$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$ $\newcommand{\sgn}{\mathrm{sgn}}$ $\newcommand{\He}{\mathrm{He}}$
Consider Schrödinger operator \begin{equation} H=h^2D^2 +V(x) \label{eq-6.3.1} \end{equation} and an energy level $E$ such that \begin{align} &V(x_E^-)=V(x_E^+)=E, && V(x)< E \iff \ x_E^-< x < x_E^+, \label{eq-6.3.2}\\[3pt] &V'(x_E^-)< 0, \quad V'(x_E^+) > 0. \label{eq-6.3.3} \end{align}
We are interested in the energy levels (i.e. eigenvalues) of $H$ close to $E$. To do this we employ the stationary theory. Consider Lagrangian manifold $\Lambda_E=\{(x,p): p^2+V(x)=E\}$ and construct function $S$ on it which everywhere except the end points $(x_E^\pm)$ can be locally expressed as function of $x$ satisfying $S_x^2 + V(x)=E$.
However globally $S(x)$ is not defined uniquely: as point $(x,p)$ circles once counterclockwise $\Lambda$ its increment is $\Delta S= \int_{\Lambda_E} p\,dx$. On the other hand,
Exercise 1. Prove that Maslov index of this path is $2\mod 4$.
Therefore argument of amplitude $u_h(x)$ is increased by $h^{-1}\int_{\Lambda_E} p\,dx +\pi$ where $\pi$ comes from the increment of amplitude $A(x)$.
Since $u_h(x)$ must be a function of $x$ we conclude that this increment is $\equiv 0\mod 2\pi\bZ$: \begin{equation} F(E):= -\frac{1}{2\pi h} \int_{\Lambda_E} p\,dx = n +\frac{1}{2}\qquad n\in \mathbb{Z}. \label{eq-6.3.4} \end{equation}
This is Bohr-Sommerfeld formula. One can prove rigorously
Example 1. As $V(x)=x^2$ (harmonic oscillator) then $F(E)=\pi E$ and $E_n= (2n+1)h$ precisely. Eigenfunctions are $h^{-\frac{1}{4}} \He_n (h^{-\frac{1}{2}}x)($ where $\He_n$ are Hermite functions.
Remark 2. What we denote by "$h$" physicists denote by "$\hbar$", and "their" $h=2\pi \hbar$ is the minimal possible action (according to N. Bohr).
Question: Do quasieigenvalues approximate eigenfunctions and do quasimodes approximate eigenfunctions?
Sometimes yes, sometimes no...
Example 2. Consider 1D Schrödinger operator with potential like this (try to formulate precise assumptions)
Simple well |
Then near level $E^*$ Bohr-Sommerfeld eigenvalues $e_n(h)$ are simple and also true eigenvalues are simple, distance between neighbouring eigenvalues is $E_{n+1}(h)-E_n(h) \asymp h$, and distance between Bohr-Sommerfeld eigenvalues is $e_{n+1}(h)-e_n(h) \asymp h$ and therefore Bohr-Sommerfeld quasieigenvalues provide a good approximation for eigenvalues \begin{gather} E_n(h)= e_n(h)+O(h^\infty). \label{eq-6.3.7} \end{gather} The same is true for corresponding eigenfumctions \begin{gather} \Psi_n(h)= \psi_n(h)+O(h^\infty). \label{eq-6.3.8} \end{gather}
Example 3. Consider 1D Schrödinger operator with potential like this (try to formulate precise assumptions)
Symmetric double well well |
Then near level $E^*$ Bohr-Sommerfeld eigenvalues $e_n(h)$ are double and corresponding quasimodes are supported in the left or right segments.
However true eigenvalues are simple but paired $E'_n(h)$ and $e''_n(h)$, corresponding to functions $\Psi'_n(x)$ and $\Psi''_n(x)$ which are even and odd with respect to $x$, distance between neighbouring eigenvalues is $E'_{n+1}(h)-E'_n(h) \asymp h$, $E''_{n+1}(h)-E''_n(h) \asymp h$ but distance between eigenvalues in the same pair is $E''_{n}(h)-E'_n(h) = O(h^\infty)$ (more precisely $O(e^{-\kappa h^{-1}})$ and there is method to calculate $\kappa>0$. Connection between left and right segments is due to tunnelling.
Distance between Bohr-Sommerfeld eigenvalues is $e_{n+1}(h)-e_n(h) \asymp h$ and therefore Bohr-Sommerfeld quasieigenvalues provide a good approximation for eigenvalues \begin{gather} E'_n(h)= e_n(h)+O(h^\infty), \qquad E''_n(h)= e_n(h)+O(h^\infty). \label{eq-6.3.9} \end{gather} However quasimodes do not provide a good approximation for eigenfunctions.
Example 4. Consider 1D Schrödinger operator with potential like this (try to formulate precise assumptions)
Well on the island |
Then near level $E^*$ Bohr-Sommerfeld eigenvalues $e_n(h)$ are simple.
But there are not true eigenvalues close to $E^*$, the spectrum here is continuous. However $e_n(h)$ are approximating resonances $E_n(h)$, $\Im E_n(h)>0$, which are studied in the Scattering theory and they correspond to quasistable states: if we sovle non-Stationary Schrödinger equation with initial data $\psi_n(h)$ then the solution $\Psi_n (t,h)$ decays albeit very slowly, "escaping to infinity" due to tunnelling.
\begin{gather} E_n(h)= e_n(h)+O(h^\infty). \label{eq-6.3.10} \end{gather}
Remark 2. One can combine these examples: f.e. in Example 4 consider well deeper than the "sea level"; then quasieigenvalues below sea level would approximate eigenvalues and quasieigenvalues above it will approximate resonances.