Second Quantization∗

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Jörg Schmalian May 19, 2016

1 The harmonic oscillator: raising and lowering operators

Lets ﬁrst reanalyze the harmonic oscillator with potential

mω2 V (x) = x2 (1) 2 where ω is the frequency of the oscillator. One of the numerous approaches we use to solve this problem is based on the following representation of the momentum and position operators: r x = ~ a† + a b 2mω b b r m ω p = i ~ a† − a . (2) b 2 b b From the canonical commutation relation

[x,b pb] = i~ (3) follows

† ba, ba = 1 † † [ba, ba] = ba , ba = 0. (4) Inverting the above expression yields

rmω i ba = xb + pb 2~ mω r † mω i ba = xb − pb (5) 2~ mω ∗Copyright Jörg Schmalian, 2016

1 † demonstrating that ba is indeed the operator adjoined to ba. We also deﬁned the operator † Nb = ba ba (6) which is Hermitian and thus represents a physical observable. It holds

mω i i Nb = xb − pb xb + pb 2~ mω mω mω 2 1 2 i = xb + pb − [p,b xb] 2~ 2m~ω 2~ 2 2 1 pb mω 2 1 = + xb − . (7) ~ω 2m 2 2 We therefore obtain 1 Hb = ω Nb + . (8) ~ 2

1 Since the eigenvalues of Hb are given as En = ~ω n + 2 we conclude that the eigenvalues of the operator Nb are the integers n that determine the eigenstates of the harmonic oscillator. Nb |ni = n |ni . (9) † Using the above commutation relation ba, ba = 1 we were able to show that √ a |ni = n |n − 1i b √ † ba |ni = n + 1 |n + 1i (10) † The operator ba and ba raise and lower the quantum number (i.e. the number of quanta). For these reasons, these operators are called creation and annihilation operators.

2 second quantization of noninteracting bosons

While the above results were derived for the special case of the harmonic oscillator there is a similarity between the result

1 E = ω n + (11) n ~ 2 for the oscillator and our expression X E{np} = εpnp (12) p for the energy of a many body system, consisting of non-interacting indistinguishable particles. While n in case of the oscillator is the quantum number label, we may alternatively argue that it is the number of oscillator quanta in the oscillator. Similarly we can consider the many body system as a collection

2 of a set of harmonic oscillators labelled by the single particle quantum number p (more generally by p and the spin). The state of the many body system was characterized by the set {np} of occupation numbers of the states (the number of particles in this single particle state). We the generalize the wave function |ni to the many body case

|{np}i = |n1, n2, ..., np, ...i (13) and introduce operators √ bap |n1, n2, ..., np, ...i = np |n1, n2, ..., np − 1, ...i † p bap |n1, n2, ..., np, ...i = np + 1 |n1, n2, ..., np + 1, ...i (14) That obey h † i bap, bap0 = δp,p0 . (15) It is obvious that these operators commute if p 6= p0. For p = p0 follows

† p bapbap |n1, n2, ..., np, ...i = np + 1bap |n1, n2, ..., np + 1, ...i = (np + 1) |n1, n2, ..., np, ...i (16) and

† √ † bapbap |n1, n2, ..., np, ...i = npbap |n1, n2, ..., np − 1, ...i = np |n1, n2, ..., np, ...i (17)

† † which gives bapbap − bapbap = 1. Thus the commutation relation follow even if the operators are not linear combinations of position and momentum. It also follows † nbp = bapbap (18) for the operator of the number of particles with single particle quantum number P † p. The total number operator is Nb = p bapbap. Similarly, the Hamiltonian in this representation is given as

X † Hb = εpbapbap (19) p which gives the correct matrix elements. We generalize the problem and analyze a many body system of particles with single particle Hamiltonian

p2 bh = b + U (r) (20) 2m b which is characterized by the single particle eigenstates

bh |φαi = εα |φαi . (21)

3 α is the label of the single particle quantum number. We can then introduce the occupation number representation with

|n1, n2, ..., nα, ...i (22)

h † i and corresponding creation and destruction operators baα, baα0 = δα,α0 . We can then perform a unitary transformation among the states X X |βi = Uβα |αi = |αi hα|βi (23) α α

