Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 48 / 3

Use the unsquared quantum fidelity convention in the question. For any matrix $X$, its trace norm is $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}=\mathrm{Tr}\sqrt{X{X}^{†}}$: the two positive matrices have the same nonzero eigenvalues, namely the squared singular values. This follows directly from the singular value decomposition.

Set $X=\sqrt{\rho }\sqrt{\sigma }$. Then $X{X}^{†}=\sqrt{\rho }\sigma \sqrt{\rho }$ and $X^{†}X=\sqrt{\sigma }\rho \sqrt{\sigma }$. Their positive square roots therefore have the same trace, provingThe argument also covers singular density operators, since zero singular values contribute zero.

For positive operators, the unique positive square root of an operator respects tensor products: $\sqrt{A\otimes B}=\sqrt{A}\otimes \sqrt{B}$. To verify this, diagonalize $A,B$; the product eigenvectors have eigenvalues $a_i{b}_j$, whose square roots are $\sqrt{a_i}\sqrt{b_j}$. Hence the operator inside the outer square root in the fidelity factors asTaking its positive square root and using $\mathrm{Tr}(C\otimes D)=\mathrm{Tr}C\mathrm{Tr}D$ gives $\boxed{F(\rho \otimes \sigma ,{\rho }^{\prime }\otimes {\sigma }^{\prime })=F(\rho ,{\rho }^{\prime })F(\sigma ,{\sigma }^{\prime })}$.

We first establish the purification-overlap formula for quantum fidelity, rather than assume the desired monotonicity. Write a canonical purification of a density operator asTracing the second factor gives $\rho$. The Schmidt decomposition shows that any other purification on a sufficiently large fixed ancillary space differs by a unitary on that space: its orthonormal ancillary vectors can be mapped to those of the canonical purification and the map extended to a unitary operator. Zero eigenvalues are handled by padding the ancillary basis.

The overlap of two canonical purifications, with a relative ancillary unitary $U$, is $\mathrm{Tr}(\sqrt{\rho }\sqrt{\sigma }{U}^T)$, with zero padding when the ancillary dimension is larger. The unitary variational formula for the trace norm isIndeed if $X=PD{Q}^{†}$ is a singular value decomposition, the trace is $\mathrm{Tr}(DW)$ with $W=Q^{†}VP$ unitary. Its modulus is at most ${\sum }_j{D}_{jj}$ because $|W_{jj}|\le 1$, and $V=Q{P}^{†}$ attains that bound. Thus the maximal absolute overlap of purifications equals $F(\rho ,\sigma )$; this also proves the needed form of Uhlmann's theorem.

Choose purifications $|\psi {⟩}_{ABE}$ and $|\varphi {⟩}_{ABE}$ attaining $F({\rho }_{AB},{\sigma }_{AB})$. Regard the very same vectors as purifications of ${\rho }_A,{\sigma }_A$ with ancillary system $BE$. Their overlap is among those maximized in the marginal purification formula, soThis proves the monotonicity of quantum fidelity under partial trace.

Write a qubit density operator as $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with Bloch vector $|\mathbf{r}|\le 1$. The depolarizing channel sends $\mathbf{r}$ to $p\mathbf{r}$. Its output eigenvalues are $(1\pm |p||\mathbf{r}|)/2$, so its Von Neumann entropy is minimized at $|\mathbf{r}|=1$. DefineThe minimum output entropy is $h(q)$, since binary entropy is symmetric under $q\mapsto 1-q$. The usual probabilistic-mixture convention has $0\le p\le 1$. The argument also holds throughout the physical qubit CPTP range $-1/3\le p\le 1$, where this $q$ still lies in $[0,1]$.

To handle multiple channel uses, let a classical message $x$ select a product input ${\rho }_x={⨂}_{j=1}^n{\rho }_{xj}$ with probability ${\pi }_x$. Memorylessness gives product outputs ${\sigma }_x={⨂}_j\Phi ({\rho }_{xj})$. Additivity of Von Neumann entropy on tensor products gives $S({\sigma }_x)\ge nh(q)$. The average output $\bar{\sigma }={\sum }_x{\pi }_x{\sigma }_x$ acts on $n$ qubits, so $S(\bar{\sigma })\le {\log }_2(2^n)=n$. Thus the Holevo quantity obeys the product-input block bound for a qubit depolarizing channelThe Holevo bound says the classical mutual information obtained by any measurement, including a joint measurement on all outputs, is at most $\chi$.

For a reliable code with $M$ equiprobable messages and decoding error $P_e$, Fano's inequality gives $H(X\mid \hat{X})\le h(P_e)+P_e{\log }_2(M-1)$. Consequently $(1-P_e){\log }_{2}M\le n[1-h(q)]+h(P_e)$. Letting the error tend to zero proves $\boxed{\mathcal{I}(\Phi )\le 1-h((1-p)/2)}$.

The bound is tight: equal use of $|0⟩,|1⟩$ and computational-basis output measurements produce a binary symmetric channel with crossover probability $q$. Ordinary classical coding achieves $1-h(q)$ bits per use.