How to calculate the log of a density matrix?

Question

In quantum information theory, calculating the log of a density operator is essential for things like the Von Neumann entropy or the entropy of entanglement. Unfortunately, this topic is considered a bit blasé and I haven't been able to find a source where the calculation is rigorously derived.

Here's what I know so far:

By the eigendecomposition rule, a density matrix $\rho$ can be written as:

$$\rho = VDV^{-1}$$

Where $D$ is a diagonal matrix with eigenvalues of $\rho$ as elements, and where $V$ is a square matrix whose columns are corresponding eigenvectors of $\rho$.

Apparently, taking the log of $\rho$ is equivalent to taking the log of each element in $D$ -- that is:

$$\log(\rho) = V\log(D)V^{-1}$$

Clearly for this to be true it must also be true that

$$ \log(VDV^{-1}) = V\log(D)V^{-1}$$

But It's not obvious to me why this should be the case. I have the feeling that I need to make use of the Taylor series expansion of $\log(\rho)$ but I'm not so confident in my ability to do this correctly. Any help is greatly appreciated.

Answer 1

By definition, a $\log$ of a matrix $M$ is a matrix $L$ such that $\exp(L)=M$. Note that I wrote a matrix because in general it may not be uniquely defined.

Recall the definition of $\exp(L)$: $$\exp(L)=\sum_{n=0}^{+\infty}\frac{L^n}{n!}$$

For a diagonal matrix its exponential is the exponential of its components, and the same is true for the log of a diagonal matrix and for powers of a diagonal matrix.

Finally, note that for any diagonal matrix $D$ and any invertible matrix $V$: $$\forall n,\left(V^{-1}DV\right)^n= V^{-1}D^nV$$

Thus: $$\begin{align}\exp\left(V^{-1}(\log D)V\right)&=\sum_{n=0}^{+\infty}\frac{\left(V^{-1}(\log D)V\right)^n}{n!}\\&=V^{-1}\left(\sum_{n=0}^{+\infty}\frac{(\log D)^n}{n!}\right)V\\&=V^{-1}\exp(\log D))V\\&=V^{-1}DV.\end{align}$$ Thus, $V^{-1}(\log D)V$ is a logarithm of $V^{-1}DV$.