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.
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