Weyl law for (non-semiclassical) Schrodinger operator

Question

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.

Answer 1

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.

Answer 2

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