- #1
Gabriel Maia
- 72
- 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.
[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.