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QM: Adding momentum

  1. Apr 11, 2006 #1
    I have a general question on finding momentum states

    Let's say I have two spin 1/2 particles, so
    and |J| ranges from [tex]|S_1 + S_2|[/tex] to [tex]|S_1 - S_2|[/tex]
    in this case J=1 is triplet and J=0 is singlet.

    Now how do you find the J=0 state? I know that
    [tex]|j=0,m_j=0>=\frac{1}{\sqrt{2}}(\uparrow_1 \downarrow_2 - \downarrow_1 \uparrow_2)[/tex]
    but how do you get this in the first place? Is it pretty much trial and error and then use operator to generate rest of the states for that particular J? Or is there an algorithm for finding state [tex]|j,m_j=j>[/tex]?
  2. jcsd
  3. Apr 12, 2006 #2


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    Well, it's all about reading & understanding Clebsch-Gordan theorem & coefficients correctly. Technically

    [tex] |j,m\rangle =\sum_{m_{1},m_{2}} \langle j_{1},m_{1},j_{2},m_{2}|j,m\rangle |j_{1},m_{1},j_{2},m_{2} \rangle [/tex]

    The coefficients are tabulated, also, you know that j=0 and m=0.

  4. Apr 12, 2006 #3

    Meir Achuz

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    If you don't have a CG table, you can generate the spin estates.
    First form J_11=\up\up. Then use the lowering operator to form the three J=1 states. J_00 will be orthogonal to J_10.
  5. Apr 12, 2006 #4
    I see. We never really went over CG table so I had to guess the first form and generate rest of the spin (or orbital angular momentum) states. Thanks for the clarification.
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