Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 37 / 4

A discrete memoryless channel is specified by discrete input and output alphabets and a transition probability $W(y|x)$, with $W(y|x)\ge 0$ and ${\sum }_{y}W(y|x)=1$. Its successive outputs are conditionally independent given the inputs, with the same transition rule at every use:Operationally, its channel capacity is the supremum of communication rates in bits per use attainable by increasingly long block codes with decoding error probability tending to zero. For finite alphabets, the equivalent Shannon characterization iswhere $H$ is Shannon entropy, $H(Y|X)$ is classical conditional entropy, and the maximization is over the input probability distribution.

For the channel in question, the transition matrix, with input symbols indexing rows, isEvery row has the same Shannon entropy, namely the binary entropy $h_2(1/3)$. Consequently $H(Y|X)=h_2(1/3)$ for every input probability distribution. Since the output has three symbols, the maximum entropy on a finite alphabet givesChoose a uniform input. Each column of $W$ sums to one, so $P_Y(y)=1/3$ for every output symbol and the upper bound is attained. FinallyThus the cyclic ternary channel capacity isachieved by a uniform input probability distribution. This is a weakly symmetric channel capacity calculation: the matrix rows are permutations and the column sums agree. The channel is not the usual ternary symmetric-error channel, since an error has only one possible changed output for each input.

There is a normalization error in the original PDF: the first term contains $I$ rather than $\rho$. For a normalized qubit density operator, that printed expression has trace $2(1-p)+p=2-p$, so it is not a quantum channel except at $p=1$. The intended depolarizing channel is the Pauli channel that leaves the input unchanged with probability $1-p$ and applies each nonidentity Pauli matrix with probability $p/3$, for $0\le p\le 1$. Its Kraus operators areand ${\sum }_{k=0}^3{K}_k^{†}{K}_k=I$, proving that it is trace-preserving and a completely positive map.

Write the input in its Bloch vector representation,Conjugation by a Pauli matrix keeps its own Bloch vector component and reverses the other two. More explicitly, ${\sigma }_k{\sigma }_j{\sigma }_k={\sigma }_j$ for $j=k$ and $-{\sigma }_j$ for $j\ne k$. HenceThe corrected quantum channel therefore givesThus its action isThe Bloch sphere is sent to a sphere of radius $|1-4p/3|$ about the origin of the Bloch ball. At $p=0$ the action is the identity. For $0<p<3/4$ the Bloch vector shrinks without changing direction. At $p=3/4$ every input becomes the maximally mixed density operator $I/2$. For $3/4<p\le 1$, the Bloch vector reverses direction and shrinks, reaching factor $-1/3$ at $p=1$. The reversal is compatible with complete positivity because this remains a probabilistic mixture of Pauli matrices. The parameter here is the total nonidentity-error probability, not the retention parameter used in some definitions of the quantum depolarizing channel.

For completeness, retain the PDF literally and call its output $F_p(\rho )$. The same calculation givesThis has no normalized-output Bloch ball interpretation as printed. If it is manually normalized, thenThat different transformation sends $p=0$ to $I/2$, rather than to the input, and agrees with the intended answer only at $p=1$. The literal and corrected conventions therefore cannot be silently identified.