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Is sixfold degeneracy the maximum degeneracy an angular momentum can have?

  1. Oct 7, 2012 #1
    If ψ is normalize-able and a function of nx, ny, nz, is the maximum energy degeneracy 6?

    I.E. There can be degeneracy at the same Energy with each state taking a different value of n, yet adding up to some (nx^2+ny^2+nz^2)=Same E, due to the linearity of the operators involved. I guess the question is, assuming that any potential involved is physical and causes bounded states, can the maximum degenerate states be 3!.
     
    Last edited: Oct 7, 2012
  2. jcsd
  3. Oct 8, 2012 #2

    mfb

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    Staff: Mentor

    Your are looking for >6 ways to get the same sum of 3 squares?

    3^2+4^2+n^2=25+n^2
    5^2+0^2+n^2=25+n^2
    +permutations

    6^2+2^2+1^2=41
    5^2+4^2+0^2=41
    +permutations

    7^2+1^2+n^2=50+n^2 for every n
    5^2+5^2+n^2=50+n^2
    +permutations (just up to 9 in per n, but without using 0^2)

    I am sure there are more solutions. No idea if the number of different ways to get the same sum is bounded.
     
  4. Oct 8, 2012 #3
    The question I'm struggling to get out is to say if you have a wave function that is bounded on three dimensions with a specific energy, will you at maximum have 3 different separated Schrodinger's Eq, and of those maximum versions have 3 quantization conditions for each dimension. Which implies that there are a maximum of 6 ways for the 3 quantization conditions to exist when all are different values (and the sum is of course the total E).

    Or am I missing an idea with the bounds and possible potentials?
     
  5. Oct 9, 2012 #4

    tom.stoer

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    I am not sure if I understand the question correctly.

    You are looking for solutions

    [tex](H-E)\,\psi_{E,\nu}(\vec{r}) = 0;\;\nu=1 \ldots N(E)[/tex]

    in D=3 dimensions. Each energy E has N-fold degeneracy where N(E) could in general depend on E.

    Is this correct?

    Example 1: for the hydrogen atom you have for each E i.e. each n:

    [tex]l=0,1, \ldots n-1[/tex]
    [tex]m=-l, \ldots +l[/tex]

    which means that each n has n²-fold degeneracy

    [tex]n=1:\;N=1[/tex]
    [tex]n=2:\;N=1+3=4[/tex]
    [tex]n=3:\;N=1+3+5=9[/tex]
    ...
    [tex]n:\;N=n^2[/tex]

    Example 2: for the 3-dim. square well potential the energy scales as

    [tex]E = n_x^2 + n_y^2 + n_z^2[/tex]

    So for each E you have N-fold degenaracy where N counts the number of ways you can write the same E as sum of three squares

    [tex]E = 3 = 1+1+1[/tex]
    [tex]E = 6 = 1+1+4 = 1+4+1 = 4+1+1[/tex]
    [tex]E = 9 = 1+4+4 = 4+1+4 = 4+4+1[/tex]

    Example 3: for the D-dim. harmonic oscillator the energy scales as

    [tex]E = n_1 + n_2 + \ldots + n_D[/tex]

    For D=3 you have

    [tex]E = 0 = 0+0+0[/tex]
    [tex]E = 1 = 0+0+1 = 0+1+0 = 1+0+0[/tex]
    [tex]E = 2 = 0+0+2 = 0+1+1 = 0+2+0 = 1+0+1 = 1+1+0 = 2+0+0[/tex]

    So the degeneracy for given E depends on the potential.

    Usually the degeneracy N(E) reflects the dimension of a representation of a symmetry group of the potential. For rotationally invariant potentials these are the representations of SO(3), for the hydrogen atom this is SO(4), for the D-dim. Harmonic oscillator this is SU(D) which is larger than SO(D).
     
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