Second Quantization

Reminder on Second Quantization Second quantization Second quantization and operators, two-body operator with Morten Hjorth-Jensen1 examples Wick’s theorem National Superconducting Cyclotron Laboratory and Department of Physics and Thouless’ theorem and analysis of the Hartree-Fock equations Astronomy, Michigan State University, East Lansing, MI 48824, USA1 using second quantization Examples on how to use bit representations for Slater Jun 27, 2017 determinants c 2013-2017, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license

Second quantization Second quantization

In Eq. (1) the operator aα† acts on the vacuum state 0 , which | i does not contain any particles. Alternatively, we could deﬁne a We introduce the time-independent operators aα† and aα which create and annihilate, respectively, a particle in the single-particle closed-shell nucleus or atom as our new vacuum, but then we need to introduce the particle-hole formalism, see the discussion to come. state ϕα. We deﬁne the fermion creation operator aα† In Eq. (2) aα† acts on an antisymmetric n-particle state and creates

aα† 0 α , (1) an antisymmetric (n + 1)-particle state, where the one-body state | i ≡ | i ϕ is occupied, under the condition that α = α , α , . . . , α . It α 6 1 2 n and follows that we can express an antisymmetric state as the product a† α1 . . . αn AS αα1 . . . αn AS (2) of the creation operators acting on the vacuum state. α| i ≡ | i

α . . . αn = a† a† ... a† 0 (3) | 1 iAS α1 α2 αn | i

Second quantization Second quantization

Using the Pauli principle It is easy to derive the commutation and anticommutation rules for a† α1 . . . αi . . . αi . . . αn AS = 0 (6) the fermionic creation operators α. Using the antisymmetry of the | i states (3) it follows that α . . . α . . . α . . . α = α . . . α . . . α . . . α (4) aα† aα† = 0. (7) | 1 i k niAS −| 1 k i niAS i i If we combine Eqs. (5) and (7), we obtain the well-known we obtain anti-commutation rule a† a† = a† a† (5) αi αk − αk αi a† a† + a† a† a† , a† = 0 (8) α β β α ≡ { α β} Second quantization Second quantization

What is the physical interpretation of the operator aα and what is the eﬀect of a on a given state α α . . . α ? Consider the α | 1 2 niAS following matrix element The hermitian conjugate of aα† is α α . . . α a α0 α0 . . . α0 (11) h 1 2 n| α| 1 2 mi aα = (aα† )† (9) where both sides are antisymmetric. We distinguish between two If we take the hermitian conjugate of Eq. (8), we arrive at cases. The ﬁrst (1) is when α α . Using the Pauli principle of ∈ { i } Eq. (6) it follows aα, aβ = 0 (10) { } α1α2 . . . αn aα = 0 (12) h | The second (2) case is when α / α . It follows that an hermitian ∈ { i } conjugation α α . . . α a = αα α . . . α (13) h 1 2 n| α h 1 2 n|

Second quantization Second quantization

For the last case, the minus and plus signs apply when the sequence α, α1, α2, . . . , αn and α10 , α20 , . . . , αn0 +1 are related to each other via even and odd permutations. If we assume that α / α we obtain Eq. (13) holds for case (1) since the lefthand side is zero due to the ∈ { i } Pauli principle. We write Eq. (11) as α α . . . α a α0 α0 . . . α0 = 0 (15) h 1 2 n| α| 1 2 n+1i α1α2 . . . αn aα α10 α20 . . . αm0 = αα1α2 . . . αn α10 α20 . . . αm0 (14) h | | i h | i when α α . If α / α , we obtain ∈ { i0} ∈ { i0} Here we must have m = n + 1 if Eq. (14) has to be trivially aα α10 α20 . . . αn0 +1 =α = 0 (16) diﬀerent from zero. | i6

and in particular | {z } a 0 = 0 (17) α| i

Second quantization Second quantization

If ααi = αi0 , performing the right permutations, the sequence { } { } The action of the operator aα from the left on a state vector is to α, α , α , . . . , αn is identical with the sequence α0 , α0 , . . . , α0 . 1 2 1 2 n+1 α This results in remove one particle in the state . If the state vector does not contain the single-particle state α, the outcome of the operation is a α1α2 . . . αn aα αα1α2 . . . αn = 1 (18) zero. The operator α is normally called for a destruction or h | | i annihilation operator. and thus The next step is to establish the commutator algebra of aα† and aβ. a αα α . . . α = α α . . . α (19) α| 1 2 ni | 1 2 ni Second quantization Second quantization

The action of the anti-commutator aα† ,a on a given n-particle If it is present we arrive at { α} state is k aα† aα α1α2 . . . αk ααk+1 . . . αn 1 = aα† aα( 1) αα1α2 . . . αn 1 | − i − | − i a† aα α α . . . αn = 0 k α 1 2 = ( 1) αα1α2 . . . αn 1 = α1α2 . . . αk ααk+1 . . . αn 1 | i − | − i | − i =α 6 aαaα† α1α2 . . . αk ααk+1 . . . αn 1 = 0 (21) | − i a a† α α . . . α = a αα α . . . α = α α . . . α (20) α α | 1 2{z n}i α | 1 2 ni | 1 2 ni =α =α =α From Eqs. (20) and (21) we arrive at 6 6 6 | {z } | {z } | {z } a† , a = a† a + a a† = 1 (22) if the single-particle state α is not contained in the state. { α α} α α α α

Second quantization Second quantization

The ﬁrst case results in

aα† aβ αβα1α2 . . . αn 2 = 0 | − i aβaα† αβα1α2 . . . αn 2 = 0 (23) | − i The action of aα† , a , with α = β on a given state yields three β 6 while the second case gives possibilities. Then ﬁrsto case is a state vector which contains both α and β, then either α or β and ﬁnally none of them. aα† aβ β α1α2 . . . αn 1 = α α1α2 . . . αn 1 | − i | − i =α =α 6 6 aβaα† β α| 1α2 .{z . . αn }1 = aβ |α βα1{zα2 . . . α} n 1 | − i | − i =α =α 6 6 = α α1α2 . . . αn 1 (24) | {z } −| | {z − i} =α 6 | {z }

Second quantization Second quantization

We can summarize our ﬁndings in Eqs. (22) and (26) as

Finally if the state vector does not contain α and β a† , aβ = δαβ (27) { α }

a† a α α . . . α = 0 with δαβ is the Kronecker δ-symbol. α β| 1 2 ni =α,β The properties of the creation and annihilation operators can be 6 summarized as (for fermions) a a† α α . . . α = a α α α . . . α = 0 (25) β α| | 1 2{z n}i β| 1 2 ni =α,β =α,β a† 0 α , 6 6 α| i ≡ | i | {z } | {z } For all three cases we have and aα† α1 . . . αn AS αα1 . . . αn AS. a† , a = a† a + a a† = 0, α = β (26) | i ≡ | i { α β} α β β α 6 from which follows

α . . . αn = a† a† ... a† 0 . | 1 iAS α1 α2 αn | i Second quantization Second quantization

The hermitian conjugate has the folowing properties A very useful operator is the so-called number-operator. Most a = (a ) . α α† † physics cases we will study in this text conserve the total number of Finally we found particles. The number operator is therefore a useful quantity which allows us to test that our many-body formalism conserves the (d, p) (p, d) aα α10 α20 . . . αn0 +1 =α = 0, in particular aα 0 = 0, number of particles. In for example or reactions it is | i6 | i important to be able to describe quantum mechanical states where particles get added or removed. A creation operator aα† adds one and | {z } particle to the single-particle state α of a give many-body state aα αα1α2 . . . αn = α1α2 . . . αn , | i | i vector, while an annihilation operator aα removes a particle from a and the corresponding commutator algebra single-particle state α.

