While trying to solve a second quantization exercise regarding a bosonic gas, I've been having trouble trying to understand the $(1-\delta_{\text{pq}})$ term in the decomposition of the following matrix element (which leads to the pair correlation function) from Gordon Baym's "Lectures on Quantum Mechanics", (p. 430, eq. 19-80):

$$ \langle \Phi |a_\text{p}^\dagger a_\text{q}^\dagger a_\text{q'} a_\text{p'}|\Phi\rangle = (1-\delta_\text{pq})\left[\delta_\text{pp'}\delta_\text{qq'}\langle \Phi |a_\text{p}^\dagger a_\text{q}^\dagger a_\text{q} a_\text{p}|\Phi\rangle+\delta_\text{pq'}\delta_\text{qp'}\langle \Phi |a_\text{p}^\dagger a_\text{q}^\dagger a_\text{p} a_\text{q}|\Phi\rangle\right]+\delta_\text{pq}\delta_\text{pp'}\delta_\text{qq'}\langle \Phi |a_\text{p}^\dagger a_\text{p}^\dagger a_\text{p} a_\text{p}|\Phi\rangle ,$$ where $| \Phi \rangle$ is a Fock state.

What I can't grasp from here is why do we get a minus sign if we are treating with bosons, which work with a commutating algebra. Also, the only term which has 3 deltas that I can relate to the original matrix element is the last summand, which is self-explanatory, but I don't understand how the rest would be.

Maybe there's another way of doing this that could be more ink-wasting but more comprehensible, I'm open to every procedure, not necessarily the way Gordon Baym does it.

  • $\begingroup$ What is $|\Phi\rangle$? $\endgroup$ Nov 29, 2021 at 7:49
  • $\begingroup$ @Jakob It's a Fock state, the question has been updated to include it. Thanks. $\endgroup$ Nov 29, 2021 at 10:31
  • $\begingroup$ But a Fock state with definite particle number?! $\endgroup$ Nov 29, 2021 at 11:23
  • $\begingroup$ Yes @Jakob, $| \Phi \rangle = |n_{\text{p}_0}, \ n_{\text{p}_1}, \ n_{\text{p}_2}, \ ... \rangle$ $\endgroup$ Nov 29, 2021 at 11:58

1 Answer 1


In the text directly above the formula Baym mentions the cases in which this amplitude is non-vanishing are $p=p',q=q'$ and $p=q',q=p'$. This follows from the fact that $\vert\Phi\rangle$ is of the form $\vert n_{p_0},n_{p_1},...\rangle$ because if we don't have one of the two cases he mentiones we are taking the inner product of two states with at least one $n_{p_i}$ being different and such states are orthogonal.

It is also pointed out that these two cases are the same if $p=q$ and we shouldn't count them twice. So we first have to make a distinction between $p=q$ which is the last term of the equation (because of the $\delta_{pq}$) and $p\ne q$ for which the first two terms are responsible because the $(1-\delta_{pq})$ is exactly 1 if $p\ne q$ and 0 if $p=q$.

So basically to derive the result one starts by writing $$\langle \Phi \vert a_p^\dagger a_q^\dagger a_{p'} a_{q'}\vert \Phi\rangle=(1-\delta_{pq}+\delta_{pq})\langle \Phi \vert a_p^\dagger a_q^\dagger a_{p'} a_{q'}\vert \Phi\rangle=...$$ It has nothing to do with there being bosons or fermions, instead one only wants to seperately look at the cases $p=q$ and $p\ne q$.


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