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.

  • $\begingroup$ you could argue that's a way to define what the log of a matrix is. So asking "why" in this case could mean to ask why it's compatible with whatever other definition you have in mind (in which case, you should probably spell out what is it) or to ask why this is a meaningful definition $\endgroup$
    – glS
    Dec 23, 2022 at 11:11

1 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 $V^{-1}DV$.


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