a† , a† = a , a = 0 a† , a = δ . { α β} { α β} { α β} αβ

Second quantization Second quantization

Let us consider an operator proportional with aα† aβ and α = β. It acts on an n-particle state resulting in The operator ˆ N = aα† aα (30) 0 α / α α ∈ { i } X a† a α α . . . α = (28) is called the number operator since it counts the number of α α| 1 2 ni   particles in a given state vector when it acts on the diﬀerent  α1α2 . . . αn α αi | i ∈ { } single-particle states. It acts on one single-particle state at the time  Summing over all possible one-particle states we arrive at and falls therefore under category one-body operators. Next we ˆ look at another important one-body operator, namely H0 and study its operator form in the occupation number representation. a† a α α . . . α = n α α . . . α (29) α α | 1 2 ni | 1 2 ni α ! X

Second quantization Second quantization Deﬁning ˆ ˆ h0(xi )ψαi (xi ) = ψα0 (xi ) αk0 h0 αk (32) We want to obtain an expression for a one-body operator which k h | | i αk0 conserves the number of particles. Here we study the one-body X ˆ operator for the kinetic energy plus an eventual external one-body we can easily evaluate the action of H0 on each product of potential. The action of this operator on a particular n-body state one-particle functions in Slater determinant. From Eq. (32) we with its pertinent expectation value has already been studied in obtain the following result without permuting any particle pair coordinate space. In coordinate space the operator reads ˆ h0(xi ) ψα1 (x1)ψα2 (x2) . . . ψαn (xn) Hˆ = hˆ (xi ) (31) 0 0 i ! i X X ˆ = α0 h0 α1 ψα (x1)ψα (x2) . . . ψαn (xn) h 1| | i 10 2 and the anti-symmetric n-particle Slater determinant is deﬁned as α X10 ˆ 1 + α0 h0 α2 ψα (x1)ψ (x2) . . . ψαn (xn) p ˆ 2 1 α20 Φ(x1, x2,..., xn, α1, α2, . . . , αn) = ( 1) Pψα1 (x1)ψα2 (x2) . . . ψαn (xn). h | | i √ − α0 n! p X2 X + ...

+ α0 hˆ α ψ (x )ψ (x ) . . . ψ (x ) (33) h n| 0| ni α1 1 α2 2 αn0 n Xαn0 Second quantization Second quantization

If we interchange particles 1 and 2 we obtain We can continue by computing all possible permutations. We rewrite also our Slater determinant in its second quantized form and skip the dependence on the quantum numbers x . Summing up hˆ (x ) ψ (x )ψ (x ) . . . ψ (x ) i 0 i α1 1 α1 2 αn n all contributions and taking care of all phases ( 1)p we arrive at i ! − X ˆ ˆ ˆ = α20 h0 α2 ψα1 (x2)ψα0 (x1) . . . ψαn (xn) H α , α , . . . , α = α0 h α α0 α . . . α h | | i 2 0| 1 2 ni h 1| 0| 1i| 1 2 ni α0 α X2 X10 ˆ ˆ + α10 h0 α1 ψα0 (x2)ψα2 (x1) . . . ψαn (xn) + α0 h α α α0 . . . α h | | i 1 h 2| 0| 2i| 1 2 ni α0 α X1 X20 + ... + ... ˆ + αn0 h0 αn ψα1 (x2)ψα1 (x2) . . . ψα (xn) (34) + α hˆ α α α . . . α h | | i n0 n0 0 n 1 2 n0 (35) α h | | i| i Xn0 Xαn0

Second quantization Second quantization In Eq. (35) we have expressed the action of the one-body operator of Eq. (31) on the n-body state in its second quantized form. This equation can be further manipulated if we use the properties of the In the number occupation representation or second quantization we creation and annihilation operator on each primed quantum get the following expression for a one-body operator which number, that is conserves the number of particles ˆ ˆ H0 = α h0 β aα† aβ (38) α1α2 . . . αk0 . . . αn = aα† aαk α1α2 . . . αk . . . αn (36) | i k0 | i h | | i Xαβ Inserting this in the right-hand side of Eq. (35) results in ˆ Obviously, H0 can be replaced by any other one-body operator

ˆ ˆ † which preserved the number of particles. The stucture of the H0 α1α2 . . . αn = α10 h0 α1 aα aα1 α1α2 . . . αn | i h | | i 10 | i operator is therefore not limited to say the kinetic or single-particle α0 X1 energy only. ˆ † + α20 h0 α2 aα aα2 α1α2 . . . αn Hˆ β h | | i 20 | i The opearator 0 takes a particle from the single-particle state α X20 to the single-particle state α with a probability for the transition ˆ + ... given by the expectation value α h0 β . ˆ h | | i + αn0 h0 αn a† aαn α1α2 . . . αn h | | i αn0 | i Xαn0 = α hˆ β a† a α α . . . α (37) h | 0| i α β| 1 2 ni Xα,β

Second quantization Second quantization

Using the commutation relations for the creation and annihilation operators we have

It is instructive to verify Eq. (38) by computing the expectation aα a† aβa† = (δαα a† aα )(δβα a† aβ), (40) ˆ 1 α α2 1 α 1 2 α2 value of H0 between two single-particle states − − which results in α hˆ α = α hˆ β 0 aα a† aβa† 0 (39) h 1| 0| 2i h | 0| ih | 1 α α2 | i αβ 0 aα a† aβa† 0 = δαα δβα (41) X h | 1 α α2 | i 1 2 and α hˆ α = α hˆ β δ δ = α hˆ α (42) h 1| 0| 2i h | 0| i αα1 βα2 h 1| 0| 2i Xαβ Two-body operators in second quantization Operators in second quantization

We can now let Hˆ act on all terms in the linear combination for Let us now derive the expression for our two-body interaction part, I α α . . . α . Without any permutations we have which also conserves the number of particles. We can proceed in | 1 2 ni exactly the same way as for the one-body operator. In the coordinate representation our two-body interaction part takes the V (x , x ) ψ (x )ψ (x ) . . . ψ (x )  i j  α1 1 α2 2 αn n following expression i

Operators in second quantization Operators in second quantization

For the other terms on the rhs we obtain similar expressions and summing over all terms we obtain We introduce second quantization via the relation

HI α1α2 . . . αn = α10 α20 vˆ α1α2 α10 α20 . . . αn aα† aα† aαl aαk α1α2 . . . αk . . . αl . . . αn | i h | | i| i k0 l0 | i α0 ,α0 X1 2 k 1 l 2 + ... = ( 1) − ( 1) − aα† aα† aαl aαk αk αl α1α2 . . . αn − − k0 l0 | i =αk ,αl + α10 αn0 vˆ α1αn α10 α2 . . . αn0 6 h | | i| i k 1 l 2 α0 ,α = ( 1) − ( 1) − α0 α0 α α . . . α X1 n0 − − | k l 1 2 ni | {z } + ... =α ,α 6 k0 l0 + α0 α0 vˆ α α α α0 . . . α0 h 2 n| | 2 ni| 1 2 ni = α1α2 . . . αk0 . . . αl0 . . . α|n {z } (47) α ,α | i X20 n0 + ... (46)

