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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.

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    $\begingroup$ 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

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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. ;)

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  • $\begingroup$ 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
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    $\begingroup$ @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

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