Intro Quantum: Expanding infinite square well

  • Thread starter camron_m21
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  • #1
camron_m21
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Homework Statement



Griffiths Intro to Quantum, problem 2.38:
A particle of mass m is in the ground state of the infinite square well. Suddenly the well expands to twice its original size: the right wall moving from a to 2a, leaving the wave function (momentarily) undisturbed. The energy of the particle is now measured.



Homework Equations



[tex]c_{n} = \sqrt{\frac{2}{a}}\int^{a}_{0} sin(\frac{n\pi}{a} x) \Psi (x,0) dx[/tex]


The Attempt at a Solution



Since the particle starts out in the ground state in a well of length a, at t=0 (right when the well expands) the wave function should be

[tex]\Psi (x,0) = \sqrt{\frac{2}{a}} sin\left( \frac{\pi x}{a} \right)[/tex]

I know this can be written as a sum of the new wave functions,

[tex]\Psi[/tex](x,0) = [tex]\sum c_{n} \psi_{n} (x)[/tex]

The problem wants the most probable result of measuring the energy, as well as the next most probable. For this, I was thinking of using [tex]|c_{n}|^{2} [/tex] as the probability of getting an energy. However, when I do this, I get pi times an integer as the argument for the sine in the answer, which gives me zero.

I'm mostly just at a loss on how to start on this, so any help would be appreciated. I'm not sure what I need to calculate to find the most and next most probable energies.
 

Answers and Replies

  • #2
camron_m21
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Scratch that, I figured it out. For anyone else stuck on this, calculate [tex]c_{n}[/tex] the regular way, but use [0,a] for the integration bounds rather than [0,2a], since the initial function at t=0 is 0 for x > a.
 
  • #3
vela
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Are you using your expression for cn above? It's not quite correct for this particular problem because it's using the eigenstates for the original well, not the expanded well.
 
  • #4
camron_m21
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Yeah, sorry about that, that was just the formula for the general infinite square well. The formula I used replaces a with 2a, except in the t=0 function.
 

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