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What is the collapse state after measuring Sx?

  1. Dec 7, 2013 #1
    1. The problem statement, all variables and given/known data

    So say I have a 1 qubit normalised state |ψ> = 1/√N(c1|z+> - c2|z->) and I make the measurement Sx and the measurement yields +hbar/2
    as |x+> is a superposition of both z+ & z- basis eigenstates does the collapsed state end up the same as the original state? Or would I need to take the inner product of <x+|ψ> to find the collapsed state? Confused. .

    2. Relevant equations



    3. The attempt at a solution

    Its not really a direct homework problem its more of an understanding thing so my attempt is above in 1.
     
  2. jcsd
  3. Dec 7, 2013 #2

    TSny

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    It doesn't matter what the original state is. If you know that the outcome of a measurement is "spin up" along the x axis, then the collapsed wave function is simply |x+>, independent of the original state |ψ>.

    The complex number <x+|ψ> is the probability amplitude that the outcome of the measurement yields |x+>. If <x+|ψ> happened to be zero, then it would have been impossible for the measurement outcome to be spin up along x.

    As long as <x+|ψ> is nonzero, then there is a nonzero probability |<x+|ψ>|2 that the outcome is spin up along x. If it does turn out that the outcome is spin up along x, then the measurement has collapsed |ψ> to |x+>
     
  4. Dec 7, 2013 #3
    Ok great, that clears things up a lot, thanks!
     
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