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

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?

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.