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In Schwabls Book "Advanced Quantum Mechanics" in the chapter for Bosons he calculates the Bosonic pair distribution function for noninteracting bosons. He said the expectation value of \begin{align*} \left\langle\phi\left|a_{\mathbf{k}}^{\dagger} a_{\mathbf{q}}^{\dagger} a_{\mathbf{q}^{\prime}} a_{\mathbf{k}^{\prime}}\right| \phi\right\rangle \end{align*} differs from zero only if $k = k′$ and $q=q′$,or $k=q′$ and $q=k′$.The case $k=q$,which,in contrast to fermions, is possible for bosons, has to be treated separately. Then he comes to the following equation: \begin{align*} \begin{array}{l}{\left\langle\phi\left|a_{\mathbf{k}}^{\dagger} a_{\mathbf{q}}^{\dagger} a_{\mathbf{q}^{\prime}} a_{\mathbf{k}^{\prime}}\right| \phi\right\rangle} \\ {=\left(1-\delta_{\mathbf{k} \mathbf{q}}\right)\left(\delta_{\mathbf{k} \mathbf{k}^{\prime}} \delta_{\mathbf{q q}^{\prime}}\left\langle\phi\left|a_{\mathbf{k}}^{\dagger} a_{\mathbf{q}}^{\dagger} a_{\mathbf{q}} a_{\mathbf{k}}\right| \phi\right\rangle+\delta_{\mathbf{k} \mathbf{q}^{\prime}} \delta_{\mathbf{q} \mathbf{k}^{\prime}}\left\langle\phi\left|a_{\mathbf{k}}^{\dagger} a_{\mathbf{q}}^{\dagger} a_{\mathbf{k}} a_{\mathbf{q}}\right| \phi\right\rangle\right)} \\ {+\delta_{\mathbf{k q}} \delta_{\mathbf{k} \mathbf{k}^{\prime}} \delta_{\mathbf{q} \mathbf{q}^{\prime}}\left\langle\phi\left|a_{\mathbf{k}}^{\dagger} a_{\mathbf{k}}^{\dagger} a_{\mathbf{k}} a_{\mathbf{k}}\right| \phi\right\rangle}\end{array} \end{align*} where I don't understand how the argument fits to the equation.

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  • $\begingroup$ I am not sure exactly what you are asking. What do you mean "how the argument fits to the equation"? $\endgroup$ – By Symmetry Apr 1 at 12:55
  • $\begingroup$ I don’t understand where exactly the argument for the case q=k‘ fits into the different terms in the equation. I mean where does the $(1-\delta)$ come from? For example. I would say one needs only the equation inside the first brackets and the last term. $\endgroup$ – Leviathan Apr 1 at 13:16
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    $\begingroup$ $(1-\delta_{kq})$ is non-zero iff $k\ne q$, in which case it is $1$. In your question you list 3 distinct cases, $k=k'\ne q=q'$, $k=q' \ne q = k'$ and $k=k'=q=q'$. If you look at the assorted delta functions in front of each term you will find they correspond to exactly these 3 cases. $\endgroup$ – By Symmetry Apr 1 at 13:34
  • $\begingroup$ Makes sense thanks :) $\endgroup$ – Leviathan Apr 1 at 13:50

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