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Spin probability

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Two particles with spin 1/2 and are in the spin state:

    [itex] |\psi> = |\uparrow_{1}>|\downarrow_{2}> [/itex]

    where [itex] |\uparrow_{1}> [/itex] is a state where particle 1 has spin up along the z-axis and
    [itex] |\downarrow_{2}>[/itex] is a state where particle 2 is spin down along the z-axis.

    If we measure the magnitude of the total spin of the two particles, what is the probability that the magnitude will be 1?

    2. Relevant equations

    [itex] Probability = |<n|\psi>|^{2} [/itex]

    3. The attempt at a solution

    I immediately thought that this problem was like a simple coin toss. 50% to get heads 50% to get tails. Since each particle has equal chance to be spin up or spin down, then the total probability of both being spin up after a measurement would be (0.5)(0.5) = 0.25 ?

    This doesn't feel right to me. I feel like it should be more complicated. =/
     
  2. jcsd
  3. Dec 11, 2013 #2

    TSny

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    Hello.

    Can you expand the state ##|\psi \rangle## in terms of basis states that have definite values of the total spin? These basis states are the "singlet" and "triplet" states.

    See here and here.
     
    Last edited: Dec 12, 2013
  4. Dec 12, 2013 #3
    Hello =)

    so the singlet state is
    [itex] |\psi_{singlet}>= 1/(√2)(|\uparrow\downarrow> - |\downarrow\uparrow>) [/itex]

    and the triplet state is

    [itex] |\psi_{triplet}>= 1/(√2)(|\uparrow\downarrow> + |\downarrow\uparrow>) [/itex]

    so,
    [itex] |\psi> = 1/2(|\psi_{singlet}> + |\psi_{triplet}>) [/itex]

    in order to get S=1, the spin state would need to be triplet. The probability would then be the square of the coefficient of the triplet state?

    [itex] |1/(2\sqrt{2})|^{2} = 0.125 [/itex]

    this doesn't make sense though, because the total probabilities don't add up to 1...
     
  5. Dec 12, 2013 #4

    TSny

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    Make sure ##|\psi \rangle## is properly normalized.
     
  6. Dec 12, 2013 #5
    Oh right, I forgot.

    After normalization, [itex] |\psi> = 1/\sqrt{2}(|\uparrow\downarrow> - |\downarrow\uparrow>) +1/\sqrt{2}(|\uparrow\downarrow> + |\downarrow\uparrow>) [/itex]

    leaving the probability to be in the triplet state (S=1) to be 1/2.

    Thanks for your help! I think I got it!
     
  7. Dec 12, 2013 #6

    TSny

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    That looks correct.
     
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