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Spin projection

  1. Nov 2, 2003 #1
    For a particle with S = 1 in a state of Sz = 1, what are the probabilities to obtain the various values of the spin component if measured along a direction which makes an angle Q relative to the z-axis?

    Is there a general method for an arbitrary value of S?
  2. jcsd
  3. Nov 2, 2003 #2
    In general, a system with spin J (and maximum projection along the z-axis of J) can be treated as 2J spin 1/2 particles. A projection of J along z would correspond to all of these pointing up.

    If we rotate by an angle &phi;, then each spin 1/2 has a probability P(+)=[cos(&phi;/2)]^2 of having projection "up" along z, and a probability P(-)=[sin(&phi;/2)]^2 of being "down". For that whole system to have projection K (where |K|<=2J), then we will have J+K spins pointing up and J-K spins pointing down. The probability of this projection is expressed in terms of P(+) and P(-), such that

    P(K)=(2J choose J+K)*P(+)^(J+K)*P(-)^(J-K)

    where "2J choose J+K" is equal to (2J)!/[(J+K)!(J-K)!]
  4. Nov 2, 2003 #3
    Sounds great!

    Is this something you thought about now, or you read/learned this trick somewhere in the past?
  5. Nov 2, 2003 #4
    Actually, it was an assignment question a few weeks ago :wink:
  6. Nov 2, 2003 #5
    BTW, this method works very well in other cases. For example, if you're deriving rotation matrices for a spin 1 particle, it is often simpler to express its basis as a combination of the standard spin 1/2 basis (spin up, spin down) using a tensor product and expand things out that way.
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