Calculate the von Neumann Entropy of a two-qubit entangled state

Question

After working through an exercise I got a confusion answer/solution that either may be because I've made a mistake or I'm not understanding von Neumann Entropy.

I have the two qubit system $$ | \psi \rangle = \cos(\theta) \rangle 1 \rangle | 0 \rangle + \sin(\theta) | 0 \rangle | 1 \rangle $$ And I want to calculate the von Neumann entropy

$$ S = - Tr ( \rho_A \ln \rho_A ) $$

I got $\rho_A$ as

$$ \rho_A = \begin{pmatrix} \cos^2 (\theta) & 0 \\ 0 & \sin^2 (\theta) \end{pmatrix} $$

Thus the von Neumann entropy as

$$ S = - Tr ( \begin{pmatrix} \cos^2 (\theta) & 0 \\ 0 & \sin^2 (\theta) \end{pmatrix} \ln \begin{pmatrix} \cos^2 (\theta) & 0 \\ 0 & \sin^2 (\theta) \end{pmatrix} ) $$ The natural log of the matrix is $$ \ln \rho_A = PDP^{-1} $$ $$ = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \ln(\cos^2 (\theta)) & 0 \\ 0 & \ln(\sin^2 (\theta)) \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$ $$ = \begin{pmatrix} \ln(\cos^2 (\theta)) & 0 \\ 0 & \ln(\sin^2 (\theta)) \end{pmatrix} $$ $$ S = - Tr ( \begin{pmatrix} \cos^2 (\theta) \cdot \ln(\cos^2 (\theta)) & 0 \\ 0 & \sin^2 (\theta) \cdot \ln(\sin^2 (\theta)) \end{pmatrix} ) $$

Perhaps I am not understanding but this answer does not make any sense? The trace of that matrix cannot be simplified further and I don't get an actual answer that doesn't depend on $\theta$. Also in this case the values of $\theta$ which make $\rho_A$ a pure state is any $\theta$ and the same can be said for values of $\theta$ which make is a maximally mixed state?

Answer 1

You are calculating the entropy of one of the marginal states and so you would not expect the answer to be independent of $\theta$, except in the case that $|\psi\rangle = |\phi_A\rangle \otimes |\phi_B\rangle$ -- this will be true whenever $\theta = \frac{k \pi}{2}$ for some $k \in \mathbb{Z}$. In this case the reduced state $\rho_A$ is also pure. However, just because $\rho_{AB} = |\psi\rangle \langle \psi |$ is a pure state, does not mean that $\rho_A = \mathrm{Tr}_B[\rho_{AB}]$ will also be pure.

The calculations you performed are all correct.