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Basic Quantum Physics - Particle(s) in a (3D) Box

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A system of 20 neutrons are confined in the region

    [tex]0<x<3L[/tex]
    [tex]0<y<L[/tex]
    [tex]0<z<2L[/tex]

    With [tex]L=10^{-15}m[/tex]

    If each eigenstate can accept 2 neutrons (corresponding to the 2 possible spin states) what is the total kinetic energy of the system?

    2. Relevant equations

    I found the total energy to be
    [tex]E = \frac{\pi^{2}\overline{h}^{2}}{2mL^{2}}\left(\frac{(n_{1})^{2}}{9}+(n_{2})^{2}+\frac{(n_{3})^{2}}{4}\right)[/tex]

    3. The attempt at a solution

    Now, I do not really understand what the question is asking.
    But my current idea is to do trial and error for a combination of ns that gives the same energy and multiply it by 20.

    Is this right?

    Thanks in advance
     
  2. jcsd
  3. Sep 3, 2010 #2
    In "relevant equations", the energy expression you have ain't for the total energy, but rather for the energy one particle, given its n1, n2, n3 state.

    What do you know about the Pauli exclusion principle?
     
  4. Sep 3, 2010 #3
    Hi bjnartowt,

    I honestly don't know about the principle.
    By the sound of it and after looking quickly at wikipedia, my understanding is that in this case, no three or more neutrons can have the same combination of ns (since each eigenstate can accept 2 neutrons).

    So I'm guessing that the question is asking for the sum of the lowest 10 energies i.e. at (n1,n2,n3 = 1,1,1
    n1,n2,n3 = 2,1,1
    n1,n2,n3 = 1,1,2
    n1,n2,n3 = 3,1,1 and so on)
    and multiply it by 2?
     
  5. Sep 3, 2010 #4
    Are you saying that only three neutrons are allowed to have the same permutation of "n"-values?
     
  6. Sep 3, 2010 #5
    I'm saying that only two neutrons can have the same permutation (combination) of n-values.
     
  7. Sep 3, 2010 #6
    Excellent: then you understand Pauli exclusion.

    Perhaps the path to the answer is a little clearer?
     
  8. Sep 3, 2010 #7
    I do not know which permutations of n-values to choose though.

    My guess is that the question is asking for the sum of the lowest 10 energies, so choose 10 permutations of n-values giving the lowest energies and sum them up. Then times this sum by 2 since each energy is occupied by 2 neutrons.

    Would this be right?
     
  9. Sep 3, 2010 #8
    You know what, I don't know which permutations of n-values to choose through either...unless someone told me "Hey, this box of 20 neutrons is at absolute-zero temperature". Then, I *think* you'd start from two neutrons in 001, two in 010, two in 100, two in 002, two in 012, etc. which I think is obeying Pauli-exclusion.

    I think that in general, if your temperature was "T", you'd use the Fermi-Dirac distribution.

    So yeah...I think I'm steering us in the right direction, but I flub little details. I hope someone else posts too.
     
  10. Sep 3, 2010 #9

    diazona

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    Homework Helper

    bjnartowt brings up a good point, that if the system is at a nonzero temperature, some of the neutrons could be excited into higher energy levels. If this is for a statistical mechanics class, then that might be the intent of the problem. But if it's just a regular quantum mechanics class, I would guess that you're supposed to assume the system is in its ground state.

    By the way, be careful to actually figure out which energy levels are the lowest ones. Don't just assume that 001, 010, 100, etc. are in order of increasing energy.
     
  11. Sep 3, 2010 #10
    It is a regular class.

    Yep, I'm fully aware of that.

    And I think n starts at 1 (111) since if one of them is 0, the wavefunction will also be equal to zero which is not true.

    Thanks for the help!:smile:
     
  12. Sep 3, 2010 #11

    diazona

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    Oops, sorry :blushing: You're right, I wasn't paying attention.
     
  13. Sep 3, 2010 #12

    Redbelly98

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    Science Advisor
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    Yes.

    BTW, the prefactor on the energy expression in Post #1 corresponds to a temperature ~1012 K, so we can safely assume the system is in the lowest allowed state:
     
    Last edited: Sep 3, 2010
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