Quantum Operators in the Fock Space and Wave Function Languages

A quantum operator acting on identical bosons can be described in terms of N–particle wave functions (the ﬁrst-quantized formalism) or in terms of the creation and annihilation operators in the Fock space (the second-quantized formalism). In these notes I explain how to translate between the two formalisms.

The Fock space formalism is explained in my notes on free ﬁelds, harmonic oscillators, and identical bosons. On pages 8–12 of those notes, I deﬁne the bosonic Fock space as the Hilbert space of states containing arbitrary numbers of identical bosons,

∞ M F = H(N identical bosons), (1) N=0

then, starting with an arbitrary basis of one-particle states |αi I build the occupation-number

basis |{nα}i for the whole Fock space, and eventually deﬁne the creation operators and the annihilation operators in terms of that occupation number basis,

† def √ 0 aˆα {nβ} = nα + 1 {nβ = nβ + δαβ} , ( √ E (2) 0 def nα {nβ = nβ − δαβ} for nα > 0, aˆα {nβ} = 0 for nα = 0.

In the present notes, I shall start by translating these deﬁnition to the language of N– † boson wave function. And then I shall use the wave-function deﬁnitions of thea ˆα anda ˆα operators to translate the more complicated one-body, two-body, etc., operators between the Fock-space and the wave-function languages.

1 Creation and Annihilation Operators in the Wave Function Language

First, a quick note on multi-boson wave functions. A wave function of an N–particle state depends on all N particles’ positions, ψ(x1, x2,..., xN ), where by abuse of notations th each xi includes all the independent degrees of freedom of the i particle,

xi = (xi, yi, zi, spini, isospini,...).

Moreover, the wave function must be totally symmetric WRT any permutations of the identical bosons,

ψ(any permutation of the x1,..., xN ) = ψ(x1,..., xN ). (3)

This Bose symmetry plays essential role in the wave-function-language action of the creation and annihilation operators.

Deﬁnitions: Let the φα(x) be the wave function of the one-particle states α which we want † to be created by thea ˆα operators and annihilated by thea ˆα operators. Then, given an

N–boson state |N, ψi with a totally symmetric wave function ψ(x1,..., x1), we construct the 0 00 totally symmetric wave functions ψ (x1,..., xN+1) and ψ (x1,..., xN−1) of the (N + 1)– 0 † 00 boson state |N + 1, ψ i =a ˆα |N, ψi and the (N − 1)–boson state |N − 1, ψ i =a ˆα |N, ψi according to:

N+1 0 1 X ψ (x1,..., xN+1) = √ φα(xi) × ψ(x1,..., 6xi,..., xN+1), (4) N + 1 i=1 √ Z 00 3 ∗ ψ (x1,..., xN−1) = N d xN φα(xN ) × ψ(x1,..., xN−1, xN ). (5)

† 0 In particular, for N = 0 the statea ˆα |0i has ψ (x1) = φα(x1), while for N = 1 the state 00 def aˆα |βi has ψ (no arguments) = hφα|ψi. Also, for N = 0 we simply deﬁnea ˆα |0i = 0.

† Let me use eqs. (4) and (5) as definitions of the creation operatorsa ˆα and the annihilation operatorsa ˆα. To verify that these definitions are completely equivalent to the definitions (2) in terms of the occupation-number basis, I am going to prove the following lemmas:

2 † Lemma 1: The creation operatorsa ˆα deﬁned according to eq. (4) are indeed the hermitian conjugates of the operatorsa ˆα deﬁned according to eq. (5).

