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Energy eigenvalue and eigen vector

  1. Sep 4, 2007 #1
    I have some question on energy eigenvalue and eigenfunction
    help plz

    A particle, mass m , exists in 3 dimensions, confined in the region
    0< x < 2L, 0 < y < 3L, 0 < z < 3L

    a) what are the energy eigenvalues and eigenfunctions of the particle?

    b) if the particel is a neutron which is confined in a volume with L=10^-15m, what are the three lowest energy eigenvalues, in MeV? what is the lowest energy eigenvalue which is degrenerate?
     
  2. jcsd
  3. Sep 5, 2007 #2

    olgranpappy

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    I think this should be posted in the homework help section, eh?
     
  4. Sep 5, 2007 #3

    malawi_glenn

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    Yes, and also show work done etc. Exactly whatis it that you dont understand? If we dont know how then we can not help you. It is against the policy of the forumus to just hand out solutions to problems. Our teachers in real life do not do so either..

    Somebody will move this post eventually, so " dusrkeric " do not make a new one.
     
  5. Sep 5, 2007 #4

    olgranpappy

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    No, it's a mere homework issue.
     
  6. Sep 5, 2007 #5
    simple formula dude...
    use equations of particle in three dimension box.
    But u don't mean that v outside bos is finite,I guess.
     
  7. Sep 6, 2007 #6
    These have pretty straightforward solutions. Please post your work/formulae that you have used. Just few hints...
    a] Use the energy eigenvalue equation for a 3-dimensional box after normalizing the eigenfunctions.
    b]This is even easier...use the particle in a 3D square box solution.
     
  8. Sep 6, 2007 #7

    Gokul43201

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    The box is not a cube - it is a cuboid. You can not use the square well energy eigenvalues.
     
  9. Sep 7, 2007 #8

    malawi_glenn

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    why not? The solution is obtained via separation of variables. And the general solution in a 1-dim box of lenght a is [tex] \sqrt{2/a} sin \dfrac{n \pi x}{a} [/tex] just substitut n, x and a to proper values and the solution for 3dim is obtained by multiply all these into one equation.

    At least we have done so here in sweden
     
    Last edited: Sep 7, 2007
  10. Sep 7, 2007 #9

    Gokul43201

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    Yes, the eigenvalues of a 3D-well will be sums of three 1D-well eigenvalues, but this does not make the box a "3D square box" (since, for example [itex]L_x \neq L_y. [/itex]) I should have specified that "you can not use the 3D square box energies."
     
  11. Sep 7, 2007 #10
    Yes it is a cuboid in the first case. However, in the second case, he has been given only one length. So it is a special case of a square box. I don't know if case a] & b] are connected. If yes, then I am wrong.
     
  12. Sep 7, 2007 #11

    olgranpappy

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    Look. The problem just has a confusing wording. He is only given one length 'L' in part b. But since it is part "b", apparently this implies that the result of part "a" is to be applied to the specific case.

    There is only one length 'L' given in part "a" as well--but the region is not a cube in part a, it is a rectangular solid of sides L, 2L, and 3L.

    I'm sure that we all understand the elementary quantum mechanics, so the point now under discussion by Reshma and Gokul is just the slightly vague wording of part "b".
     
  13. Sep 8, 2007 #12

    Gokul43201

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    That's right (gramps).

    Reshma, in part b, you are not (explicitly) given the length of the sides; you are given the value of L. The lengths of the sides are still (2L,3L,3L).
     
    Last edited: Sep 8, 2007
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