$\newcommand\ket[1]{|#1\rangle}$
$\newcommand\bra[1]{\langle #1|}$
$\newcommand\mean[1]{\langle #1\rangle}$
Let us consider an entangled state $\mathcal{H} = \mathcal{H_1}\otimes\mathcal{H_2}$. In this state, one defines the density operator $\rho$ as $\rho=\sum_i\rho_i\ket{i}\bra{i}$. It is a well-known result that the mean value of $A_1\otimes\mathcal{I}_2$ (where $\mathcal{I}_2$ is the identity in $\mathcal{H}_2$) would be given by
\begin{equation}
\mean{\mathcal{A}_1\otimes\mathcal{I}_2} = \mathrm{Tr}\left[\rho\left(\mathcal{A}_1\otimes\mathcal{I}_2\right)\right] = \mean{\mathcal{A}_1}_{\rho_1} = \mathrm{Tr}_{\mathcal{H}_1}\left[\rho_1\mathcal{A}_1\right]
\end{equation}
where $\rho_1 = \mathrm{Tr}_{\mathcal{H}_2}\rho$ is the reduced density matrix on the subsystem $\mathcal{H}_1$ and $\mathrm{Tr}_{\mathcal{H}_i}$ is the partial trace over the Hilbert space $\mathcal{H}_i$. Let us consider $\rho_1$ and $\rho_2$.
One defines the Von Neumann entropy as
\begin{equation}
S(\rho) = -\mathrm{Tr}\left[\rho \log\rho\right]
\end{equation}
One of the many-properties of this entropy is that – for any entangled state, $S(\rho_1\otimes\rho_2) = S(\rho_1)+S(\rho_2)$.
I have proved it the following way. I define ${\ket{1:\alpha,2:a}}$ as the basis of $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$. In this basis, one writes
\begin{align}
S(\rho_1\otimes\rho_2) = – \mathrm{Tr}\left[\rho_1\otimes\rho_2\; ln\left(\rho_1\otimes\rho_2\right)\right] &= -\sum_{\alpha,a}\bra{1:\alpha,2:a}\rho_1\otimes\rho_2\ket{1:\alpha,2:a}\log(\rho_1\otimes\rho_2)\\
\end{align}
I use now the fact that the logarithm of a product is equal to the sum of the logarithms: $\log(a b) = \log(a)+\log(b)$. Am I allowed to generalize this to the tensor product of two states? Using this property, I find that
\begin{equation}
S(\rho_1\otimes\rho_2) = -\sum_\alpha \bra{1:\alpha}\rho_1\ket{1:\alpha} + \left(-\sum_\beta \bra{2:a}\rho_2\ket{2:a}\log(\rho_2)\right) = S(\rho_1)+S(\rho_2)
\end{equation}
It feels like cheating, yet it seems to work. Any insights?
Additionally, I have trouble interpreting this result. Thanks for your input!
Best Answer
Remarks on your calculations $\newcommand{Tr}{\operatorname{Tr}}$
The last equality does not make sense, as $\log(\rho_1\otimes \rho_2)$ is an operator : it should come before the ket (see my calculations below).
Given a hermitian operator $\hat \Lambda$ and a function $f$, we can define $f(\Lambda)$ the following way : if $|i\rangle$ is an orthonormal basis such that $\hat\Lambda|i\rangle = \lambda_i|i\rangle$, then we define $f(\lambda)$ as the linear operator such that : $$f(\hat\Lambda) |i\rangle = f(\lambda_i) |i\rangle$$ (It turns out that this operator does not depend on the eigenbasis we chose.)
Since $f(\lambda) = \lambda \log(\lambda)$ is well-defined on $[0,+\infty)$ (by setting $f(0) = 0$, we can define $\hat \Lambda \log \hat \Lambda$ if $\hat \Lambda$ is positive semi-definite (which is true for density matrices).
Using bases of eigenvectors
Let $|i\rangle_1$ and $|x\rangle_2$ be orthonormal bases of eigen-vectors of $\rho_1$ and $\rho_2$ respectively, with : $$\rho_1 |i\rangle_1 = p_i|i\rangle_1 \qquad \text{and}\qquad \rho_2|x\rangle_2 = q_x |x \rangle_2$$
Then, $|i\rangle_1 \otimes |x\rangle_2$ is a basis of $\mathcal H_1\otimes \mathcal H_2$, so we can use it to compute the trace : \begin{align} S(\rho_1\otimes \rho_2) &= -\Tr(\rho \log\rho) \\ &= -\sum_{i,x}\langle i|_1 \otimes \langle x|_2(\rho_1\otimes \rho_2\log(\rho_1\otimes \rho_2)) |i\rangle_1\otimes |x\rangle_2 \\ &= -\sum_{i,x} p_i q_x \log (p_i q_x) \\ \end{align} From this point on, we are only dealing with real numbers, so we get : \begin{align} S(\rho_1\otimes \rho_2) &= -\sum_{i} p_i \log (p_i)-\sum_{x} q_x \log ( q_x) \\ &= S(\rho_1) + S(\rho_2) \end{align}
What is $\log( \hat A\otimes \hat B)$ ?
In the intermediate steps of the calculation above, we showed that if $\hat A$ and $\hat B$ are positive definite hermitian operators on $\mathcal H_1$ and $\mathcal H_2$, then : $$\log (\hat A \otimes \hat B) = \log(\hat A)\otimes \mathbb I_2 + \mathbb I_1 \otimes\log(\hat B) $$
where $\mathbb I_1$ and $\mathbb I_2$ are the identity operators on $\mathcal H_1$ and $\mathcal H_2$.
Using this formula, we can redo the calculations directly, without using a basis : \begin{align} S(\rho_1\otimes \rho_2) &= -\Tr(\rho_1\otimes \rho_2 \log(\rho_1 \otimes \rho_2) )\\ &= -\Tr(\rho_1\otimes \rho_2 \cdot( \log(\rho_1) \otimes \mathbb I_2 + \mathbb I_1 \otimes \log \rho_2) )\\ &= -\Tr ( (\rho_1 \log \rho_1) \otimes \rho_2 + \rho_1 \otimes (\rho_2 \log \rho_2)) \\ &= -\Tr ( \rho_1 \log \rho_1))\Tr( \rho_2) - \Tr(\rho_1)\Tr(\rho_2 \log \rho_2)) \\ &= S(\rho_1) + S(\rho_2) \end{align}