Electron on Helium Surface

An electron close to the surface of liquid helium experiences an attractive force due to the electrostatic polarization of the helium and a repulsive force due to the exclusion principle(hard core). To a reasonable approximation for the potential when helium fills the space where $z<0$:

$V(z)={\begin{cases}-{\frac {Q^{2}e^{2}}{z}}&{\mbox{if}}\qquad z>0\\\infty &{\mbox{if}}\qquad else\end{cases}}$

Note: the potential is infinite when $z<=0$ because the cannot penetrate the helium surface.

(a) Solve the Schrödinger equation. Find the Eingenenergies and Eigenvalues.

(b) An electric field is turned on at t=0 which produces the perturbation:

$V(z)={\begin{cases}eE_{0}ze^{t/\tau }&{\mbox{if}}\qquad t>=0\\0&{\mbox{if}}\qquad else\end{cases}}$

If the electron is initially in its ground state, find the probability makes a transition to its first excited state for times $t>>\tau$.

Solution...

(a) Solve the Schrödinger equation.

$H=-{\frac {h^{2}}{2m}}\nabla ^{2}+V(z)$

The Schrodinger equation for when $z>0$ is:

$\left[{\frac {h^{2}}{2m}}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)-{\frac {Q^{2}e^{2}}{z}}\right]\psi (\mathbf {r} )=E\psi (\mathbf {r} )$

Using separation of variables:

$\psi (x,y,z)=X(x)Y(y)Z(z)$

For X and Y we get place waves.

$X(x)Y(y)=Ae^{i(k_{x}x+k_{y}y)}$

This corresponds to motion parallel to the helium surface.

For z-component the Schordinger equation becomes:

<math>

 \left[ 
   \frac{h^2}{2m}\frac{\partial ^2}{\partial z^2} - \frac{Q^2e^2}{z} 
 \right] Z(z) =