The states |βi are in general not the eigenstates of the single particle Hamil- tonian (they only are if Uβα = hα|βi = δαβ). We can nevertheless introduce creation and destruction operators of these states, that are most naturally de- ﬁned as: X baβ = hβ|αi baα (24) α and the corresponding adjoined equation

† X ∗ † baβ = hβ|αi baα. (25) α

This transformation preserves the commutation relation (see below for an example). We can for example chose the basis β as the eigenbasis of the potential. Then holds in second quantization

X † Ub = hβ |U (r)| βi aβaβ (26) β and we can transform the result as

4 The commutation relation is then δα,α0

h 0 i X ∗ 0 h † i ψb(r) , ψb(r ) = φα (r) φα0 (r ) baα, baα0 α,α0 X ∗ 0 X 0 = φα (r) φα (r ) = hr|αi hα|r i α α = hr|r0i = δ (r − r0) (29) and it follows Ub = d3rU (r) ψb† (r) ψb(r) (30) ˆ Similarly holds for the kinetic energy 2 ~ 3 3 0 2 0 † 0 Tb = − d rd r r ∇ r ψb (r) ψb(r ) 2m ˆ 2 ~ 3 3 0 † 2 0 0 = − d rd r ψb (r) ∇ δ (r − r ) ψb(r ) 2m ˆ 2 ~ 3 † 2 = − d rψb (r) ∇ ψb(r) (31) 2m ˆ Thus we ﬁnd X † H = εαbaαbaα α 2 2 3 † ~ ∇ = d rψb (r) − + U (r) ψb(r) (32) ˆ 2m With the help of the ﬁeld operators ψb(r) and ψb† (r) is it possible to bring the many body Hamiltonian in occupation number representation into the same form as the Hamiltonian of a single particle.

2.1 Example 1: a single particle We consider the most general wave function of a single spinless boson:

3 † |ψαi = d rφα (r) ψb (r) |0i (33) ˆ where |0i is the completely empty system. Let the Hamiltonian be 2 2 3 † ~ ∇r H = d rψb (r) − + U (r) ψb(r) (34) ˆ 2m It follows 2 2 3 3 0 † ~ ∇r 0 † 0 H |ψαi = d r d r ψb (r) − + U (r) φα (r ) ψb(r) ψb (r ) |0i ˆ ˆ 2m 2 2 3 3 0 † ~ ∇r 0 † 0 = d r d r ψb (r) − + U (r) φα (r ) ψb (r ) ψb(r) |0i ˆ ˆ 2m 2 2 3 3 0 † ~ ∇r 0 0 + d r d r ψb (r) − + U (r) φα (r ) δ (r − r ) |0i(35) ˆ ˆ 2m

5 The ﬁrst term disappears since ψb(r) |0i = 0 for the empty state. Performing the integration over r0 gives

2 2 3 † ~ ∇r H |ψαi = d rψb (r) − + U (r) φα (r) |0i ˆ 2m 2 2 3 ~ ∇r † = d r − + U (r) φα (r) ψb (r) |0i (36) ˆ 2m Thus, we need to ﬁnd the eigenvalue of and eigenfunction of

2∇2 −~ + U (r) φ (r) = ε φ (r) (37) 2m α α α to obtain 3 † H |ψαi = εα d rphiα (r) ψb (r) |0i = εα |ψαi . (38) ˆ Thus, for a single particle problem we recover the original formulation of the "ﬁrst quantization". The function φ (r) in Eq.33 is therefore the wave function of the single particle problem. † P ∗ † † 3 † Using ψb (r) = φα (r) baα follows baα = d rφα (r) ψb (r) and our above α ´ wave function is nothing but

† |ψαi = baα |0i (39) Applying the Hamiltonian to the wave function in this basis is obviously giving the same answer.

