• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

198 submissions , 156 unreviewed
4,911 questions , 2,087 unanswered
5,312 answers , 22,547 comments
1,470 users with positive rep
806 active unimported users
More ...

  Fourier transform of the G-lesser function.

+ 1 like - 0 dislike

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.

I have tried to redo everything carefully from scratch and I have found for example the relation

$$ G^{<} ( \mathbf{p}, \omega, \mathbf{p}', \omega' ) = 2\pi V \delta(\omega - \omega') \delta_{\mathbf{p}, \mathbf{p}'} G^{<} ( \mathbf{p}, \omega). $$

However, I cannot really get the factor $1/V$ which was obtained in the book.

Closed by author request
asked Feb 12, 2020 in Theoretical Physics by JezuzStardust (5 points) [ revision history ]
closed Feb 26, 2020 by author request

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights