Perturbations of the Landau Hamiltonian: Asymptotics of Eigenvalue Clusters

Abstract

We consider the asymptotic behavior of the spectrum of the Landau Hamiltonian plus a short-range continuous potential. The spectrum of the operator forms eigenvalue clusters. We obtain a Szegő limit theorem for the eigenvalues in the clusters as the cluster index and the field strength B tend to infinity with a fixed ratio ${\mathcal E}$. The answer involves the averages of the potential over circles of radius $\sqrt{{\mathcal E}/2}$ (classical orbits). After rescaling, this becomes a semiclassical problem where the role of Planck’s constant is played by 2/B. We also discuss a related inverse spectral result.

Appendices

A. The Weyl Quantization of Radial Functions

For the benefit of the reader, we include here some results on the Weyl quantization of radial functions that shed light on the material in Sect. 4. The result in Eq. (A.13) has been originally shown in [3] [Proposition 4.1]; we include a derivation here for completeness. See also [21], §4.

If a(x, p) is a symbol in ${\mathbb R}^{2n}$, its Weyl quantization is the operator $a^W(x, \hbar D)$ with kernel

$$\begin{aligned} {\mathcal K}_a(x,y) = \frac{1}{(2\pi \hbar )^n} \int e^{i\hbar ^{-1}(x-y)\cdot p} a\left( \frac{x+y}{2}, p\right) \, \mathrm{d}p. \end{aligned}$$

The corresponding bilinear form $Q_a(f,g) = \langle a^W(x,\hbar D)(f),\overline{g}\rangle$ is

$$\begin{aligned} \forall f,\,g\in {\mathcal S}({\mathbb R}^n)\qquad Q_a(f,g)&= \frac{1}{(2\pi \hbar )^n} \iiint e^{i\hbar ^{-1}(x-y)\cdot p} a \left( \frac{x+y}{2}, p\right) \,\nonumber \\&\quad \times f(y)\, g(x) dp\,dx\,dy. \end{aligned}$$

(A.1)

It is not hard to see that

$$\begin{aligned} Q_a(f,g) = \iint a(u,p)\, {\mathcal G}(f,g)(u,p)\, du\, dp, \end{aligned}$$

(A.2)

where

$$\begin{aligned} {\mathcal G}(f,g)(u,p) = \frac{1}{(\pi \hbar )^n}\int e^{2i\hbar ^{-1}v\cdot p} f(u-v)\,g(u+v)\, \mathrm{d}v. \end{aligned}$$

(A.3)

Let $a\in {\mathcal S}({\mathbb R}^2)$ be a radial function, that is

$$\begin{aligned} a(x,p) = \rho (r), \quad r=\sqrt{x^2+p^2}. \end{aligned}$$

To simplify notation let $A:=a^W(x,\hbar D)$. By the equivariance of Weyl quantization with respect to the action of the symplectic (metaplectic) group, A commutes with the quantum harmonic oscillator ${\mathcal Z}= -\hbar ^2d^2/dx^2 + x^2$ and, by simplicity of the eigenvalues of the latter, the eigenfunctions $e_n$ of ${\mathcal Z}$ are also eigenfunctions of A. Our goal is to compute the corresponding eigenvalues. We follow the argument in [5].

One can show (starting with section 13.1 of [1], for example) that if one defines the functions $g_n(x)$ by the generating function

$$\begin{aligned} \pi ^{-1/4}\, e^{-x^2/2 +xt-t^2/4} =\sum _{n=0}^\infty \frac{t^n}{\sqrt{2^n n!}}\,g_n(x) \end{aligned}$$

(A.4)

then the $g_n$ are orthonormal in $L^2({\mathbb R})$ and satisfy

$$\begin{aligned} -g_n''(x) + x^2 g_n(x) = (2n+1)\,g_n(x). \end{aligned}$$

For our problem, we need the eigenfunctions of ${\mathcal Z}$, so we need to re-scale the variable x. Define

$$\begin{aligned} e_n(x) := \hbar ^{-1/4}\,g_n(x/\sqrt{\hbar }). \end{aligned}$$

(A.5)