X † † H |ψαi = εα0 baα0 baα0 baα |0i = εa |ψαi (40) α0 3 Second quantization of interacting bosons

Next we analyze the formulation of particle-particle interactions within the second quantization. We consider a two body interaction Vb that has, by deﬁnition, matrix elements that depend on the states of two particles. Thus the expression for a single particle where

X 0 † Ub = hα |U| α i baαbaα0 (41) α,α0 will be determined by a matrix elements of the kind:

0 0 3 3 0 ∗ ∗ 0 0 0 hαγ |V | α γ i = d rd r φ (r) φ (r ) V (r, r ) φ 0 (r ) φ 0 (r) . (42) ˆ α γ α γ In general there will be a two particle basis |αγi where the interaction is diagonal

Vb |αγi = Vαγ |αγi (43)

6 where Vαγ = hαγ |V | αγi. In this basis we can proceed just like for the interac- P † tion Ub, where the operator was given by α hα |U| αi baαbaα. In case of a two particle interaction we have contributions if there are two particles, one in state α the other in state γ. This the interaction must be 1 X Vb = Vαγ Pbαγ (44) 2 αγ where Pbαβ is the operator which counts the number of pairs of particles in 1 the states |αi and |γi. The prefactor 2 takes into account that each pair is considered only once. If |αi = |γi, the number of pairs is nα (nα − 1), while for |αi 6= |γi it is nαnγ , where the nα are the occupation numbers of those states. It follows

Pbαγ = nbαnbγ − δαγ nbα † † † † = aαaγ aαaγ = aαaγ aγ aα (45) and we ﬁnd

1 X † † 1 X † † Vb = Vαγ a a aγ aα = hαγ |V | αγi a a aγ aα (46) 2 α γ 2 α γ αγ αγ

Transforming this expression into an arbitrary basis |µi = P |αi hα|µi , we insert α the operators in the new basis

† X † baα = hλ|αi baλ λ X baα = hα|λi baλ (47) λ and it follows

1 X † † Vb = hλµ |V | ρνi a a aρaν (49) 2 λ µ λµρν If for example

hr, r0 |V | r00r000i = v (r − r0) δ (r00 − r0) δ (r000 − r) (50) for an interaction that only depends on the distance between the two particles, it follows 1 Vb = d3rd3r0v (r − r0) ψb† (r) ψb† (r0) ψb(r0) ψb(r) . (51) 2 ˆ

7 4 Second quantization of noninteracting fermions 4.1 The fermionic "harmonic oscillator" When we introduced the second quantized representation for bosons we took advantage of the fact that the eigenstates of a free bose system X E = εαnα (52) α could be expressed in terms of the set {nα} of occupation numbers. In case of bosons these occupation numbers were allowed to take all integer values na = 0, 1, ··· , ∞, reminiscent of the quantum number of the harmonic oscillator. The latter then led to the introduction of creation and annihilation operators of the † bosons, where the occupation number operator of a given state was nbα = baαbaa. The Hamiltonian was then written as X Hb = εαnbα (53) α

Obviously, this approach cannot be used to describe fermions where nα = 0 or 1. In case of fermions, the single particle quantum state always includes the spin, for example α = (k, σ). We need to ﬁnd the fermion analog to the harmonic oscillator, i.e. a state that only allows for the two occupations nα = 0 or 1. We want to express the Hamiltonian for a single quantum state as

bh = εnb (54) This is easily done with the help of a (2 × 2) matrix representation (note, these matrices have nothing to do with the spin of the system). If we introduce

1 0 |0i = and |1i = (55) 0 1 for the empty and occupied state, it holds

0 0 n = (56) b 0 1

We can equally introduce lowering and raising operators

ba |0i = 0 and ba |1i = |0i (57) as well as † † ba |1i = 0 and ba |0i = |1i . (58) It follows easily that this is accomplished by

0 1 0 0 a = and a† = . (59) b 0 0 b 1 0

8 † As in case of bosons, ba is the adjoined operator of ba. The action of these operators of a state with arbitrary occupation is then

ba |ni = n |n − 1i = n |1 − ni † ba |ni = (1 − n) |n + 1i = (1 − n) |1 − ni (60) If we now determine a†a it follows 0 1 0 1 0 0 a†a = = (61) b b 1 0 0 0 0 1 and we have, just as for bosons,

† nb = ba ba. (62) However, an important diﬀerence is of course that now holds

† † baba + ba ba = 1 (63) in addition we immediately see

† † ba ba = baba = 0 (64) Fermionic creation and annihilation operators do not commute, they anticom- mute:

a, a† = 1 b b + a†, a† = [a, a] = 0. (65) b b + b b + Note, we could have introduced equally

0 1 0 0 a = − and a† = − . (66) b 0 0 b 1 0

† with the only change that now ba |1i = − |0i and ba |0i = − |1i and all other results remain unchanged.

