# Coherent Information and Entanglement Breaking channels

The book by John Watrous, "The Theory of Quantum Information" is an exciting read for anyone wanting to research quantum information theory. The following question presumes some background covered in the book, which I will do my best to explain.

Question: If $$\mathcal{X}_0, \mathcal{X}_1, \mathcal{Y}_0, \mathcal{Y}_1$$ are complex Euclidean spaces (Finite dimensional Hilbert spaces on complex field) and $$\Phi_0 \in C(\mathcal{X}_0,\mathcal{Y}_0)$$ $$\Phi_1 \in C(\mathcal{X}_1,\mathcal{Y}_1)$$ are two quantum channels, with $$\Phi_0$$ being entanglement breaking, then show that $$\begin{equation} I_C(\Phi_0\otimes\Phi_1) = I_C(\Phi_1). \end{equation}$$

Background: Here, the space of channels $$C(\mathcal{X}_0, \mathcal{Y}_0)$$ consists of linear, completely positive trace-preserving (CPTP) maps on $$L(\mathcal{X}_0)$$ (linear operators on $$\mathcal{X}_0$$) returning operators in $$L(\mathcal{Y}_0)$$. For any channel $$\Phi$$ and state $$S$$ (states are non-negative operators with unit trace), the coherent information $$I_C(\Phi;S) = H(\Phi(S)) - H((\Phi\otimes I)(vec(\sqrt{S})vec(\sqrt{S})^\dagger))$$ and $$I_C(\Phi) = \max_S I_C(\Phi;S)$$, where the maximum is taken with respect to all states. For any operator $$A = \sum_{i,j}a_{ij}|e_i\rangle\langle f_j|$$, $$vec(A) = \sum_{i,j}a_{ij}|e_i\rangle \otimes |f_j\rangle$$.

A separable state is a state of the form $$\sum_i p_i P_i\otimes Q_i$$, where $$P_i$$ and $$Q_i$$ are states with $$\sum_i p_i = 1$$ with $$p_i\ge 0$$. States that do not have this kind of representation are called entangled. An entanglement breaking channel is one whose output state is always separable even if the input is entangled.

There is a theorem that may prove useful: The following are equivalent.

1. $$\Phi$$ is entanglement breaking (EB).
2. For any finite-dimensional Hilbert space $$\mathcal{Z}$$, $$\Phi\otimes I_\mathcal{Z}$$ is separable where $$I_\mathcal{Z}$$ is the identity map in $$\mathcal{Z}$$.
3. There exist states $$\rho_i$$ and non-negative operators $$M_i$$ such that $$\sum_{i}M_i = I$$ (called a POVM) such that $$\Phi(S) = \sum_i \rho_i Tr(SM_i)$$.

I was able to show $$I_C(\Phi_0 \otimes\Phi_1) \ge I_C(\Phi_0) + I_C(\Phi_1)$$ for any two channels $$\Phi_0$$ and $$\Phi_1$$. The book does prove that $$I_C(\Phi) \le 0$$ for $$\Phi$$ EB. But I've been unable to make any progress beyond that.

I will be happy to offer any clarification if required. This problem has been bugging me for quite some time now and I cannot see a way through. This problem would interest students and professors working in Quantum Information Theory.

Cross-posted on math.SE

• My suggestion is that you first try to solve the problem for the case that $\Phi_0$ is quantum-to-classical. – John Watrous Jul 4 at 17:09
• Wow! I did not expect a response from the author. Thanks for the hint. I will try it for the q-c case. – K Gautam Shenoy Jul 4 at 17:29
• Just a remark that if $\Phi_0$ is EB then you must have $I_C(\Phi_1)=0$ and so you already have the inequality $I_C(\Phi_0\otimes \Phi_1)\geq I_C(\Phi_1)$. Perhaps you can derive a contradiction by supposing the strict inequality holds? – Condo Aug 2 at 18:27