Variational method in a finite square well

jcsimon89
Messages
1
Reaction score
0
I am trying to prove that there is always one bound state for a finite square well using variational method, and I am stuck. I've tried using e^(-bx^2) as my trial wave function, but I am left with E(b)=(hbar^2)b/2m - V, where V is the depth of the well. In this equation, taking the derivative in terms of b and setting it to zero doesn't lead to a critical value of b, am I doing something wrong?

Thanks,
Jacob
 
Physics news on Phys.org
I think you evaluated your potential wrong. If your well goes from 0 to a, then you need to evaluate
<br /> -V \int_0^a \psi^*(x) \psi(x) dx<br />
which should give you error functions.
 
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!
Back
Top