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Relation between the matrix elements of the density matrix

  1. Apr 8, 2017 #1
    Hi. I must prove that, in general, the following relation is valid for the elements of a density matrix

    [tex] \rho_{ii}\rho_{jj} \geq |\rho_{ij}|^{2}. [/tex]

    I did it for a 2x2 matrix. The density matrix is given by

    [tex] \rho = \left[ \begin{array}{cc} \rho_{11} & \rho_{12} \\ \rho^{\ast}_{12} & \rho_{22} \end{array}\right]. [/tex]

    Now, the trace of the square of the density matrix is

    [tex] \rho_{11}^2 + \rho_{22}^{2} + 2|\rho_{12}|^{2} \leq 1 \\ (\rho_{11} + \rho_{22})^2 + 2|\rho_{12}|^{2} - 2\rho_{11}\rho_{22} \leq 1[/tex]

    Because the sum of the diagonal elements of the density matrix is 1 we have that

    [tex] |\rho_{12}|^{2} \leq \rho_{11}\rho_{22} [/tex]

    This is what I've done so far but I have no idea how to prove it in the general way of

    [tex] \rho_{ii}\rho_{jj} \geq |\rho_{ij}|^{2}. [/tex]

    If, for example, the density matrix is a 3x3 matrix this means that this inequality is valid for any two elements of the diagonal. Do you know how can I approach this? Thank you.
     
  2. jcsd
  3. Apr 13, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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