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Particle in a box question

  1. Feb 21, 2006 #1
    "Particle in a box" question

    The solution of Schrodingers equation for the simple "particle in a box" problem is:

    E = (n^2)(h^2) / 8m(L^2)

    where L = the length of the box.

    See http://en.wikipedia.org/wiki/particle_in_a_box

    Say that the initial energy of the particle is E1.
    If at a later time the length of the box is changed (a moveable wall), then there are now a completely new set of possible energies available, for n=1,2,3 etc. What if none of these new allowed energies equals the original energy of the particle?

    What happens to the energy difference?
     
  2. jcsd
  3. Feb 21, 2006 #2

    quasar987

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    Nobody can answer?!
     
  4. Feb 21, 2006 #3

    Physics Monkey

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    It depends on how slowly the walls of the box are moved.
     
  5. Feb 21, 2006 #4

    quasar987

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    Suppose they are abruptly removed. I.e. they instantly vanish. The particle state is now in a superposition of the eigenstates of the new box and will settle into one of them upon measurement. In any case, the energy upon measurement wil be different than originally.
     
  6. Feb 22, 2006 #5
    Particle in a box

    In my original question I was assuming that the energy of the particle was
    conserved after changing the size of the box.

    Was I wrong in assuming this. Maybee changing the size of the box does work on the particle? Like compressing a gas if there were many particles.

    Is it the energy state n that is conserved?

    Please confirm.
     
  7. Feb 23, 2006 #6

    Physics Monkey

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    Hi allanm1,

    Yes, if you move the walls of the box slowly then the particle will always be in the ground state. In this case you must clearly have done work on the system since the energy of the system is changing.

    Hope this helps.
     
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