The Weyl law for a semiclassical Schrodinger operator $$ A_h\ := \ -h^2\Delta+V(x) $$ on an $d$-dimensional complete Riemannian manifold $M$ says that the number $N(A_h,1)$ of eigenvalues of $A_h$ which are smaller than 1 has asymptotic behavior $$ N(A_h,1)\ \sim \ \frac1{(2\pi h)^d}\ \text{Vol}\ {\Large(}(x,\xi)\in T^*M:\ |\xi|^2+V(x)\le 1{\Large)}, \quad h\to 0.\ \ \ (\ast) $$ I am interested in a non-semiclassical Schrodinger operator $$ A\ := \ -\Delta+V(x).$$ I believe that the number $N(A,\lambda)$ of eigenvalues of $A$ smaller than $\lambda$ has a similar asymptotic $$ N(A,\lambda) \sim \ \frac1{(2\pi)^d} \text{Vol}\ {\Large(}(x,\xi)\in T^*M:\ |\xi|^2+V(x)\le \lambda{\Large)}, \qquad \lambda\to \infty. \quad(\ast\ast) $$ This is, of course, true on a compact manifolds, due to the classical Weyl's law. It is also not difficult to verify (**) for operator $-\Delta+|x|^n$ on $\mathbb{R}^d$ (where $A_h$ and $A$ can be related by rescaling of $x$).

So my question: is (**) true and, if it is true, where can I find it?

PS. Victor Ivrii in his review article https://arxiv.org/abs/1608.03963 mentions (page 5) that, using Birman-Schwinger principle, one can obtain a Weyl law for $N(A,\lambda)$ from (*). But I don't see how it can be done.

  • 1
    $\begingroup$ Victor Ivrii is somewhat responsive to emails. One could try emailing the question to him for a response. $\endgroup$ Apr 14 '19 at 6:44
  • 2
    $\begingroup$ I don't think completeness alone will suffice to say anything this general. E.g. there are complete riemannian manifolds with purely continuous spectrum (such as the upper half-plane $\mathbb{H}^2$ has spectrum $[1/4,\infty)$.) If you're not considering closed manifolds, and you don't assume something about how your potential $V(x)$ behaves at infinity I don't think you can hope to get a Weyl Law. You might look at H\"{o}rmander's "spectral function of an elliptic operator" since he proves a weyl law very generally, though at the cost of some FIO theory being necessary. $\endgroup$ Apr 14 '19 at 17:24
  • 1
    $\begingroup$ @HadrianQuan Clearly, I assume that $V(x)$ grows at infinity, so that the volume in (**) is finite. I did look at Hormander. I don't think the result I need follows from it. $\endgroup$ Apr 16 '19 at 15:21
  • 1
    $\begingroup$ Apologies for not realizing you meant this. I think you're right, as the principal symbol in classical calculus he considers will not include the potential $V(x)$ (I suppose this is the point of considering semiclassical operators!) I am glad you were able to find some references. $\endgroup$ Apr 16 '19 at 16:54

We are talking about a non-compact Riemannian manifold, right? Then Weyl's law may be incorrect, at least out of the box. Let us consider $H = -\Delta$ (so $V(x)=0$) in the domain $X \subset \mathbb{R}^d$ with the Dirichlet or Neumann boundary condition (the case of the closed manifold is very similar to the Neumann case).

Let non-compact part of $X$ be $\{x\colon x_1 > c, \ x'\in \rho(x_1) \Omega\}$ with $x'=(x_2,\ldots, x_d)$, $\Omega$ a compact domain with a smooth boundary in $\mathbb{R}^{d-1}$ and $\rho \to 0$ as $x_1\to \infty$.

Observe that depending on the rate of decay of $\rho$ the volume of $X$ could be finite or infinite (and the area of $\partial X$ also could be finite or infinite).

The literal Weyl's law says: if $X$ has a finite volume then the spectrum is discrete and Weyl's law holds and if $X$ has an infinite volume then there is an essential spectrum.

However the reality is more nuanced:

Dirichlet Laplacian In this case spectrum is discrete for sure; simply in the case of the infinite volume the correct asymptotic expansion is different. Let $\rho (t)= t^{-\mu}$, $\mu >0$. Volume is finite for $\mu (d-1)> 1$ and infinite otherwise.

  1. Literal Weyl's law holds for $\mu (d-1)> 1$.
  2. For $\mu (d-1)=1$ (logarithmic divergence) $N(\lambda)\asymp \lambda ^{(d-1)/2}\ln (\lambda)$ and is given by the same formula but with integration over $x_1\colon \rho(x_1)>\lambda^{-1/2}$.
  3. For $\mu (d-1)<1$ (power divergence) $$ N(\lambda)\sim \sum_{j} n_j(\lambda) \tag{D} $$ where $n_j(\lambda)$ is a Weyl's expression for $1$-dimensional Schrödinger operator $\mathsf{h}_j = -\partial_1^2 +\nu_j \rho(x_1)^{-2}$ and $\nu_j >0$ are eigenvalues of $(d-1)$-dimensional Dirichlet Laplacian in $\Omega$.

Neumann Laplacian Then things change drastically. There is an essential spectrum unless $\rho\to 0$ really fast. To understand why look at (D) but instead of a Dirichlet Laplacian in $\Omega$ one should consider a Neumann Laplacian, and $\nu_1=0$. Does it mean essential spectrum?

  1. If $\rho =x_1^{-\mu}$ or even $\rho =\exp (-c x_1)$ then yes.
  2. But if $\rho = \exp(-x_1^{k+1})$ with $k>0$ then no, and $$ N(\lambda) \sim N^W(\lambda) + N^c(\lambda) \tag{N} $$ where $N^W(\lambda)$ is a Weyl expression for an original operator, and $N^c(\lambda)$ is a Weyl expression for $1$-dimensional Schrödinger $\mathsf{h}_0= -\partial_1^2 + W(x_1)$ with a potential $W =\frac{1}{4} (\partial_{x_1} \log \rho (x_1))^2$ and depending on $k$ either the first or the second term in (N) may prevail.

For Dirichlet/Neumann case in more general setting see subsections 3.2.2/3.2.3 of 100 years of Weyl's law either in arXiv or in Bull Math Sci

  • 3
    $\begingroup$ As someone mentioned, the best way to reach me is by email $\endgroup$ Aug 17 '19 at 10:25

I talked to several experts. Here is the punchline: apparently, only the case of $\mathbb{R}^n$ can be found in the literature (I would be glad to be wrong and would highly appreciate a reference). For $\mathbb{R}^n$ the asymptotic (**) holds if the potential does not oscillate too much and otherwise is reasonable. Probably the first result in this direction is in

de Wet, J. S.; Mandl, F. On the asymptotic distribution of eigenvalues. Proc. Roy. Soc. London. Ser. A. 200, (1950). 572–580.

where only the cases on $n=1,2,3$ are considered and the potential $V(x)$ is supposed to grow "fast enough". One of the best results is in

Levendorskii. Spectral asymptotics with a remainder estimate for Schr ̈odinger operators with slowly growing potentials. Proc. Roy. Soc. Edinburgh Sect. A, 126(4):829–836, 1996

where very slowly growing potentials are allowed. Another interesting paper is

G. V. Rozenbljum. Asymptotic behavior of the eigenvalues of the Schr ̈odinger operator. Mat. Sb. (N.S.), 93 (135):347–367, 487, 1974. (in Russian)

where very singular potentials are allowed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.