Lemma 2: The operators (4) and (5) obey the bosonic commutation relations

† † † [âα, aˆβ] = 0, [âα, aˆβ] = 0, [âα, aˆβ] = δαβ . (6)

Lemma 3: Let φαβ···ω(x1, x2 ..., xN ) be the N–boson wave function of the state

1 |α, β, ··· , ωi = √ aˆ† ··· aˆ† aˆ† |0i (7) T ω β α

† where the creation operatorsa ˆα act according to eq. (4) while T is the number of trivial permutations between coincident entries of the list (α, β, . . . , ω) (for example, α ↔ β when Q α and β happen to be equal). In terms of the occupation numbers nγ, T = γ nγ!. Then

distinct permutations of (α,β,...,ω) 1 X φαβ···ω(x1, x2 ..., xN ) = √ φα˜(x1) × φ ˜(x2) × · · · × φω˜ (xN ) D β (˜α,β,...,˜ ω˜) (8) all N! permutations of (α,β,...,ω) 1 X = √ φα˜(x1) × φ ˜(x2) × · · · × φω˜ (xN ), T × N! β (˜α,β,...,˜ ω˜)

where D = N!/T is the number of distinct permutations. In other words, the state (7) is precisely the symmetrized state of N bosons in individual states |αi , |βi ,..., |ωi.

Together, the lemmas 1–3 establish that the deﬁnitions (4) and (5) of the creation and annihilation operators completely agree with the deﬁnitions (2)of the same operators in terms of the occupation number basis.

3 † Proof of Lemma 1: To prove that the operatorsa ˆα anda ˆα are hermitian conjugates of each other, we need to compare their matrix elements and verify that for any two states |N, ψi and |N,e ψei in the Fock space we have

† ∗ hN,e ψe| aˆα |N, ψi = hN, ψ| aˆα |N,e ψei . (9)

† Since thea ˆα always lowers the number of particles by 1 while thea ˆα always raises it by 1, it is enough to check this equation for Ne = N − 1 — otherwise, we get automatic zero on both sides of this equation.

00 00 Let ψ (x1,..., xN−1) be the wave function of the state |N − 1, ψ i =a ˆα |N, Ψi according to eq. (5). Then, on the LHS of eq. (9) we have

00 hN − 1, ψe| aˆα |N, ψi = N − 1, ψe|N − 1, ψ Z Z 3 3 ∗ 00 = d x1 ··· d xN−1 ψe (x1,..., xN−1) × ψ (x1,..., xN−1) Z Z 3 3 ∗ = d x1 ··· d xN−1 ψe (x1,..., xN−1) × √ Z 3 ∗ × N d xN φα × ψ(x1,..., xN ) √ Z Z 3 3 ∗ ∗ = N d x1 ··· d xN ψe (x1,..., xN−1) × φα(xN ) × ψ(x1,..., xN ). (10) 0 0 † Likewise, let ψe (x1,..., xN ) be the wave function of the state N, ψe =a ˆα |N − 1, ψei according to eq. (4). Then the matrix element on the RHS of eq. (9) becomes Z Z † 3 3 ∗ 0 hN, ψ| aˆα |N − 1, ψei = d x1 ··· d xN ψ (x1,..., xN ) × ψe (x1,..., xN ) Z Z 3 3 ∗ = d x1 ··· d xN ψ (x1,..., xN ) ×

N 1 X × √ φα(xi) × ψe(x1,... 6xi,..., xN ) N i=1 N Z Z 1 X 3 3 ∗ = √ d x1 ··· d xN ψ (x1,..., xN ) × N i=1 × φα(xi) × ψe(x1,... 6xi,..., xN ). (11) By bosonic symmetry of the wavefunctions ψ and ψe, all terms in the sum on the RHS are

4 equal to each other. So, we may replace the summation with a single term — say, for i = N — and multiply by N, thus

Z Z † N 3 3 ∗ hN, ψ| aˆ |N − 1, ψei = √ × d x1 ··· d xN ψ (x1,..., xN )×φα(xN )×ψe(x1,... xN−1). α N (12) By inspection, the RHS of eqs. (10) and (12) are complex conjugates of each other, thus

† ∗ hN − 1, ψe| aˆα |N, ψi = hN, ψ| aˆα |N − 1, ψei . (9)

This completes the proof of Lemma 1.

