Solving Infinite Square Well: Eigen Functions & Solutions

mr_whisk
Messages
7
Reaction score
0
Hi all,

Sorry if this question is not very challenging, but I am driving myself to confusion...

I happily derived the eigen functions for an infinite square well spanning from 0 to L and found them to be:

20276c56b7727f9948f8a7cacfc05d52.png


...in agreement with wikipedia.

However my course notes derive solutions for an inf sq well between -a and +a.

Surely L=2a, no?

In the notes n=1,3,5... produces cosine solutions and n=2,4,6... produce sine solutions.

Why is there now cosine solutions?

Please help, I'm going crazy :)
 
Physics news on Phys.org
You can get the solutions on [-L/2 , L/2] from the solutions on [0, L] by replacing x by x + L/2 (i.e. moving the solutions to the left by L/2). For example, sin(πx/L) becomes sin((π/L)(x + L/2)) = sin(πx/L + π/2) = cos(πx/L).
 
Thankyou dx,

That's brilliant, it all makes sense :)
 
Draw a graph of the ground state (n = 1) solution for both cases. You'll see that the two graphs are identical except for the origin (x = 0) being in different places relative to the walls of the well. Likewise for n = 2, etc.
 
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