Pair correlation function for non-interacting spinless bosons

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.

• What is $|\Phi\rangle$? Nov 29 '21 at 7:49
• @Jakob It's a Fock state, the question has been updated to include it. Thanks. Nov 29 '21 at 10:31
• But a Fock state with definite particle number?! Nov 29 '21 at 11:23
• Yes @Jakob, $| \Phi \rangle = |n_{\text{p}_0}, \ n_{\text{p}_1}, \ n_{\text{p}_2}, \ ... \rangle$ Nov 29 '21 at 11:58

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 $$p\ne q$$.