Proof of Lemma 2: Let’s start by verifying that the creation operators deﬁned according † † to eq. (4) commute with each other. Pick any two such creation operatorsa ˆα anda ˆβ, and 000 pick any N-boson state |N, ψi. Consider the (N + 2)-boson wavefunction ψ (x1,..., xN+2) 000 † † of the state |N + 2, ψ i =a ˆαaˆβ |N, ψi. Applying eq. (4) twice, we immediately obtain

φα(xi) × φ (xj) × 000 1 X β ψ (x1,..., xN+2) = p (N + 1)(N + 2) i,j=1,...,N+2 × ψ(x1,..., xN+2 except xi, xj). i6=j (13) On the RHS of this formula, interchanging α ↔ β is equivalent to interchanging the summa- † † tion indices i ↔ j — which has no eﬀect on the sum. Consequently, the statesa ˆαaˆβ |N, ψi † † anda ˆβaˆα |N, ψi have the same wavefunction (13), thus

† † † † aˆαaˆβ |N, ψi =a ˆβaˆα |N, ψi . (14)

Since this is true for any N and any totally-symmetric wave function ψ, this means that the † † creation operatorsa ˆα anda ˆβ commute with each other.

Next, let’s pick any two annihilation operatorsa ˆα anda ˆβ deﬁned according to eq. (5) and show that they commute with each other. Again, let |N, ψi be an arbitrary N-boson

5 state . For N < 2 we have

† aˆαaˆβ |N, ψi = 0 =a ˆβaˆα |N, ψi , (15) so let’s focus on the non-trivial case of N ≥ 2 and consider the (N − 2)-boson wavefunction 0000 0000 ψ of the state |N − 2, ψ i =a ˆαaˆβ |N, ψi. Applying eq. (5) twice, we obtain

∗ ∗ Z Z φ (xN ) × φ (xN−1) × 0000 p 3 3 α β ψ (x1,..., xN−2) = N(N − 1) d xN d xN−1 × ψ(x1,..., xN−2, xN−1, xN ). (16) On the RHS of this formula, interchanging α ↔ β is equivalent to interchanging the integrated-over positions of the N th and the (N − 1)th boson in the original state |N, ψi. Thanks to bosonic symmetry of the wave-function ψ, this interchange has no eﬀect, thus

aˆαaˆβ |N, ψi =a ˆβaˆα |N, ψi . (17)

Therefore, when the annihilation operators deﬁned according to eq. (5) act on the totally- symmetric wave functions of identical bosons, they commute with each other.

† Finally, let’s pick a creation operatora ˆβ and an annihilation operatora ˆα, pick an arbitrary N-boson state |N, ψi, and consider the diﬀerence between the states

5 † 6 † N, ψ =a ˆβaˆα |N, ψi and N, ψ =a ˆαaˆβ |N, ψi . (18)

Suppose N > 0. Applying eq. (5) to the wave function ψ and then applying eq. (4) to the result, we obtain

N 5 1 X 00 ψ (x1,..., xN ) = √ φβ(xi) × ψ (x1,..., 6xi,..., xN ) N i=1 (19) N Z X 3 ∗ = φβ(xi) × d xN+1 φα(xN+1) × ψ(x1,..., 6xi,..., xN , xN+1). i=1

6 On the other hand, applying ﬁrst eq. (4) and then eq. (5), we arrive at

√ Z 6 3 ∗ 0 ψ (x1,..., xN ) = N + 1 d xN+1 φα(xN+1) × ψ (x1,..., xN , xN+1)

Z N+1 3 ∗ X = d xN+1 φα(xN+1) × φβ(xi) × ψ(x1,..., 6xi,..., xN+1) i=1

 N  X Z φ (x ) × ψ(x ,..., 6x ,..., x , x ) 3 ∗  β i 1 i N N+1  = d xN+1 φ (xN+1) ×   α  i=1  + φβ(xN+1) × ψ(x1,..., xN )

N Z X 3 ∗ = φβ(xi) × d xN+1 φα(xN+1) × ψ(x1,..., 6xi,..., xN , xN+1) i=1 Z 3 ∗ + ψ(x1,..., xN ) × d xN+1 φα(xN+1) × φβ(xN+1)

5 = ψ (x1,..., xN ) hh compare to eq. (19) ii

+ ψ(x1,..., xN ) × φα|φβ . (20) Comparing eqs. (19) and (20), we see that

6 5 ψ (x1,..., xN ) − ψ (x1,..., xN ) = ψ(x1,..., xN )× φα|φβ = ψ(x1,..., xN )×δαβ , (21)

where φα|φβ = δαβ by orthonormality of the 1-particle basis {φγ(x)}γ. In Dirac notations, eq. (21) amounts to † † aˆαaˆβ − aˆβaˆα |N, ψi = |N, ψi × δαβ . (22)

Thus far, we have checked this formula for all bosonic states |N, ψi except for the vacuum |0i. To complete the proof, note that

† aˆα |0i = 0 =⇒ aˆβaˆα |0i = 0, (23) while † aˆαaˆβ |0i =a ˆα 1, φβ = φα|φβ × |0i = δαβ × |0i , (24)

7 hence

† † aˆαaˆβ − aˆβaˆα |0i = δαβ × |0i . (25)

Altogether, eqs. (22) and (25) verify that

† [ˆaα, aˆβ] |Ψi = δαβΨ (26)

† for any state Ψ in the bosonic Fock space, hence the operatorsa ˆα anda ˆβ deﬁned according † to eqs. (4) and (5) indeed obey the commutation relation [ˆaα, aˆβ] = δα,β. This completes the proof of Lemma 2. √ Proof of Lemma 3: Let me start with a note on the normalization factor 1/ T in eq. (7). We need this factor to properly normalize the multi-boson states in which some bosons may be in the same 1-particle mode. For example, for the two particle states,

1 |α, βi =a ˆ† aˆ† |0i when α 6= β, but |α, αi = √ aˆ† aˆ† |0i . (27) β α 2 α α

In terms of the occupation numbers, the properly normalized states are

† nα † nα ! O aˆα Y aˆα |{nα}αi = |nαi = √ |0i = √ |vacuumi . (28) mode α α nα! α nα!

√ hence the factor 1/ T in eq. (7).

Now let’s work out the wave functions of the states (7) by successively applying the creation operators according to eq. (4):

† 1. For N = 1, states |αi =a ˆα |0i have wave functions φα(x).

√ † † 2. For N = 2, states T |α, βi =a ˆβaˆα |0i have wavefunctions

√ 1 T × φαβ(x1, x2) = √ φβ(x1)φα(x2) + φβ(x2)φα(x1) . (29) 2

8 √ † † † 3. For N = 3, states T |α, β, γi =a ˆγaˆβaˆα |0i have

 1  φγ(x1) × √ φβ(x2)φα(x3) − φβ(x3)φα(x2)  2    √ 1  1   + φ (x ) × √ φ (x )φ (x ) − φ (x )φ (x )  T × φαβγ(x1, x2, x3) = √  γ 2 β 1 α 3 β 3 α 1  3  2     1  + φγ(x3) × √ φβ(x1)φα(x2) − φβ(x2)φα(x1) 2

6 permutations of (x1,x2,x3) 1 X = √ φγ(x˜1)φβ(x˜2)φα(x˜3) 3! (x˜1,x˜2,x˜3)

6 permutations of (α,β,γ) 1 X = √ φα˜(x1)φ ˜(x2)φγ˜(x3). 3! β (˜α,β,˜ γ˜) (30)

Proceeding in this fashion, acting with a product of N creation operators on the vacuum

we obtain a totally symmetrized product of the 1-particle wave functions φα(x) through √ † † φω(x). Extrapolating from eq.(30), the N-particle state T |α, . . . , ωi =a ˆω ··· aˆα |0i, has the totally-symmetrized wave function

all permutations of (α,...,ω) √ 1 X T × φα···ω(x1,..., xN ) = √ φα˜(x1) × · · · × φω˜ (xN ). (31) N! (˜α,...,ω˜) √ Dividing both sides of this formula by the T factor, we immediately arrive at the second line of eq. (8).

