I am coming up with that question as I simply cannot satisfy myself with the frustrating fear that it might not be possible to show that a Hamiltonian corresponding to a particle in a box with periodic boundary conditions has pure-point spectrum and that most of its eigenvalues have finite-dimensional eigenspaces.

Particle confined in a box with periodic boundary conditions

So let us be clear about the setup of the problem. Suppose we have a particle which is confined to a box which is subject to periodic boundary conditions. Inside the box, we have a potential which also satisfies the periodic boundary conditions. So we simply have a single-particle Hamiltonian $$ H = \frac{\mathbf p^2}{2 m} + V(\mathbf r) ,$$ where $V(\mathbf r)$ as above mentioned satisfies the periodic boundary conditions and where we are lurking for a solution $\psi(\mathbf r)$ which also satisfies the periodic boundary conditions.

Clearly, we all know the standard example of a rectangular box containing a free particle (no potential). We all know the solutions and due to the boundary conditions can convince ourselves that the spectrum of the Hamiltonian $\sigma(H)$ is indeed pure-pont spectrum. Taking a look at a particular eigenvalue $E \in \sigma(H)$, we will also quickly notice that the corresponding eigenspace $\mathcal H_E$ is finite-dimensional.

This all seems to be perfectly intuitive. Now, however, what happens when allowing for an arbitrary but bounded potential $V(\mathbf r)$?

Intuition proposes that nothing should have changed with respect to the pure-point nature of the spectrum $\sigma(H)$ and the corresponding dimensionality of the eigenspaces $\mathcal H_E$.

Yet it remains to be proven, if it indeed is the case, which I am not sure of.

Thoughts so far:

I am aware of a Theorem for self-adjoint, bounded, compact operators which looks pretty much like what one would like to end up with as a result.

(Spectral theorem for compact operators). Suppose the operator $K$ is self-adjoint and compact. Then the spectrum of $K$ consists of an at most countable number of eigenvalues which can only cluster at $0$. Moreover, the eigenspace to each nonzero eigenvalue is finite dimensional [...]

  • Theorem 6.6, Gerald Teschl, Mathematical Methods in Quantum Mechanics

However in general compact operators seem to be quite rarely encountered in general quantum mechanics.

Nonetheless, my hope was based on $H$ potentially being a compact operator due to the restriction to a box with period boundary conditions.

  • 1
    $\begingroup$ What do you mean by $H$ being compact? That it has a bounded spectrum? Also, have you considered that the periodic boundary conditions force the momentum operator to take discrete values, regardless of the form of $V(\mathbf{r})$? Because, assuming $V(\mathbf{r})$ is bounded from above and below, you can use perturbation theory to transition from the $V=0$ case using standard techniques. $\endgroup$ Mar 12, 2018 at 18:13
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    $\begingroup$ Try to show that H has a compact resolvent . $\endgroup$
    – jjcale
    Mar 12, 2018 at 18:18
  • $\begingroup$ @SeanE.Lake I see, quite an interesting about perturbation theory. How would one argue in that case exactly, like slowly turning up on the potential? I am a bit confused as without box this appears a bit tricky to me, as switching on a infinitesimal periodic potential in that case would lead immediately to band gapps for instance, I suppose...? $\endgroup$
    – Lars D.
    Mar 13, 2018 at 1:02
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    $\begingroup$ @LarsD. As long as $-\infty<V_{\mathrm{min}} \le V(\mathbf{r}) \le V_{\mathrm{max}}<\infty$ you can introduce a parameter in front $V$, say $\lambda$, that guarantees the shift in eigenvalues and eigenvectors is actually infinitesimal, and therefore continuous. The only wrinkle I can see is going to be related to getting the degenerate perturbation theory right. $\endgroup$ Mar 13, 2018 at 1:18
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    $\begingroup$ @Lars D. From the spectral decomposition of normal operators (see e.g. Rudin, functional analysis, chapter 13) it follows that H and its resolvent have the same eigenvectors and there functional dependency carries over to the spectrum, i.e. $\sigma((z-H)^{-1} ) = (z - \sigma(H))^{-1}$ $\endgroup$
    – jjcale
    Mar 13, 2018 at 21:25

1 Answer 1


Your hamiltonian is not compact as it has an unbounded spectrum. It is true, however, that the inverse (i.e. the Green function or resolvent) of a Schroedinger Hamiltonian in a finite box and with a bounded potential is a compact operator. I'm not sure where I learned this though...

  • $\begingroup$ That's equivalent to the spectrum being gapped no? $\endgroup$ Mar 12, 2018 at 20:12
  • $\begingroup$ Hah, you're right. That point about unbounded spectrum was actually on my mind some days ago but apparently I forgot. :P The information you provided helps me already quite a bit. I assume you however cannot recall right now where you learnt that from? xD $\endgroup$
    – Lars D.
    Mar 13, 2018 at 1:12
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    $\begingroup$ @Lars D. Probably by browsing Read and Simon "Analysis of Operators". I never read that book thoroughly though. $\endgroup$
    – mike stone
    Mar 13, 2018 at 13:28
  • $\begingroup$ @mikestone I got the book and actually found a Theorem about box-systems that states what I would like to prove. I still lack the basis to thoroughly understand the line of argumentation but will spend the next days struggling with it. Thank you for the reference, it's much appreciated. :) $\endgroup$
    – Lars D.
    Mar 14, 2018 at 1:07

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