Then, for each $\hbar$, $e_n$ is $L^2$-normalized and

$$\begin{aligned} \begin{array}{lcl} \hbar ^{1/4}\,\left[ -\hbar ^2 \frac{d^2 \ }{dx^2} e_n(x) + x^2 e_n(x)\right] &{} = &{} -\hbar \,g_n''(x/\sqrt{\hbar }) + \hbar \left( \frac{x}{\hbar }\right) ^2 g_n(x/\sqrt{\hbar }) \\ &{} = &{} \hbar (2n+1) g_n(x/\sqrt{\hbar }). \end{array} \end{aligned}$$

(A.6)

In other words, the normalized eigenfunctions $e_n$ are given by the generating function

$$\begin{aligned} G_t(x):=(\pi \hbar )^{-1/4}\, e^{-x^2/2\hbar +xt/\sqrt{\hbar }-t^2/4} =\sum _{n=0}^\infty \frac{t^n}{\sqrt{2^n n!}}\,e_n(x,\hbar ), \end{aligned}$$

(A.7)

where the notation emphasizes that $e_n$ also depends on $\hbar$.

We now use this generating function to compute the eigenvalues of A. Note that

$$\begin{aligned} Q_a(G_t, G_t) = \langle A(G_t),G_t\rangle = \sum _{n=0}^\infty \frac{t^{2n}}{2^n n!}\, \lambda _n, \end{aligned}$$

where $\lambda _n = \langle A(e_n),e_n\rangle$ is the eigenvalue of A corresponding to $e_n$. Computing using (A.2) and (A.3):

$$\begin{aligned} {\mathcal G}(G_t, G_t)(x,p) = \frac{e^{-t^2/2}}{(\pi \hbar )^{3/2}} \int e^{2\hbar ^{-1}ivp}\, e^{-x^2/\hbar }\,e^{-v^2/\hbar }\, e^{2xt/\sqrt{\hbar }}\, \mathrm{d}v \end{aligned}$$

$$\begin{aligned} \qquad \qquad \quad = \frac{e^{-t^2/2-x^2/\hbar + 2xt/\sqrt{\hbar }}}{(\pi \hbar )^{3/2}} \int e^{2\hbar ^{-1}ivp}\,e^{-v^2/\hbar }\,\mathrm{d}v \end{aligned}$$

$$\begin{aligned} = \frac{1}{\pi \hbar }\,e^{-t^2/2-(x^2+p^2)/\hbar + 2xt/\sqrt{\hbar }}, \end{aligned}$$

and therefore,

$$\begin{aligned} \sum _{n=0}^\infty \frac{t^{2n}}{2^n n!}\, \lambda _n = \frac{e^{-t^2/2}}{\pi \hbar }\, \int _{{\mathbb R}^2} a(x,p) \,e^{-(x^2+p^2)/\hbar + 2xt/\sqrt{\hbar }}\, \mathrm{d}x\,\mathrm{d}p. \end{aligned}$$

(A.8)

Next, we use that a is radial and integrate in polar coordinates. The key integral is

$$\begin{aligned} \int _0^{2\pi } e^{2tr\cos (\theta )/\sqrt{\hbar }}\, \mathrm{d}\theta = 2\pi I_0(2tr/\sqrt{\hbar }), \end{aligned}$$

(A.9)

where $I_0$ is the modified Bessel function of order zero. At this point, we can conclude that

$$\begin{aligned} \sum _{n=0}^\infty \frac{t^{2n}}{2^n n!}\, \lambda _n = \frac{2e^{-t^2/2}}{\hbar }\, \int _0^\infty \rho (r) e^{-r^2/\hbar }\,I_0(2tr/\sqrt{\hbar })\, r\mathrm{d}r. \end{aligned}$$

(A.10)

Now, it is known that, for any $u\in {\mathbb R}$,

$$\begin{aligned} I_0(s) = e^u\sum _{k=0}^\infty \frac{(-u)^k}{k!}\, L_k(s^2/4u), \end{aligned}$$

(A.11)

where the $L_k$ are the Laguerre polynomials (in particular the right-hand side is independent of u). If we take $u=t^2/2$, (A.11) gives us that

$$\begin{aligned} I_0(s) = e^{t^2/2}\sum _{k=0}^\infty \frac{(-t^2/2)^k}{k!} L_k(s^2/2t^2). \end{aligned}$$

