Quantum Wave Function: Infinite Sheet of Charge & Pinhole

GRDixon
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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.
 
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GRDixon said:
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
 
eaglelake said:
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
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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