Suppose we want to numerically solve the following Schrodinger equation on a bounded open set $\Omega \subset \mathbb R^n$:
$$Hu = (-\Delta + V(x)) u = \lambda u, \qquad \left.u\right\vert_{\partial \Omega} = 0,\qquad V \in L^2(\Omega,\mathbb R), \not \equiv 0$$
I will discretize the Hamiltonian as a finite-dimensional operator $H_N$ acting on an orthonormal basis $(\phi_j)_{j=1}^N$ (where $\phi_j$ are Dirichlet eigenfunctions of $-\Delta$ on $\Omega$ with eigenvalues $\mu_j$). This is called the expansion method. Now I have:
$$H_{N_{nm}} = \langle \phi_n, H \phi_m\rangle = \mu_n\delta_{nm} + \langle \phi_n, V \phi_m\rangle$$
Alternatively, we can write this as
$$H_N = P_N H ,\qquad P_N = \text{projection onto }\operatorname{span}(\phi_j)_{j=1}^N$$
Note that both $H$ and $H_N$ are self-adjoint. Furthermore, by convention we can define $H_N$ as an operator acting in $H^1_0(\Omega)$ by defining:
$$H_N := P_N H P_N$$
Now, referencing to Mathematical Methods in Quantum Mechanics by G. Teschl, I am able to prove $H_N$ converges to $H$ in the strong resolvent sense, which in turn gives us:
$$\lim_{N\to \infty} \sigma(H_N) \supseteq \sigma(H)$$
I am, however, a bit confused why it's so hard to show either $H_N \to H$ in the norm resolvent sense (which would give us equality in the spectral limit above), or go from the above spectral containment then show it is indeed an equality (perhaps by contradiction). Do I have reason to believe that:
$$\lim_{N\to\infty} \sigma(H_N) = \sigma(H)$$
isn't necessarily true? If so, could I put additional requirements on $V$ in order for this to be true? I suppose the proof might be similar to the analog for a finite-difference method, but I couldn't find any literature on that either.
Best Answer
We have for eigenfunctions $u_k^{(N)}$ of $H_N$ the following,
\begin{equation} (1) \qquad H_N u_k^{(N)} = P_N(-\Delta u_k^{(N)}) + P_N (V u_k^{(N)}) = -\Delta u_k^{(N)} + P_N( V u_k^{(N)}) = \lambda_k^{(N)} u_k^{(N)}, \end{equation}
since $P_N$ projects onto eigenfunctions of $-\Delta$. Theorem 2.2.3 of A. Henrot's Extremum Problems for Eigenvalues of Elliptic Operators states:
"Let $L_n$ be a sequence of uniformly elliptic operators defined on an open set $D$ by $$ L_n u:=-\sum_{i, j=1}^N \frac{\partial}{\partial x_i}\left(a_{i j}^n(x) \frac{\partial u}{\partial x_j}\right)+a_0^n(x) u . $$ We assume that, for fixed $i, j$, the sequence $a_{i, j}^n$ is bounded in $L^{\infty}$ and converge almost everywhere to a function $a_{i, j}$; we also assume that the sequence $a_0^n$ is bounded in $L^{\infty}$ and converges weakly-* in $L^{\infty}$ to a function $a_0$. Let $L$ be the (elliptic) operator defined on $D$ as in (3.2) by the functions $a_{i, j}$ and $a_0$. Then each eigenvalue of $L_n$ converges to the corresponding eigenvalue of $L$."
For (1), the operator $\left. H_N \right\vert_{\{u_k^{(N)}\}_{k=1}^N}$ is of the form $L_n$ in the theorem, where $a_0^n = P_n V$ has the appropriate convergence conditions, hence the eigenvalues of $H_N$ converge to those of $H$. We obtain the desired result: $$\lim_{N\to \infty} \sigma(H_N) = \sigma(H)$$
The results are now in "A generalized expansion method for computing Laplace-Beltrami eigenfunctions on manifolds" from Jackson C. Turner, Elena Cherkaev, and Dong Wang: https://arxiv.org/abs/2210.10982