Consider a non-relativistic electron moving above a large, flat grounded conductor while it is attracted by its image charge, but cannot penetrate the conductor's surface.
What is the Hamiltonian of the electorn and the BC its wavefunction must satifsfy?
What is its ground state energy and its average distance above the conductor?
Classically I can write $H = T + V = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) + \frac{e^2}{4 \pi \epsilon_0} \frac{1}{2\left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2\right)^\frac{1}{2} }$
Now I would guess that the wavefunction would have to obey $\psi(x,y,z\leq 0) = 0$ and $\psi( |\vec{r}| \rightarrow \infty) \rightarrow 0$
Writing down the time indep. Schrodinger equation, however (with the QM Hamilotnian)
$E \psi(\vec{r}) = \left( \frac{- \hbar^2}{2m_e} \nabla^2 + V(\vec{r}) \right) \psi(\vec{r}) $
I am unable to find a solution since I can't use spherical symmetry, and most of the Sch. equation I have dealt with were usually for free particles (e.g. particle in box) etc. Any help would be appreicated. I'm assuming once I get the wf as a function of z I can compute its expectation value to find the average distance?
The potential depends only on the distance to the conductor (z) not x and y. Hence it is a free particle in the x and y dimensions leaving you with the single z dimension. I believe the z equation can be related to the zero angular momentum radial functions for the hydrogen atom - recall the effective 1-D radial equation for u(r)=r R(r) is essentially a 1-d problem with and effective potential that has a centrifugal barrier term, which vanishes for zero angular momentum. Take care with the identification of r - here the distance is 2z between the charge and its image.