Operators in second quantization Operators in second quantization

Inserting this in (46) gives Here we let 0 indicate that the sums running over α and β run over all single-particle states, while the summations γ and δ run † † P HI α1α2 . . . αn = α10 α20 vˆ α1α2 aα aα aα2 aα1 α1α2 . . . αn over all pairs of single-particle states. We wish to remove this | i h | | i 10 20 | i α ,α X10 20 restriction and since + ... αβ vˆ γδ = βα vˆ δγ † † (49) = α10 αn0 vˆ α1αn aα aα aαn aα1 α1α2 . . . αn h | | i h | | i h | | i 10 n0 | i α ,α X10 n0 we get + ... = α α vˆ α α a† a† a a α α . . . α 20 n0 2 n α α αn α2 1 2 n αβ vˆ γδ aα† aβ† aδaγ = βα vˆ δγ aα† aβ† aδaγ (50) h | | i 20 n0 | i h | | i h | | i α ,α αβ αβ X20 n0 X X + ... = βα vˆ δγ a† a† a a (51) h | | i β α γ δ 0 αβ = αβ vˆ γδ a† a† a a α α . . . α (48) X h | | i α β δ γ| 1 2 ni α,β,γ,δX where we have used the anti-commutation rules. Operators in second quantization Operators in second quantization

Changing the summation indices α and β in (51) we obtain With this expression we can now verify that the second αβ vˆ γδ aα† aβ† aδaγ = αβ vˆ δγ aα† aβ† aγaδ (52) ˆ h | | i h | | i quantization form of HI in Eq. (53) results in the same matrix αβ αβ X X between two anti-symmetrized two-particle states as its From this it follows that the restriction on the summation over γ corresponding coordinate space representation. We have and δ can be removed if we multiply with a factor 1 , resulting in 2 ˆ 1 α1α2 HI β1β2 = αβ vˆ γδ 0 aα aα aα† a† aδaγa† a† 0 . h | | i 2 h | | ih | 2 1 β β1 β2 | i 1 αβγδ HˆI = αβ vˆ γδ a† a† aδaγ (53) X 2 h | | i α β (54) αβγδX where we sum freely over all single-particle states α, β, γ og δ.

Operators in second quantization Operators in second quantization

Using the commutation relations we get The vacuum expectation value of this product of operators becomes aα aα a† a† a aγa† a† 2 1 α β δ β1 β2 † † † = aα aα aα† a† (aδδγβ a† aδa† aγa† ) 0 aα2 aα1 aα† aβaδaγaβ aβ 0 2 1 β 1 β2 − β1 β2 h | 1 2 | i = (δ δ δ δ ) a a a a† = aα aα aα† a† (δγβ δδβ δγβ a† aδ aδa† δγβ + aδa† a† aγ) γβ1 δβ2 δβ1 γβ2 0 α2 α1 α† β 0 2 1 β 1 2 − 1 β2 − β1 2 β1 β2 − h | | i = (δγβ1 δδβ2 δδβ1 δγβ2 )(δαα1 δβα2 δβα1 δαα2 ) (56) = aα2 aα1 aα† a† (δγβ1 δδβ2 δγβ1 a† aδ − − β − β2 δδβ δγβ + δγβ a† aδ + aδa† a† aγ) (55) − 1 2 2 β1 β1 β2

Operators in second quantization Operators in second quantization

The two-body operator can also be expressed in terms of the Insertion of Eq. (56) in Eq. (54) results in anti-symmetrized matrix elements we discussed previously as 1 ˆ 1 HÎ = αβ vˆ γδ a† a† aδaγ α1α2 HI β1β2 = α1α2 vˆ β1β2 α1α2 vˆ β2β1 2 h | | i α β h | | i 2 h | | i − h | | i αβγδ α α vˆ β β + α α vˆ β β X −h 2 1| | 1 2i h 2 1| | 2 1i 1 = [ αβ vˆ γδ αβ vˆ δγ ] aα† aβ† aδaγ = α1α2 vˆ β1β2 α1α2 vˆ β2β1 4 h | | i − h | | i h | | i − h | | i αβγδ = α α vˆ β β . (57) X h 1 2| | 1 2iAS 1 = αβ vˆ γδ a† a† aδaγ (58) 4 h | | iAS α β αβγδX Operators in second quantization Particle-hole formalism Second quantization is a useful and elegant formalism for constructing many-body states and quantum mechanical operators. One can express and translate many physical processes into simple 1 1 pictures such as Feynman diagrams. Expecation values of The factors in front of the operator, either 4 or 2 tells whether we use antisymmetrized matrix elements or not. many-body states are also easily calculated. However, although the We can now express the Hamiltonian operator for a many-fermion equations are seemingly easy to set up, from a practical point of system in the occupation basis representation as view, that is the solution of Schroedinger’s equation, there is no particular gain. The many-body equation is equally hard to solve, 1 irrespective of representation. The cliche that there is no free lunch H = α tˆ+u ˆ β a† aβ + αβ vˆ γδ a† a† aδaγ. (59) h | ext| i α 4 h | | i α β brings us down to earth again. Note however that a transformation α,β αβγδ X X to a particular basis, for cases where the interaction obeys specific This is the form we will use in the rest of these lectures, assuming symmetries, can ease the solution of Schroedinger’s equation. that we work with anti-symmetrized two-body matrix elements. But there is at least one important case where second quantization comes to our rescue. It is namely easy to introduce another reference state than the pure vacuum 0 , where all single-particle | i states are active. With many particles present it is often useful to introduce another reference state than the vacuum state 0 . We | i will label this state c (c for core) and as we will see it can reduce | i considerably the complexity and thereby the dimensionality of the many-body problem. It allows us to sum up to infinite order specific many-body correlations. The particle-hole representation is one of these handy representations. Particle-hole formalism Particle-hole formalism

If we use Eq. (60) as our new reference state, we can simplify In the original particle representation these states are products of considerably the representation of this state the creation operators aα† acting on the true vacuum 0 . Following i | i c α α . . . α α = a a ... a a Eq. (3) we have 1 2 n 1 n α† 1 α† 2 α† n 1 α† n 0 (63) | i ≡ | − i − | i α α . . . α α = a a ... a a n + n 1 2 n 1 n α† 1 α† 2 α† n 1 α† n 0 (60) The new reference states for the 1 and 1 states can then be | − i − | i − α α . . . α α α = a a ... a a a written as 1 2 n 1 n n+1 α† 1 α† 2 α† n 1 α† n α† n+1 0 (61) | − i − | i n n α α . . . α = a† a† ... a† α α . . . α α α = ( ) a c ( ) α 1 2 n 1 α1 α2 αn 1 0 (62) 1 2 n 1 n n+1 1 α† n+1 1 n+1 c (64) | − i − | i | − i − | i ≡ − | i n 1 n 1 α1α2 . . . αn 1 = ( 1) − aαn c ( 1) − αn 1(65)c | − i − | i ≡ − | − i

Particle-hole formalism Particle-hole formalism The first state has one additional particle with respect to the new The physical interpretation of these new operators is that of vacuum state c and is normally referred to as a one-particle state so-called quasiparticle states. This means that a state defined by | i or one particle added to the many-body reference state. The the addition of one extra particle to a reference state c may not | i second state has one particle less than the reference vacuum state necesseraly be interpreted as one particle coupled to a core. We c and is referred to as a one-hole state. When dealing with a new define now new creation operators that act on a state α creating a | i reference state it is often convenient to introduce new creation and new quasiparticle state annihilation operators since we have from Eq. (65) aα† c = α , α > F | i | i aα c = 0 (66) bα† c = (70) | i 6 | i ( 1 aα c = α− , α F since α is contained in c , while for the true vacuum we have | i | i ≤ | i a 0 = 0 for all α. where F is the Fermi level representing the last occupied α| i The new reference state leads to the definition of new creation and single-particle orbit of the new reference state c . | i annihilation operators which satisfy the following relations The annihilation is the hermitian conjugate of the creation operator

