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In Fundamentals of Many-body Physics by W. Nolting, 1e, the author arrives at the following formula for the electron-electron contribution to the Hamiltonian of Jellium:

$$ \hat{\mathcal{H}}_{ee}=\frac{2\pi q^2}{\alpha^2 V}\left(\hat{N}^2-\hat{N}\right) $$

where $q$ is the electronic charge, $\hat{N}$ is the particle number operator, and $\alpha$ is a small parameter needed to make the integrals converge. The term involving $\hat{N}^2$ exactly cancels other divergent contributions to the Hamiltonian. Regarding the term linear in $\hat{N}$, the author says:

The [term linear in $\hat{N}$] leads to an energy per particle which vanishes in the thermodynamic limit.

I understand $2\pi q^2/\alpha^2 V$ (ie. the energy per particle) vanishes if we first send $V\rightarrow\infty$, then send $\alpha\rightarrow 0$. But if we commute the limits, it will diverge. How do we know $V\rightarrow\infty$ then $\alpha\rightarrow 0$ is the correct order in which to take the limits?

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To quote the standard reference on the Jellium:$^1$

The physical hamiltonian is recovered by letting $\kappa \to 0$ after going to the thermodynamic limit $L\to \infty$. The procedure is justified a posteriori since the physical quantities obtained by taking the limits in this order are finite and independent of $\kappa$.

But in principle you don't have to introduce the regularization, although it helps in deriving the Fourier transform of the Coulomb potential, see e.g. this PSE post. References 2 and 3 derive the Jellium Hamiltonian (in the thermodynamic limit) without the "issue" of the order of the limits. In reference 4, the author, after proceeding just as Nolting, states that

The reader may feel uneasy about the results we obtained; they rely on the mathematical artifact of introducing an exponential damping term, and on the sequence in which the limits are taken. The reader may rest assured that the results are correct. In fact, the same results are obtained without introducing the exponential term.

and then derives the results in the same spirit as references 2 and 3.

Very roughly, the idea is to work with a finite volume and periodic boundary conditions. This is also done in reference 1, but there the Fourier coefficients (in the series of the Coulomb potential) are replaced by the Fourier transform, making the regularization necessary due to the $k=0$ term. Note that this replacement is fine in the thermodynamic limit, see e.g. this PSE post.

Instead, e.g. reference 3 makes this replacement only for the $k\neq 0$ terms (eq. $(10.1)$ therein), and then shows that the $k=0$ contributions (finite for finite volume) in the total Hamiltonian vanish in the thermodynamic limit.

References:

  1. Quantum Theory of the Electron Liquid. G. Giuliani and G. Vignale. Section 1.3.2, p. 14
  2. Many-Body Problems and Quantum Field Theory. An Introduction. P. Martin and F. Rothen. Section 4.2.1, p.133
  3. Many-Particle Theory. E. Gross and E. Runge. Chapter 10, p.79
  4. Feynman Diagram Techniques in Condensed Matter Physics. R. Jishi. Section 4.1, p. 68
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    $\begingroup$ Absolutely shocking, Sir. I have a signed copy of Vignale; wonderful man. But when you said that there is no need to introduce the regularisation for Fourier transforming the Coulomb potential, did you mean Pablo's answer in that link of yours? I'll have to get your two references... $\endgroup$
    – naturallyInconsistent
    Commented Jul 18 at 3:48
  • $\begingroup$ @naturallyInconsistent No, not really. I meant the following. You can go through the derivation of Jellium Hamiltonian using PBC in a finite volume and leaving the Fourier coefficients as they are. The $v_{q=0}$ coefficient will cancel out from the various terms, except something like $-\frac{N}{2V}v_{q=0}$ (which corresponds to the second term in the question). One can show that this vanishes in the TDL (if seen per particle). Then you arrive at form of the Hamiltonian with the $q=0$ term missing. Now you can replace the coefficient by the Fourier transform. $\endgroup$
    – Tobias Fünke
    Commented Jul 18 at 5:40
  • $\begingroup$ Well but yes, using the regularization to arrive at the form of the FT of the Coulomb potential is the most common way, in my experience. Pablo's answer looks fine, from my naive point of view, too. ---My point above was that one does not need the regularization for the Jellium derivation, because $v_{q=0}$, the series coefficient, is finite (and, as explained, will cancel out). How to derive the Fourier transform of the Coulomb potential is another thing. $\endgroup$
    – Tobias Fünke
    Commented Jul 18 at 5:57
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    $\begingroup$ I suppose the generic trick is to punch out the $q=0$ term in all the terms that have it, and analytically cancel them, and then the rest of the stuff are all well-defined. $\endgroup$
    – naturallyInconsistent
    Commented Jul 18 at 6:13

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