The triangle inequality or Araki-Lieb inequality of the von Neumann entropy is $$ S(A,B)\ge|S(A)-S(B)| $$ this is proven by introducing a system $R$ which purifies systems $A$ and $B$. Applying subadditivity obtains $S(R)+S(A)\ge S(A,R)$.

$ABR$ is in a pure state $\implies $ $S(A,R)=S(B)$ and $S(R)=S(A,B)$. Therefore,

\begin{align} S(R)+S(A)&=S(A,B)+S(A)\ge S(A,R)=S(B)\\ &\implies S(A,B)+S(A)\ge S(B)\\ &\implies S(A,B)\ge S(B)-S(A) \end{align}

In Page 516, Quantum Computation and Quantum Information by Nielsen and Chuang, Exercise 11.16 is given as

Let $ρ^{AB}=\sum_i λ_i|i\rangle\langle i|$ is a spectral decomposition for $ρ^{AB}$. Show that $S(A,B)=S(B)−S(A)$ if and only if the operators $ρ^A_i≡\text{tr}_B(|i\rangle\langle i|)$ have a common eigenbasis, and the $ρ^B_i≡\text{tr}_A(|i\rangle\langle i|)$ have orthogonal support.

How do I approach the problem, and are there any physical explanations for the equality conditions?

My Attempt

Klein inequality:$$S(\rho||\sigma)=-S(\rho)-\text{tr}(\rho\log\sigma)= 0\implies \rho=\sigma\,.$$ Setting $\rho=\rho^{AR}$ and $\sigma=\rho^A\otimes\rho^R$ then \begin{align} S(\rho^{AR}||\rho^A\otimes\rho^R)&=-S(\rho^{AR})-\text{tr}(\rho^{AR}(\log\rho^A\otimes I_R+I_A\otimes\log\rho^R))\\ &=-S(\rho^{AR})-\text{tr}((\rho^{AR}(\log\rho^A\otimes I_R)+-\text{tr}((I_A\otimes\log\rho^R))\\ &=-S(\rho^{AR})-\text{tr}(\rho^A\log\rho^A)-\text{tr}(\rho^R\log\rho^R)\\ &=-S(\rho^{AR})+S(\rho^A)+S(\rho^R)\\ &=-S(AR)+S(A)+S(R)\\ \end{align} We used $$\log(\rho^A\otimes\rho^R)=\log\rho^A\otimes I_R+I_A\otimes\log\rho^R$$ $$\text{tr}(\rho^{AR}(\sigma^A\otimes I))=tr(\rho^A\sigma^A)$$ Please check Partial trace over a product of matrices - one factor is in tensor product form for proof.

Therefore, $$S(\rho^{AR}||\rho^A\otimes\rho^R)=0\implies S(A,R)= S(A)+S(R)\implies \rho^{AR}=\rho^A\otimes\rho^R$$

$$ S(A,R)=S(A)+S(R)\implies S(A,B)=S(B)−S(A)\implies ρ^{AR}=ρ^A\otimes ρ^R\\ \rho^A=tr_B(\rho^{AB})=\sum_i λ_itr_B(|i\rangle\langle i|)\\ \rho^B=tr_A(\rho^{AB})=\sum_i λ_itr_A(|i\rangle\langle i|)\\ $$

  • $\begingroup$ what are the last equations in the post for? Are you asking about an "equality condition" in those equation or something else? Also, > text is really for quotes, not to highlight what you're asking (in fact, it achieves the exact opposite, making the most important part of the post less evident) $\endgroup$
    – glS
    Commented Mar 4, 2023 at 21:27
  • $\begingroup$ @glS Thanks for responding. The last three lines of equations are my attempt to show the given statement "$S(A,B)=S(B)−S(A)$ if and only if the operators $ρ^A_i≡tr_B(|i\rangle\langle i|)$ have a common eigenbasis, and the $ρ^B_i≡tr_A(|i\rangle\langle i|)$ have orthogonal support." $\endgroup$
    – Sooraj S
    Commented Mar 5, 2023 at 5:32
  • $\begingroup$ @glS I am asking about the equality condition of the triangle inequality for the von Neumann entropy, ie., $S(A,B)\ge |S(A)-S(B)|$ $\endgroup$
    – Sooraj S
    Commented Mar 5, 2023 at 11:33
  • $\begingroup$ I feel like there might be a typo in the statement, or maybe I'm not understanding it correctly. As a counterexample, $\rho=\frac12(|\Psi\rangle\!\langle\Psi|+|\tilde\Psi\rangle\!\langle\tilde\Psi|)$ with $|\Psi\rangle=\frac1{\sqrt2}(|00\rangle+|11\rangle)$ and $|\tilde\Psi\rangle=\frac1{\sqrt2}(|22\rangle+|33\rangle)$. This is symmetric upon changing $A$ and $B$, thus $S(A)=S(B)$, but also $S(A,B)=1\neq S(B)-S(A)$, although it satisfies the assumptions b/c $\rho^A_1=\rho^B_1=\frac12(P_0+P_1)$ and $\rho^A_2=\rho^B_2=\frac12(P_2+P_3)$ $\endgroup$
    – glS
    Commented Mar 7, 2023 at 12:34
  • $\begingroup$ @glS I believe the condition "$\rho^A_i$ have a common eigenbasis and $\rho^B_i$ have orthogonal support" should include all the vectors $|i\rangle$ forming the complete basis. In your counter example, $\{|i\rangle\}$ should include $|e_3\rangle=(|00\rangle-|11\rangle)/\sqrt2, |e_4\rangle = \ldots$. And in this case, $\rho^B_1 = \rho^B_3$ and they do not have orthogonal support. $\endgroup$
    – Guangliang
    Commented May 23, 2023 at 21:25


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