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 1


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

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

  • $\begingroup$ 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 $RB$ $\endgroup$
    – forky40
    Commented May 5 at 17:35
  • $\begingroup$ 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. $\endgroup$
    – rnva
    Commented May 6 at 4:20

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