Assume a potential of the form \begin{eqnarray}V(x) &=& \frac{1}{2}m\omega^2x^2,~-x_0\leq x\leq x_0,\\&=& 0,\hspace{2.5cm}{\rm otherwise} \end{eqnarray} where $x_0$ is a finite positive number. Surely tunneling is possible in this case and the particle can escape to $x\to\pm\infty$. How to qualitatively draw the stationary states of this problem (of course, without solving it in detail). In particular, how different will the stationary state in this case be from those of the actual harmonic oscillator potential $V(x)=\frac{1}{2}m\omega^2x^2, ~\forall x$? If someone could draw the stationary states that will be helpful.
Can it be solved on a computer to obtain the stationary states?
Best Answer
Because tunneling can happen at all accessible energies, the spectrum will be purely continuous, i.e., there will be no stationary states in the Hilbert space of square integrable wave functions. The solutions of $H\psi=E\psi$ corresponding to the continuous spectrum have real $E$ and are concentrated near $x=0$, with sinusoidal tails outside $[-x_0,x_0]$. To construct them numerically, pick an energy $E$ and solve a 2-point boundary-value problem in $[-x_0,x_0]$, with boundary conditions at $x=\pm x_0$ matching those of a free solution with energy $E$.
With outgoing boundary conditions at infinity, the Schrödinger equation will have Gamov states (= Siegert states) with complex eigenvalues, corresponding to resonances, unstable states decaying into the continuum. Gamov states satisfy the equation $H\psi=E\psi$ with complex $E$; the imaginary part is the inverse lifetime of the state. These states form a discrete set and are the analogues of the harmonic oscillator wave functions.