# Operators and periodic boundary conditions

Background:

In Ref. 1, a system of $$N$$ (identical) fermions is considered. The system is enclosed in a cubic box of volume $$\Omega=L^3$$ and periodic boundary conditions are employed, that is (I'll change and ease the notation a bit): $$\langle x_1\ldots x_j+L_i\ldots x_N|\Psi\rangle = \langle x_1\ldots x_j\ldots x_N|\Psi\rangle \tag{E.2}$$ for all $$j=1,\ldots,N$$ and $$i=1,2,3$$. It is then stated that:

Although these conditions seem natural they are not trivial in the presence of many-body interactions. This is because the many-body interaction is invariant under the simultaneous translation of all particles, i.e., $$v(x − x^\prime) = v((x + L_i ) − (x^\prime + L_i ))$$ but not under the translation of a single particle. The Hamiltonian does not therefore have a symmetry compatible with the boundary conditions (E.2). To solve this problem we replace the two-body interaction by $$v(x − x^\prime ) = \frac{1}{\Omega} \sum\limits_k\, \tilde v_k\,e^{ik(x-x^\prime)} \quad , \tag{E.3}$$ [...] With this replacement and the BvK boundary conditions $$(\mathrm{E.2})$$ the eigenvalue equation for the Hamiltonian $$H$$ becomes well-defined (of course the spatial integrations in $$(\mathrm{E.1})$$ must be restricted to the box).

The relevant equation of the Hamiltonian is $$H= \int\mathrm dx\, \psi^\dagger(x) \, \left(-\nabla/2 -V(x)\right)\,\psi(x)\,\tag{E.1} + \frac{1}{2} \int \mathrm dx\,\mathrm dy \,v(x,y)\, \psi^\dagger(x)\,\psi^\dagger(y)\, \psi(y)\,\psi(x) \quad ,$$

where, by the usual abuse of notation, $$v(x,y)=v(x-y)$$.

Question:

I wonder what exactly the authors mean by saying that The Hamiltonian does not therefore have a symmetry compatible with the boundary conditions $$(\mathrm{E.2})$$ and why the replacement is necessary.

My understanding is that the boundary conditions do not fix the domain of $$H$$, but instead mean that the single-particle Hilbert space is $$\mathfrak h=L^2(\mathbb T^3)$$ (neglecting spin) instead of $$L^2(\Omega)$$ and the corresponding $$N$$-particle space is $$H_N:= \wedge^N \mathfrak h$$, i.e. the $$N$$-fold antisymmetric tensor product. This is for example used (for a non-interacting system) in Ref. 2.

So why should the lack of a symmetry (I guess the authors mean that the Hamiltonian is not invariant under the application of the (discrete) translation operator) of the Hamiltonian be problematic? Do the authors claim that $$H$$ in $$(\mathrm{E.1})$$ without the replacement of $$v$$ (but with integration restricted to the volume $$\Omega$$) is not an operator on $$H_N$$?

I guess/have the feeling that one can also rephrase the question for a non-interacting system (even of a single-particle) with a non-periodic external potential. See also this related MathSE question and answer.

References:

Ref. 1: Nonequilibrium many-body theory of quantum systems. Stefanucci and Leuuwen. Cambridge University Press. Appendix E, page 529.

Ref. 2: Mathematical Quantum Theory. Lecture notes 2019. Marcello Porta. Section 9.4.1, page 103. A PDF can be found here.

Your intuition is right. The salient detail can be understood by studying a single particle on a ring, with Hilbert space $$L^2(\mathrm S^1)$$.

The first thing to do is consider how we write down a function $$\psi:\mathrm S^1\rightarrow \mathbb C$$, for which there are at least two approaches.

1. We could define two overlapping coordinate charts on $$\mathrm S^1$$ - say, $$\theta : \mathrm S^1 \rightarrow (-\pi,\pi)$$ and $$\phi: \mathrm S^1\rightarrow (0,2\pi)$$ - and then study $$\psi_\theta := \psi \circ \theta^{-1}$$ and $$\psi_\phi := \psi \circ \phi^{-1}$$.
2. We could define the equivalence relation $$\sim$$ on the interval $$[0,2\pi]$$ such that $$\forall x, x\sim x$$ and additionally $$0\sim 2\pi$$. From there, we could model $$\mathrm S^1 = [0,2\pi]/\sim$$.

Here we take the second approach. Let $$q:[0,2\pi] \rightarrow \big([0,2\pi]/\sim\big) \equiv \mathrm S^1$$ be the canonical quotient map corresponding to our (nearly trivial) equivalence relation.

• Note that given any function $$f:\mathrm S^1\rightarrow \mathbb C$$ we may define a function $$\tilde f:[0,2\pi] \rightarrow \mathbb C$$ via $$\tilde f := f\circ q$$, and notice that $$\tilde f(0)=\tilde f(2\pi)$$.
• Similarly, given any $$\tilde f:[0,2\pi]\rightarrow \mathbb C$$ such that $$\tilde f(0)=\tilde f(2\pi)$$, we may define a map $$f:\mathrm S^1\rightarrow \mathbb C$$ via $$f = \tilde f \circ q^{-1}$$, which is well-defined precisely because of the periodicity requirement.

