Relation between the matrix elements of the density matrix

[Gabriel Maia]

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

$$\rho_{ii}\rho_{jj} \geq |\rho_{ij}|^{2}.$$

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

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

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

$$\rho_{11}^2 + \rho_{22}^{2} + 2|\rho_{12}|^{2} \leq 1 \\ (\rho_{11} + \rho_{22})^2 + 2|\rho_{12}|^{2} - 2\rho_{11}\rho_{22} \leq 1$$

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

$$|\rho_{12}|^{2} \leq \rho_{11}\rho_{22}$$

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

$$\rho_{ii}\rho_{jj} \geq |\rho_{ij}|^{2}.$$

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.

 

[mark726]

Thank you for sharing your progress in proving the relation \rho_{ii}\rho_{jj} \geq |\rho_{ij}|^{2} for a 2x2 density matrix. Your approach is correct and can be extended to prove the relation for any size density matrix.

To prove the relation for a general density matrix, we can start by considering the trace of the square of the density matrix. This will give us:

Tr(\rho^2) = \sum_{i=1}^{n} \rho_{ii}^2 + 2\sum_{i<j}^{n} |\rho_{ij}|^2

Where n is the dimension of the density matrix. We can rewrite this as:

Tr(\rho^2) = \sum_{i=1}^{n} \rho_{ii}^2 + \sum_{i<j}^{n} |\rho_{ij}|^2 + \sum_{i>j}^{n} |\rho_{ij}|^2

Since the trace of the density matrix is equal to 1, we can rewrite the first term as:

Tr(\rho^2) = \sum_{i=1}^{n} (1 - \sum_{j\neq i}^{n} \rho_{ij}^2) + \sum_{i<j}^{n} |\rho_{ij}|^2 + \sum_{i>j}^{n} |\rho_{ij}|^2

Expanding the squares and rearranging the terms, we get:

Tr(\rho^2) = n - \sum_{i,j=1}^{n} \rho_{ij}^2 + 2\sum_{i<j}^{n} |\rho_{ij}|^2

Using the Cauchy-Schwarz inequality, we know that:

|\rho_{ij}|^2 \leq \rho_{ii}\rho_{jj}

Substituting this in the above equation, we get:

Tr(\rho^2) \leq n - \sum_{i,j=1}^{n} \rho_{ij}^2 + 2\sum_{i<j}^{n} \rho_{ii}\rho_{jj}

Since the trace of the density matrix is equal to 1, we can rewrite this as:

1 \leq n - \sum_{i,j=1}^{n}