Consider the one dimensional Schrödinger hamiltonian $\mathcal{H}=-\frac{\hbar^2}{2} \frac{d^2}{dx^2} + V(x)$.

Suppose that $V:\mathbb{R} \rightarrow \mathbb{R}^+$ is a continuous and confining potential $\displaystyle \lim_{\lvert x\rvert \to +\infty} V(x)=+\infty$

It is well known that $\mathcal{H}$ has pure discrete spectrum $\lambda_1< \cdots \leq \lambda_n$ with $\lambda_n \to +\infty$ as $n\to + \infty$.

Furthermore, if $V$ is a polynomial of even degree $2s$, then the eigenvalues obey the asymptotic formula

$$\lambda _k \sim {C_{s,\hbar}\, k}^{\frac{2s}{s+1}}$$

I want to know if the above asymptotic formula holds if, instead of having a polynomial potential as above, I consider a "pseudo-polynomial" in the sense that $V$ is another function on a segment $[a,b]\subset \mathbb{R}$ (like $\sin(x)$) and outside $\mathbb{R}\setminus[a,b]$ its an even polynomial such that the result is a continuous piecewise function.


1 Answer 1


The answer is yes. Suppose your $V$ equals to a polynomial $P$ when $|x|$ is large. Then there is a constant $c$ such that we have $P-c<V<P+c$, which implies that $\lambda_k^\prime-c<\lambda_k<\lambda^\prime_k+c$, where $\lambda_k^\prime$ are the eigenvalues of the potential $P$. So $\lambda_k$ and $\lambda_k^\prime$ have the same asymptotics.

EDIT. I suppose it is clear (from physical interpretation, or from the analogy with matrices) that bigger potential gives bigger eigenvalues, but for a formal proof of this one can refer on Sturm's comparison theorem.


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