A Proving Subadditivity of Entropy for Uncorrelated Systems in Pure States

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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.
 
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ergospherical said:
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.)
 
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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!).
 
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