While reading Altland and Simons book (Condensed matter field theory p.52), I came across the following problem. We consider the simplest electron gas model with interaction term

$ H_{int} = \int d^d r a^{\dagger}_{\sigma} (r) V(r) a_{\sigma} (r) + \frac{1}{2} \int d^d r \int d^d r’ V_{ee} a^{\dagger} _{\sigma} (r) a^{\dagger}_{\sigma’} (r’)a_{\sigma’} a_{\sigma} (r) $

Where the first term is the interaction between the electrons and the background positive ions, the second term is the electron interactions.Then we could do Fourier transformation to represent the interaction in the momentum space, where the electron-electron interaction is

$H_{ee}= \frac{1}{2V} \sum_{k,k’,q} V_{ee} (q) a^{\dagger} _{k-q \sigma} a^{\dagger}_{k’+q \sigma’} a_{k’\sigma’} a_{k\sigma}$

And the electron-ion interaction:

$ H_{ie} = \sum_{k,q} V_{ie}(q) a^{\dagger}_{k +q \sigma} a_{k \sigma} $

The author then says that after doing Fourier transformation, the zero momentum term of the electron interaction is cancelled by the electron-background ion interaction. To see this we could calculate the interaction when $q= 0$. We have

$H_{ee} (q=0)= \frac{1}{2V} \sum_{k,k’} V_{ee} (0)n_{k \sigma} n_{k’ \sigma’}$

At this point I still don’t know why the positive background cancels the zero momentum interaction between the electrons, can anybody show how this result is derived?


The $q=0$ interactions corresponds to summing up all the charges of the electrons in the system (you sum over all $k$, without any spatial modulation) $$ \sum_{k} n_k = \sum_{r} n_r = N$$ therefore this is just a constant, which is independent of any dynamics or spatial configurations of the Hamiltonian. As we know that the total charge of the system itself is neutral, we can even further generalize that this interaction contributes zero.

Contrast that with some non-zero component $q \neq 0$, where the spatial organization of the charges will leave some areas more positively and other areas more negatively charged, and this will contribute to the energy via the Coulomb interaction.

  • $\begingroup$ Thanks, but can you be more clear? I also saw a lot of “neutral charge” argument in the book, but how is this argument relates to the cancellation of the zero momentum electron interaction with the electron-ion interactions. I also add the electron-ion interaction calculation in the question to be more clear. $\endgroup$
    – Jiahao Fan
    Feb 21 '20 at 17:16
  • $\begingroup$ More specifically, in my expression for electron-electron interaction above, I can only see the term is proportional to the “square” of the electron density rather than the density itself. $\endgroup$
    – Jiahao Fan
    Feb 21 '20 at 17:24
  • $\begingroup$ I'm sorry but I don't understand what's unclear to you in my answer. For $q=0$ we get $\sum_{k,k'} n_k n_{k'} = \sum_k n_k \sum_{k'} n_{k'} = N^2$ while for nonzero $q$ $\sum_{k,k'} a^{\dagger}_{k-q}a_k \sum_{k'} a^{\dagger}_{k'+q}a_{k'} \neq N^2$ where $N$ is the total number of electrons in the system. $\endgroup$
    – user245141
    Feb 22 '20 at 8:43

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