Finally, the top line of eq. (8) obtains from the bottom line by adding up coincident terms. Indeed, if some one-particle states appear multiple times in the list (α, . . . , ω), then permuting coincident entries of this list has no eﬀect. Altogether, there T such trivial permutations. By group theory, this means that out of N! possible permutations of the list, there are only D = N!/T distinct permutations. But for each distinct permutations, there are T coincident terms in the sum on the bottom line of eq. (8). Adding them up gives us the top line of eq. (8).

9 This completes the proof of Lemma 3.

† Altogether, the three lemmas verify that the operatorsa ˆα anda ˆα deﬁned according to eqs. (4) and (5) are indeed the creation and the annihilation operators in the bosonic Fock space.

One-body and Two-body Operators in the Wave Function and the Fock Space Languages

The one-body operators are the additive operators acting on one particle at a time, for example the net momentum or net kinetic energy of several bosons. In the wave-function language (AKA, the ﬁrst-quantized formalism), such operators act on N–particle states according to N ˆ(1) X ˆ th Anet = A(i particle) (32) i=1

where Aˆ is some kind of a single-particle operator. For example,

N ˆ (1) X the net momentum operator Pnet = pˆi , (33) i=1 N 2 (1) X pˆ the net kinetic energy operator Kˆ = i , (34) net 2m i=1 N ˆ (1) X the net potential energy operator Vnet = V (xˆi). (35) i=1

In the Fock space language (AKA, the second-quantized formalism), such net one-body operators take form

ˆ(2) X ˆ † Anet = hα| A |βi × aˆαaˆβ (36) α,β

ˆ where the matrix elements Aα,β = hα| A |βi are taken in the one-particle Hilbert space.

Theorem 1: Although eqs. (32) and (36) look very diﬀerent from each other, they describe exactly the same net one-body operator.

ˆ Let’s start by relating the matrix elements on the LHS of this formula to the Aα,β = hα| A |βi. For N = 1 the relation is very simple: Since the states |αi make a complete basis of the 1-particle Hilbert space, for any 1-particle states hψe| and |ψi

In the N–particle Hilbert space we have a similar formula for the matrix elements of the Aˆ acting on particle #i, the only modiﬁcation being integrals over the coordinates of the other particles,

th hN, ψe| Aˆ1(i ) |N, ψi = Z Z Z 3 3 3 X 3 ∗ = ··· d x1 ··· /////d xi ··· d xN Aαβ × d x˜i ψe (x1,..., x˜i,..., xN )φα(x˜i) α,β Z 3 ∗ × d xi φβ(xi)ψ(x1,..., xi,..., xN ) Z Z X 3 3 3 ∗ = Aαβ × ··· d x1 ··· d xN d x˜i ψe (x1,..., x˜i,..., xN ) × φα(x˜i) α,β ∗ × φβ(xi) × ψ(x1,..., xi,..., xN ). (39) For symmetric wave-functions of identical bosons, we get the same matrix element regardless of which particle #i we are acting on with the operator Aˆ, hence for the net A operator (32),

Z Z ˆ(1) X 3 3 3 3 hN, ψe| Anet |N, ψi = N × Aαβ × ··· d x1 ···d xN−1 d xN d xÑ (40) α,β ∗ ψe (x1,..., xN−1, xÑ ) × φα(xÑ ) ∗ × φβ(xN ) × ψ(x1,..., xN−1, xN ).