b c = 0 (67) bα = (b† )†, α| i α b† , b† = b , b = 0 { α β} { α β} resulting in b† , b = δ (68) { α β} αβ aα† α > F aα α > F We assume also that the new reference state is properly normalized bα† = bα = (71) ( ( aα α F aα† α F c c = 1 (69) ≤ ≤ h | i Particle-hole formalism Particle-hole formalism With the new creation and annihilation operator we can now construct many-body quasiparticle states, with ˆ one-particle-one-hole states, two-particle-two-hole states etc in the We express the one-body operator H0 in terms of the quasi-particle same fashion as we previously constructed many-particle states. We creation and annihilation operators, resulting in can write a general particle-hole state as ˆ ˆ ˆ ˆ H0 = α h0 β bα† bβ + α h0 β bα† bβ† + β h0 α bβbα 1 1 1 h | | i h | | i h | | i β1β2 . . . βnp γ− γ− . . . γn− b† b† ... b† bγ† bγ† ... bγ† c αβ>F | 1 2 h i ≡ β1 β2 βnp 1 2 nh | i X αX > F h i >F F β F ≤ ≤ (72) | {z } | {z } + α hˆ α β hˆ α b† bβ (77) We can now rewrite our one-body and two-body operators in terms h | 0| i − h | 0| i α α F αβ F of the new creation and annihilation operators. The number X≤ X≤ operator becomes The ﬁrst term gives contribution only for particle states, while the last one contributes only for holestates. The second term can Nˆ = a† a = b† b + n b† b (73) α α α α c − α α create or destroy a set of quasi-particles and the third term is the α α>F α F X X X≤ contribution from the vacuum state c . | i where n is the number of particle in the new vacuum state c . c | i The action of Nˆ on a many-body state results in

1 1 1 1 1 1 N β β . . . β γ− γ− . . . γ− = (n +n n ) β β . . . β γ− γ− . . . γ− | 1 2 np 1 2 nh i p c − h | 1 2 np 1 2 nh i (74) Here n = n + n n is the total number of particles in the p c − h quasi-particle state of Eq. (72). Note that Nˆ counts the total Particle-holenumber of particles formalism present Particle-hole formalism

Nqp = bα† bα, (75) α Before we continue with the expressionsX for the two-body operator, The two-particle operator in the particle-hole formalism is more wegives introduce us the number a nomenclature of quasi-particles we will use as forcan the be restseen of by this computing text. It complicated since we have to translate four indices αβγδ to the is inspired by the notation1 1 used in1 quantum chemistry. We reserve1 1 1 possible combinations of particle and hole states. When performing N = β β . . . β γ− γ− . . . γ− = (n +n ) β β . . . β γ− γ− . . . γ− theqp labels| 1 i,2j, k,...np for1 hole2 statesnh i and ap, b, ch,...| 1 for2 statesnp above1 2 nh i the commutator algebra we can regroup the operator in ﬁve (76) F , viz. particle states. This means also that we will skip the diﬀerent terms where nqp = np + nh is the total number of quasi-particles. ˆ constraint F or > F in the summation symbols. Our operator H0 ≤ ˆ ˆ(a) ˆ(b) ˆ(c) ˆ(d) ˆ(e) reads now HI = HI + HI + HI + HI + HI (79)

Particle-hole formalism Particle-hole formalism The ﬁrst line stands for the creation of a two-particle-two-hole state, while the second line represents the creation to two ˆ(b) one-particle-one-hole pairs while the last term represents a The next term HI reads contribution to the particle single-particle energy from the hole (b) 1 states, that is an interaction between the particle states and the Hˆ = ab Vˆ ci b†b†b†bc + ai Vˆ cb b†bi bbbc (81) I 4 h | | i a b i h | | i a hole states within the new vacuum state. The fourth term reads Xabci ˆ(d) 1 ˆ ˆ This term conserves the number of quasiparticles but creates or H = ai V jk ba†b† b†bi + ji V ak b† bj bi ba + I 4 h | | i k j h | | i k ˆ(c) aijk removes a three-particle-one-hole state. For HI we have X 1 ai Vˆ ji b†b† + ji Vˆ ai ji Vˆ ia bj ba (83). (c) 1 4 h | | i a j h | | i − h | | i Hˆ = ab Vˆ ij b†b†b†b† + ij Vˆ ab babbbj bi + aij I 4 h | | i a b j i h | | i X Xabij 1 1 The terms in the ﬁrst line stand for the creation of a particle-hole ai Vˆ bj b†b†bbbi + ai Vˆ bi b†bb. (82) 2 h | | i a j 2 h | | i a state interacting with hole states, we will label this as a Xabij Xabi two-hole-one-particle contribution. The remaining terms are a particle-hole state interacting with the holes in the vacuum state. Finally we have

(e) 1 1 1 Hˆ = kl Vˆ ij b†b†bl bk + ij Vˆ kj b† bi + ij Vˆ ij I 4 h | | i i j 2 h | | i k 2 h | | i Xijkl Xijk Xij (84) The first terms represents the interaction between two holes while the second stands for the interaction between a hole and the remaining holes in the vacuum state. It represents a contribution to single-hole energy to first order. The last term collects all contributions to the energy of the ground state of a closed-shell system arising from hole-hole correlations. Summarizing and defining a normal-ordered Hamiltonian Summarizing and defining a normal-ordered Hamiltonian

A 1 P ˆ ΦAS (α1, . . . , αA; x1,... xA) = ( 1) P ψα (xi ), √A − i ap† , aq = δpq, p, q αF Pˆ i=1 ≤ X Y n o ap, aq† = δpq, p, q > αF which is equivalent with α1 . . . αA = aα† ... aα† 0 . We have also | i 1 A | i n o with i, j,... α , a, b, . . . > α , p, q,... any ≤ F F − ap† 0 = p , ap q = δpq 0 | i | i | i | i a a Φ = Φ , a† Φ = Φ i | 0i | i i a| 0i | i δpq = ap, aq† , and n o and a† Φ = 0 a Φ = 0 i | 0i a| 0i 0 = a† , a = a , a = a† , a† p q { p q} p q Φ =n α . .o . α , α , . . . ,n α αo | 0i | 1 Ai 1 A ≤ F

Summarizing and deﬁning a normal-ordered Hamiltonian Summarizing and deﬁning a normal-ordered Hamiltonian

One- and two-body operators The one-body operator is defined as We can also define a three-body operator Fˆ = p fˆ q a† a h | | i p q 1 pq Vˆ = pqr vˆ stu a† a† a†a a a X 3 h | 3| iAS p q r u t s 36 pqrstu while the two-body opreator is defined as X with the antisymmetrized matrix element 1 Vˆ = pq vˆ rs AS a† a† as ar 4 h | | i p q pqr vˆ stu = pqr vˆ stu + pqr vˆ tus + pqr vˆ ust pqr vˆ sut pqr vˆ tsu pqr vˆ uts . pqrs h | 3| iAS h | 3| i h | 3| i h | 3| i − h | 3| i − h | 3| i − h | 3| i X (85) where we have defined the antisymmetric matrix elements

pq vˆ rs = pq vˆ rs pq vˆ sr . h | | iAS h | | i − h | | i

Hartree-Fock in second quantization and stability of HF Hartree-Fock in second quantization and Thouless’ theorem solution We wish now to derive the Hartree-Fock equations using our second-quantized formalism and study the stability of the Let us give a simple proof of Thouless’ theorem. The theorem equations. Our ansatz for the ground state of the system is states that we can make a linear combination av particle-hole approximated as (this is our representation of a Slater determinant excitations with respect to a given reference state c . With this | i in second quantization) linear combination, we can make a new Slater determinant c0 | i which is not orthogonal to c , that is | i Φ = c = a†a† ... a† 0 . | 0i | i i j l | i c c0 = 0. h | i 6 We wish to determine uˆHF so that E HF = c Hˆ c becomes a local 0 h | | i minimum. To show this we need some intermediate steps. The exponential product of two operators exp Aˆ exp Bˆ is equal to exp (Aˆ + Bˆ) In our analysis here we will need Thouless’ theorem, which states × only if the two operators commute, that is that an arbitrary Slater determinant c0 which is not orthogonal to n | i ˆ ˆ a determinant c = aα† 0 , can be written as [A, B] = 0. | i i | i Yi=1

