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.
$\begingroup$
$\endgroup$
1
-
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>. $\endgroup$– RammusFeb 1, 2022 at 12:50
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
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:
- Factor (or "diagonalize") the matrix: $A = U\Lambda U^\dagger$, so that $U$ is a unitary transformation and $\Lambda$ is a diagonal matrix.
- Apply the function $f(x)$ on each element of $\Lambda$. Call the resulting matrix $f(\Lambda)$.
- 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:
- https://en.wikipedia.org/wiki/Logarithm_of_a_matrix
- https://en.wikipedia.org/wiki/Analytic_function_of_a_matrix
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? $\endgroup$– User101Feb 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. $\endgroup$– jecadoFeb 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 $\endgroup$ Feb 2, 2022 at 16:00