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


1 Answer 1


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 $p_iS(|i\rangle\langle i | + S(\rho_i)) = p_iS(\rho_i)$? $\endgroup$ Commented Apr 21, 2020 at 14:09
  • $\begingroup$ Sure, $| i\rangle \langle i |$ is a pure state $\endgroup$
    – Danylo Y
    Commented Apr 21, 2020 at 14:15
  • $\begingroup$ That all makes sense, thank you so much! $\endgroup$ Commented Apr 21, 2020 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.