I have a question about the deduction of interacting term in Jellium model. In the text book Condensed Matter Field Theory ed.2 Alexander Altland, Ben Simons, pg.52. Author gives the expression of e-e interaction without deduction details:

$$ V_{ee}=\sum_{k_1,k_2,q} a^\dagger_{k_1-q \sigma_1} a^\dagger_{k_2+q \sigma_2} \frac {e^2}{q^2}a_{k_2 \sigma_2}a_{k_1 \sigma_1} \tag 1 $$

I try to get this form via inserting the Fourier expansions of each term into the "classical" Coulomb interaction expression

$$ V_{ee}=\int dr_1dr_2 \ a^\dagger_{r_1 \sigma_1} a_{r_1 \sigma_1} \frac {e^2}{r_{12}} \ a^\dagger_{r_2 \sigma_2} a_{r_2 \sigma_2} \tag 2 $$


\begin{align} a^\dagger_{r} & = \sum_{k} e^{-ikr}a^\dagger_k \\ a_r & = \sum_k e^{ikr}a_k \\ \frac {1}{r} & =\sum_q e^{iqr}\frac {1}{q^2} \end{align}

When I have done this, I got an expression with different order of operators. This is mine: $$ V_{ee}=\sum_{k_1,k_2,q} a^\dagger_{k_1-q \sigma_1} a_{k_1 \sigma_1} \frac {e^2}{q^2} a^\dagger_{k_2+q \sigma_2} a_{k_2 \sigma_2} \tag 3 $$

When $q \neq k_1-k_2$ or $\sigma_1 \neq \sigma_2$, eq.3 can be re-ordered thus becomes identical to eq.1. However, if $k_1=k_2+q$ and $\sigma_1=\sigma_2$, the re-order cannot be done as before since $[a_{k_1 \sigma_1},a_{k_1 \sigma_1}^\dagger]=1$. Therefore, I cannot reach eq. 1?

So, what's wrong? Any suggestions are appreciated.


1 Answer 1


Suppose that you have two quartic terms with different orderings, like $a^\dagger a^\dagger a a$ and $a^\dagger a a^\dagger a $, assuming that total momenta and Lorentz spins etc. of the oscillators sum to zero. As you say, if you try to re-order one of them you "wrong" terms via the commutator $[a^\dagger,a]=1$, like

$$g \sum \ a^\dagger a^\dagger a a \sim g \sum a^\dagger a a^\dagger a + g \sum a^\dagger a.$$

The "wrong" term you get just renormalizes the coupling in front of the mass term $$\int dr\, a_r^\dagger a_r$$ that you have in your Hamiltonian. So it's nothing to worry about.


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