I guess this topic may have been addressed somewhere but I cannot really find a reference myself, so I ask here.

For each $N \in \mathbb{N}$, consider the Laplacian $\Delta$ on the interval $[-N,N]$ subject to the periodic boundary conditions. Then, it is well-known that it is a self-adjoint operator on $L^2[-N,N]$ with discrete spectrum. Yet, the "gaps" between neighboring eigenvalues become "smaller" with larger $N$.

Now, the Laplacian $\Delta$ on whole $\mathbb{R}$ has the continuous spectrum $(-\infty,0]$.

My question is that: \begin{equation} \text{ In what sense does the Laplacian } \Delta \text{ on } [-N,N] \text{ converges to one on the whole real line as } N \to \infty? \end{equation} \begin{equation} \text{ Moreover, does the boundary conditions on } [-N,N] \text{ matter as long as } \Delta \text{ remains self-adjoint?} \end{equation}

Could anyone please clarify for me?

I guess this topic may have been addressed somewhere but I cannot really find a reference myself, so I ask here.

For each $N \in \mathbb{N}$, consider the Laplacian $\Delta$ on the interval $[-N,N]$ subject to the periodic boundary conditions. Then, it is well-known that it is a self-adjoint operator on $L^2[-N,N]$ with discrete spectrum. Yet, the "gaps" between neighboring eigenvalues become "smaller" with larger $N$.

Now, the Laplacian $\Delta$ on whole $\mathbb{R}$ has the continuous spectrum $(-\infty,0]$.

My question is that:

In what sense does the Laplacian $\Delta$ on $[-N,N]$ converge to one on the whole real line as $N \to \infty$?

Moreover, does the boundary condition on $[-N,N]$ matter as long as $\Delta$ remains self-adjoint?

Could anyone please clarify for me?