# Bosonic Pair Distribution Function

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.

• I am not sure exactly what you are asking. What do you mean "how the argument fits to the equation"? – By Symmetry Apr 1 at 12:55
• 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. – Leviathan Apr 1 at 13:16
• $(1-\delta_{kq})$ 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. – By Symmetry Apr 1 at 13:34
• Makes sense thanks :) – Leviathan Apr 1 at 13:50