2
$\begingroup$

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

$\endgroup$
0

1 Answer 1

2
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.