Consider a Schrödinger operator $H:=-\Delta+V$ on $\mathbb R$, where $V$ is such that $H$ has a purely discrete spectrum $-\infty<\lambda_1\leq\lambda_2\leq\cdots$ converging to $+\infty$. Do there exist reasonably convenient criteria on $V$ that can guarantee that the eigenvalues are all simple (i.e., $\lambda_1<\lambda_2<\cdots$)?

If we instead consider $H$ on an bounded or half-line interval $[a,b]$ (i.e, at least $a$ or $b$ finite), with Dirichlet boundary conditions, then there are well-known (and relatively simple) results guaranteeing multiplicity-one spectra under very general conditions. However, I can't seem to find any information on the full-space case.


You don't need to assume anything. Only the absolutely continuous spectrum of a whole line operator (assumed to be in the limit point case at both endpoints, for convenience, though that, too, could be relaxed) can have multiplicity greater than $1$.

The result is usually attributed to I.S. Kac, On the multiplicity of the spectrum of a second order differential operator, Sov. Math. 3 (1962), 1035--1039. It follows from the form of the standard spectral representation of such whole line operators (using a $2\times 2$ matrix valued spectral measure) and is discussed in several textbooks.

In your situation, though, none of this is needed. In limit point situations (so for example if you just assume that $V(x)\ge -C$ or even only $V(x)\ge -Cx^2$), there is only one solution that is in $L^2$ at a given endpoint, so trivially all eigenvalues are simple.

  • $\begingroup$ Thanks for your answer, I didn't suspect this property to hold in such generality! $\endgroup$ – user78270 Aug 26 '18 at 23:56
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    $\begingroup$ @user78270: I should also add (in case my first paragraph is confusing) that if you have a limit circle endpoint, then the spectrum can not be bounded below, so you really don't need to assume anything in addition to what you already have. $\endgroup$ – Christian Remling Aug 27 '18 at 15:07

It is sufficient to assume $V \in L^\infty_{loc}(\mathbb R)$ and $\lim_{x\to\infty} V(x) = \infty$. This result is quoted in, for example, these lecture notes by A. Pankov, section 7.

  • $\begingroup$ That's great, thanks! The result in the lecture notes (and its referenced proof) is precisely what I was looking for! $\endgroup$ – user78270 Aug 26 '18 at 15:30

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