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Angular momentum eigenstates

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]|0,0\rangle [/itex] be the simultaneous eigenstate of [itex]\mathbf{J}^2[/itex] and [itex]J_z[/itex] with eigenvalues 0 and 0. Find
    J_x|0,0\rangle \quad\quad J_y |0,0\rangle \quad\quad [J_x,J_y]|0,0\rangle

    2. The attempt at a solution
    It seemed reasonable to write [itex]J_x[/itex] and [itex]J_y[/itex] in terms of ladder operators
    J_{+}=J_x + iJ_y
    J_{-}=J_x -i J_y
    and then to have them operate on the states (for the last one I would just use the x,y,z commutation relations). But I was looking at the normalization constant out front
    J_{+}|j,m\rangle = \hbar\sqrt{(j+m+1)(j-m)} |j,m+1\rangle
    J_{-}|j,m\rangle = \hbar \sqrt{(j-m+1)(j+m)}|j,m-1\rangle
    but given the initial state, these seem to be zero... Please tell me I'm doing this wrong, zero is such a unsatisfactory answer....

  2. jcsd
  3. Nov 1, 2011 #2


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    Science Advisor
    Gold Member

    Because you are in a j=0 state, you can't actually raise or lower the m eigenvalue since m ranges between -j and +j so in this case it must stay 0.

    So, it would seem like indeed you get 0 for everything.
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