11 Now consider matrix elements of the Fock-space operator (36). In light of eq. (5), the 00 state |N − 1, ψ i =a ˆβ |N, ψi has wave-function

√ Z 00 3 ∗ ψ (x1,..., xN−1) = N d xN φβ(xN ) × ψ(x1,..., xN−1, xN ). (41)

00 Likewise, the state N − 1, ψe =a ˆα |N, ψei has wave-function

√ Z 00 3 ∗ ψe (x1,..., xN−1) = N d xÑ φα(xÑ ) × ψe(x1,..., xN−1, xÑ ). (42)

Consequently,

† 00 00 hN, ψe| aˆαaˆβ |N, ψi = N − 1, ψe N − 1, ψ Z Z 3 00∗ 00 = ··· d x1 ··· xN−1 ψe (x1,..., xN−1) × ψ (x1,..., xN−1) Z Z √ Z 3 3 ∗ = ··· d x1 ··· xN−1 N d xÑ φα(xÑ ) × ψe (x1,..., xN−1, xÑ ) × √ Z 3 ∗ × N d xN φβ(xN ) × ψ(x1,..., xN−1, xN ). (43) Comparing this formula to the integrals in eq. (40), we see that

Quod erat demonstrandum.

???

Now consider the two-body operators — the additive operators acting on two particles at a time. For example, the net two-body potential

i,j=1,...,N (1) 1 X Vˆ = V (xˆ − xˆ ). (45) |rmnet 2 i j i6=j

More generally, in the wave-function language we start with some operator Bˆ in a two- particle Hilbert space, make it act on all (i, j) pairs of particles in the N-particle Hilbert

12 space, and total up the pairs,

i,j=1,...,N (1) 1 X Bˆ = Bˆ(ith and jth particles). (46) net 2 i6=j

In the Fock space language, such a two-body operator becomes

ˆ(2) 1 X ˆ † † Bnet = 2 (hα| ⊗ hβ|)B(|γi ⊗ |δi)a ˆαaˆβaˆδaˆγ . (47) α,β,γ,δ

Note: in this formula, it is OK to use the un-symmetrized 2-particle states hα| ⊗ hβ| and

|γi⊗|δi, and hence the un-symmetrized matrix elements of the Bˆ2. At the level of the second- ˆ(2) † † † † quantized operator Bnet, the Bose symmetry is automatically provided bya ˆαaˆβ =a ˆβaˆγ and ˆ aˆδaˆγ =a ˆγaˆδ, even for the un-symmetrized matrix elements of the two-particle operator B.

For example, the two-body potential V (xˆ1 −xˆ2) in the un-symmetrized momentum basis has matrix elements

0 0 −3 ( p ⊗ p )Vˆ (|p i ⊗ |p i) = L δ 0 0 × W (q) 1 2 1 2 p1+p2,p1+p2 0 0 where q = p1 − p1 = p2 − p2 (48) Z −iqx and W (q) = dx e V2(x), hence

(2) X X Vˆ = 1 L−3 W (q) aˆ† aˆ† aˆ aˆ . (49) net 2 p1+q p2−q p2 p1 q p1,p2

Theorem: Again, for any two-particle operator Bˆ eqs. (46) and (47) deﬁne exactly the same net operator Bˆnet. Indeed, the operators deﬁned according to the two equations have the same matrix elements between any two N–boson states,

Proof: This works similarly to the previous theorem, except for more integrals. Let

ˆ Bαβ,γδ = hα| ⊗ hβ| B2 |γi ⊗ |δi (51) be matrix elements of a two-body operator Bˆ2 between un-symmetrized two-particle states.

13 Then for generic two-particle states hψe| and |ψi — symmetric or not — we have ˆ X hψe| B2 |ψi = Bαβ,γδ × hψe| |αi ⊗ |βi × hγ| ⊗ hδ| |ψi α,β,γ,δ ZZ X 3 3 ∗ = Bαβ,γδ × d x˜1 d x˜2 ψe (x˜1, x˜2)φα(x˜1)φβ(x˜2) (52) α,β,γ,δ ZZ 3 3 ∗ ∗ × d x1 d x2 φγ(x1)φδ(x2)ψ(x1, x2).

