How do I compute the Relative Entropy between pure and mixed states?

Question

Let

$$        rho = begin{bmatrix} .7738 & -.0556  -.0556 & .0040 end{bmatrix} , 
        sigma = begin{bmatrix} .9454 & -.2273  -.2273 & .0546 end{bmatrix} $$

As you can see $rho$ is an operator of mixed states and $sigma$ is a density operator from a pure state. I can calculate entropy of them individually. But can I calculate the relative entropy between them? I am not sure about what it would mean. Anyways, considering the definition of relative entropy:

$$S(rho || sigma) = mathrm{tr}(rho  log (rho)) - mathrm{tr}(rho log ( sigma))$$

I know that I can calculate the entropy of $sigma$ from it's eigenvalues. But here I can't use the eigenvalue approach, can I? I have to take the logarithm I think. But there is no logarithm for $sigma$ in matlab. What can I do in this sort of cases?

DaftWullie · Answer

As @NorbertSchuch said in a comment, matlab has a function for taking the logarithm of a matrix: logm. In general, there is a standard method for calculating the function $f(sigma)$ of a matrix $sigma$. You first diagonalise the matrix:
$$
sigma=UDU^dagger,
$$
where $U$ is a unitary and $D$ is diagonal. We then say
$$
f(sigma)=Uf(D)U^dagger,
$$
where $f(D)$ simply involves calculating the function $f$ on just the diagonal elements of the matrix.

Note this means that in your particular case, since $sigma=|psiranglelanglepsi|$ corresponds to a pure state, one of the eigenvalues is 0, so unless $rho$ is an identical pure state, the answer you get will be $infty$.

How do I compute the Relative Entropy between pure and mixed states?

One Answer

Add your own answers!

Ask a Question