# Prove that the coherent information of an antidegradable channel is equal to zero

I want to show that antidegradable channels have zero coherent information, based on Exercise 13.5.6 in [1]. So the solution should use the following relationship: For Hilbert spaces $$R, B, E$$, a pure state $$|\phi\rangle_{RBE} \in RBE$$ obeys (Exercise 11.6.6 in [1]): $$I(R\rangle B)_\phi = \frac{1}{2}(I(R:B)_\phi - I(R:E)_\phi) \tag{1}$$ where $$I(\cdot \rangle \cdot)$$ denotes coherent information.

My work so far:

Given an antidegradable channel $$\mathcal{N}$$ with isometric extension $$V: A \rightarrow BE$$, we start with an entangled state $$|\psi\rangle_{RA} \in RA$$ and apply $$V$$ to system $$A$$, resulting in $$|\phi\rangle_{RBE}$$. Then we use antidegradability to show that $$I(R\rangle B)_\phi = 0$$ in (1).

Say that $$U: E\rightarrow B' E'$$ is the isometric extension of our antidegradable $$\mathcal{N}$$. Antidegradability means that for reduced states $$\rho_E = \text{tr}_{RB}|\phi\rangle\langle \phi|_{RBE}\quad\text{ and } \quad\rho_B = \text{tr}_{RE}|\phi\rangle\langle \phi|_{RBE},$$ $$U$$ always obeys $$\text{tr}_{E'}(U \rho_E U^\dagger) = \rho_B'$$. Applying the isometric extension of the antidegrading map to $$|\phi\rangle_{RBE}$$ prepares the state $$|\chi\rangle_{RBB'E'}:= (I_{RB}\otimes U)|\phi\rangle_{RBE}$$

With this we can compute \begin{align} I(R:B)_\phi - I(R:E)_\phi &\leq I(R:B)_\phi - I(R:B')_\chi \qquad\qquad\qquad\qquad \text{data processing inequality} \tag{1} \\&= H(B)_\phi - H(RB)_\phi + H(RB')_\chi - H(B')_\chi \\&= - H(RB)_\phi + H(RB')_\chi \qquad\qquad\qquad\qquad \text{by antidegradability} \\&= -H(E)_\phi + H(BE')_\chi \qquad\qquad\qquad\qquad\, \text{entropy equal across bipartitions} \end{align}

Then I do not know how to proceed. I thought maybe one could argue that applying the map $$V^\dagger: E'B' \rightarrow E$$ to subsystem $$E'B$$ should recover a state with entropy $$H(E)$$, since $$\rho_B = \rho_{B'}$$ (using antidegradability). But $$V^\dagger$$ isn't necessarily an isometry and so I can't make sense of applying it to the system $$BE'$$ in this way.

[1 Wilde, Mark. From Classical to Quantum Shannon Theory (2016). arXiv:1106.1445

## 1 Answer

Let the channel be $$\mathcal{N}_{A\rightarrow B}$$ and its complementary channel be $$\mathcal{N}^c_{A\rightarrow E}$$.

By Property 13.5.1 of [1], we have that $$Q(\mathcal{N})\geq 0.$$

Consider the map $$\mathcal{D}_{E\rightarrow B}$$ which satisfies $$(\mathcal{D}\circ\mathcal{N}^c)(\rho_A) = \mathcal{N}(\rho_A)$$ for any input state $$\rho_A$$. This exists because the channel $$\mathcal{N}$$ is anti-degradable. By the data processing inequality, you have

$$I(R:E)_{\mathcal{N}^c(\rho_A)} \geq I(R:B)_{(\mathcal{D}\circ\mathcal{N}^c)(\rho_A)} = I(R:B)_{\mathcal{N}(\rho_A)}.$$

This implies that

$$Q(\mathcal{N}) = I(R:B)_{\mathcal{N}(\rho_A)} - I(R:E)_{\mathcal{N}^c(\rho_A)} \leq 0.$$

• I don't see why $I(R:B)_{(\mathcal{D}\circ\mathcal{N}^c)(\rho_A)} = I(R:B)_{\mathcal{N}(\rho_A)}$ is true though? This is equivalent to the second-to-last line of my derivation, so I understand that it should be true. I just don't see why the antidegrading map $\mathcal{D}$ should leave $RB'$ (you just called this $RB$) in the same state as $RB$ Commented May 5 at 17:35
• The reason this equality holds is because, by definition, the degradable channel is one for which there exists a map $\mathcal{D}$ such that $\mathcal{D}\circ\mathcal{N}^c = \mathcal{N}$. Note that this is an equality of channels so you get equality of output states too when the input state is the same.
– rnva
Commented May 6 at 4:20