In Exercise 11.25, Page 522, Entropy and information, Quantum Computation and Quantum Information by Nielsen and Chuang, it is required to show that the concavity of the conditional entropy may be deduced from strong subadditivity by introducing an auxiliary system $R$ into the problem.

My understanding about the concavity of the conditional entropy is that, if $\rho_1$ and $\rho_2$ are quantum bipartite states of the system $AB$ and $\lambda\in[0,1]$ then $$ S(A|B)_{\lambda\rho_1+(1-\lambda)\rho_2}\ge\lambda S(A|B)_{\rho_1}+(1-\lambda)S(A|B)_{\rho_2} $$ The strong subadditivity is given by the inequality, $S(A,B,C)+S(B)\le S(A,B)+S(B,C)$

We can possibly use the properties: $$ \rho\text{ is a pure state}\iff S(\rho)=0\\ AB\text{ is a pure state}\implies S(A)=S(B)\\ S(A|B)=S(A,B)-S(B)\\ $$

If we introduce an auxiliary system $R$ to purify our quantum system $ABC$ such that $\rho^{ABCR}$ is a pure state, then $$ S(A,B,C)=S(R) $$

It'd be helpful if one can point me in the right direction to obtain the concavity condition of the relative quantum entropy.

Note: Concavity of Conditional Entropy can be shown from the Joint Convexity of Relative Entropy

My Attempt

We have $\rho^{AB}=\lambda\rho_1^{AB}+(1-\lambda)\rho_2^{AB}$ and it is required to prove that, $$ S(A|B)_{\lambda\rho_1+(1-\lambda)\rho_2}\ge\lambda S(A|B)_{\rho_1}+(1-\lambda)S(A|B)_{\rho_2}\\ S(A|B)_{\rho^{AB}}\ge\sum_i p_iS(A|B)_{\rho^{AB}_i} $$

Lemma: Suppose $p_i$ are probabilities, $|i\rangle$ are orthogonal states for a system $A$, and $\rho_i$ is any set of density operators for another system $B$, then $$ S(\sum_ip_i\rho_i\otimes |i\rangle\langle i|)=H(p_i)+\sum_i p_iS(\rho_i) $$ proof: $$ \rho_{AB}=\sum_ip_i\rho_i\otimes |i\rangle\langle i|\quad\&\quad \rho_i|\lambda_i^j\rangle=\lambda_i^j|\lambda_i^j\rangle\\ (\rho_i\otimes|i\rangle\langle i|)(|\lambda_i^j\rangle\otimes |i\rangle)=\rho_i|\lambda_i^j\rangle\otimes|i\rangle=\lambda_i^j(|\lambda_i^j\rangle\otimes|i\rangle)\\ \rho_{AB}(|\lambda_i^j\rangle\otimes|i\rangle)=(\sum_ip_i\rho_i\otimes |i\rangle\langle i|)(|\lambda_i^j\rangle\otimes |i\rangle)=p_i\lambda_i^j(|\lambda_i^j\rangle\otimes |i\rangle)\\ $$ $\implies$ The eigenvectors of $\rho_{AB}=\sum_ip_i\rho_i\otimes |i\rangle\langle i|$ are $|i\rangle\otimes|\lambda_i^j\rangle$ with eigenvalue $p_i\lambda_i^j$. \begin{align} S(\sum_ip_i\rho_i\otimes |i\rangle\langle i|)&=-\sum_{i,j} p_i\lambda_i^j\log (p_i\lambda_i^j)\\ &=-\sum_{i,j}p_i\lambda_i^j(\log p_i+\log\lambda_i^j)\\ &=-\sum_ip_i\log p_i\sum_j\lambda_i^j-\sum_i p_i\sum_j\lambda_i^j\log\lambda_i^j\\ &=-\sum_i p_i\log p_i-\sum_i p_i\sum_j\lambda_i^j\log\lambda_i^j\\ &=H(p_i)+\sum_i p_iS(\rho_i) \end{align}

