I once read that the first eigenvalue of a Schrödinger operator always is simple, together with an easy proof of it. But I cannot remember where. Does anybody know a reference?
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asked May 20, 2012 at 16:57
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2 Answers
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Say that $V$ be bounded by below. Up to the addition of a large enough constant, you may assume that $V\ge0$. Then argue as follow.
The operator $L=-\Delta+V$ (where $V(x)$ is the potential) satisfies the maximum principle: if $f\ge0$ and $f\not\equiv0$, then the solution $u$ of $-\Delta u+Vu=f$ exists, is unique and satisfies $u>0$. Then apply the Krein--Rutman Theorem to $L^{-1}$ ; this is the infinite-dimensional version of Perron-Frobenius Theorem, the latter applying to positive matrices. You find that the spectral radius is an eigenvalue, a simple one, associated with a positive eigenfunction.
answered May 20, 2012 at 17:02
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$\begingroup$ Excuse me, for the maximum principle, do we assume $V\geq 0$? $\endgroup$ Commented Jun 3, 2014 at 23:47
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$\begingroup$ @user38600. Add a constant to $V$, and do the same argument. $\endgroup$ Commented Nov 15, 2021 at 7:12
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Barry Simon's book, "Functional Integration and Quantum Physics", should fit the bill.
