6

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.

Tobias Fünke
  • 8,777

1 Answers1

5

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.

Albatross
  • 72,909