Explanation on anticommutation relations

Question

Setup

Given two states: $|K\rangle=a_i^+a_j^+|\rangle$ and $|L\rangle=a_k^+a_l^+|\rangle$.
Evaluating the overlap: $\langle K|L\rangle=\langle|a_ja_ia_k^+a_l^+|\rangle$
Introducing: $a_ia_k^+=\delta_{ik}-a_k^+a_i$
Replacing 3 in 2 and solving: $\langle K|L\rangle=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+|\rangle=\delta_{ik}\langle|a_ja_l^+|\rangle-\langle|a_ja_k^+a_ia_l^+|\rangle$

Question

Fourth term has two terms, how come is obtained:

$\langle K|L\rangle=\delta_{ik}\delta_{jl}\langle|\rangle-\delta_{ik}\langle|a_i^+a_j|\rangle-\delta_{il}\langle|a_ja_k^+|\rangle+\langle|a_ja_k^+a_l^+a_i|\rangle$

For the underlying discussion of CARs, see http://physics.stackexchange.com/q/77384/2451 , http://physics.stackexchange.com/q/68445/2451 , http://physics.stackexchange.com/q/17893/2451 and links therein. — Qmechanic, Apr 18 '15 at 23:54
Yes, in those links give the background of second quantization. But, over there do not response the posted question. — Another.Chemist, Apr 18 '15 at 23:58
Use the rule 3. to put the four operators in the order of the last term of your result. Each time you swap two operators you get an additional term. BTW, I think that the second term of your result is wrong. — GCLL, Apr 19 '15 at 00:04
@GCLL I have updated the post. But, I still don't see how is obtained the last term — Another.Chemist, Apr 19 '15 at 00:30
You are near. The rule (3) is valid for generic values of the indexes, so you can apply it again to swap the two operators on the right in the second term. — GCLL, Apr 19 '15 at 00:36
But ... how many times are mandatory to swap and introduce, over and over again? What are the reason behind that repeated operation? — Another.Chemist, Apr 19 '15 at 00:41
You have messed up your indices somewhere along the line in the last line. For example, your second term in the last line has two "i" indices and no "l" index. — hft, Apr 19 '15 at 00:45

score 1 · Answer 1 · answered Apr 19 '15 at 00:53

Introducing: $a_ia_k^+=\delta_{ik}-a_k^+a_i$

Replacing 3 in 2 and solving: $\langle K|L\rangle=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+|\rangle=\delta_{ik}\langle|a_ja_l^+|\rangle-\langle|a_ja_k^+a_ia_l^+|\rangle$

$\langle K|L\rangle=\langle|a_ja_ia_k^+a_l^+|\rangle$ $$ =\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+\rangle $$ $$ =\langle a_ja_l^+\rangle\delta_{ik}-\langle a_ja_k^+a_ia_l^+\rangle $$ $$ =\delta_{ik}\delta_{jl}-\langle a_l^+a_j\rangle\delta_{ik}-\langle a_ja_k^+a_ia_l^+\rangle $$ $$ =\delta_{ik}\delta_{jl}-\langle a_l^+a_j\rangle\delta_{ik}-\langle a_ja_k^+(\delta_{il}-a_l^+a_i)\rangle $$ $$ =\delta_{ik}\delta_{jl}-\langle a_l^+a_j\rangle\delta_{ik}-\langle a_ja_k^+\rangle\delta_{il} +\langle a_ja_k^+a_l^+a_i\rangle $$

GCLL · Accepted Answer · 2015-04-19T01:29:17.833

Using the rule (3) you can sort the four operators in the order you prefer, by swapping terms. After each swapping, you obtain a piece with four operators, plus an additional one with two operators. And so on.

In your case, by swapping the second and the third term,

$$a_j a_i a_k^+ a_l^+ = - a_j a_k^+ a_i a_l^+ + \delta_{ki} a_j a_l^+ $$

and by swapping the third and the fourth in the first addendum

$$a_j a_i a_k^+ a_l^+ = - \delta_{il} a_j a_k^+ + a_j a_k^+ a_l^+ a_i + \delta_{ki} a_j a_l^+ $$

Finally swapping the terms in the third addendum

$$a_j a_i a_k^+ a_l^+ = - \delta_{il} a_j a_k^+ + a_j a_k^+ a_l^+ a_i + \delta_{ki} \delta_{jl} + \delta_{ki} a_l^+ a_j $$

which is the final result.

A motivation for doing this kind of manipulation could be, for example, to order each term in such a way that the destruction operators are on the right side. In this particular case one obtain

$$a_j a_i a_k^+ a_l^+ = a_k^+ a_l^+ a_j a_i - \delta_{jl} a_k^+ a_i + \delta_{il} a_k^+ a_j + \delta_{jk} a_l^+ a_i - \delta_{ik}a_l^+ a_j +\delta_{ik}\delta_{jl} - \delta_{jk}\delta_{il}$$

This is called "normal order". Note that, when the operator acts on the vacuum state, only the last two addenda survives.

Explanation on anticommutation relations

2 Answers2