Eigenstates of the Laplacian and boundary conditions

Question

Consider the following setting. I have a box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$, for some $L> 0$. In physics, this is usually the case in statistical mechanics or some problems in quantum mechanics. The free Hamiltonian $H_{0}$ is just $H_{0} = -\Delta$. Suppose $\Omega$ is considered with periodic boundary conditions. In this case, plane waves $$f_{p}(x) = L^{-d/2}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{d}$$ are eigenstates of $H_{0}$.

I am a bit confused with the role of boundary conditions in this setting. The functions $f_{p}$ are periodic, so these are only eigenstates of $H_{0}$ because of the boundary conditions assumed. How these eigenfunctions change if we assume other boundary conditions, say, Neumann or Dirichlet? I always take for granted that plane waves are eigenstates of $H_{0}$, and this is usually what is considered in most problems in physics, but maybe I am just very used to periodic boundary conditions and I don't even notice that other options are possible.

Answer 1

If we use Dirichlet ($f(x)= 0$ if $x$ is on the box's boundary) or von Neumann ($\nabla f$ is orthogonal to the normal vector of the boundary) boundary conditions , then only some linear combination of plane waves of the same frequency can be eigenstates.

In dimension $d=1$, only the sine remains for Dirichlet boundary condition : $$f^{(d=1, \text{Dirichlet})}_p(x) = \mathcal N\sin(px)$$ with $p\in \frac{2\pi}{L}\mathbb Z$ and energy $E = p^2$ (and $\mathcal N$ a normalization constant). and only the cosine for von Neumann condition : $$f^{(d=1, \text{von Neumann})}_p(x) = \mathcal N\cos(px)$$ We get the solutions in arbitrary $d$ by taking tensor product of the $d=1$ solutions. Explicitly, for pure Dirichlet boundary condition, with $p \in \frac{2\pi}{L}\mathbb Z^d$, $$f_p(x) = \mathcal N \prod_{i=1}^d \sin(p_i x_i)$$ is an eigenstate of $H_0$ with energy is $E = |p|^2$.

We could of course do the same with cosine for pure von Neumann condition, or with a mix of sines and cosines for a mix of Dirichlet and von Neumann conditions.

Even in $d=1$ there are more possibilities :

• if we impose Dirichlet at one end (say $x=0$) and von Neumann at the other ($x=L$), this changes the allowed values of $p$ : the solutions are $f_p(x) = \sin(px)$ to satisfy the $x=0$ condition. The $x=L$ conditions imposes : $$p \in \frac{2\pi}{L}(\frac{1}{2}+ \mathbb Z)$$
• if we impose a mixed boundary condition at $x=0$, like $\cos(\theta)f(0) + \sin(\theta)f'(0)= 0$, then the solution is $f_p(x) = \sin(x-\theta)$. The $x=L$ boundary condition will determine the allowed values of $p$.