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Degeneracy of each of energies - Quantum

  1. Nov 18, 2013 #1
    Hello, I dont understand something in this exercice and i have another question:

    1. The problem statement, all variables and given/known data
    Use separation of variables in Cartesian coordinates to solve the infinite cubical well (or "particle in a box"):

    V (x, y, z) = { O,if x, y, z are all between 0 and a;

    (a) Find the stationary state wave functions and the corresponding energies.

    (b) Cali the distinct energies E 1, E1, E3, ... , in order of increasing energy. Find E1, E2, E3 , E4 , Es, and E6. Determine the degeneracy of each ofthese energies (that is, the number of different states that share the same energy). Recall (Problem 2.42) that degenerate bound states do not occur in one dimension, but
    they are common in three dimensions.

    (c) What is the degeneracy of E 14, and why is this case interesting?

    II/ I was wondering for the lowering and increasing operators a- and a+
    When we have a-a+=1/(hω) H-1/2
    H is considered as an operator ?

    2. Relevant equations

    3. The attempt at a solution

    a. The energy : E=h2π2/(2ma2) (nx2+ny2+nz2)
    with ki=ni2π2/a2

    b. The thing that I dont get is that according to the correction that we can find here:
    Why the ground state is nx=ny=nz=1 so n=3?
    Whyt not for n=nx2+ny2+nz2=1
    the combinations 1,0,0 0,1,0 0,0,1 ?

    Thanks !
  2. jcsd
  3. Nov 19, 2013 #2


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    Staff: Mentor

    What happens to the wave function when ##n_\alpha=0## (##\alpha \in \{x,y,z\}##)?
  4. Nov 19, 2013 #3
    When what ?
  5. Nov 19, 2013 #4


    User Avatar

    Staff: Mentor

    When any of ##n_x##, ##n_y## or ##n_z## is equal to zero.

    You ask why we can't take, for instance, ##n_x = 1##, ##n_y = 0##, ##n_z = 0## as the ground state. What does the wave function for that state look like?
  6. Nov 19, 2013 #5
    Oh right, it's equal to 0 everywhere, and this is not possible !
    Thank you !
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