c0 = exp C a†a c | i  ai a i  | i a>F i F X X≤    Thouless’ theorem Thouless’ theorem If the operators do not commute, we need to resort to the We note that Baker-Campbell-Hauersdorf. This relation states that C a†a C a† a c = 0, ai a i bi b i | i Cˆ = Aˆ Bˆ, i a>F b>F exp exp exp Y X X with and all higher-order powers of these combinations of creation and n annihilation operators disappear due to the fact that (ai ) c = 0 1 1 1 | i Cˆ = Aˆ + Bˆ + [Aˆ, Bˆ] + [[Aˆ, Bˆ], Bˆ] [[Aˆ, Bˆ], Aˆ] + ... when n > 1. This allows us to rewrite the expression for c0 as 2 12 − 12 | i

From these relations, we note that in our expression for c0 we | i c0 = 1 + Cai aa†ai c , have commutators of the type | i ( ) | i Yi aX>F

[aa†ai , ab† aj ], which we can rewrite as

and it is easy to convince oneself that these commutators, or higher c0 = 1 + Cai a†ai a† a† ... a† 0 . powers thereof, are all zero. This means that we can write out our | i a | i1 i2 in | i i ( a>F ) new representation of a Slater determinant as Y X The last equation can be written as 2

New operators Showing that c˜ = c0 1 + Cai a†ai a† ... 0 = a† + Cai a† 0 . × | i 2 a| 2 i i2 | i i a | i If we define a new creation operator We need to showa>F that c˜!= c . We needi also toa> assumeF that! the X | i | 0i Y X new state is not orthogonal to c , that is c c˜ = 0. From this(87) it | i h | i 6 bi† = ai† + Cai aa†, (88) follows that aX>F in in we have c c˜ = 0 a ... a f a† f a† 0 , h | i h | in i1  i1p p  int t  | i pX=i1 tX=i1     c0 = bi† 0 = ai† + Cai aa† 0 , | i | i | i which is nothing but the determinant det(fip) which we can, using i i a>F ! Y Y X the intermediate normalization condition, normalize to one, that is meaning that the new representation of the Slater determinant in second quantization, c , looks like our previous ones. However, det(fip) = 1, | 0i this representation is not general enough since we have a restriction meaning that f has an inverse defined as (since we are dealing with on the sum over single-particle states in Eq. (88). The orthogonal, and in our case unitary as well, transformations) single-particle states have all to be above the Fermi level. The question then is whether we can construct a general representation 1 f f − = δ , of a Slater determinant with a creation operator ik kj ij Xk ˜ bi† = fipap† , and p 1 X fij− fjk = δik . j where fip is a matrix element of a unitary matrix which transforms X our creation and annihilation operators a† and a to b˜† and b˜. WrappingThese new it operators up define a new representation of a Slater Thouless’ theorem determinant as Using these relations we canc˜ then= defineb˜† 0 . the linear combination of This means that we can actually write an ansatz for the ground | i i | i creation (and annihilation as well)i operators as state of the system as a linear combination of terms which contain Y the ansatz itself c with an admixture from an infinity of ∞ ∞ | i 1 1 1 one-particle-one-hole states. The latter has important consequences f − b˜† = f − fipa† = a† + f − fipa† . ki i ki p k ki p when we wish to interpret the Hartree-Fock equations and their i i p=i1 i p=in+1 X X X X X stability. We can rewrite the new representation as Defining 1 c0 = c + δc , ckp = fki− fip, | i | i | i i F X≤ where δc can now be interpreted as a small variation. If we | i we can redefine approximate this term with contributions from one-particle-one-hole (1p-1h) states only, we arrive at ∞ ∞ 1 ak† + fki− fipap† = ak† + ckpap† = bk† , i p=in+1 p=in+1 X X X c0 = 1 + δCai aa†ai c . | i ! | i our starting point. We have shown that our general representation Xai of a Slater determinant In our derivation of the Hartree-Fock equations we have shown that ˜ c˜ = bi† 0 = c0 = bi† 0 , δc Hˆ c = 0, | i | i | i | i h | | i Yi Yi which means that we have to satisfy with ∞ † † ˆ bk = ak + ckpap† . c δCai aa†ai H c = 0. p=i h | | i Xn+1 Xai n o With this as a background, we are now ready to study the stability of the Hartree-Fock equations. Hartree-Fock in second quantization and stability of HF Hartree-Fock in second quantization and stability of HF solution solution The variational condition for deriving the Hartree-Fock equations ˆ guarantees only that the expectation value c H c has an extreme The norm of c is given by (using the intermediate normalization h | | i | 0i value, not necessarily a minimum. To figure out whether the condition c0 c = 1) extreme value we have found is a minimum, we can use second h | i 2 3 quantization to analyze our results and find a criterion for the c0 c0 = 1 + δC + O(δC ). h | i | ai | ai above expectation value to a local minimum. We will use Thouless’ a>F i F X X≤ theorem and show that The expectation value for the energy is now given by (using the c Hˆ c h 0| | 0i c Hˆ c = E , Hartree-Fock condition) c c ≥ h | | i 0 h 0| 0i c0 Hˆ c0 = c Hˆ c + δC ∗ δC c a†a Haˆ † a c + h | | i h | | i ai bj h | i a b j | i with ab>F ij F ≤ c0 = c + δc . X X | i | i | i 1 1 Using Thouless’ theorem we can write out c0 as δCai δCbj c Haˆ †ai a† aj c + δC ∗ δC ∗ c a†aba†aaHˆ c +... | i 2! h | a b | i 2! ai bj h | j i | i ab>F ij F ab>F ij F X X≤ X X≤

c0 = exp δC a†a c (89) | i  ai a i  | i a>F i F X X≤    1 = 1 + δCai a†ai + δCai δCbj a†ai a† aj + ...  a 2! a b  a>F i F ab>F ij F  X X≤ X X≤  Hartree-Fock in second quantization and stability of HF(90) Hartree-Fock in second quantization and stability of HF   solutionwhere the amplitudes δC are small. solution

We have already calculated the second term on the right-hand side 1 1 † c a†ab a†aa VˆN c = c VˆN a†ai a† aj c of the previous equation 2!h | { j }{ i } | i 2!h | { a }{ b } | i c a†a Hˆ a† a c = δC ∗ δC p hˆ q c a†a a† a a† a c which is nothing but h | { i a} { b j } | i ai bj h | 0| ih | { i a}{ p q}{ b j } | i pq ijab X X 1 ˆ 1 (91) c VN aa†ai a† aj c ∗ = ( ij vˆ ab )∗δCai∗ δCbj∗ 2!h | { }{ b } | i 2 h | | i 1 Xijab + δC ∗ δC pq vˆ rs c a†a a† a† a a a† a c , ai bj h | | ih | { i a}{ p q s r }{ b j } | i 4 pqrs or X Xijab 1 ( ab vˆ ij )δC ∗ δC ∗ (92) 2 h | | i ai bj Xijab resulting in where we have used the relation 2 2 E δC + δC (ε ε ) aj vˆ bi δC ∗ δC . 0 ai ai a i ai bj a Aˆ b = ( b Aˆ† a )∗ | | | | − − h | | i h | | i h | | i Xai Xai Xijab due to the hermiticity of Hˆ and Vˆ.

