# What is the Von Neumann entropy of $\rho = \sum_ip_i|i\rangle\langle i| \otimes \rho_i$?

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 $$S(\rho) = \sum _i p_iS(\rho_i) - \sum_ip_i\text{ln}p_i = \sum _i p_iS(\rho_i) + S(\overline{p})$$
• Does $S(|i\rangle \langle i |) = 0$? What is the reason that $p_iS(|i\rangle\langle i | + S(\rho_i)) = p_iS(\rho_i)$? – the mmmPodcast Apr 21 '20 at 14:09
• Sure, $| i\rangle \langle i |$ is a pure state – Danylo Y Apr 21 '20 at 14:15