Fourier transform of the G-lesser function

Question

I am studying Kadanoff & Baym's book Quantum Statistical Mechanics and I am stuck a one point.

The are considering a system of non-interacting particles, (let's say fermions to not having to write both signs), and are then considering the G-lesser function:

$$ G^{<} (1, 1') = i \left< \psi^\dagger (1') \psi(1) \right>, $$

where $ 1 = \mathbf{r}_1, t_1 $ and similarly for $1'$.

Since the Hamiltonian has rotational and translational symmetry they argue that the Green's function above only depends on $| \mathbf{r}_1 - \mathbf{r}_{1'} |$. Also, since the Hamiltonian is time independent the Green's function should only depend on the time difference $t_1 - t_1'$. All this seems fine and I think I have sucessfully convinced myself of these facts by considering e.g. the translation operator.

However, they then define the Fourier transform as

$$ G^{<} ( \mathbf{p}, \omega) = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} G^{<}(\mathbf{r}, t), $$

where we now use $ \mathbf r = \mathbf r_1 - \mathbf r_2$ and similarly for $t$. Now come the claim that I cannot really see. They say that, due to the invariances I talked about above, we have

$$ G^{<}(\mathbf{p} , \omega) = \int dt \frac{e^{i\omega t}}{V} \left< \psi^\dagger(\mathbf{p}, 0) \psi(\mathbf{p}, t) \right> ,$$

where $V$ is the volume of the system. Can someone please explain how this follows from the above? If I naively try to calculate this I instead get

$$ G^{<} ( \mathbf{p}, \omega) = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} G^{<}(\mathbf{r}, t) \\ = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} \left< \psi^\dagger(\mathbf{0}, 0) \psi(\mathbf{r}, t ) \right>, $$

which would only give the Fourier transform of the annihilation operator.

Answer 1

The way forward, I think, is to go in the other direction. Start from the r-h-s, and plug in the FT for $\psi(\mathbf{p})$ and $\psi^{\dagger}(\mathbf{p})$. You will get two integrals, one over $r_1$ and the other over $r_2$

$$ \langle \psi^{\dagger}(\mathbf{p})\psi(\mathbf{p}) \rangle = \int\! d^3r_1 d^3r_2 e^{-i\mathbf{p(r_1-r_2)}} \langle \psi^{\dagger}(\mathbf{r}_1)\psi(\mathbf{r}_2) \rangle$$

Now you can do the substitution $r_1-r_2=r$, $r_1+r_2 = 2R$, and as the correlation function will depend only on $r$, you can integrate over $R$ immediately and get the factor $V$.

Edit: detailed calculation. The time part is completely independent so I will ignore it. Let's start with the r-h-s of the expression that the authors get and work our way from there $$ \langle \psi^{\dagger}(\mathbf{p})\psi(\mathbf{p}) \rangle = \int\! d^3r_1 d^3r_2 e^{-i\mathbf{p(r_1-r_2)}} \langle \psi^{\dagger}(\mathbf{r}_1)\psi(\mathbf{r}_2) \rangle = \int\! d^3R d^3r e^{-i\mathbf{p r}} G^<(\mathbf{r}) = V \int\! d^3r G^<(\mathbf{r}) = V G^<(\mathbf{p})$$ where we used the fact that the Jacobian of the transformation is $1$ and that $\int\! d^3R = V$. The last equation is simply the definition of the FT of the Green function when written in relative coordinates. So we get

$$ G^<(\mathbf{p}) = \frac{1}{V}\langle \psi^{\dagger}(\mathbf{p})\psi(\mathbf{p})\rangle$$