What information does the Von Neumann entropy give for mixed states?

Von Neumann entropy is defined as $$S=-\mathrm{Tr}\left(\rho \ln\rho\right)$$ It can be used to measure the entanglement between two sub-systems, provided that the total system is in pure state.

1) What information does the Von Neumann entropy give if the total system is in a mixed state? As NorbertSchuch briefly explained in the post How useful is entanglement entropy? it should "measure classical entropy of the subsystem, classical correlations with the other system, and entanglement". Can someone provide an example and a deeper explanation?

2) If one wants to compute the entanglement between two sub-systems of a system which is in a mixed state, what measure should be employed? I guess that such measure should, of course, filter out the classical (i.e. purely statistical) disorder. Is it correct?

• Regarding your second question, for mixed states the Peres-Horodecki criterion provides a necessary but not sufficient criterion for a state to be separable. In short, it states that if the partial transpose of the density operator has negative eigenvalues then the state cannot be separable. For certain dimensions, such as 2x2 and 2x3, this criterion is also sufficient, so can be used as a definite test of entanglement in these cases. – Oliver Lunt Dec 17 '17 at 15:22
• Related to my previous comment, the negativity $\mathcal{N}$ provides an entanglement measure that can be applied to mixed states. In words, it is the sum of the absolute values of the negative eigenvalues of $\rho^{T_{A}}$. Of course, since the P-H criterion is not generally sufficient, $\mathcal{N}$ can be 0 even for entangled states, so it is not a perfect entanglement measure. As far as I know, the main reason it receives attention is because it is easy to compute, unlike many other more sophisticated entanglement measures. – Oliver Lunt Dec 17 '17 at 15:38

First of all you have to distinguish the classical uncertainty form the quantum one. The density matrix can be written as $$\rho=\sum w_i|\alpha_i\rangle\langle\alpha_i|$$ where the $w_i$ can be the classical probabilities, if you can't say exactly where the state is in the Hilbert space, or the quantum ones, if you don't want (or can't) write the state as a definite vector of the Hilbert space.

Classical

As an example take a two level system and consider the case in which you classically can't say in which one of the two states the system is. In that case your density matrix will be $$\rho=\frac{1}{2}|0\rangle\langle0|+\frac{1}{2}|1\rangle\langle1|$$ and your entropy will be $S=-Tr(\rho\log\rho)=\log2$. This entropy is a measure of your classic uncertainty about the state and has nothing to do with quantum uncertainty.

Quantum

Now consider a system composed as the two two level systems of above each one with his Hilber space $\mathcal{H}_1$ and $\mathcal{H}_2$. If you want to write the total Hilbert space you have to do the tensor product and you will have some vectors that can't be write in the separate Hilbert spaces. A famous example is one of the Bell's states $$|\psi\rangle=\frac{1}{\sqrt{2}}\left(|0\rangle\otimes|1\rangle+|1\rangle\otimes|0\rangle\right)~.$$ While a total state can be always written in base where his state is one of the base itself and so its density matrix will be like $$\rho=\begin{pmatrix}1&...&0\\.&...&.\\0&...&0\end{pmatrix}$$ and its entropy $\rho=-Tr(\rho\log\rho)=0$, this is not true for a single subsystem of the Bell's state for which the density matrix will be obtained with the trace operator on the Hilber sbace of the other subsystem acting on the total density matrix: $$\rho_{A}=Tr_B(\rho)=\frac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix}~.$$

In this case the Von Neumann entropy will be $S=-Tr(\rho\log\rho)=\log2$. This entropy measures the quantum correlations between the two subsystems and are always different from zero if the state you are considering is somehow interacting with something else (another subsystem or an environment).

From the classical and the maximally entangled scenario you can see that there are not differences between the two density matrices and so, given ad arbitary density matrix you do not have a criterion to distinguish the two cases and the Von Neumann entropy will result the same. It can be show that distinguish an entangled or non-entangled density matrix is a NP-hard problem known as Quantum Separability Problem.

• Thanks a lot for this very detailed explanation. Can you please explain what does the Von Neumann entropy of a reduced sub-system give, if the total system is in a mixed state? Is it a mixing of such concepts as entanglement and classical correlation? – AndreaPaco Dec 17 '17 at 19:11
• Since a total system is always pure by definition (in the sense that is writable as a vector of the total Hilber space) it can be mixed only from the classical point of view. Aside form that yes, the entropy of a reduced density matrix contains classical and quantum correlations. – NicoPranzo Dec 18 '17 at 18:08