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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?

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    $\begingroup$ In matlab, use "logm". $\endgroup$ – Norbert Schuch Sep 28 at 11:43
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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 $\infty$.

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You are running into problems because $\rho$ is not a density operator. A mixed state density operator has $\text{tr}(\rho^2) < 1$, but even a mixed state density operator must have $\text{tr}(\rho)=1$. This is necessary because $\text{tr} (\rho) = \sum \limits_i p_i \, \text{tr}\left(\vert \psi_i \rangle \langle \psi_i \vert \right) = \sum \limits_i p_i = 1$.

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  • $\begingroup$ You're right, but again, how would I calculate the logarithm of sigma? $\endgroup$ – Hasan Iqbal Sep 28 at 0:48
  • $\begingroup$ Entropy in this context measures deviation from a pure state so the entropy of $\sigma$ is zero. The lack of disorder in a pure state density matrix manifests as idempotence, $\sigma^2 = \sigma$. $\endgroup$ – ChainedSymmetry Sep 28 at 1:22

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