I want to compute the following trace $$ \operatorname{tr}(\exp(-\sum_k\varepsilon_k a_k^\dagger a_k)a_i a_j) $$ with bosonic operators $a_i,a_j$. I think the result will be proportional to $\delta_{i,j}$ but I have no idea how to do this precisely. Any help is appreciated.
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1 Answer
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On the trace, the operator $\exp(-\sum_k\varepsilon_ka^\dagger_k a_k)$ is diagonal while for, any state $|n\rangle$, $\langle n|a^2|n\rangle=0$, so that sums up to zero.
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$\begingroup$ Thanks for your answer. But wouldn't this imply that correlations of the form $\langle a_i a_j\rangle$ always vanish and we are back in the fermionic regime? $\endgroup$– julianCommented Jan 14, 2018 at 17:13
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$\begingroup$ Not exactly. What determines being bosons or fermions is the positivity of the energy and so, the way $a^\dagger,\ a$ commutes or anti-commutes each other. $\endgroup$– JonCommented Jan 14, 2018 at 17:17