Hartree-Fock in second quantization and stability of HF Hartree-Fock in second quantization and stability of HF solution solution With these deﬁnitions we write out the energy as

2 2 HF HF c0 H c0 = 1 + δC c H c + δC (ε ε ) + A δC ∗ δC + h | | i | ai | h | | i | ai | a − i ai,bj ai bj ai ! ai ijab We deﬁne two matrix elements X X X (93)

Aai,bj = aj vbiˆ 1 1 3 −h | i Bai∗ ,bj δCai δCbj + Bai,bj δCai∗ δCbj∗ + O(δCai ), 2 2 ijab ijab and X X B = ab vˆ ij (94) ai,bj h | | i both being anti-symmetrized. which can be rewritten as

2 3 c0 H c0 = 1 + δCai c H c + ∆E + O(δCai ), h | | i | | ! h | | i Xai and skipping higher-order terms we arrived

The condition 1 ˆ We have defined ∆E = χ M χ 0 1 2h | | i ≥ ∆E = χ Mˆ χ 2h | | i for an arbitrary vector with the vectors T T χ = [δC δC ∗] χ = [δC δC ∗] and the matrix means that all eigenvalues of the matrix have to be larger than or equal zero. A necessary (but no sufficient) condition is that the ∆ + AB Mˆ = , matrix elements (for all ai ) B ∆ + A ∗ ∗ (εa εi )δabδij + Aai,bj 0. with ∆ = (ε ε )δ δ . − ≥ ai,bj a − i ab ij This equation can be used as a first test of the stability of the Hartree-Fock equation.

Operators in second quantization Operators in second quantization Assume that we have at our disposal n diﬀerent single-particle orbits α0, α2, . . . , αn 1 and that we can distribute among these − In the build-up of a shell-model or FCI code that is meant to tackle orbits N n particles. Hˆ ≤ large dimensionalities is the action of the Hamiltonian on a A Slater determinant can then be coded as an integer of n bits. As Slater determinant represented in second quantization as an example, if we have n = 16 single-particle states α0, α1, . . . , α15 and N = 4 fermions occupying the states α3, α6, α10 and α13 we α1 . . . αn = a† a† ... a† 0 . | i α1 α2 αn | i could write this Slater determinant as The time consuming part stems from the action of the Hamiltonian ΦΛ = a† a† a† a† 0 . on the above determinant, α3 α6 α10 α13 | i The unoccupied single-particle states have bit value 0 while the 1 † occupied ones are represented by bit state 1. In the binary notation α t + u β aα† aβ + αβ vˆ γδ aα† aβaδaγ aα† aα† ... aα† n 0 .  h | | i 4 h | | i  1 2 | i we would write this 16 bits long integer as Xαβ αβγδX   A practically useful way to implement this action is to encode a α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 α12 α13 α14 α15 Slater determinant as a bit pattern. 0001001000100100 which translates into the decimal number

23 + 26 + 210 + 213 = 9288.

We can thus encode a Slater determinant as a bit pattern.

Operators in second quantization Operators in second quantization We assume again that we have at our disposal n different single-particle orbits α0, α2, . . . , αn 1 and that we can distribute − among these orbits N n particles. The ordering among these ≤ states is important as it defines the order of the creation operators. With N particles that can be distributed over n single-particle We will write the determinant states, the total number of Slater determinats (and defining thereby ΦΛ = a† a† a† a† 0 , the dimensionality of the system) is α3 α6 α10 α13 | i n in a more compact way as dim( ) = . H N Φ = 0001001000100100 . 3,6,10,13 | i The total number of bit patterns is 2n. The action of a creation operator is thus

a† Φ , , , = a† 0001001000100100 = a† a† a† a† a† 0 , α4 3 6 10 13 α4 | i α4 α3 α6 α10 α13 | i which becomes

a† a† a† a† a† 0 = 0001101000100100 . − α3 α4 α6 α10 α13 | i −| i Operators in second quantization Operators in second quantization

Consider the action of aα† 2 on various slater determinants: Similarly aα† Φ = aα† 00111 = 0 00111 2 00111 2 | i × | i † † a† Φ , , , = a† 0001001000100100 = a† a† a† a† a† 0 , aα2 Φ01011 = aα2 01011 = ( 1) 01111 α6 3 6 10 13 α6 | i α6 α3 α6 α10 α13 | i | i − × | i aα† Φ = aα† 01101 = 0 01101 2 01101 2 | i × | i which becomes aα† Φ01110 = aα† 01110 = 0 01110 2 2 2 a† (a† ) a† a† 0 = 0! | i × | i α4 α6 α10 α13 aα† Φ = aα† 10011 = ( 1) 10111 − | i 2 10011 2 | i − × | i This gives a simple recipe: aα† Φ10101 = aα† 10101 = 0 10101 2 2 | i × | i aα† Φ = aα† 10110 = 0 10110 If one of the bits bj is 1 and we act with a creation operator 2 10110 2 | i × | i on this bit, we return a null vector aα† 2 Φ11001 = aα† 2 11001 = (+1) 11101 | i × | i l † † If b = 0, we set it to 1 and return a sign factor ( 1) , where l aα2 Φ11010 = aα2 11010 = (+1) 11110 j − | i × | i is the number of bits set before bit j. What is the simplest way to obtain the phase when we act with one annihilation(creation) operator on the given Slater determinant representation?

Operators in second quantization Operators in second quantization

We have an SD representation The action

ΦΛ = a† a† a† a† a† 0 , aα0 Φ0,3,6,10,13 = 0001001000100100 , α0 α3 α6 α10 α13 | i | i in a more compact way as can be obtained by subtracting the logical sum (AND operation) of Φ0,3,6,10,13 and a word which represents only α0, that is Φ0,3,6,10,13 = 1001001000100100 . | i 1000000000000000 , | i The action of from Φ = 1001001000100100 . 0,3,6,10,13 | i a† aα Φ , , , , = a† 0001001000100100 = a† a† a† a† a† 0 , This operation gives 0001001000100100 . α4 0 0 3 6 10 13 α4 | i α4 α3 α6 α10 α13 | i | i Similarly, we can form aα† 4 aα0 Φ0,3,6,10,13, say, by adding which becomes 0000100000000000 to a Φ , ﬁrst checking that their | i α0 0,3,6,10,13 logical sum is zero in order to make sure that orbital α4 is not a† a† a† a† a† 0 = 0001101000100100 . − α3 α4 α6 α10 α13 | i −| i already occupied.

Operators in second quantization Exercises It is trickier however to get the phase ( 1)l . One possibility is as Exercise 1 − follows This exercise serves to convince you about the relation between two Let S be a word that represents the 1 bit to be removed and different single-particle bases, where one could be our new 1 − all others set to zero. Hartree-Fock basis and the other a harmonic oscillator basis. Consider a Slater determinant built up of single-particle orbitals ψλ, In the previous example S = 1000000000000000 1 | i with λ = 1, 2,..., A. The unitary transformation Define S2 as the similar word that represents the bit to be added, that is in our case ψa = Caλφλ, S = 0000100000000000 . λ 2 | i X Compute then S = S S , which here becomes 1 − 2 brings us into the new basis. The new basis has quantum numbers S = 0111000000000000 a = 1, 2,..., A. Show that the new basis is orthonormal. Show that | i the new Slater determinant constructed from the new single-particle wave functions can be written as the determinant based on the Perform then the logical AND operation of S with the word previous basis and the determinant of the matrix C. Show that the containing old and the new Slater determinants are equal up to a complex Φ , , , , = 1001001000100100 , constant with absolute value unity. (Hint, C is a unitary matrix). 0 3 6 10 13 | i Starting with the second quantization representation of the Slater which results in 0001000000000000 . Counting the number of | i determinant 1 bits gives the phase. Here you need however an algorithm for n − bitcounting. Several efficient ones available. Φ = a† 0 , 0 αi | i Yi=1 use Wick’s theorem to compute the normalization integral Φ Φ . h 0| 0i Exercises Exercises Exercise 2 Calculate the matrix elements Exercise 3 α α Fˆ α α h 1 2| | 1 2i Show that the onebody part of the Hamiltonian and Hˆ = p hˆ q a† a , ˆ 0 h | 0| i p q α1α2 G α1α2 pq h | | i X with can be written, using standard annihilation and creation operators, α α = a† a† 0 , | 1 2i α1 α2 | i in normal-ordered form as ˆ ˆ F = α f β aα† aβ, h | | i Hˆ = p hˆ q a† a + i hˆ i . αβ 0 h | 0| i p q h | 0| i X pq X n o Xi α fˆ β = ψ∗ (x)f (x)ψ (x)dx, h | | i α β Explain the meaning of the various symbols. Which reference Z 1 vacuum has been used? Gˆ = αβ gˆ γδ a† a† aδaγ, 2 h | | i α β αβγδX and