Substituting back into (A.10), we obtain

$$\begin{aligned} \sum _{n=0}^\infty \frac{t^{2n}}{2^n n!}\, \lambda _n = \frac{2}{\hbar } \sum _{k\ge 0} \int _0^\infty \rho (r) e^{-r^2/\hbar }\,\frac{(-t^2)^k}{2^k k!} L_k(2r^2/\hbar )\, r\mathrm{d}r. \end{aligned}$$

(A.12)

Equating coefficients of like powers of t we conclude that $\forall n$

$$\begin{aligned} \lambda _n = \frac{(-1)^k2}{\hbar }\int _0^\infty \rho (r) e^{-r^2/\hbar }\,L_n(2r^2/\hbar )\, r\mathrm{d}r. \end{aligned}$$

If we now let $u=r^2$, we finally get

$$\begin{aligned} \boxed { \lambda _n = \frac{(-1)^n}{\hbar } \int _0^\infty \rho (\sqrt{u}) e^{-u/\hbar }\,L_n(2u/\hbar )\, \mathrm{d}u. } \end{aligned}$$

(A.13)

Although we do not need it for the proof of our main theorem, we note the following:

Theorem A.1

Let (as in the main body of the paper)

$$\begin{aligned} \hbar = \frac{\mathcal E}{2n+1},\quad \mathcal {E} \text { fixed},\quad n = 1,2, \ldots , \end{aligned}$$

Then, maintaining the previous notation, as $n\rightarrow \infty$

$$\begin{aligned} \lambda _n = \rho (\sqrt{{\mathcal E}}) + O(\hbar ). \end{aligned}$$

Proof

By the functional calculus the operator $\rho ({\mathcal Z}^{1/2})$ is an $\hbar$ pseudo-differential operator with principal symbol $\rho (r)$, that is, with the same principal symbol as $a^W$. Therefore,

$$\begin{aligned} \lambda _n = \langle a^W(e_n),e_n\rangle = \langle \rho ({\mathcal Z}^{1/2})(e_n),e_n\rangle + O(\hbar ) = \rho (\sqrt{{\mathcal E}}) + O(\hbar ). \end{aligned}$$

$\square$

In view of (A.13), we immediately obtain:

Corollary A.2

Let

$$\begin{aligned} \psi _n(u) := \frac{(-1)^n}{\hbar }e^{-u/\hbar } L_n \left( \frac{2u}{\hbar } \right) \end{aligned}$$

so that $\lambda _n = \int _0^\infty \rho (\sqrt{u}) \psi _n (u)\, \mathrm{d}u$. Then, if $\hbar$ and n are related as above, the sequence $(\psi _n)$ tends weakly to the delta function at ${\mathcal E}$.

It is instructive to consider directly the behavior of the functions $\psi _n$. As we will see, there is an oscillatory and a decaying region of $\psi _n$ (similar to the Airy function). For a fixed n, $\psi _n$ has n zeros. As n increases, where do the zeros concentrate? According to [8], the zeros of $L_n$ are real and simple.

Let us denote by $\lambda _{n,k}$ the zeros of $L_n$. According to [7] (restricting to the case $\alpha =0$), the zeros $\lambda _{n,k}$ are in the oscillatory region

$$\begin{aligned} 0< x < \nu := 4n+2 \end{aligned}$$

and satisfy the following inequalities and asymptotic approximation:

Theorem A.3

([7]). The first zero $\lambda _{n,1}$ satisfies

$$\begin{aligned} 0 < \lambda _{n,1} \le \frac{3}{2n+1}, n=1,2,\ldots . \end{aligned}$$

Theorem A.4

([7]). For a fixed m, the zeros of $L_n$ satisfy

$$\begin{aligned} \lambda _{n,n-m+1} = \nu +2^{1/3} a_m \nu ^{1/3}+\frac{1}{5} 2^{4/3} a_m^{2}\nu ^{-1/3}+O(n^{-1}),\text { as }n \rightarrow \infty , \end{aligned}$$

where $a_m$ is the m-th negative zero of the Airy function, in decreasing order.