Similarly, in the Hilbert space of N > 2 particles — identical bosons or not — the operator

Bˆ2 acting on particles #i and #j has matrix elements

th th hN, ψe| Bˆ2(i , j ) |N, ψi = Z Z X 3 3 3 3 = Bαβ,γδ × ··· d x1 ··· /////d xi ··· d/////xj ··· d xN α,β,γ,δ ZZ 3 3 ∗ d x˜i d x˜j ψe (x1,..., x˜i,..., x˜j,..., xN )φα(x˜i)φβ(x˜j)

ZZ 3 3 ∗ ∗ × d xi d xj φγ(xi)φδ(xj)ψ(x1,..., xi,..., xj,..., xN ) (53) For identical bosons — and hence totally symmetric wave-functions ψ and ψe — such matrix

elements do not depend on the choice of particles on which Bˆ2 acts, so we may just as well let i = N − 1 and j = N. Consequently, the net Bˆ operator (21) has matrix elements

(1) N(N − 1) hN, ψe| Bˆ |N, ψi = × hN, ψe| Bˆ2(N − 1,N) |N, ψi net 2 N(N − 1) X (54) = × B × I 2 αβ,γδ αβ,γδ α,β,γ,δ where Z Z 3 3 Iαβ,γδ = ··· d x1 ··· d xN−2

ZZ 3 3 ∗ d xÑ−1 d xÑ ψe (x1,..., xN−2, xÑ−1, xÑ )φα(xÑ−1)φβ(xÑ ) (55)

ZZ 3 3 ∗ ∗ × d xN−1 d xN φγ(xN−1)φδ(xN )ψ(x1,..., xN−2, xN−1, xN )

Now let’s compare these formulae to the Fock space formalism. Applying eq. (5) twice,

14 we ﬁnd that the (N − 2)–particle state

000 N − 2, ψ =a ˆδaˆγ |N, ψi (56) has wave function

ZZ 000 p 3 3 ∗ ∗ ψ (x1,..., xN−2) = N(N − 1) d xN−1 d xN φ (xN−1)φ (xN ) γ δ (57) × ψ(x1,..., xN−2, xN−1, xN ).

Likewise, the (N − 2)–particle state

000 N − 2, ψe =a ˆβaˆα |N, ψei (58)

has wave function

ZZ 000 p 3 3 ∗ ∗ ψe (x1,..., xN−2) = N(N − 1) d xN−1 d xN φ (xÑ−1)φ (xÑ ) β α (59) × ψe(x1,..., xN−2, xÑ−1, xÑ ).

Taking Dirac product of these two states, we obtain

† † 000 000 hN, ψe| aˆαaˆβaˆδaˆγ |N, ψi = N − 2, ψe N − 2, ψ Z Z 3 3 000∗ 000 = ··· d x1 ··· d xN−2 ψe (x1,..., xN−2) × ψ (x1,..., xN−2)

= N(N − 1) × Iαβ,γδ (60)

where Iαβ,γδ is exactly the same mega-integral as in eq. (55). Therefore,

ˆ(2) where the second equality follows directly from the eq. (47) for the Bnet operator. Quod erat demonstrandum.

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15 I would like to conclude these notes with a couple of simple theorems about the commutators of the net one-body and two-body operators.

ˆ ˆ ˆ ˆ(2) ˆ(2) ˆ(2) Theorem 1: Let A, B, and C be some one-particle operators, and let Anet, Bnet, and Cnet be the corresponding net one-body operators in the fock space according to eq. (36).

