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Is the singlet state just a special case of mixture?

  1. Oct 5, 2008 #1
    If the answer is no, why?

    Is it possible to show that, given the state spaces we have for individual particles, the singlet state is not a mixture of product states? If yes, how?
    Last edited: Oct 5, 2008
  2. jcsd
  3. Oct 6, 2008 #2


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    I guess you are talking about the 2-particle state:
    |up>|down> - |down> |up>

    and you wonder whether this is any different from a mixture of, say, 50% |up>|down> and 50% |down> |up>.

    The short answer is that the difference is exactly what violates the Bell inequalities. The mixture wouldn't, the singlet state does. At least in an ideal experiment.

    The long answer is that the density matrix is different. In the basis of the two-particle space:

    { |up> |up> ,
    |up> |down>,
    |down> |up>,
    |down>|down> }

    the singlet state has a non-diagonal matrix, while the mixture has a diagonal matrix. Note that the diagonal elements of both are the same (0, 1/2, 1/2, 0).
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