Random unitary channel

A random unitary channel has $T(\rho )={\sum }_j{p}_j{U}_j\rho U_j^{†}$. Its Choi matrix is a convex combination of maximally entangled pure states.

On a $d$-dimensional Hilbert space, let $X|j⟩=|j+1\mathrm{mod}d⟩$ and $Z|j⟩=e^{2\pi ij/d}|j⟩$. The $d^2$ unitaries $W_{k,m}=X^k{Z}^m$ form an orthogonal operator basis under the Hilbert-Schmidt inner product.

Averaging conjugation by all Heisenberg-Weyl operators completely depolarizes a system:Summing over phases $Z^m$ kills off-diagonal matrix elements, and averaging shifts $X^k$ makes all diagonal elements equal. Acting on subsystem $B$ of a bipartite operator gives $({\mathrm{Tr}}_{B}A)\otimes I_B/d$. This identity turns joint convexity of quantum relative entropy into monotonicity under partial trace.

A one-qubit Pauli channel applies $I,{\sigma }_x,{\sigma }_y,{\sigma }_z$ with probabilities $p_0,p_x,p_y,p_z$. It maps a Bloch vector diagonally by independently contracting or reversing its three components.

A bit-flip channel applies the Pauli X gate with probability $p$ and does nothing otherwise. If the classical error outcome is not recorded, its output is the convex mixture ${\mathcal{E}}_p(\rho )=(1-p)\rho +pX\rho X$, not a coherent superposition of the two alternatives.

// Target: quantum-information-theory.bigb

For pure input $|\psi ⟩$, squared quantum fidelity after the bit-flip channel is $(1-p)+p|⟨\psi |X|\psi ⟩{|}^2$. The last factor is the square of the input's Bloch $x$ component. Rotational symmetry of the uniform pure-qubit average gives $⟨r_x^2⟩=1/3$, because $r_x^2+r_y^2+r_z^2=1$. Hence the average squared fidelity is $1-2p/3$. Averaging an unsquared fidelity or over a different ensemble gives a different quantity.

// Target: quantum-mechanics.bigb

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.

The phase-flip channel is ${\Lambda }_p(\rho )=(1-p)\rho +p{\sigma }_z\rho {\sigma }_z$. It multiplies the $x$ and $y$ components of the Bloch vector by $1-2p$ and leaves the $z$ component fixed.

Independent successive applications of the phase-flip channel $D_{ϵ}(\rho )=(1-ϵ)\rho +ϵZ\rho Z$ preserve diagonal entries and multiply off-diagonal entries by $(1-2ϵ{)}^n$. Hence $D_{ϵ}^n=D_{{ϵ}_n}$ with the displayed parameter, for nonnegative integers $n$. If $0<ϵ<1/2$, the limit is the completely dephasing channel, equivalently a nonselective projective measurement in the computational basis. A time-interval law can be iterated this way only under the memoryless composition assumption.

A dephasing channel suppresses off-diagonal entries of a density matrix in a preferred basis while preserving quantum state populations. Complete dephasing is identical to a projective measurement in that basis with its outcome discarded.

For the Bell correlations versus a classical mixture example, completely dephasing Alice's qubit in the computational basis maps ${\rho }_{\mathrm{Bell}}$ to ${\rho }_{\mathrm{mix}}$. It preserves their perfect $Z$ agreement and each reduced density matrix, yet destroys the $X$ correlation. This is why testing only populations can miss damage to a quantum memory: the stored quantum coherence may have disappeared while every preferred-basis measurement still looks correct. The joint Von Neumann entropy rises from zero to one bit, although each local entropy stays fixed.

An environment interaction preserving the computational populations multiplies a qubit's upper off-diagonal density entry by $\overline{c}$. A real phase-flip channel representation exists for every input precisely when $c$ is real, with probability $(1-c)/2$. For $c=r{e}^{i\phi }$, the channel is a phase-flip channel of probability $(1-r)/2$ followed by the diagonal unitary operator $\mathrm{diag}(1,e^{i\phi })$. The coherent phase cannot be discarded just because the populations are unchanged.