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Rigid Walled Box (Quantum Mechanics)

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

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


    3. The attempt at a solution

    a) The energy of the system as a function of x should obey Schrodinger'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!!!!
     
  2. jcsd
  3. Oct 27, 2014 #2

    jtbell

    User Avatar

    Staff: Mentor

    No, you have three rigid walled boxes (bounded by the "rigid interior walls"), each containing one particle.
     
  4. Oct 29, 2014 #3
    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 Emin=3*[(π2ħ2)/(2m)]*[27/(4L2)]=(27π2ħ2)/(8mL2).



     
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