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.

  • 2
    $\begingroup$ 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>. $\endgroup$
    – Rammus
    Commented Feb 1, 2022 at 12:50

1 Answer 1


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. ;)

  • $\begingroup$ That looks cool! Could you suggest some reference(s) regarding the above 3 steps? $\endgroup$
    – User101
    Commented Feb 1, 2022 at 21:27
  • $\begingroup$ 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. $\endgroup$
    – jecado
    Commented Feb 1, 2022 at 22:27
  • 1
    $\begingroup$ @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 $\endgroup$ Commented Feb 2, 2022 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.