Let $\overline{p}$ be a probability distribution on $\{1,....,d\}$. Then let $\rho = \sum_ip_i|i\rangle\langle i| \otimes \rho_i$.

How should I take the Von-Neumann entropy of $\rho$? I know that Von-Neumann entropy is additive under the tensor product. So

$S(\rho) = S(\sum_ip_i|i\rangle\langle i| \otimes \rho_i) = S(p_i\sum_i|i\rangle\langle i|) + S(\sum_i\rho_i)$

How can I break this down further? My goal is to prove $S(\rho) = S(\overline{p}) + \sum _i p_iS(\rho_i)$ but I would just like help on how to work with the two terms I've broken $S(\rho)$ into


Operator $\rho$ is not a tensor product, it's a sum of tensor products $$ p_1|1\rangle\langle 1| \otimes \rho_1 + p_2|2\rangle\langle 2| \otimes \rho_2 + \dots + p_d|d\rangle\langle d| \otimes \rho_d. $$ This is not the same as $$ \big(\sum_ip_i|i\rangle\langle i|\big) \otimes \big(\sum_i\rho_i\big), $$ so your expansion isn't correct.

Also in general $S(A+B)\neq S(A)+S(B)$, but in this situation the supports of $|i \rangle\langle i|\otimes \rho_i$ and $|j \rangle\langle j|\otimes \rho_j$ are orthogonal, so we can write $$ S(\rho) = S(p_1|1\rangle\langle 1| \otimes \rho_1) + \dots + S(p_d|d\rangle\langle d| \otimes \rho_d) $$ Here $p_i|1\rangle\langle 1| \otimes \rho_i$ is not a density matrix because it's scaled, i.e. its trace equals $p_i<1$, so technically $S$ is not defined. But for such matrices we also can define expression $S(M) = -\sum_i \lambda_i\text{ln}\lambda_i$, where $\lambda_i$ are eigenvalues of $M$. It's easy to check that for $c>0$ and density matrix $\rho$ we have $S(c\rho) = cS(\rho) - c\text{ln}c$. So we can write $$ S(p_i|i\rangle\langle i| \otimes \rho_i) = p_iS(|i\rangle\langle i| \otimes \rho_i)-p_i\text{ln}p_i = $$ $$ = p_i\big(S(|i\rangle\langle i|) + S(\rho_i)\big)-p_i\text{ln}p_i = p_iS(\rho_i)-p_i\text{ln}p_i $$ After the summation we will have $$ S(\rho) = \sum _i p_iS(\rho_i) - \sum_ip_i\text{ln}p_i = \sum _i p_iS(\rho_i) + S(\overline{p}) $$

  • $\begingroup$ Does $S(|i\rangle \langle i |) = 0$? What is the reason that $p_iS(|i\rangle\langle i | + S(\rho_i)) = p_iS(\rho_i)$? $\endgroup$ Apr 21 '20 at 14:09
  • $\begingroup$ Sure, $| i\rangle \langle i |$ is a pure state $\endgroup$
    – Danylo Y
    Apr 21 '20 at 14:15
  • $\begingroup$ That all makes sense, thank you so much! $\endgroup$ Apr 21 '20 at 14:18

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