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Degeneracy for different energy states in Infinite cubic well

  1. Dec 2, 2012 #1
    Alright, I'm back with yet another question...

    So the prof was explaining that the energy in an infinite cubical well is E((h22)/2ma2))(nx2+ny2+nz2)

    Which is all well and good, and he gave us the example of:

    ψ1,2,1 = E = 6((h22)/2ma2))

    And with little explanation mixed it up once switching one of the 1's with the 2. After some searching through my book I found out that this is the first energy level or E1, and has 3-fold degeneracy.

    Simply enough, E2 is ψ1,2,2 with 3 -fold degeneracy also..

    But after looking online, I find that E3 is not ψ2,2,2.

    How do you derive/find out what to use for the quantum numbers? As I found out, ψ2,2,2 is E4.

    Is there a pattern? Or a formula? Any help is appreciated.
     
  2. jcsd
  3. Dec 2, 2012 #2

    mfb

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    Just sort by increasing energy:

    111 (energy 3*const)
    112 (energy 6*const)
    122 (energy 9*const)
    113 (energy 11*const)
    222 (energy 12*const)
     
  4. Dec 2, 2012 #3

    Vanadium 50

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    E3 has quantum numbers 1,1,3, for a total energy of 1+1+9 = 11.

    E4 has quantum numbers 2,2,2 for a total energy of 4+4+4 = 12.

    There is no pattern, but there is a formula - the very first one you wrote down. Plug in all the n's and then rank them by energies.
     
  5. Dec 2, 2012 #4
    Oh. Wow I was making that way too difficult. I.. don't even know why I didn't think of it that way. Geeze.

    Well thanks for clarifying! hehe. I guess I should stop studying Q.M at 4 in the morning.
     
  6. Dec 3, 2012 #5

    tom.stoer

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    The interesting question is whether one can find two energies with

    mx² + my² + mz² = nx² + ny² + nz²

    with different ordered triples (mx, my, mz) and (nx, ny, nz)
     
  7. Dec 3, 2012 #6
    I think you should be able to...
    that's exactly is what degeneracy is, right? Finding similar energy states for different wave functions.
    So I think it would still count as degeneracy even though your values for m and n (in your example) are different. Like: (3,3,3) = 27... and (5,1,1) would also be 27.
    I suppose it would work ... but hey just when you think you know something about QM, you don't. ;)
     
  8. Dec 3, 2012 #7

    tom.stoer

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    The intersting thing is that for different ordered triples the wave functions are not related by a symmetry transformation; for (1,1,2), (1,2,1) and (2,1,1) it's permutation symmetry, but for (3,3,3) and (5,1,1) there is no (no obvious ?) symmetry. One may find a hint when looking from geometrical aspects; the equation reads

    nx² + ny² + nz² = Z²

    which means that one tries to find integral lattice points with identical radius Z
     
    Last edited: Dec 3, 2012
  9. Dec 3, 2012 #8
    So, beyond that it has not permutation symmetry (which, yes that is odd), does it make E14 special? Is there something particular about that energy state that makes it have those particular triples?
     
  10. Dec 3, 2012 #9

    tom.stoer

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    Mathematically it means that there are two inequivalent ways to write Z² as a sum of three squares; physically I don't see anything special
     
  11. Dec 3, 2012 #10
    Oh, alight. On the inside I'm a little sad. I was half expecting the physical meaning to be something odd and hard to grasp; like figuring out how many times you'd have to throw a baseball at a wall before it tunneled through, just something odd.
     
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