Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There is a well-known theorem, firstly obtained by Denes Petz, in quantum information theory, which is described as follows:

$\mathbf{Theorem.}$ Let $\rho$ and $\sigma$ be two states on $\mathcal H$, $\Phi$ be a quantum channel defined over $\mathcal H$. If $\mathrm{supp}(\rho)\subseteq\mathrm{supp}(\sigma)$, then \begin{eqnarray} S(\rho||\sigma) = S(\Phi(\rho)||\Phi(\sigma))\quad\text{if and only if}\quad \Phi^\dagger_\sigma\circ\Phi(\rho) = \rho, \end{eqnarray} where \begin{eqnarray} \Phi^\dagger_\sigma(*) = \sigma^{1/2}\Phi^\dagger(\Phi(\sigma)^{-1/2} * \Phi(\sigma)^{-1/2}) ) \sigma^{1/2}. \end{eqnarray}

This theorem deals with the saturation of the monotonicity inequality of relative entropy.

My question is: Given arbitrary two states $\rho,\sigma$ on $\mathcal H$, can we estimate the fidelity $F(\rho,\Phi^\dagger_\sigma\circ\Phi(\rho))$ between two states $\rho$ and $\Phi^\dagger_\sigma\circ\Phi(\rho)$? Here for $\Phi^\dagger_\sigma$, you are referred to the above definition.

The notations mentioned above are explained as follows:

(1) $\mathcal H$ is a finite-dimensional Hilbert space.

(2) a state means a positive semi-definite operator on $\mathcal H$ with a unite trace such as $\rho,\sigma$.

(3) $\Phi$ is a trace-preserving completely positive map, i.e. quantum channel.

(4) $\mathrm{Ad}_M$ is a conjugate map, defined by $\mathrm{Ad}_M(X) = MXM^\dagger$.

(5) the notation $\mathrm{supp}(\rho)$ is the support space of the state $\rho$.

(6) The relative entropy between two states $\rho$ and $\sigma$ on $\mathcal H$ is defined as: $$ S(\rho||\sigma) = \mathrm{Tr}(\rho(\log\rho - \log\sigma)), $$ where $\mathrm{supp}(\rho)\subseteq \mathrm{\sigma}(\sigma)$.

(7) The fidelity between two states $\rho$ and $\sigma$ is defined by $$ F(\rho,\sigma) = \mathrm{Tr}(\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}). $$

(8) The operator power is taken on the support of this operator.

Another questions:

In fact, we can also estimate the relative entropies of both states: $$ S(\rho||\Phi^\dagger_\sigma\circ\Phi(\rho))\qquad S(\Phi^\dagger_\sigma\circ\Phi(\rho)||\rho). $$ Or the trace-distance $\|\rho - \Phi^\dagger_\sigma\circ\Phi(\rho))\|_1$.

share|improve this question
    
The resolution of this problem can give a good estimate of conditional mutual information. –  Lin Zhang Apr 13 '13 at 23:55
    
I find a work [Koenraad M.R. Audenaert, On the asymmetry of the relative entropy, arXiv:1304.0409], deal with the amount of asymmetry by providing a sharp upper bound in terms of two parameters the trace norm distance between the two states, and the smallest of the smallest eigenvalues of both states. This amounts to give the estimate of the following quantity: $$ |S(\rho||\Phi^\dagger_\sigma\circ\Phi(\rho)) - S(\Phi^\dagger_\sigma\circ\Phi(\rho)) ||\rho)|. $$ –  Lin Zhang Apr 15 '13 at 1:40
    
Aren't there exact formulas for all the quantities above? Why do we need any estimation? –  Peter Shor Apr 19 '13 at 21:05
    
In fact, I mean all the quantities above are known. On the one hand, I am interested in a univeral lower bound of $S(\rho||\sigma) - S(\Phi(\rho)||\Phi(\sigma))$. Since if the lower bound is obtained, we can use it to lower bounding the quantum conditional mutual information, defined by $$ I(A:C|B)_\rho := S(\rho_{AB}) + S(\rho_{BC}) - S(\rho_{ABC}) - S(\rho_B). $$ That is because the conditional mutual information can be represented by the difference of two relative entropy. On the other hand, I would like to consider the perturbation of Markov chain states, i.e. $I(A:C|B)_\rho =0$ –  Lin Zhang Apr 21 '13 at 5:18
    
Along with the above line, I proposed an approach---self-commutator--- to this problems. The following link can be visited: arxiv.org/abs/1212.5023v2 arxiv.org/abs/1010.1750 arxiv.org/abs/1210.3181 arxiv.org/abs/1210.4720v3 –  Lin Zhang Apr 21 '13 at 5:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.