# Eigenfunctions of Schrödinger Operators on the boundary

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.

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## 2 Answers

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.)

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What do you mean by "test functions". Smooth functions with compact support (in our case this would mean all smooth functions)? –  supersnail Jun 22 '13 at 8:12
Yes, test functions are meant to be smooth with compact support. On a compact smooth space, yes, it'd be all smooth functions. On a merely piecewise-smooth thing (which a boundary might be), it's less clear what we might want at the "corners", e.g., on the boundary of a square? –  paul garrett Jun 22 '13 at 17:32

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 $\phi$.

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Thank you for the comment. Actually I wasn't looking for a physical reason (there is no ambient space in my situation). –  supersnail Jun 22 '13 at 7:26