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If one takes the Fourier transform w.r.t. space of the following Green function

$$G(x,t;x',t')=-i\langle T\{\psi(x,t)\psi^\dagger(x',t')\}\rangle$$

one ends up with an expression like

$$G(k,t;k',t')=-i\langle T\{ a^{\phantom{\dagger}}_k(t)a^\dagger_{k'}(t')\}\rangle$$

where the $a$ operators create/destroy a particle in the respective momentum states.

My questions are the following: I know that in homogeneous space momentum is conserved and it should be $k=k'$ in the above expression.

Is there a rigorous argument to see this by which one can avoid setting $k=k'$ by hand?

What happens for Green functions when space is no longer homogeneous (e.g. in the presence of impurities or in a crystal)? Is the FT of the GFs still a "good" object?

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Momentum conservation is a consequence of shift invariance. (Thanks to Noether's theorem we know that continuous symmetries are related to conserved quantities.) Shift invariance means that the Green function can only depend on the relative displacement. In other words, (ignoring $t$) we would have $$ G(x,x') = G(x-x') . $$ In this sense $G(x)$ is the autocorrelation function of the field. Now there is another theorem, the Wiener-Khinchin theorem that states that the Fourier transform of the autocorrelation function is the power spectral density $$ {\cal F}\{G(x)\} = S(k) . $$ The power spectral density is the modulus squared of the spectrum of the field $$ S(k) = |{\cal F}\{ \psi(x) \}|^2 . $$ So we see that it contains only one $k$, which implies that in effect $k=k'$.

Now what about the $t$ that we ignored? Nature is also shift invariant in time, which leads to energy conservation. Hence, a similar analysis can be followed for the temporal degrees of freedom.

When space is not homogeneous, so that the shift invariance is lost, then momentum conservation would also be lost. Then one would not be able to set these quantities equal.

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