So the takeaway is that the set of all functions from $$\mathrm S^1\rightarrow \mathbb C$$ is in one-to-one correspondence with the set of periodic functions from $$[0,2\pi]\rightarrow \mathbb C$$.

Having made this (possibly trivially obvious) point, we can proceed to construct $$L^2(\mathrm S^1)$$ as follows: $$L_0^2 := \left\{ f :[0,2\pi] \rightarrow \mathbb C \ \bigg| \ f(0)=f(2\pi) \text{ and } \int_0^{2\pi} |f(x)|^2 \mathrm dx <\infty\right\}$$ $$\forall f,g \in L_0^2 : f \sim_{L^2}g \iff \int_0^{2\pi} |f(x)-g(x)|^2 = 0$$ $$L^2(\mathrm S^1) := L_0^2 / \sim_{L^2}$$

The key thing to note is that the periodicity requirement is baked into the definition of the Hilbert space from the start. Every function in $$L_0^2$$ is periodic, and so any representative of an equivalence class in $$L^2(\mathrm S^1)$$ is also periodic. To put it differently, if you have a square-integrable function $$f:[0,2\pi]\rightarrow \mathbb C$$ such that $$f(0)\neq f(2\pi)$$, then it does not$$^\ddagger$$ correspond to an element of $$L^2(\mathrm S^1)$$.

As a result, given a multiplication operator $$V:L^2(\mathrm S^1) \rightarrow L^2(\mathrm S^1)$$ $$\big(V\psi\big)(x) = v(x)\psi(x)$$ we must have that $$v(0)=v(2\pi)$$; otherwise the resulting vector would generically not belong to $$L^2(\mathrm S^1)$$, and $$V$$ would not be an operator on that space.

The extension to higher-dimensional spaces and to multiparticle systems is then straightforward. When we study a particle on a compact space without boundary, we usually talk about functions defined on some compact region $$R\in\mathbb R^n$$ and then "glue" the edges of $$R$$ together via some equivalence relation(s). The result of this process - if we carry it through formally - is that periodicity is a fundamental prerequisite for the entire relevant Hilbert space, and that an object which maps a periodic function to a non-periodic function is not an acceptable operator.

$$^\ddagger$$This is in contrast to $$L^2([0,2\pi])$$, which does not have that requirement. In that case, we may impose periodicity on the domain of some operator, for example, but the full Hilbert space is not so restricted.

• Dear J. Murray, thank you very much for your detailed and throughout answer! I need some time to read it carefully. Besides that, do you happen to know a good textbook on e.g. solid state physics discussing these mathematical things a bit more? IMHO most textbooks in that subject are really not careful, don't mention the Hilbert space etc. at all... Commented Sep 18, 2022 at 14:52
• I think I've understood it. There is one point, however a trivial one I guess, which I do fail to see at the moment: You defined the equivalence relation through $x\sim x$ and $0 \sim 2\pi$. Is this equivalent to $x\sim x+2\pi$? In other words, for the construction of $L_0^2$ you consider functions with $f(0)=f(2\pi)$. Don't we need $f(x+2\pi)=f(x)$ or is this implied from your definition? Thanks in advance. Commented Sep 18, 2022 at 15:05
• @JasonFunderberker Thanks Jason - upon close inspection, I suspect you'll find most of what I wrote fairly trivial. For a few notes on the construction and properties of $L^2(\mathrm S^1)$, you might look here, though this doesn't discuss the multiplication operator issue. Unfortunately I'm not familiar with many resources which treat solid-state physics at this level of precision - Hall's Quantum Theory for Mathematicians is my go-to resource for the fundamentals, which can be extrapolated from there. Commented Sep 18, 2022 at 15:07
• @JasonFunderberker I suppose you could look at it that way, though I defined $\sim$ to be a relation on the set $[0,2\pi]$ rather than $\mathbb R$, so the only value of $x$ for which $x\sim x+2\pi$ makes sense is $x=0$. If you wanted to consider periodic functions on the entire real line then you could do so, but you'd also want to make sure that when defining the $L^2$ inner product, you only integrate over a single period. Commented Sep 18, 2022 at 15:09
• @JasonFunderberker I don't think it's a big problem. One could simply say that $\big(T_a\psi\big)(x) = \psi([x+a]\text{ mod }2\pi)$ or something equivalent. Ultimately, that's what you'd be doing when working with $2\pi$-periodic functions on $\mathbb R$ anyway, bearing in mind that you'll be specifically focusing on (e.g. integrating over) the interval $[0,2\pi]$ at any given time. But in any case, I think this would be a matter of personal preference. Commented Sep 19, 2022 at 5:11