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Ladder operators to prove eigenstates of total angular momen

  1. Sep 11, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider the following state constructed out of products of eigenstates of two individual angular momenta with ##j_1 = \frac{3}{2}## and ##j_2 = 1##:
    $$
    \begin{equation*}
    \sqrt{\frac{3}{5}}|{\tiny\frac{3}{2}, -\frac{1}{2}}\rangle |{\tiny 1,-1}\rangle + \sqrt{\frac{2}{5}}|{\tiny\frac{3}{2},-\frac{3}{2}}\rangle|{\tiny1,0}\rangle
    \end{equation*}
    $$

    (a) Show that this is an eigenstate of the total angular momentum. What are the values of ##j## and ##m_j## for this state?

    2. Relevant equations
    $$
    \begin{equation}
    J^2|\Psi\rangle=\hbar ^2j(J+1)|\Psi\rangle
    \end{equation}
    $$

    $$
    \begin{equation}
    J_\pm=J_x\pm iJ_y
    \end{equation}
    $$

    $$
    \begin{equation}
    J^2 = (J_1)^2 + (J_2)^2 + 2(J_1\cdot J_2)
    \end{equation}
    $$


    $$
    \begin{equation}
    J_1 \cdot J_2 = J_{1x}J_{2x} + J_{1y}J_{2y} + J_{1z}J_{2z}
    \end{equation}
    $$
    3. The attempt at a solution

    I want to use (2) to write (4) in terms of the ladder operators so I can prove the left hand side of (8) does work out to be the right hand side. Re-writing (4) gives me:


    $$
    \begin{equation}
    J_1 \cdot J_2 = J_{1\pm}J_{2\pm} \mp i J_{2y}J_{1\pm} \mp i J_{1y}J_{2\pm} + J_{1z}J_{2z}
    \end{equation}
    $$

    I know how ##J_{\pm}## and ##J_z## act on ##|\Psi\rangle## however I don't know how ##J_y## acts on ##|\Psi\rangle## Can someone tell what I can do to the ##J_y## terms?
     
  2. jcsd
  3. Sep 11, 2015 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can solve (2) for both ##J_x## and ##J_y## by computing ##J_+ \pm J_-##.
     
  4. Sep 12, 2015 #3
    Ah great. Thanks!
     
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