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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_0 = \frac{hbar^2}{mQ^2e^2} }

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

The electric field introduces a perturbation to the hamiltonian:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_t = eE_0ze^{t/\tau} }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} 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^2E_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{align}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle &= \langle zR{20} | z | zR_{10} \rangle \\ &= \int_{0}^{\infty} dz z^3 e^{ frac{3}{2} frac{z}{a_0} } \left( 2 - \frac{z}{a_0} \right) \left[ \frac{1}{ \left( \int_{0}^{\infty} dz z^2 e^{\frac{-2z}{a_0}} \right)^{1/2} \left( \int_{0}^{\infty} dz z^2 \left(2 - \frac{z}{a_0} \right)^2 e^{\frac{-z}{a_0}} \right)^{1/2} } \right] \end{align}}

where the quantity inside the brackets is the normalization for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_1 } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_2 } , respectively.

let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \frac{z}{a_0} }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \langle \psi_{n=2}(z) | z | \psi_{n=1}(z) \rangle &\rightarrow a_0 \int_{0}^{\infty} dx x^3 e^{ /frac{3}{2} x } \left( 2 - x \right) \left[ \frac{1}{ \left( \int_{0}^{\infty} dx x^2e^{-2x} \right)^{1/2} \left( \int_{0}^{\infty} dx x^2 \left( 2 - x \right)^2 e^{ -x } \right)^{1/2} } \right] \\ &= a_0 \frac{32 \sqrt{2}}{81} \end{align}}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{1 \rightarrow 2} = \frac{e^2E_0^2}{\hbar^2} \frac{1}{ \omega_{21}^2 + \frac{1}{\tau^2} } \frac{2^{11}}{3^{8}} }