Thermodynamic limit and periodic boundary conditions

Question

Background: Concerning the use of periodic boundary conditions for systems of $N$ (identical) particles, Ref. 1 considers a cube $\Lambda$ with volume $L^3$ and notes regarding the interaction potential of the form $V(x_1-x_2)$:

It is often useful, for calculations, to express $V$ in the plane-wave basis $|k\sigma\rangle$ with periodic boundary conditions. The Fourier transform of $V(x)$ in a cubic box $\Lambda$ is given by $$ \tilde V_L(k)=\int_\Lambda \mathrm dx\, e^{-ikx} \, V(x) \tag{3.111}$$ with corresponding Fourier series $$V_L(x) = \frac{1}{L^3} \sum\limits_k e^{ikx}\, \tilde V_L(k) \quad .\tag{3.112} $$ Note that $V_L(x)$ is a periodic function with period $L$ and is only equal to $V(x)$ if $x$ is in $\Lambda$. In the limit of infinite volume, however, $\tilde V_L(k)$ approaches the Fourier transform $\tilde V(k)$ of $V(x)$ in the whole space, $$ \lim\limits_{L\to\infty}\tilde V_L(k) = \int \mathrm d x\, e^{-ikx}\, V(x) = \tilde V(k) \tag{3.113}$$ and $$\lim\limits_{L\to\infty} V_L(x)=V(x) \quad .\tag{3.114} $$

The authors then proceed to express the two-body interaction in second quantization and again note that $V_L(x_1-x_2)$ is not equal to $V(x_1-x_2)$, only if $x_1-x_2 \in \Lambda$. But again they argue that in the limit of infinite volume, we recover the original potential. All of this more or less makes sense to me, but I have one doubt.

Question: It seems that it matters what cube exactly we take. Indeed, I see that if we take $\Lambda=[-L/2,L/2]^3 \subset \mathbb R^3$, then the limits in $(3.113)$ and $(3.114)$ yield the desired results. More generally, we can take some "sequence" of cubes which "converge" to $\mathbb R^3$, a notion which I think can be made rigorously.

However, by taking $\Lambda=[0,L]$ instead, I cannot see that this is true in general. First, it does not reproduce the Fourier transform in equation $(3.113)$. Moreover, for such an interval (I think for any asymmetric interval), the periodically extended interaction potential (cf. Ref. 4) $V_L$ is not symmetric anymore, which should be the case for identical particles.

But the authors do nowhere assume some special choice of $\Lambda$. Other References (see below) proceed and argue in the same manner.

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References:

1. Many-Body Problems and Quantum Field Theory: An Introduction. Martin and Rothen. Section 3.2 page 111 and section 4 pages 135-137.

2. Nonequilibrium many-body theory of quantum systems. Stefanucci and Leuuwen. Appendix E, page 529.

3. Response Theory of the Electron-Phonon Coupling. Starke and Schober. Sections 2.2.1 and A.5. arxiv link.

4. Theoretical Solid State Physics In two Volumes. Volume 1. Albert Haug. Chapter II.B, section 24 (a), page 200.

5. Closely related PSE post: Operators and periodic boundary conditions

Answer 1

You could construct $V_L$ in a first naive way without using a unit cells. Starting with $V$ defined on $\mathbb R^3$, take a Bravais lattice $\Lambda$ in real space (sorry for the conflict, but this is standard notation). The idea is by taking the lattice $L\Lambda=\{Lx|x\in\Lambda\}$ with $L\to\infty$, you should recover the infinite space limit.

There are two ways of viewing the construction of $V_L$. In Fourier space, this is done by sampling the Fourier transform: $$ \tilde V_L(k) = \sum_{l\in \frac{1}{L}\Lambda^*} \tilde V(l)\delta(k-l) $$ where $\frac{1}{L}\Lambda^*$ is the dual lattice of $L\Lambda$. Note in particular that the dependence in $L$ is inverted when going to dual space, and that the sampling lattice gets finer as $L\to\infty$ which is why: $$ \tilde V_L \to \tilde V $$

In real space, this is done by the Poisson summation formula: $$ V_L(x) = \sum_{y\in L\Lambda}V(x-y) $$ Intuitively, if $V$ decays fast, you'll only have one term left. You can prove that under more general assumptions over $V$: $$ V_L\to V $$

You can check that that both approach are related by Fourier transform.

However, if you know a bit of sampling, it's best to filter out high frequencies before the sampling in order to avoid aliasing. This is how you recover your method and the unit cell comes into play. The idea is to use a brick-wall filter over a unit cell of $\Lambda$ (real space) $\Omega$, to that $V$ becomes $V_{L\Omega}:=V1_{L\Omega}$ (indicator function). This is why when you calculate the Fourier transform, you only integrate over the finite domain. In Fourier space, $\tilde V_{L\Omega}$ is a smoothed out version of $\tilde V$.

Having done this filtering, you can now proceed to sampling in dual space, or equivalently periodising in real space. It turns out the thanks to the filtering, the next step loses no information thanks to the the higher dimensional version of the Sannon-Wittaker interpolation formula in dual space or by direct inspection in real space. $V_L$ is no given by: $$ V_L(x) = \sum_{y\in L\Lambda}V_{L\Omega}(x-y) \\ \tilde V_L(k) = \sum_{l\in \frac{1}{L}\Lambda^*} \tilde V_{L\Omega}(l)\delta(k-l) $$

In order to recover the correct limiting behavior $V_L\to V$, one approach is to impose that $L\Omega\to\mathbb R^3$. Indeed, $V_L$ and $V$ coincide on $L\Omega$, so this is sufficient. $L\Omega\to\mathbb R^3$ is equivalent to $\Omega$ being a neighborhood of the origin. This is why your counter example failed since the origin was at the boundary.

The fact that $\Omega$ is a unit cell is necessary and sufficient for the sampling/periodisation to be invertible. However, it is not entirely relevant if all you are interested in is the limit. You can choose $\Omega$ to be rather arbitrary and relax the condition of being a unit cell of $\Lambda$. The key argument is that $V_L,V$ coincide in a neighbourhood of the origin. A natural generalisation would be that the complementary set of $(\Lambda-\{0\})+\Omega$ is a neighbourhood of the origin. This guarantees that the periodisation does not generate an overlap at the origin.

Btw, you can generalise the approach by considering different filters instead of brick wall filters.

Hope this helps.