Quantum depolarizing channel

A qubit depolarizing channel shrinks every Bloch vector by the same factor: ${\Lambda }_p(\rho )=p\rho +(1-p)I/2$. In this retention-parameter convention it is completely positive for $-1/3\le p\le 1$; the usual mixture of unchanged and fully randomized states uses $0\le p\le 1$.

For the quantum depolarizing channel applying the identity with probability $1-p$ and each nonidentity Pauli operator with probability $p/3$, the entanglement-assisted classical capacity is the displayed number of bits per use. Here $H_4$ is Shannon entropy of the four listed probabilities. Its unassisted product-state capacity is $1-h_2(2p/3)$. Both vanish at $p=3/4$. If $q$ instead denotes replacement by $I/2$, then $p=3q/4$ and complete depolarization is at $q=1$. The original entanglement-assisted capacity paper gives the depolarizing-channel formula.

With Bloch vector retention $\eta \to 0$, the unassisted product-state capacity of a quantum depolarizing channel is ${\eta }^2/(2\ln 2)+O({\eta }^4)$. Its entanglement-assisted capacity is $3{\eta }^2/(2\ln 2)+O({\eta }^3)$, obtained by expanding Shannon entropy around the uniform four-outcome distribution. Consequently their limiting ratio is three, independent of the linear noise-parameter convention.

The displayed qubit quantum channel is completely positive for $0\le q\le 4/3$. Indeed its Pauli channel probabilities are $1-3q/4$ for the identity and $q/4$ for each nonidentity Pauli operator. In the range $q>1$ the coefficient $1-q$ is negative, so the displayed expression is not a convex mixture of the unchanged and maximally mixed states, even though the Pauli representation is a valid random unitary channel. Its Bloch vector is contracted and reversed by the factor $1-q$. The Holevo capacity of a qubit depolarizing channel remains $1-h_2(q/2)$ throughout this interval.

If the identity is applied with probability $p$ and the three nonidentity Pauli operators each with probability $(1-p)/3$, the Bloch vector retention factor is $(4p-1)/3$. Thus complete depolarization occurs at $p=1/4$, not at zero. The Holevo capacity of a qubit depolarizing channel becomes $1-h_2((2p+1)/3)$ in this identity-probability parametrization.

For the retention parameter $p$, the Holevo capacity isEvery output has Von Neumann entropy at least the value on a pure input, $h_2((1+p)/2)$. Two equiprobable orthogonal pure inputs attain this minimum entropy individually and have maximally mixed average output, attaining the bound. Additivity of Holevo capacity makes the unassisted classical capacity of a quantum channel equal to this value. King's depolarizing-channel capacity paper proves additivity with an arbitrary other channel.

For $\Phi (\rho )=p\rho +(1-p)I/2$, every output qubit has Von Neumann entropy at least $h((1-p)/2)$. A product input therefore has output entropy at least $nh((1-p)/2)$. The average output has entropy at most $n$, so its Holevo quantity is bounded by the displayed expression. This allows correlated classical message labels and joint output measurements, but requires product quantum inputs. The Holevo bound and Fano's inequality turn it into the corresponding asymptotic rate bound.