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.

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

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

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

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 Schroedinger equation becomes:

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

This has the same form as the hydrogin atom with l=0 (s-wave). Since similar equations have similar solutions, the solution to the z-component is:

$Z(z)=zR_{n0}(z)$

where

$R_{10}=\left({\frac {1}{a_{0}}}\right)^{3/2}ze^{-{\frac {z}{a_{0}}}}$

$R_{20}=\left({\frac {1}{2a_{0}}}\right)^{3/2}\left(2-{\frac {z}{a_{0}}}\right)e^{-{\frac {z}{2a_{0}}}}$

The total wave function and energies are:

$\psi =Ae^{i(k_{x}x+k_{y}y)}zR_{n0}(z)$

$E_{k_{x}K_{y}K_{z}}={\frac {\hbar ^{2}}{2m}}\left(k_{x}^{2}+k_{y}^{2}\right)-{\frac {Q^{4}e^{4}m}{2\hbar ^{2}n^{2}}}$

where n = 1,2,... is the quantum number for the z-direction and the bohr radius has become

$a_{0}={\frac {hbar^{2}}{mQ^{2}e^{2}}}$

(b) Turn on electric field at t=0.

The electric field introduces a perturbation to the hamiltonian:

$V_{t}=eE_{0}ze^{t/\tau }$

From expression 2.1.10 in Time Dependent Perturbation Section of the PHY5646 page:

${\begin{aligned}P_{1\rightarrow 2}(t\rightarrow \infty )&=|\langle n|\psi (t)\rangle |^{2}\\&=\left|{\frac {1}{i\hbar }}\int _{0}^{\infty }dt'e^{{\frac {i}{\hbar }}(E_{2}-E_{1})t'}\langle \psi _{n=2}|V_{t'}|\psi _{n=1}\rangle \right|^{2}\\&={\frac {e^{2}E_{0}^{2}}{\hbar ^{2}}}{\frac {1}{\omega _{21}^{2}+{\frac {1}{\tau }}}}\left|\langle \psi _{n=2}(z)|z|\psi _{n=1}(z)\rangle \right|^{2}\end{aligned}}$

${\begin{aligned}\langle \psi _{n=2}(z)|z|\psi _{n=1}(z)\rangle &=\langle zR{20}|z|zR_{10}\rangle \\&=\int _{0}^{\infty }dzz^{3}e^{frac{3}{2}frac{z}{a_{0}}}\left(2-{\frac {z}{a_{0}}}\right)\left[{\frac {1}{\left(\int _{0}^{\infty }dzz^{2}e^{\frac {-2z}{a_{0}}}\right)^{1/2}\left(\int _{0}^{\infty }dzz^{2}\left(2-{\frac {z}{a_{0}}}\right)^{2}e^{\frac {-z}{a_{0}}}\right)^{1/2}}}\right]\end{aligned}}$

where the quantity inside the brackets is the normalization for $\psi _{1}$ and $\psi _{2}$, respectively.

let $x={\frac {z}{a_{0}}}$

${\begin{aligned}\langle \psi _{n=2}(z)|z|\psi _{n=1}(z)\rangle &\rightarrow a_{0}\int _{0}^{\infty }dxx^{3}e^{/frac{3}{2}x}\left(2-x\right)\left[{\frac {1}{\left(\int _{0}^{\infty }dxx^{2}e^{-2x}\right)^{1/2}\left(\int _{0}^{\infty }dxx^{2}\left(2-x\right)^{2}e^{-x}\right)^{1/2}}}\right]\\&=a_{0}{\frac {32{\sqrt {2}}}{81}}\end{aligned}}$

So the probablity of a transition from ground state to the first excited state is:

$P_{1\rightarrow 2}={\frac {e^{2}E_{0}^{2}}{\hbar ^{2}}}{\frac {1}{\omega _{21}^{2}+{\frac {1}{\tau ^{2}}}}}{\frac {2^{11}}{3^{8}}}$