Eigenfunctions of Schrödinger Operators on the boundary

Question

Hello,

let's consider a compact and connected Riemannian manifold with the Schrödinger Operator $L=-\Delta +V:dom(H)\subset L^2(M)\rightarrow L^2(M)$ whereas $dom(L):=\lbrace f\in C^{\infty}(M,\mathbb{R}) \vert f_{\vert \partial M}=0 \rbrace$ and $V\in L^2(M)$ bounded.

The spectrum of the Friedrichs extension of $L$ consists of a discrete set of Eigenvalues $(\lambda_i)_{i=1}^{\infty}$ with corresponding eigenfunctions $\phi_i$, which form an $L^2$-orthonormal basis.

I want to know, why the eigenvalues $\phi_i$ vanish on the boundary , i.e. $\phi_i (x)=0 \forall x\in\partial M$. I hope, you can explain it to me.

Regards.

Answer 1

In addition to Michael Renardy's apt comment about the physical sense of the situation, one can also see the boundary vanishing from the characterization of the Friedrichs extension. Namely, by construction, the domain of the Friedrichs extension is inside the +1 Levi-Sobolev space closure of the original domain. With a piecewise smooth $\partial M$, this means that integrations of boundary values of eigenfunctions against test functions on $\partial M$ itself will all vanish. (These functionals are in the -1 Levi-Sobolev space.)

Answer 2

If you are looking for a physical reason, you should ask the question why the eigenfunctions are confined to M in the first place. Usually, the reasoning for this is that V is infinite outside of M. Since the term $V\phi$ appears in the Schroedinger equation, an infinite V should go with a zero $\phi$.