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Spin problem and question

  1. Sep 24, 2006 #1
    I'm preparing my QM exam (problem part) and I have stumbled upon one I can't quite "digest" properly. It goes like this

    Two particles with spin s=1 are given and in total state [tex]|S=2,M=1>[/tex]. What is probability that by simultaneous measurement of both particle's spin projections onto some axis we obtain same result?

    Back on exercise classes few years ago TA gave some ridiculoulsy large solution I'm reading now and I dont understand it completely. I thought of next. Problem's question is equivalent to next (I hope): What is probability that by measuring difference between particle's spin projections onto some axis we obtain zero as a result? We first set up spin difference operator[tex]\hat{s}_{diff}=\hat{s}_1\otimes\hat{I}_2-\hat{I}_1\otimes\hat{s}_2[/tex] (it has propper eigenvaules [tex]s_1-s_2[/tex] and gives orthogonal decomposition of total state space so I guess it is spin difference operator). Probability asked for is [tex]v=<2,1|\hat{P}_0|2,1>[/tex], where [tex]\hat{P}_0[/tex] projects to spin difference operator's subspace of eigenvalue 0. Now I represent it in noncorrelated basis [tex]|s_1,m_1>\otimes|s_2,m_2>[/tex]. Since both particles have same spin, they will be represented by same matrices in this basis. As a consequence, spin difference operator will be zero (in any basis) and accordingly, I suppose, all projectors in it's subspaces will be zero. Hence probability is zero (this is same result TA got). Now, this reasoning would hold for any particles of identical spin.

    Q1: Is my solution OK?

    Q2: Does this mean that it is actually impossible for two particles of identical spin to have same spin projections onto some axis?
     
    Last edited: Sep 24, 2006
  2. jcsd
  3. Sep 25, 2006 #2
    Still no reply? Well anyway...

    First: What worries me is that no one noticed I made a mistake (or someone noticed and decided not to post). Spin projection difference operator is not zero (contribution of nonocommutativity of Kronecker product). I will try to get around this.

    Second: Q2 still remains
     
  4. Sep 25, 2006 #3
    Doesn't this belong in the "homework" category?
     
  5. Sep 25, 2006 #4

    CarlB

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    Of course the answer to Q2 is "no", Bose statistics allow as many as you like in the same spin state.

    By the way, I'm sort of surprised this didn't get sent over to the homework section.

    Carl
     
  6. Sep 29, 2006 #5

    reilly

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    S=2 implies 5 total angular momentum states. Count how many can have the same m or sz components, and you have a good start. My guess is that with careful counting you can avoid Clebsch-Gordon coefficients or a Gram-Schmidt procedure.

    Regards,
    Reilly Atkinson
     
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