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Density matrix problem

  1. May 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Find condition for which ##\hat{\rho}## will be pure state density operator?
    ##\hat{\rho} = \begin{bmatrix}
    1+a_1 & a_2 \\[0.3em]
    a_2^* & 1-a_1
    \end{bmatrix}##


    2. Relevant equations
    In case of pure state ##Tr(\hat{\rho}^2)=Tr(\hat{\rho})=1##.



    3. The attempt at a solution
    Using that condition I got
    ##\hat{\rho}^2 = \begin{bmatrix}
    (1+a_1)^2+|a_2|^2 & (1+a_1)a_2+a_2(1-a_1) \\[0.3em]
    (1+a_1)a_2^*+a_2^*(1-a_1) & (1-a_1)^2+|a_2|^2
    \end{bmatrix}##
    and from that
    ## 2|a|^2+2a_1^2=-1## which can not be true. Because ##a_1## must be real, condition to ##\hat{\rho}## is hermitian.
     
  2. jcsd
  3. May 21, 2014 #2
    I don't quite understand. That density operator you stated, it's not a pure state density operator, and that's why The trace of the square of rho will not be 1. It's correct. I afraid I did not understand your question.
     
  4. May 21, 2014 #3
    Question is find condition put on ##a_1## and ##a_2## for which ##\hat{\rho}## is pure state density operator.
     
  5. May 21, 2014 #4
    The problem occurs because you have not normalized the original density operator yet.
     
  6. May 21, 2014 #5
    Tnx a lot Fightfish. I did not see that trace of given matrix is not ##1##. Density matrix must be
    ##\hat{\rho} =\frac{1}{2} \begin{bmatrix}
    1+a_1 & a_2 \\[0.3em]
    a_2^* & 1-a_1
    \end{bmatrix}##
     
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