The strong subadditivity is, $$ S(A,B,X)+S(B)\le S(A,B)+S(B,X)\\ S(A,B,X)-S(B,X)\le S(A,B)-S(B)\\ S(A|B)\ge S(A|B,X)\\ $$ $$ S(A|B)\ge S(A|B,X)=S(A,B,X)-S(B,X)\\ $$ $$ \rho^{AB}=\sum_i p_i\rho_i^{AB}\implies\rho^{ABX}=\sum_i p_i\rho_i^{AB}\otimes|i\rangle\langle i|\\ \implies \rho^{BX}=tr_A\big(\sum_i p_i\rho_i^{AB}\otimes|i\rangle\langle i|\Big)=\sum_i p_itr_A\Big(\rho_i^{AB}\otimes|i\rangle\langle i|\Big)=\sum_i p_i\Big(\rho_i^{B}\otimes|i\rangle\langle i|\Big) $$

Using lemma, $$ S(A,B,X)=S(\rho^{ABX})=S(\sum_i p_i\rho_i^{AB}\otimes|i\rangle\langle i|)=H(p_i)+\sum_i p_iS(\rho_i^{AB})\\ S(B,X)=S(\sum_i p_i\rho_i^B\otimes |i\rangle\langle i|)=H(p_i)+\sum_i p_iS(\rho_i^{B}) $$ Substituting into the result of the strong subadditivity gives, \begin{align} S(A|B)\ge S(A|B,X)&=S(A,B,X)-S(B,X)\\ &=S(\rho^{ABX})-S(\rho^{BX})\\ &=H(p_i)+\sum_i p_iS(\rho_i^{AB})-H(p_i)-\sum_i p_iS(\rho_i^{B})\\ &=\sum_i p_iS(\rho_i^{AB})-\sum_i p_iS(\rho_i^{B})\\ &=\sum_i p_i\Big(S(\rho_i^{AB})-S(\rho_i^{B})\Big)\\ &=\sum_i p_iS(A|B)_{\rho^{AB}_i}\\ \end{align}


2 Answers 2


You start with a two bipartite states $\rho_1$ and $\rho_2$ on a system $AB$. Their convex mixture with parameter $\lambda \in [0,1]$ is another quantum state $$\rho_{AB} = \lambda \rho_1 + (1-\lambda)\rho_2$$ on $AB$. Operationally, this state corresponds to preparing $\rho_1$ with probability $\lambda$ and $\rho_2$ with probability $(1-\lambda)$ and then forgetting which state you prepared.

If we don't forget what state we prepared then we can describe the state of the system in the following way. We have a classical register $X$ where we record which state we prepared $1$ or $2$. Then the state of the whole system $XAB$ is $$ \rho_{XAB} =\lambda |1\rangle\langle 1|\otimes \rho_1 + (1-\lambda)|2\rangle \langle 2 |\otimes \rho_2. $$ Notice if we trace out the system $X$ (forget about the preparation) we recover the convex mixture above.

Now for your question, I will give you a hint but let you finish the details. By strong subadditivity you know that $$ H(A|B) \geq H(A|BX). $$ I would then try to think about how you can simplify the RHS knowing that $X$ is a classical system.

  • $\begingroup$ Thanks for your hint. I have tried to follow the proof in my attempt section in the post. Could you please have a look? $\endgroup$
    – Sooraj S
    Apr 13, 2023 at 17:39
  • $\begingroup$ Looks fine to me $\endgroup$
    – Rammus
    Apr 13, 2023 at 18:13

A concise rewriting of what has already been said: consider $\rho=\sum_i p_i \rho_i$ with $\sum_i p_i=1$, $p_i\ge0$. Here $\rho_i$ are bipartite states, with the spaces labelled with $A$ and $B$. We want to prove that $$S(A|B)_\rho \ge \sum_i p_i S(A|B)_{\rho_i}.$$ Start observing that the strong subadditivity of the von Neumann entropy is equivalent to the statement $S(A|B)_\rho \ge S(A|BX)_\rho$, for any $\rho\equiv\rho_{ABX}$, and spaces $A,B,X$. Apply this to the state $$\rho_{ABX}=\sum_i p_i (\rho_i\otimes |i\rangle\!\langle i|).$$ This gives $\rho_{AB}=\sum_i p_i \rho_i$, and $$S(A|BX) = \sum_i \operatorname{Prob}(X=i) S(A|B, X=i)_\rho = \sum_i p_i S(A|B)_{\rho_i},$$ which concludes the proof.


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