# Why von Neumann entropy requires diagonalization and linear entropy doesn't?

The linear entropy for a state $$\rho$$ is defined as $$S_L = 1 - Tr[\rho^2]$$, while as von Neumann entropy as $$S_{N} = -Tr[\rho \ln \rho]$$. According to quantiki, the computation of $$S_{N}$$ requires diagonalization but $$S_L$$. But this is not clear to me why.

• Well how would you compute $\rho \log \rho$ and $\rho^2$? <Fun fact: technically you wouldn't actually need diagonalization to compute the logarithm of full rank density matrices, you could also do it via a power series>. Feb 1, 2022 at 12:50

The linear entropy $$S_L = 1 - {\rm Tr}[\rho^2]$$ certainly doesn't need diagonalization because $$\rho^2$$ is straightforward to calculate, as is its trace.

The von Neumann entropy $$S_N=-{\rm Tr}[\rho \ln\rho]$$ seems to require diagonalization because of the factor $$\ln \rho$$. Taking logarithms is not a "natural" operation for a matrix, and it's not at all clear at first glance what it should even mean.

But, physicists find it useful to say that any scalar function $$f(x)$$ can be applied to any matrix $$A$$ in the following way:

1. Factor (or "diagonalize") the matrix: $$A = U\Lambda U^\dagger$$, so that $$U$$ is a unitary transformation and $$\Lambda$$ is a diagonal matrix.
2. Apply the function $$f(x)$$ on each element of $$\Lambda$$. Call the resulting matrix $$f(\Lambda)$$.
3. Define the matrix $$f(A) \equiv U f(\Lambda) U^\dagger$$.

Essentially we have applied the scalar function to each element of $$A$$ when represented in its eigenbasis.

So, the very definition of $$\ln \rho$$, and therefore $$S_N$$, implicitly includes a diagonalization.

But as @Rammus points out in a comment, the joys of calculus do let us sometimes use alternate means of calculation such as Taylor series. Actually, it seems the function $$f(A)$$ is often (maybe usually!) defined in terms of the Taylor series. See for example:

I was a little embarrassed to find that even a mathematical physics textbook defines it that way. My physics teachers are perhaps a little iconoclastic. ;)

• That looks cool! Could you suggest some reference(s) regarding the above 3 steps? Feb 1, 2022 at 21:27
• That is an excellent question. In fact I've just checked three different sources and they all contradict me, defining $f(A)$ in terms of the Taylor series! ^_^ I'll edit the end of my answer to acknowledge this and give a couple links. Feb 1, 2022 at 22:27
• @jecado don't be so quick to dismiss your physics teachers: you can have operator functions that are not analytic! They are then defined by their action on eigenstates. When $A|\psi_i\rangle=\lambda_i|\psi_i\rangle$, we define operators $f(A)$ by $f(A)|\psi_i\rangle=f(\lambda_i)|\psi_i\rangle$, plain and simple. Then if $f$ happens to have a Taylor series expansion we can do other things too Feb 2, 2022 at 16:00