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?

Answer 1

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 $\sigma=|\psi\rangle\langle\psi|$ 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 $\infty$.