Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 34 / 3

The PDF's first Pauli term contains an undefined ${\rho }_x$. For the stated depolarizing channel, use ${\sigma }_x\rho {\sigma }_x$, consistently with the other two terms. This is the minimal typographical correction needed to define the intended Pauli channel.

Write $\rho =\frac{1}{2}(I+\mathbf{s}\cdot \mathbf{\sigma })$ using its Bloch vector. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k{\sigma }_j={\sigma }_k$ for $j=k$ and $-{\sigma }_k$ for $j\ne k$. Thus conjugating by each nonidentity Pauli matrix leaves one Bloch component unchanged and reverses the other two. Summing the three conjugates givesSubstitution yieldsso the Pauli-mixture parametrization of qubit depolarization, with $p$ here the total error probability, givesThere is also a genuine range error in the printed claim. For its full $0<p<1$ assumption, $0<q<4/3$, and $0<q<1$ holds exactly when $0<p<3/4$. For example $p=9/10$ gives $q=6/5$; on $|0⟩⟨0|$ the output is $\mathrm{diag}(2/5,3/5)$, with negative $z$ Bloch component. No $q\in (0,1)$ could produce this output from that input, since its component would be $1-q>0$.

The corrected channel is still completely positive in the larger range: its original Pauli probabilities are $1-p,p/3,p/3,p/3$. This is depolarizing noise weight above one; the formula with $q>1$ is not a convex mixture with weights $1-q,q$, but it has a valid random unitary channel representation. The subsequent parts can therefore be solved both in the printed $q\in (0,1)$ regime and in the actual parameter range.

From the derived formula,so the action on the Bloch vector isFor $0<q<1$, every direction is contracted by the same positive factor. The pure-state Bloch sphere becomes a concentric sphere of radius $1-q$, and the Bloch ball contracts towards the maximally mixed density operator $I/2$. The lack of a preferred direction explains the name depolarizing channel.

At $q=1$ all inputs become $I/2$. For the allowed Pauli-error range $1<q<4/3$, the factor is negative, so there is also an antipodal reversal; the radius is $q-1\le 1/3$. Thus it would be incorrect to describe the whole printed $p\in (0,1)$ range as positive radial contraction.

For the quantum state ensemble $\mathcal{E}=\{p_i,{\rho }_i\}$ with average $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, the Holevo quantity isThe second equality follows by expanding the trace definition: ${\sum }_i{p}_i\mathrm{Tr}{\rho }_i\log \overline{\rho }=\mathrm{Tr}\overline{\rho }\log \overline{\rho }$. All positive-weight states are supported within $\overline{\rho }$, so the expression is finite.

For a CPTP map $\Lambda$, the output average is $\Lambda (\overline{\rho })$, with unchanged probabilities $p_i$. The data-processing inequality for quantum relative entropy givesThus the Holevo quantity cannot increase under a quantum channel.

The underlying monotonicity can also be seen directly from Strong subadditivity of Von Neumann entropy. Attach the classical label $X$, and realize $\Lambda$ by a Stinespring representation of a completely positive map with output $B$ and discarded environment $E$. The isometry preserves $I(X:BE)$, whose input value is $\chi (\mathcal{E})$. The difference from the output value iswhich is precisely strong subadditivity. This explains why discarding an environment loses accessible ensemble information.

Here “operation” means the unconditional channel, including the classical outcome when a measurement is retained. Conditioning on a selected outcome can increase the Holevo quantity. For instance, take equally likely vectors $\sqrt{a}|0⟩\pm \sqrt{1-a}|1⟩$, with $1/2<a<1$. Their input Holevo quantity is $h_2(a)<1$. The filter $K=\mathrm{diag}(\sqrt{(1-a)/a},1)$ succeeds on either state with probability $2(1-a)$ and produces the orthogonal states $|+⟩,|-⟩$ after normalization. The success-conditioned quantity is $1$. This does not violate the unconditional statement.

The Holevo-Schumacher-Westmoreland theorem gives the classical coding formulaIts ensemble-coding statement makes every rate below a one-use output Holevo quantity achievable with product input encodings and collective output decoding. Under this product-input restriction, the capacity isFor any such input ensemble, rates below its output Holevo quantity are achievable with asymptotically vanishing decoding error. Maximization gives the product-state classical capacity. The converse follows from the Holevo bound, entropy bounds on product outputs, and Fano's inequality.

Define the binary entropy $h_2(x)=-x{\log }_{2}x-(1-x){\log }_2(1-x)$. For an input Bloch radius $r\le 1$, the output eigenvalues are $(1\pm (1-q)r)/2$. Its entropy isOn $[1/2,1]$, $h_2$ is decreasing, so this is minimized by pure inputs $r=1$. Throughout the completely positive interval $0\le q\le 4/3$, the minimum equals $h_2(q/2)$, using symmetry $h_2(x)=h_2(1-x)$. The average output entropy is at most $1$, because it is a qubit. Hence every ensemble satisfiesTwo equally likely orthogonal pure inputs attain this bound: their average output is $I/2$, and each individual output has entropy $h_2(q/2)$. ThereforeThis gives the Holevo capacity of a qubit depolarizing channel in the noise-weight convention used here.

For completeness, correlated classical code labels do not evade this upper bound. Each product input to $n$ uses has product output entropy at least $n{h}_2(q/2)$, whereas the average output entropy is at most $n$. Its Holevo quantity is consequently at most $n[1-h_2(q/2)]$. The Holevo bound and Fano's inequality rule out a reliable rate larger than this value. Combined with HSW achievability, this proves the capacity under the stated product-input restriction. It also remains valid in the extended physical parameter range arising from the original $p$ assumption.