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.

  • $\begingroup$ I assume that in their formula the annihilation operator at $\bf{p}$ is also at time $t$ and not $t=0$? $\endgroup$
    – user245141
    Feb 10, 2020 at 14:35
  • $\begingroup$ Thank you, you are correct. I have updated the question to correct this. $\endgroup$ Feb 10, 2020 at 14:39

1 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

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

  • $\begingroup$ Unfortunately this does not work out. I do not get the factor $1/V$ in the end when doing this. If I do everything carefully from the start I get the following relation between the Green's functions: $ G^{<} (\mathbf{p}, \omega, \mathbf{p}', \omega') = 2 \pi V \delta(\omega - \omega') \delta_{\mathbf{p}, \mathbf{p}'} G^{<} (\mathbf{p}, \omega)$. If I carefully do everything you suggest from the start (switching to Wigner coordinates, etc) I do not get the volume factor. $\endgroup$ Feb 12, 2020 at 10:04
  • 1
    $\begingroup$ I've elaborated the calculation. Is there still something unclear? $\endgroup$
    – user245141
    Feb 12, 2020 at 12:10
  • $\begingroup$ Thank you for clarifying. The problem I have is that I believe that $\psi(\mathbf{p}) = \frac{1}{\sqrt{V}} \int d^3 r e^{i \mathbf{p}\cdot \mathbf{r} } \psi(\mathbf{r})$, and similarly for $\psi^\dagger$. This introduces a factor $1/V$ in your first equation, which removes the factor $V$. If this is correct, I am back to the same problem again. $\endgroup$ Feb 12, 2020 at 12:53
  • $\begingroup$ This is a strange way to define the FT, and I think that it is a result of mixing discreet FT, where the operators are dimensionless and indeed $c_p = \sum c_x e^{ikx}/\sqrt{V}$ and continuous FT, where $\psi(r)$ has dimensions of $1/\sqrt{\rm{distance}^3}$ and $\psi(p)$ dimensions of $1/\sqrt{\rm{momenta}^3}$ and $\psi(p) = \int\! d^3 r e^{ipr} \psi(r)$ (up to $2\pi$ factors) $\endgroup$
    – user245141
    Feb 12, 2020 at 13:12
  • 1
    $\begingroup$ yes you are correct (and previously I had a typo). But when you want to go to the continuum limit you want to make $\psi(p)$ dimensionfull (as $\{ \psi(p), \psi(p') \}$ becomse the dimensionfull dirac-delta instead of the dimensionless kronecker). So you define $\psi(p) \to \sqrt{V}\psi(p)$. And then you get $\psi(x) = 1/V \sum_p e^{-ipr} \psi(p)$ and can take $V\to \infty$ and make the sum an integral (with $1/V = d^3p$) $\endgroup$
    – user245141
    Feb 12, 2020 at 13:36

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