Consider the single qubit quantum depolarizing channel, given by

$$T(\rho) = (1- p)\rho + p \frac{\mathbb{I}}{2}. $$

For an $n$ qubit state $\rho$, according to Definition 6.1 of this paper, the channel satisfies a strong data processing inequality, given by:

$$S\left(T^{\otimes n}(\rho) ~||~ \frac{\mathbb{I}}{2^{n}}\right) \leq (1-\alpha)S\left(\rho ~||~ \frac{\mathbb{I}}{2^{n}}\right).$$

$S$ is the von Neumann entropy and the relative entropy $S(A~||~B)$ is given by

$$S(A~||~B) = \text{Tr}(A (\log A - \log B)). $$

Now, consider the quantum amplitude damping channel $\mathcal{A}$, as defined here: $$\mathcal{A}_\gamma(\rho)=\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}\rho\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}+\begin{pmatrix}0&\sqrt{\gamma}\\0&0\end{pmatrix}\rho\begin{pmatrix}0&0\\\sqrt{\gamma}&0\end{pmatrix}.$$

Is there a known data processing inequality for the same? For example, something like:

$$S\left(\mathcal{A}^{\otimes n}(\rho) ~||~ |0^{n}\rangle \langle 0^{n}| \right) \leq (1-\alpha)S\left(\rho ~||~ |0^{n}\rangle \langle 0^{n}|\right)?$$

(Note that the fixed point of the amplitude damping channel is not the maximally mixed state over $n$ qubits but the all zeros state.)


1 Answer 1


An easy (but perhaps cheating) answer: The relative entropy $S(A||B)$diverges when $A$ has support over the kernel of $B$ (e.g., wikipedia). Now, the kernel of $B=|0\rangle^{\otimes n}\langle 0|^{\otimes n}$ is everything other than $|0\rangle^{\otimes n}$, i.e., $\mathbb{I}-|0\rangle^{\otimes n}\langle 0|^{\otimes n}$! For the inequality to be meaningful, we want the right-hand side to be finite, so we only need to consider cases with $$0=\text{Tr}[\rho(\mathbb{I}-|0\rangle^{\otimes n}\langle 0|^{\otimes n})]=1-\langle 0|^{\otimes n}\rho|0\rangle^{\otimes n};$$ otherwise, the RHS is infinite and the inequality won't tell us anything. This means that the inequality is only useful for $\rho=|0\rangle^{\otimes n}\langle 0|^{\otimes n}$, which then trivially transforms as $\mathcal{A}^{\otimes n}(\rho)=\rho$ and the inequality is simply an equality with $\alpha=0$.

Now, are there cases in which the RHS is infinite while the LHS is finite? The inequality wouldn't be useful in the sense of data processing, but it would at least make us feel more comfortable with the inequality. Indeed, we can begin with $\rho\neq|0\rangle^{\otimes n}\langle 0|^{\otimes n}$ such that the RHS diverges and end with $\mathcal{A}^{\otimes n}(\rho)=|0\rangle^{\otimes n}\langle 0|^{\otimes n}$ by simply setting the amplitude damping constant to yield maximal damping $\gamma= 1$. Then, the LHS is looking at the relative entropy between identical states, so it gives the reasonable inequality $0<(1-\alpha)\infty$.

In summary, while I don't think the present inequality is useful, I do believe that it is correct.


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