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Lieb's theorem states that, let $X$ be a matrix and $0\le t\le 1$ then the function $f(A,B)\equiv tr(X^\dagger A^tXB^{1-t})$ is jointly concave in positive matrices $A$ and $B$, ie., $$ tr\Big[X^\dagger\big(\lambda A_1+(1-\lambda)A_2\big)^tX\big(\lambda B_1+(1-\lambda)B_2\big)^{1-t}\Big]\ge\lambda tr\big( X^\dagger A_1^t XB_1^{1-t}\big)+(1-\lambda)tr\big(X^\dagger A^t_2XB_2^{1-t}\big) $$

Lemma: Let $R_1,R_2,S_1,S_2,T_1,T_2$ be positive operators such that $[R_1,R_2]=[S_1,S_2]=[T_1,T_2]=0$, and $R_1\ge S_1+T_1$ $,R_2\ge S_2+T_2$ then for all $0\le t\le 1$, \begin{align} R_1^tR_2^{1-t}\ge S_1^tS_2^{1-t}+T_1^tT_2^{1-t}\tag{A6.8}\label{A6.8} \end{align} is true as a matrix inequality.

which is proved as given in Show that if $\mu,\eta\in I\implies \dfrac{\mu+\eta}{2}\in I$ and $\{0,1,1/2\}\subset I$ implies $[0,1]\subset I$, but I think the proof is on the assumption that the operators are standard matrices.

The Lieb's theorem is proved, as given in my reference, Page 648, Appendix 6: Proof of Lieb’s theorem, Quantum Computation and Quantum Information by Nielsen and Chuang using the lemma by taking the superoperators, $$ S_1(X)=\lambda A_1X\\ S_2(X)=\lambda XB_1\\ T_1(X)=(1-\lambda)A_2X\\ T_2(X)=(1-\lambda)XB_2\\ R_1(X)=S_1+T_1\\ R_2(X)=S_2+T_2\\ $$ where $S_1$ and $S_2$, $T_1$ and $T_2$,$R_1$ and $R_2$ are said to be commute, and $A_1,A_2,B_1,B_2$ are positive operators.

My understanding is that from the definition of the given superoperators, $$ S_1(S_2(X))=S_1(\lambda XB_1)\lambda A_1.\lambda XB_1=\lambda^2A_1XB_1\\ S_2(S_1(X))=S_2(\lambda A_1X)=\lambda.\lambda A_1XB_1=\lambda^2 A_1XB_1 $$ which are equal.

Is it valid to apply the lemma to these superoperators (linear operators on matrices), as it seems to be proven by taking the operators are matrices ?

Sooraj S
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  • If the $A_i$s and $B_i$s are positive semidefinite and $\lambda\in[0,1]$, the $R_i$s, $S_i$s and $T_i$s will be positive (with respect to Frobenius inner product) and hence the lemma should be applicable. – user1551 Apr 02 '23 at 00:52
  • @user1551 I can see that, for example, $A_1$ is positive$\implies tr(X^\dagger A_1(X))\ge 0$ then $tr(X^\dagger S_1(X))=tr(X^\dagger .\lambda A_1X)=\lambda tr(X^\dagger A_1X)\ge 0\implies S_1$ is positive w.r.t the Frobenius inner product, right ? – Sooraj S Apr 02 '23 at 03:03
  • @user1551 It may be a stupid doubt, but I am confused about the commutation relation. In lemma we take $[S_1,S_2]=0$ by treating $S_1,S_2$ as matrices. But in Leib's therem it is defined as $S_1(X)=\lambda A_1X,S_2(X)=\lambda XB_1$ in terms of its action on any matrix $X$. Looks like $S_1,S_2$ can't be represented as usual matrices. Thats where my doubt comes from. – Sooraj S Apr 02 '23 at 03:15
  • Nothing except the phrase “is true as a matrix inequality” in the lemma is specific to matrices. The inequality itself is true as an one between positive operators subject to PSD partial ordering. There is no need to mention any matrix. But of course, in the context of the lemma, the inner product space is finite-dimensional. So, you may view the operators as matrices. In this case, identify $X, S_1$ and $S_1(X)$ with the column vector $\operatorname{vec}(X)$, the matrix $I\otimes\lambda A_1$, and the matrix-vector multiplication $(I\otimes\lambda A_1)\operatorname{vec}(X)$ respectively. – user1551 Apr 02 '23 at 15:39

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