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Spin wavefunction confusion

  1. May 22, 2010 #1
    My lecturer writes:

    The spin wavefunctions are symmetric on exchange of spins for the spin 3/2 states. These states include:

    [tex]|\uparrow \uparrow \uparrow \rangle [/tex]

    and [tex]|\uparrow \uparrow \downarrow \rangle + |\uparrow \downarrow \uparrow \rangle + |\downarrow \uparrow \uparrow \rangle [/tex]

    How is the second wavefunction a state for a spin 3/2 particle? I thought the spin is 1/2 + 1/2 - 1/2 = 1, so the measured spin can be 1, 0 or -1?
     
  2. jcsd
  3. May 22, 2010 #2
    Start with the spin 1 states that you get from adding two spin 1/2 particles and then add the third, standard excersice in QM

    |++>

    |+-> + |-+>

    |-->

    are the three spin 1 states you can build from adding two spin 1/2 particles-

    The second state you wrote is the |S, S_z> = |3/2, 1> state
     
  4. May 23, 2010 #3
    How are they spin 1 states though? How do you figure that out from those states?
     
  5. May 23, 2010 #4
    Have you done adding angular momenta in your QM class yet? yes or no?
     
  6. May 23, 2010 #5
    We did it briefly, just in terms of quantum numbers though, so S = s1 + s2...|s1-s2|, we didn't relate it to the spin wavefunctions like the ones you have mentioned.
     
  7. May 23, 2010 #6
    ok, there are three spin-1 states - do you agree?

    do you also agree that |+-> + |-+> has S_z = 0?

    and total spin

    S^2 = (S_1 + S_2)^2 on that state gives s(s+1) = 1(1+1) = 2

    as eigenvalue.

    S^2 on |+-> + |-+> gives 0, right?
     
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