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A question on reduced density matrix

  1. Oct 18, 2011 #1
    I have worked out a problem on reduced density matrix, but my final expression looks kind of weird to me. I would appreciate someone can point out where I have gone wrong.

    1. The problem statement, all variables and given/known data

    The problem goes like this: Consider a two spin 1/2 particles. Their Hilbert space is spanned bu |1>|1>, |1>|0>, |0>|1> and |0>|0>. Supposed the two particles are in a pure state:

    |ψ> = (1/√N) (|1>|1> + |1>|0> + |0>|0>)

    Determine the normalisation N and the reduced density matrix after averaging over the 2nd spin. Show that the reduced density matrix corresponds to a mixed state.

    2. Relevant equations
    3. The attempt at a solution

    For the normalisation part, I put

    <ψ|ψ> = 1/N (1 + 1 + 1) = 1

    Therefore, N=3.

    Density matrix, ρ = |ψ><ψ|
    = (1/N) (|1>|1> + |1>|0> + |0>|0>)(<1|<1| + <1|<0| + <0|<0|)
    = (1/N) (|1>|1><1|<1| + |1>|1><1|<0| + |1>|1><0|<0|
    + |1>|0><1|<1| + |1>|0><1|<0| + |1>|0><0|<0|
    + |0>|0><1|<1| + |0>|0><1|<0| + |0>|0><0|<0|)

    To find the reduced density matrix, I used


    which will leave me with only those terms where the second ket is in the same state as the second bra (i.e. both are 1 or 0).

    This leaves me with: [itex]ρ_{1}[/itex] = (1/N) (2|1><1| + |1><0| + |0><1| + |0><0|)

    I can see that this is a mixed state. The part I find weird is the 2nd and the 3rd term, i.e. |1><0| and |0><1|. For a density matrix of a single state (which the reduced density matrix is supposed to represent in a way), I thought it does not make sense for the ket and the bra to be in different states. Furthermore, I have earlier calculated that N=3. If my final expression is correct, the coefficient of all the four terms would have exceeded 1.

    Can someone point out to me where I went wrong?
  2. jcsd
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