αβ gˆ γδ = ψ∗ (x )ψ∗(x )g(x , x )ψ (x )ψ (x )dx dx h | | i α 1 β 2 1 2 γ 1 δ 2 1 2 ZZ Compare these results with those from exercise 3c). Exercises Exercises Exercise 4 Exercise 5 Show that the twobody part of the Hamiltonian The aim of this exercise is to set up specific matrix elements that will turn useful when we start our discussions of the nuclear shell 1 HÎ = pq vˆ rs a† a† as ar , model. In particular you will notice, depending on the character of 4 h | | i p q pqrs the operator, that many matrix elements will actually be zero. X Consider three N-particle Slater determinants Φ , Φa and Φab , | 0 | i i | ij i can be written, using standard annihilation and creation operators, where the notation means that Slater determinant Φa differs from | i i in normal-ordered form as Φ by one single-particle state, that is a single-particle state ψ is | 0i i 1 1 replaced by a single-particle state ψa. It is often interpreted as a Hˆ = pq vˆ rs a† a† a a + pi vˆ qi a† a + ij vˆ ij . I h | | i p q s r h | | i p q h | | i so-called one-particle-one-hole excitation. Similarly, the Slater 4 pqrs 2 X n o Xpqi n o Xij determinant Φab differs by two single-particle states from Φ | ij i | 0i Explain again the meaning of the various symbols. and is normally thought of as a two-particle-two-hole excitation. We assume also that Φ represents our new vacuum reference This exercise is optional: Derive the normal-ordered form of the | 0i threebody part of the Hamiltonian. state and the labels ijk ... represent single-particle states below the Fermi level and abc ... represent states above the Fermi level, ˆ 1 so-called particle states. We define thereafter a general onebody H3 = pqr vˆ3 stu ap† aq† ar†auat as , 36 pqr h | | i normal-ordered (with respect to the new vacuum state) operator as Xstu

Fˆ = p f β a† a , and specify the contributions to the twobody, onebody and the N h | | i p q pq scalar part. X n o with p f q = ψ∗(x)f (x)ψ (x)dx, Exercises Exercises: Usingh sympy| | i to computep matrixq elements Z Exercise 6 and a general normal-ordered two-body operator Write a program which sets up all possible Slater determinants 1 given N = 4 eletrons which can occupy the atomic single-particle GˆN = pq g rs AS a† a† as ar , 4 h | | i p q s s p s p d pqrs states defined by the 1 , 2 2 and 3 3 3 shells. How many X n o single-particle states n are there in total? Include the spin degrees Exercisewith for example7 the direct matrix element given as of freedom as well. We include here a python program which may Compute the matrix element aid in this direction. It uses bit manipulation functions from pq g rs = ψ (x )ψ (x )g(x , x )ψ (x )ψ (x )dx dx http://wiki.python.org/moin/BitManipulation. p∗ 1 q∗ ˆ2 1 2 r 1 s 2 1 2 h | | i α1α2α3 G α0 α0 α0 , import math ZZ h | | 1 2 3i g """ withusing Wick’sbeing theoreminvariant and under express the interchange the two-body of the operator coordinatesG in the of A simple Python class for Slater determinant manipulation twooccupation particles. number The single-particle (second quantization) states ψi representation.are not necessarily Bit-manipulation stolen from: eigenstates of fˆ. The curly brackets mean that the operators are http://wiki.python.org/moin/BitManipulation normal-ordered with respect to the new vacuum reference state. """ How would you write the above Slater determinants in a second # bitCount() counts the number of bits set (not an optimal function) quantization formalism, utilizing the fact that we have defined a def bitCount(int_type): new reference state? """ Count bits set in integer """ Use thereafter Wick’s theorem to find the expectation values of count=0 while(int_type): ˆ int_type &= int_type-1 Φ0 FN Φ0 , count +=1 h | | i return(count) and ˆ Φ0GN Φ0 . # testBit() returns a nonzero result, 2**offset, if the bit at ’offset’ is one. h | i Find thereafter def testBit(int_type, offset): ˆ a mask=1 << offset Φ0 FN Φi , return(int_type& mask) >> offset h | | i and # setBit() returns an integer with the bit at ’offset’ set to 1. Φ Gˆ Φa , h 0| N | i i def setBit(int_type, offset): mask=1 << offset Finally, find return(int_type| mask) Φ Fˆ Φab , h 0| N | ij i # clearBit() returns an integer with the bit at ’offset’ cleared. and ˆ ab def clearBit(int_type, offset): Φ0 GN Φij . mask=~(1 << offset) h | | i return(int_type& mask) What happens with the two-body operator if we have a transition # toggleBit() returns an integer with the bit at ’offset’ inverted, 0 -> 1 and 1 -> 0. probability of the type

def toggleBit(int_type, offset): Φ Gˆ Φabc , mask=1 << offset h 0| N | ijk i return(int_type^ mask) where the Slater determinant to the right of the operator diﬀers by # binary string made from number more than two single-particle states? def bin0(s): return str(s) ifs<=1 else bin0(s>>1)+ str(s&1)

def bin(s, L=0): ss= bin0(s) ifL>0: return ’0’*(L-len(ss))+ ss else: return ss

class Slater: """ Class for Slater determinants """ def __init__(self): self.word= int(0)

def create(self, j): print "c^+_"+ str(j)+ " |"+ bin(self.word)+ "> = ", # Assume bit j is set, then we return zero. s=0 # Check if bit j is set. isset= testBit(self.word, j) if isset ==0: bits= bitCount(self.word&((1<

print str(s)+ " x |"+ bin(self.word)+ ">" returns def annihilate(self, j): print "c_"+ str(j)+ " |"+ bin(self.word)+ "> = ", # Assume bit j is not set, then we return zero. s=0 # Check if bit j is set. isset= testBit(self.word, j) if isset ==1: bits= bitCount(self.word&((1<" returns

