How do I show the separability of this density matrix?

Question

I am stuck since a longer time regarding this exercise where I need to work with a density matrix of the given form
$$displaystyle rho_{AB}(X)= frac{1}{N+text{tr}X^2}left( {begin{array}{cc}
    I & X 
   X & X^2 
  end{array} } right)$$
where $X$ is an $Ntimes N$ hermitian matrix in the quantum system $B$ with the Hilbert space $mathbb{C}^N$, $I$ is the identity matrix and $rho_{AB}$ a state on the quantum system $AB$ with Hilbert space $mathbb{C}^2 otimes mathbb{C}^N$.
Question: How do I show now, that $rho_{AB}(X)$ is separable by using the spectral theorem on $X$?

Adam Zalcman · Accepted Answer

Let $|irangle$ be the eigenbasis of the Hilbert space of system B which diagonalizes $X$
$$
X = sum_{i=1}^N x_i |iranglelangle i|
$$
where $x_i in mathbb{R}$ and define
$$
|psi_krangle = frac{1}{sqrt{1 + x_k^2}} (|0rangle + x_k |1rangle)
$$
$$
rho_k = |psi_kranglelanglepsi_k| = frac{1}{1 + x_k^2}begin{pmatrix}
1 & x_k 
x_k & x_k^2
end{pmatrix}.
$$
Now, note that
$$
rho_{AB}(X) = sum_{i=1}^N lambda_i rho_i otimes |iranglelangle i|tag1
$$
where
$$
lambda_i = frac{1 + x_i^2}{N + mathrm{tr}X^2}
$$
and the right hand side of $(1)$ is separable.

In order to see the equality $(1)$, consider each of the four blocks of the matrix $rho_{AB}(X)$ in turn. The top left block is
$$
langle 0|rho_{AB}(X)|0rangle = frac{1}{N + mathrm{tr}X^2} I.
$$
The bottom right block is
$$
langle 1|rho_{AB}(X)|1rangle = frac{x_i^2}{N + mathrm{tr}X^2} sum_{i=1}^N |iranglelangle i| = frac{1}{N + mathrm{tr}X^2} sum_{i=1}^N x_i|iranglelangle i| x_i|iranglelangle i| = frac{1}{N + mathrm{tr}X^2} X^2.
$$
The top right block is
$$
langle 0|rho_{AB}(X)|1rangle = frac{x_i}{N + mathrm{tr}X^2} sum_{i=1}^N |iranglelangle i| = frac{1}{N + mathrm{tr}X^2} sum_{i=1}^N x_i|iranglelangle i| = frac{1}{N + mathrm{tr}X^2} X
$$
and similarly for the bottom left block.

How do I show the separability of this density matrix?

One Answer

Add your own answers!

Ask a Question