Quantum Wave Function: Infinite Sheet of Charge & Pinhole

[GRDixon]

Imagine an infinite, positive, uniform sheet of charge with a pinhole in it. A negative particle oscillates back and forth through the pinhole and in the +-x direction. The magnitude of the force on it is constant in time (although the force reverses direction when the particle passes through the pinhole). Can anyone tell me what the formula for Psi(x) would be? Thanks. PS, I'm 72 years old, and this is not a homework problem. I just haven't found this particular potential in any of my limited supply of quantum mechanics texts.

 

[eaglelake]

Imagine an infinite, positive, uniform sheet of charge with a pinhole in it. A negative particle oscillates back and forth through the pinhole and in the +-x direction. The magnitude of the force on it is constant in time (although the force reverses direction when the particle passes through the pinhole). Can anyone tell me what the formula for Psi(x) would be? Thanks. PS, I'm 72 years old, and this is not a homework problem. I just haven't found this particular potential in any of my limited supply of quantum mechanics texts.

I assume you want to find the energy eigenfunctions in the position representation, which means we must solve the energy eigenequation, aka Schrodinger's time independent equation - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 \psi (x)}}{{dx^2 }} + V(x)\psi (x) = E\psi (x). From Gauss Law, we know that the electric field outside of an infinite charged sheet is constant, so that V(x) = ax. This problem then is equivalent to a particle in a "vee" shaped potential well. This solution to Schrodinger's equation is in terms of Airy functions. The two constants of intergration are then obtained by matching \psi (x) with the two decaying exponentials outside the well. Very tedious, unless you use a computer.
Best wishes

 

[GRDixon]

I assume you want to find the energy eigenfunctions in the position representation, which means we must solve the energy eigenequation, aka Schrodinger's time independent equation - \frac{{\hbar ^2 }}{{2m}}\frac{{d^2 \psi (x)}}{{dx^2 }} + V(x)\psi (x) = E\psi (x). From Gauss Law, we know that the electric field outside of an infinite charged sheet is constant, so that V(x) = ax. This problem then is equivalent to a particle in a "vee" shaped potential well. This solution to Schrodinger's equation is in terms of Airy functions. The two constants of intergration are then obtained by matching \psi (x) with the two decaying exponentials outside the well. Very tedious, unless you use a computer.
Best wishes

Many Thanks. GRD