Let us now denote by $\mu _{n,k}$ the zeros of $\psi _n(u)$, so that $\mu _{n,k}= \frac{\hbar }{2}\lambda _{n,k}$. Substituting $\hbar = \mathcal E/(2n+1)$, Theorem A.3 implies that the first zero satisfies

$$\begin{aligned} \mu _{n,1} \le \frac{3}{2 \mathcal E} \hbar ^2. \end{aligned}$$

On the other hand, the last zero satisfies

$$\begin{aligned} \mu _{n,n} = \mathcal E+(\mathcal E/2)^{1/3}a_1 \hbar ^{2/3} + \frac{(\mathcal E)^{-1/3} a_1^2}{5}\hbar ^{4/3} + O(\hbar ^2), \text { as } \hbar \rightarrow 0. \end{aligned}$$

This implies that the first zero is close to 0 while the last one is close to $\mathcal E$ as $\hbar \rightarrow 0$. In fact, if we define

$$\begin{aligned} N_n(x) = \big | \left\{ k \in \{ 1,2,\ldots , n\} | \lambda _{n,k} \le x \right\} \big | , x\in \mathbb R, \end{aligned}$$

it can be shown ([8]) that

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{1}{n} N_n(4 nx) =\frac{2}{\pi } \int _0^{x}t^{-1/2} (1-t)^{1/2}\mathrm{d}t, \; \;\text { for } 0 \le x \le 1. \end{aligned}$$

Fig. 2

Graph of $\psi _n$ in the interval [0, 5]. Here, $n=100$, $\mathcal E = 3$

We note that $\lambda _{n,k} \le 4 n x$ if and only if $\mu _{n,k} \le \mathcal E x \left( 1-\frac{1}{2n+1} \right)$. This implies that

$$\begin{aligned}&\lim _{n\rightarrow \infty } \frac{1}{n} \left| \left\{ k : \mu _{n,k} \le z \left( 1-\frac{1}{2n+1} \right) \right\} \right| \\&\quad = \frac{2}{\pi } \int _0^{z/\mathcal E}t^{-1/2} (1-t)^{1/2}\mathrm{d}t, \; \; 0 \le z \le \mathcal E. \end{aligned}$$

We note that the integral on the right-hand side is equal to one for $z=\mathcal E$. In particular, this shows that the zeros of $\psi _n$ “cover” the entire oscillatory region $[0,\mathcal E]$, asymptotically for n large.

Choosing $n=100$ and $\mathcal E = 3$, the corresponding graph of $\psi _n$ in the interval [0, 5] is shown in Fig. 2. We can corroborate numerically that the zeros of $\psi _n$ are located in the oscillatory region $[0,\mathcal E]$. We can easily see that $L_n$ is always locally decreasing near the origin and locally increasing/decreasing around the last zero for n even/odd. As a result, the last critical point of $\psi _n$ is always a local maximum.

B. The Remainder in Taylor’s Theorem

For completeness, we include here the elementary derivation of the expression for the remainder in Taylor’s theorem that we used in the proof of Theorem 4.6. Let us start with a smooth one-variable function f and write

$$\begin{aligned} f(t)&= f({\mathcal E}) + \int _0^1 \frac{d\ }{\mathrm{d}u}f(ut + (1-u){\mathcal E})\, \mathrm{d}u \\&= f({\mathcal E})+ (t-{\mathcal E})\int _0^1 \,f'(ut + (1-u){\mathcal E})\, \mathrm{d}u . \end{aligned}$$

So if we let

$$\begin{aligned} g(t):= \int _0^1 f'(ut + (1-u){\mathcal E})\, \mathrm{d}u, \end{aligned}$$

(B.1)

then g is smooth and $f(t) = f({\mathcal E}) + (t-{\mathcal E})g(t)$. Repeating the argument with f replaced by g, we obtain that

$$\begin{aligned} g(t) = g({\mathcal E}) + (t-{\mathcal E})R(t), \end{aligned}$$

where

$$\begin{aligned} R(t) = \int _0^1 g'(vt + (1-v){\mathcal E})\, \mathrm{d}v. \end{aligned}$$

Since $g({\mathcal E}) = f'({\mathcal E})$, substituting we obtain $f(t) = f({\mathcal E}) + (t-{\mathcal E})f'({\mathcal E}) + (t-{\mathcal E})^2 R(t)$, as desired. Finally, we compute the remainder R(t). Using (B.1),

$$\begin{aligned} g'(x) = \int _0^1 uf''(ux + (1-u){\mathcal E})\, \mathrm{d}u, \end{aligned}$$

and therefore

$$\begin{aligned} R(t) = \int _0^1 \int _0^1 uf''[u(vt + (1-v){\mathcal E})+ (1-u){\mathcal E}] \mathrm{d}u\, \mathrm{d}v. \end{aligned}$$