# First eigenvalue of Schrödinger operator is simple

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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## 2 Answers

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.

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Excuse me, for the maximum principle, do we assume $V\geq 0$? –  user38600 Jun 3 at 23:47

Barry Simon's book, "Functional Integration and Quantum Physics", should fit the bill.

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