, 383, 384, 385, Symmetric triplet state, 187 405, 520, 543Ð545, 662, 667 Split-band regime, 473 T Spontaneous magnetization, 17, 160, 176, 181, Taylor expansion, 48, 102, 115, 140, 141, 268 182, 183, 184, 227, 236Ð240, 242, 246, Tensor 247, 252, 253, 254, 268, 269Ð270, 283, operator, 52Ð53, 54, 64, 66, 83 286, 291, 298, 299, 305, 317, 319, 326, of rank, 52, 66, 83 354Ð359, 363Ð371, 385, 403, 405, 410, Term, 27, 28, 280 411, 415, 419, 420, 451, 464, 468, 484, Thermal energy, 157, 161, 179, 295, 359 637 Thermodynamic expectation value, 332, 351, Spontaneous ordering, 175, 464 353, 359, 372, 412, 413, 435, 541 Square lattice, 254, 255 Thermodynamic limit, 109, 146, 239, 242, 243, Standard components, 52, 53, 54, 64, 66, 78, 245, 246, 247, 252Ð254, 255, 264, 266, 83, 84 271, 290, 414, 633, 638 State with finite lifetime, 552 Thermodynamic potential, 19, 21, 243, 252, Static paramagnetic susceptibility, 461, 463, 359Ð361 484 Third law of thermodynamics, 173, 244 Static susceptibility, 14, 405Ð409, 419, Time evolution operator, 493 461Ð464, 468, 470, 481, 482Ð485, 487 Time ordering operator, 529 Stationary state, 552, 553, 561, 740 Total angular momentum operator, 37 Statistical operator, 516, 559, 586 Transfer function, 241 Step function, 97, 143, 205, 517, 529, 532, 560 Transfer integral, 214, 216, 219 Stoke’s theorem, 133 Transfer matrix, 241, 243 752 Index

Transfer matrix element, 213 Valence mixture, 217, 317 Transfer matrix method, 240 van Vleck paramagnetism, 166Ð171 Transition Variation ansatz, 190, 192 metals, 16, 137, 210, 387, 389, 395, 402 Variation method, 190 operator, 523, 524, 525, 526 Vector operator, 48, 49, 52, 53, 54, 78, 82, 83 Translational symmetry, 85, 171, 239, 265, Vector potential, 9, 10, 67, 68, 72, 82, 104 297, 332, 353, 384, 396, 413, 417, 435, Vertex, 256, 257, 442 452, 465, 466, 510, 522, 647, 665, 701 Virtual hopping, 431 Transposition operator, 198, 199, 492 Transverse susceptibility, 520 W Triangular inequality, 413 Wall, 247, 249, 250 Tunnelling probability, 389 Wannier functions, 508 Two-component theory, 38, 41 Wave packet, 124, 492 Two dimensional Ising model, 245Ð270 Weak-coupling behaviour, 441, 475, 479, 481, Two-electron system, 186, 187, 195, 199, 423, 482, 484 426 Weak ferromagnetism, 398 Two-magnon states, 330 Weight of the path, 261 Two-particle density of states, 524 Weiss ferromagnet, 180Ð184, 228, 298, 305, Two-peak structure, 434 397, 401, 621 Two-site Hubbard-model, 421, 424, 487, 488 Weiss model, 180Ð184, 230, 298, 395 Two-site model, 420Ð427, 432, 451, 456, 464, WignerÐEckart theorem, 45Ð56, 59, 60, 61, 64, 487, 488 65, 66, 78, 162, 167, 168, 233, 280 Tyablikov approximation, 353, 356 WignerÐSeitz cell, 281, 598

U X Uniaxial ferromagnet, 282 XY-model, 233 Upper sub-band, 425, 456 Z V Zeeman energy, 68, 156 Vacuum state, 323, 495, 501, 504, 511 Zeeman term, 58, 72, 106, 280, 281, 292 Valence electrons, 388 Zero-bandwidth limit, 417, 487