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Can someone show me how the first order coherence function $G^1(r,r')\equiv \left \langle \hat{\Psi}(r)\hat{\Psi}(r') \right \rangle $ for a system of bosons is related to the momentum distribution function $n(p)$, the diagonal elements of the density matrix in momentum representation, via:

$G^1(r,r')=\frac{1}{V}\int dp n(p)exp\left [ \frac{i}{\hbar}p(r-r') \right ]$

My problem with the derivation given here (eq.2.27) is that I don't know where the delta function $\delta(p-p')$ assumed to reduce the integral over $p'$ comes from.

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up vote 2 down vote accepted

I believe, that the derivation is wrong...

If you assume a translationally invariant state, such that $G^1(r, r') = G^1(r - r')$ then you can get the result. Rewrite the exponential as $p r - p' r' = p( r- r') + r'(p - p')$. Since, in this case, the left-hand-side of Eq. (2.27) can only depend on $r - r'$ it must be such that $p = p'$ from the second term. This gives you the delta-function in $(p - p')$ and you get the stated result. To be more precise, the delta-function comes from $\langle \psi(p)^\dagger \psi(p')\rangle$ which can only depend on the relative momentum and is therefore proportional to $\delta(p - p')$ - by translational invariance.

I do not believe the result holds for states without translational invariance (such as a small, trapped ultra-cold quantum gas).

See for example Buus & Flensberg ( appendix A.5 (the pages happens to be available in the amazon preview).

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