ˆ ˆ ˆ h ˆ(2) ˆ(2)i ˆ(2) if [A, B] = C then Anet, Bnet = Cnet . (62)

Theorem 2: Now let Aˆ be a one-particle operator while Bˆ and Cˆ are two-particle operators. ˆ(2) ˆ(2) ˆ(2) Let Anet, Bnet, and Cnet be the corresponding net operators in the Fock space according to eqs. (36) and (47).

h ˆ st ˆ nd ˆi ˆ h ˆ(2) ˆ(2)i ˆ(2) if A(1 ) + A(2 ) , B = C then Anet, Bnet = Cnet . (63)

Proof of Theorem 1: The theorem follows from the commutator

† † † † [ˆaαaˆβ, aˆγaˆδ] = δβ,γaˆαaˆδ − δα,δaˆγaˆβ (64) which you should have calculated in homework set#3, problem 4(a). Indeed, given

ˆ(2) X ˆ † Atot = hα| A |βi aˆαaˆβ (65) α,β and

ˆ(2) X ˆ † Btot = hγ| B |δi aˆγaˆδ , (66) γ,δ

16 we immediately have

h ˆ(2) ˆ(2)i X ˆ ˆ † † Atot, Btot = hα| A |βi hγ| B |δi [ˆaαaˆβ, aˆγaˆδ] α,β,γ,δ hh using eq. (64) ii X ˆ ˆ † † = hα| A |βi hγ| B |δi δβ,γaˆαaˆδ − δα,δaˆγaˆβ α,β,γ,δ X † X ˆ ˆ X † X ˆ ˆ = aˆαaˆδ × hα| A |γi hγ| B |δi − aˆγaˆβ × hγ| B |αi hα| A |βi α,δ β=γ β,γ α=δ (67) X † ˆ ˆ X † ˆ ˆ = aˆαaˆδ hα| AB |δi − aˆγaˆβ hγ| BA |βi α,δ β,γ hh renaming summation indices ii X † ˆ ˆ ˆ ˆ = aˆαaˆβ × hα| AB |βi − hα| BA |βi α,β X † ˆ ˆ ˆ ˆ(2) = aˆαaˆβ × hα| [A, B] = C |βi ≡ Ctot . α,β

Quod erat demonstrandum.

Proof of Theorem 2: Similarly to the theorem 1, this theorem also follows from a commutator you should have calculated in homework set#3, problem 4(a), namely

† † † † † † † † † † † [ˆaµaˆν, aˆαaˆβaˆγaˆδ] = δναaˆµaˆβaˆγaˆδ + δνβaˆαaˆµaˆγaˆδ − δµγaˆαaˆβaˆνaˆδ − δµδaˆαaˆβaˆγaˆν . (68)

Indeed, in the Fock space

ˆ(2) X ˆ † Atot = hµ| A |νi aˆµaˆν (36) µ nu and

ˆ(2) 1 X ˆ † † Btot = 2 hα ⊗ β| B |γ ⊗ δi aˆαaˆβaˆγaˆδ , (47) α,β,γ,δ

17 h ˆ(2) ˆ(2)i so the commutator Anet, Bnet is a linear combination of the commutators (68). Speciﬁcally,

ˆ(2) ˆ(2) 1 X ˆ ˆ † † † [Atot, Btot] = 2 hµ| A |νi hα ⊗ β| B |γ ⊗ δi [ˆaµaˆν, aˆαaˆβaˆγaˆδ] µ,ν,α,β,γ,δ hh using eq. (68) ii 1 X † † X ˆ ˆ = 2 aˆµaˆβaˆγaˆδ × hµ| A |νi hν ⊗ β| B |γ ⊗ δi µ,β,γ,δ ν 1 X † † X ˆ ˆ + 2 aˆαaˆµaˆγaˆδ × hµ| A |νi hα ⊗ ν| B |γ ⊗ δi α,µ,γ,δ ν (69) 1 X † † X ˆ ˆ − 2 aˆαaˆβaˆνaˆδ × hα ⊗ β| B |µ ⊗ δi hµ| A |νi α,β,ν,δ µ 1 X † † X ˆ ˆ − 2 aˆαaˆβaˆγaˆν × hα ⊗ β| B |γ ⊗ µi hµ| A |νi α,β,γ,ν µ hh renaming summation indices ii 1 X † † = 2 aˆαaˆβaˆγaˆδ × Cα,β,γ,δ , α,β,γ,δ where