# Do some testing:

phi= Slater() phi.create(0) phi.create(1) phi.create(2) phi.create(3)

s= phi.annihilate(2) s= phi.create(7) s= phi.annihilate(0) s= phi.create(200) Exercises: Using sympy to compute matrix elements Exercises: Using sympy to compute matrix elements The last exercise can be solved using the symbolic Python package called SymPy. SymPy is a Python package for general purpose We can expand the above Python code by defining one-body and symbolic algebra. There is a physics module with several interesting two-body operators using the following SymPy code submodules. Among these, the submodule called secondquant, # This code sets up a two-body Hamiltonian for fermions from sympy import symbols, latex, WildFunction, collect, Rational contains several functionalities that allow us to test our algebraic from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO manipulations using Wick’s theorem and operators for second # setup hamiltonian quantization. p,q,r,s= symbols(’p q r s’,dummy=True) from sympy import* f= AntiSymmetricTensor(’f’,(p,),(q,)) from sympy.physics.secondquant import* pr= NO((Fd(p)*F(q))) v= AntiSymmetricTensor(’v’,(p,q),(r,s)) i, j= symbols(’i,j’, below_fermi=True) pqsr= NO(Fd(p)*Fd(q)*F(s)*F(r)) a, b= symbols(’a,b’, above_fermi=True) Hamiltonian=f*pr+ Rational(1)/Rational(4)*v*pqsr p, q= symbols(’p,q’) print "Hamiltonian defined as:", latex(Hamiltonian) print simplify(wicks(Fd(i)*F(a)*Fd(p)*F(q)*Fd(b)*F(j), keep_only_fully_contracted=True))Here we have used the AntiSymmetricTensor functionality, together The code defines single-particle states above and below the Fermi with normal-ordering defined by the NO function. Using the latex level, in addition to the genereal symbols pq which can refer to any option, this program produces the following output type of state below or above the Fermi level. Wick’s theorem is implemented between the creation and annihilation operators Fd p 1 qp fq ap† aq vsr ap† aq† ar as and F, respectively. Using the simplify option, one can lump − 4 n o n o together several Kronecker-δ functions.

Exercises: Using sympy to compute matrix elements Exercises: Using sympy to compute matrix elements We can continue along these lines and define a normal-ordered We can now use this code to compute the matrix elements between Hamiltonian with respect to a given reference state. In our first two two-body Slater determinants using Wick’s theorem. step we just define the Hamiltonian from sympy import symbols, latex, WildFunction, collect, Rational, simplify from sympy import symbols, latex, WildFunction, collect, Rational, simplify from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies,from sympy.physics.secondquant NO, evaluate_deltas import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO, evaluate_deltas # setup hamiltonian # setup hamiltonian p,q,r,s= symbols(’p q r s’,dummy=True) p,q,r,s= symbols(’p q r s’,dummy=True) f= AntiSymmetricTensor(’f’,(p,),(q,)) f= AntiSymmetricTensor(’f’,(p,),(q,)) pr= NO((Fd(p)*F(q))) pr= Fd(p)*F(q) v= AntiSymmetricTensor(’v’,(p,q),(r,s)) v= AntiSymmetricTensor(’v’,(p,q),(r,s)) pqsr= NO(Fd(p)*Fd(q)*F(s)*F(r)) pqsr= Fd(p)*Fd(q)*F(s)*F(r) Hamiltonian=f*pr+ Rational(1)/Rational(4)*v*pqsr #define the Hamiltonian c,d= symbols(’c, d’,above_fermi=True) Hamiltonian=f*pr+ Rational(1)/Rational(4)*v*pqsr a,b= symbols(’a, b’,above_fermi=True) #define indices for states above and below the Fermi level index_rule={ expression= wicks(F(b)*F(a)*Hamiltonian*Fd(c)*Fd(d),keep_only_fully_contracted=True, simplify_kronecker_deltas’below’: ’kl’, =True) expression= evaluate_deltas(expression) ’above’: ’cd’, expression= simplify(expression) ’general’: ’pqrs’ print "Hamiltonian defined as:", latex(expression) } Hnormal= substitute_dummies(Hamiltonian,new_indices=True, pretty_indices=index_rule) The result is as expected, print "Hamiltonian defined as:", latex(Hnormal) b b a a ab which results in δac f δad f δbc f + δbd f + v . d − c − d c cd q 1 sr f a† ap + v a†a†apaq p q 4 qp s r

Exercises: Using sympy to compute matrix elements Exercises: Using sympy to compute matrix elements

In our next step we define the reference energy E0 and redefine the Hamiltonian by subtracting the reference energy and collecting the We can now go back to exercise 7 and define the Hamiltonian and coefficients for all normal-ordered products (given by the NO the second-quantized representation of a three-body Slater function). determinant. from sympy import symbols, latex, WildFunction, collect, Rational, simplify from sympy import symbols, latex, WildFunction, collect, Rational, simplify from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies,from sympy.physics.secondquant NO, evaluate_deltas import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO, evaluate_deltas # setup hamiltonian # setup hamiltonian p,q,r,s= symbols(’p q r s’,dummy=True) p,q,r,s= symbols(’p q r s’,dummy=True) f= AntiSymmetricTensor(’f’,(p,),(q,)) pr= Fd(p)*F(q) v= AntiSymmetricTensor(’v’,(p,q),(r,s)) v= AntiSymmetricTensor(’v’,(p,q),(r,s)) pqsr= NO(Fd(p)*Fd(q)*F(s)*F(r)) pqsr= Fd(p)*Fd(q)*F(s)*F(r) Hamiltonian=Rational(1)/Rational(4)*v*pqsr #define the Hamiltonian a,b,c,d,e,f= symbols(’a,b, c, d, e, f’,above_fermi=True) Hamiltonian=f*pr+ Rational(1)/Rational(4)*v*pqsr #define indices for states above and below the Fermi level expression= wicks(F(c)*F(b)*F(a)*Hamiltonian*Fd(d)*Fd(e)*Fd(f),keep_only_fully_contracted=True, simplify_kronecker_deltas=True) index_rule={ expression= evaluate_deltas(expression) ’below’: ’kl’, expression= simplify(expression) ’above’: ’cd’, print latex(expression) ’general’: ’pqrs’ } resulting in nine terms (as expected), Hnormal= substitute_dummies(Hamiltonian,new_indices=True, pretty_indices=index_rule) E0= wicks(Hnormal,keep_only_fully_contracted=True) δ v cb δ v cb+δ v cb δ v ac δ v ac +δ v ac +δ v ab+δ v ab δ v ab Hnormal= Hnormal-E0 − ad ef − ae fd af ed − bd ef − be fd bf ed cd ef ce fd − cf ed w= WildFunction(’w’) Hnormal= collect(Hnormal, NO(w)) Hnormal= evaluate_deltas(Hnormal) print latex(Hnormal) which gives us

i q 1 ii 1 ii 1 sr f + f a† ap v v + v a†a†aqap, − i p q − 4 ii − 4 ii 4 qp r s again as expected, with the reference energy to be subtracted. Exercises: Derivation of Hartree-Fock equations Exercises: Derivation of Hartree-Fock equations Exercise 9 Consider the ground state Φ of a bound many-particle system of | i fermions. Assume that we remove one particle from the single-particle state λ and that our system ends in a new state Φ . Deﬁne the energy needed to remove this particle as Exercise 8 | ni What is the diagrammatic representation of the HF equation? E = Φ a Φ 2(E E ), λ |h n| λ| i| 0 − n n n HF X αk u αi + [ αk αj vˆ αi αj αk αj v αj αi ] = 0 −h | | i h | | i − h | | i where E and E are the ground state energies of the states Φ j=1 0 n | i X and Φ , respectively. | ni (Represent ( uHF ) by the symbol X .) − − − − Show that E = Φ a† [a , H] Φ , λ h | λ λ | i where H is the Hamiltonian of this system. If we assume that Φ is the Hartree-Fock result, ﬁnd the

relation between Eλ and the single-particle energy ελ for states λ F and λ > F , with ≤ ε = λ tˆ+u ˆ λ , λ h | | i and λ uˆ λ = λβ vˆ λβ . h | | i h | | i β F X≤ We have assumed an antisymmetrized matrix element here. Discuss the result. The Hamiltonian operator is deﬁned as 1 H = α tˆ β a† aβ + αβ vˆ γδ a† a† aδaγ. h | | i α 2 h | | i α β Xαβ αβγδX