What is $\underset{\rho\ \text{pure}}{\text{max}}S(\mathcal{N}^{\otimes N}(\rho))$?

Denote $S$ as the Von Neumann entropy defined $S=-\text{tr}(\rho\ln\rho)$, $\mathcal{N}$ a quantum channel (i.e., a linear, trace preserving, completely positive map) and $\rho$ is a $n$ qubit pure density matrix. I.e., there exists some $2^n$-dimensional vector $\psi$ such that $\rho=\psi^{\dagger}\psi$. Some important properties of such a density matrix: $\rho^2=\rho$ (projector), $\rho^{\dagger}=\rho$ (hermitian), $\text{Tr}(\rho)=1$ (normalization), $\rho\geq 0$ (positive), and $\text{tr}(\rho^2)=1$ (purity).

It might not be possible to give an answer to this for general $\mathcal{N}$. If not, how about the specific case where $\mathcal{N}$ is the so called depolarizing channel?

$\mathcal{N}(\rho)=\lambda\rho+\frac{(1-\lambda)}{2}I$ with $-\frac{1}{3}\leq\lambda\leq 1$

or, equivalently,

$\mathcal{N}(\rho)=(1-p)\rho +\frac{p}{3}X\rho X+\frac{p}{3}Y\rho Y+\frac{p}{3}Z\rho Z$

I haven't been able to solve this problem. I thought possibly that for an even $N$ , $N$ bell pairs might work, but it doesn't seem to be so.

I have proven that $\underset{\rho\ \text{pure}}{\text{max}}S(\mathcal{N}_{BF}^{\otimes N}(\rho))=Nh(p)$ where $\mathcal{N}_{BF}$ is the bitflip channel and $h(p)$ is the entropy of a single qubit passing through a bit flip channel.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.