4.2 Many fermi states To generalize the single fermi result to many fermions we the any body wave function in occupation number representation

|n1, n2, . . . , nα,...i (67)

† We then need to analyze the creation and annihilation operators baα and baafor the individual states, respectively. At ﬁrst glance it is natural to introduce(1 − n) |n + 1i

baa |n1, n2, . . . , nα,...i = nα |n1, n2, . . . , nα − 1,...i † baa |n1, n2, . . . , nα,...i = (1 − nα) |n1, n2, . . . , nα + 1,...i (68)

9 (Note, these equations will turn out to be incorrect!) This implies however that

a , a† = 1 (69) bα bα + while for diﬀerent states α 6= α0 follows

h † i † aα, aα0 = 2aαaα0 (70) b b + b b

h † i 0 a result that follows from baα, baα0 = 0 for α 6= α . If we now want to transform from one basis to another, with X |li = |αi hα|li (71) α Just like in case of bosons the new operators should be linear combinations of the old ones, which yields X bal = hl|αi baα (72) α and the corresponding adjoined equation

† X ∗ † bal = hl|αi baα. (73) α We now require h †i al, al = 1 (74) b b + which leads to X 0 h † i 1 = hl|αi hα |li aα, aα0 (75) b b + α,α0 For a complete set of states hl|αi this is only possible if

h † i aα, aα0 = δα,α0 (76) b b + i.e. for α 6= α0 the anticommutator and not the commutator must vanish. We conclude that Eq.68 cannot be correct. Jordan and Wigner realized that a small change in the definition of these operators can fix the problem. To proceed we need to order the quantum numbers in some arbitrary but fixed way. We then introduce

α−1 X να = nα (77) α0=1 as the number of occupied states that precede the α-th state. We can then introduce 0 1 0 0 a = (−1)να and a† = (−1)να (78) bα 0 0 b 1 0

10 The matrix acts on the occupation of the α-th state. As shown above, a prefactor −1 in the deﬁnition of these operators causes no problem. It then follows

It obviously holds that a , a† = 1 (80) bα bα + We next analyze (assume α0 prior to α)

† να baa0 baa |n1, . . . , nα0 , . . . , nα,...i = (−1) (1 − nα) baa0 |n1, . . . , nα0 ,..., 1 − nα,...i να+να0 = (−1) nα0 (1 − nα) |n1,..., 1 − nα0 ,..., 1 − nα,...i

On the other hand:

† να0 † baabaa0 |n1, . . . , nα0 , . . . , nα,...i = (−1) nα0 baa |n1,..., 1 − nα0 , . . . , nα,...i να+να0 −1 = (−1) nα0 (1 − nα) |n1,..., 1 − nα0 ,..., 1 − nα,...i

It then follows

† † να+να0 baa0 baa + baabaa0 = (−1) (nα0 (1 − nα) − nα0 (1 − nα)) = 0 (81) The same holds of course if we assume α0 to occur after α. Thus, we ﬁnd h † i aα, aα0 = δα,α0 (82) b b + as desired, yielding after a unitary transformation

h † i X 0 0 h † i al, al0 = hl|αi hα |l i aα, aα0 = δl,l0 (83) b b + b b + α,α0 i.e. the anticommutation relation of fermionic operators is independent on the speciﬁc representation. The Hamiltonian of noninteracting fermions is then

X † H = εαbaαbaα (84) α which in case of free particles reads

X † H = εkbak,σbak,σ (85) k,σ

1 where k goes over all momentum values and σ = ± 2 .

