Rigid Walled Box (Quantum Mechanics)

[robotpie3000]

Homework Statement

I have just started my undergraduate quantum mechanics lectures and I am currently stuck in this question:

A rigid-walled box that extends from -L to L is divided into three sections by rigid
interior walls at -x to +x, where x<L . Each section contains one particle in its ground
state.
(a) What is the total energy of the system as a function of x?
(b) Sketch E(x) versus x.
(c) At what value of x is E(x) a minimum?

Homework Equations

E_n=(n²π²ħ²)/(2mL²)
EΨ(x,t) = iħ(∂Ψ(x,t))/(∂t)

The Attempt at a Solution

[/B]
a) The energy of the system as a function of x should obey Schrödinger's Equation, so EΨ(x,t) = iħ(∂Ψ(x,t))/(∂t), but I'm not sure how to find a solution to Ψ(x,t) that may help me move forward.

b) After reading my textbook for a while, I have decided to draw the total energy as a straight line parallel to the horizontal x axis.

c) I was thinking since there are 3 particles within the rigid walled box, the minimum energy of the system would be E₃=(9π²ħ²)/(2m(2L²)), since the length of the box is 2L.

Help is much appreciated!

 

[jtbell]

there are 3 particles within the rigid walled box

No, you have three rigid walled boxes (bounded by the "rigid interior walls"), each containing one particle.

 

[robotpie3000]

So I've been thinking about this for a while and I think I've found the answer.

There are three regions: [-L, -x],[-x, x] and [x, L]. So the length of each region would be L-x, 2x and L-x respectively. If we let a₁=1/(L-x)², a₂=1/(2x)², and a₃=1/(L-x)², and use the inequality relationship ((a₁² + a₂² + a₃²)/n)^1/2 ≥ (a₁ + a₂ + a₃)/n, we can show that ([1/(L-x)²] + [1/(2x)²] + [1/(L-x)²])/3 ≥ 9/4L².

E is a minimum if L-x = 2x, which means x=L/3, and E_min=3*[(π²ħ²)/(2m)]*[27/(4L²)]=(27π²ħ²)/(8mL²).