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Degeneracy of a 2-dimensional isotropic Harmonic Oscillator

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    The Hamiltonian is given by:
    [tex] H = \frac{1}{2} \sum_{i=1,2}[p_i^2 + q_i^2] [/tex]
    We define the following operators:
    [tex] J = \frac{1}{2} (a_1^+ a_1 + a_2^+ a_2) [/tex]
    [tex] J_1 = \frac{1}{2} (a_2^+ a_1 + a_1^+ a_2) [/tex]
    [tex] J = \frac{i}{2} (a_2^+ a_1 - a_1^+ a_2) [/tex]
    [tex] J = \frac{1}{2} (a_1^+ a_1 - a_2^+ a_2) [/tex]

    I have shown previosly that [tex] \textbf{J}^2=J_1^2 + J_2^2 + J_3^2 = J(J+1) [/tex]

    The question: Using this result (the one above), discuss what is the degeneracy of the eigenvalues of H.
    2. Relevant equations
    [tex] H = 2J + 1 [/tex]
    [tex] \textbf{J}^2=J_1^2 + J_2^2 + J_3^2 = J(J+1) [/tex]

    3. The attempt at a solution
    I thought just saying that [tex] [H, \textbf{J}^2] = 0 [/tex] would be enough to say that the eigenvalues of H are degenerate with respect to the eigenstates of [tex] \textbf{J}^2 [/tex]. But I am not sure if this is enough, can I calculate the degeneracy of the eigenvalues of H just using the equations found above?
     
  2. jcsd
  3. Oct 24, 2015 #2
    One makes an analogy to angular momentum. So one does something like
    $$J_{\pm} = J_1 \pm i J_2 $$
    and find that
    $$[J_+, J_3] = J_+ $$
    $$[J_-, J_3] = -J_- $$
    One also finds that ##[J_3, J^2] = 0## so one can give the eigenvalues of ##J^2## and ##J_3## a separate quantum number, so the eigenvalues of the two operators are ##j(j+1)## and ##m##, respectively.
    So from this we obtain $$J_3J_+\mid j m \rangle = (m+1) J_+\mid j m \rangle $$
    $$ J_+\mid j m \rangle = \alpha_{jm+}\mid j m+1 \rangle$$
    and similiarly for ##J_-##. Also since ##[J_\pm, J^2] = 0##, one gets that ##J_\pm\mid j m \rangle## has the same energy eigenvalue as ##\mid j m \rangle##
    To find the degeneracy one calculates ##\left| J_\pm\mid j m \rangle\right|^2 = |\alpha_{jm\pm}|^2## and notice that if one tries to exceed a particular range of ##m## one just gets ##\alpha=0##. So there is a finite range of ##m## with the same energy. Let me know if that helps.
     
    Last edited: Oct 24, 2015
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