A Proving Subadditivity of Entropy for Uncorrelated Systems in Pure States

[ergospherical]

Two systems A & B (with orthonormal basis $\{|a\rangle\}$ and $\{|b\rangle\}$) are uncorrelated, so the combined density operator $\rho_{AB} = \rho_A \otimes \rho_B$. Assume the combined system is in a pure state $\rho_{AB} = |\psi \rangle \langle \psi |$ where $|\psi \rangle = \sum_{a,b} c_{ab} |a \rangle |b \rangle$. The reduced density operator for A is $\rho_A = \mathrm{tr}_{H_{B}} (\rho_{AB}) = \sum_{a,a',b} c_{ab} \overline{c_{a'b}} |a \rangle \langle a'|$, and similarly for B. Now to show $S(\rho_{AB}) = S(\rho_A) + S(\rho_B)$,
\begin{align*}
S(\rho_{AB}) &= -\mathrm{tr}_{H_A \otimes H_B} (\rho_{AB} \ln \rho_{AB}) \\
&= \sum_{a,b} \langle a| \langle b| (\rho_{AB} \ln \rho_{AB}) |a \rangle |b \rangle
\end{align*}How to proceed with the trace of the logarithm? Cheers.

 

[Demystifier]

How to proceed with the trace of the logarithm?

Work in basis in which $\rho$ is diagonal! (There is also the Renyi entropy trick, but you don't need it here.)

 

[vanhees71]

You don't need any specific basis. Using the product states for taking the trace in the OP together with the product state $\hat{\rho}_{AB}$ is sufficient (you don't need to assume anything about the $\hat{\rho}_A$ and $\hat{\rho}_B$ either!).