11 5 Interacting fermions

To incorporate interaction eﬀects is now rather similar to the case of bosons. We start from the two particle basis |αγi where the interaction is diagonal

Vb |αγi = Vαγ |αγi . (86)

Here Vαγ = hαγ |V | αγi. In this basis we can proceed just like for bosons. In case of a two particle interaction we have contributions if there are two particles, one in state α the other in state γ. This the interaction must be 1 X Vb = Vαγ Pbαγ (87) 2 αγ where Pbαβ is the operator which counts the number of pairs of particles in 1 the states |αi and |γi. The prefactor 2 takes into account that each pair is considered only once. It follows again

Pbαγ = nbαnbγ − δαγ nbα † † † = aαaαaγ aγ − δαγ aαaα † † † † = −aαaγ aαaγ + aαδαγ aγ − δαγ aαaα † † = aαaγ aαaγ (88) and we ﬁnd 1 X † † Vb = Vαγ a a aγ aα (89) 2 α γ αγ just as in case of bosons. Transforming this expression into an arbitrary basis P P |µi = |αi hα|µi where baµ = hµ|αi baα it holds α α

1 X † † Vb = hλµ |V | ρνi a a aρaν (90) 2 λ µ λµρν If for example hr, r0 |V | r00r000i = v (r − r0) δ (r00 − r0) δ (r000 − r) (91) for an interaction that only depends on the distance between the two particles, it follows 1 Vb = d3rd3r0v (r − r0) ψb† (r) ψb† (r0) ψb(r0) ψb(r) . (92) 2 ˆ

5.1 Example 1: free electron gas We want to derive the ground state wave function of the free electron gas. The Hamiltonian of an individual electron is 2∇2 h = −~ (93) 2m

12 which leads to the single particle eigenvalues

2k2 ε = ~ . (94) k 2m The Hamiltonian of the many fermion system is then

X † H = εkbak,σbak,σ. (95) k,σ

The ground state wave function is the state where all single particle states with energy εk < EF (96) are occupied and all states above the Fermi energy are empty. ,

2 3π2 hNi2/3 E = ~ (97) F 2m V was determined earlier. The ground state wave function is then

Y † |Ψ0i = bak,σ |0i (98) k,σ(εk

This state is normalized:

* + Y † hΨ0|Ψ0i = 0 ak,σa 0 b bk,σ k,σ(εk

* + Y † = 0 1 − a ak,σ 0 bk,σb k,σ(εk

k,σ(εk

X † H |Ψ0i = εkbak,σbak,σ |Ψ0i (100) k,σ

It follows immediately that

† Y † bak,σbak,σ baq,σ |0i = θ (EF − εk) |Ψ0i . (101) q,σ(εq

Either k is among the states below the Fermi surface or it isn’t. This yields

H |Ψ0i = E0 |Ψ0i (102)

13 with X E0 = θ (EF − εk) εk (103) k,σ The states that contribute to the ground state energy are all those with an energy 2 2 ~ k below EF . Since εk = 2m is implies that the magnitude of the momentum must be smaller than a given value

2k2 2k2 ~ < ~ F = E (104) 2m 2m F

Here kF is the so called Fermi momentum. All momentum states inside the sphere of radius kF are occupied. Those outside are empty. Our above result for the Fermi energy yields

2 1/3 kF = 3π ρ (105) where ρ = hNi /V is the electron density.

5.2 Example 2: two particles The natural state of two noninteracting particles is

† † |ψα,α0 i = baαbaα0 |0i (106) Applying the Hamiltonian X † H = εαbaαbaα (107) α to this wave function gives

The eigenvalue is Eα,α0 = εα + εα0 . The wave function can also be written as

1 3 3 0 0 † † 0 |ψα,α0 i = √ d rd r φα (r) φα0 (r ) ψb (r) ψb (r ) |0i (109) 2 ˆ Since a labelling of the particles is not necessary within the second quantization, 0 there is no need to symmetrize φα (r) φα0 (r ) in this formulation. To determine the wave function in real space we analyze

0 0 Ψαα0 (r, r ) = hr, r |ψα,α0 i (110)

14 It holds

|r, r0i = ψb† (r) ψb† (r0) |0i (111) such that hr, r0| = h0| ψb(r0) ψb(r) (112) and we can analyze

0 1 3 00 3 000 00 000 hr, r |ψα,α0 i = √ d r d r φα (r ) φα0 (r ) (113) 2 ˆ × h0| ψb(r0) ψb(r) ψb† (r00) ψb† (r000) |0i (114)

Inserting this yields

0 1 0 0 Ψαα0 (r, r ) = √ (φα (r) φα0 (r ) − φα (r ) φα0 (r)) (115) 2 This is of course the correct result for the wave function of two indistinguishable fermions.