Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 60 (Quantum Information Theory)

Use the unsquared convention for quantum fidelity. For density operators on a common finite-dimensional Hilbert space,Here $\Vert A{\Vert }_1=\mathrm{Tr}\sqrt{A^{†}A}$ is the trace norm, and all square roots are the positive operator square roots. The two displayed expressions agree because $\sqrt{\rho }\sqrt{\sigma }$ and its adjoint have the same singular values. This convention has $0\le F\le 1$; some literature squares this quantity, but that convention is not used here.

For normalized pure states, the rank-one operator $\sqrt{\rho }\sqrt{\sigma }=⟨\phi |\psi ⟩|\phi ⟩⟨\psi |$ has a single nonzero singular value. Therefore

Uhlmann's theorem identifies quantum fidelity with the largest absolute overlap of purifications of a density operator. Choose a common reference Hilbert space $R$ of dimension at least that of the original system. Thenwhere $|u⟩,|v⟩\in {\mathcal{H}}_A\otimes {\mathcal{H}}_R$ have reduced density operators ${\rho }_A,{\sigma }_A$. One purification may be fixed in advance: the maximum is over the other, with the freedom to apply a unitary on the reference system. This is the unitary freedom of purification. Enlarging the reference by unused dimensions does not change the maximum.

Take maximizing purifications $|u⟩,|v⟩$ of ${\rho }_{AB},{\sigma }_{AB}$ on $ABR$. Their overlap has magnitude $F({\rho }_{AB},{\sigma }_{AB})$. Regard these same vectors as purifications of ${\rho }_A,{\sigma }_A$ with reference system $BR$. They are candidates in the larger optimization, so Uhlmann's theorem givesThis is monotonicity of quantum fidelity under partial trace: discarding a system cannot make two states more distinguishable according to their quantum fidelity.

Embed the given purifications into a common reference Hilbert space $R$, enlarging it if necessary, and add a flag register $X$ with orthonormal basis $\{|x⟩\}$. A flagged purification of a quantum ensemble isOrthogonality of the flags gives $⟨\Psi |\Psi ⟩={\sum }_x{p}_x=1$. Moreover,Thus it is a purification of a density operator. The extra flag is essential: simply superposing the purifications without orthogonal labels would generally leave unwanted cross terms.

For each $x$, choose purifications $|u_x⟩$ of ${\rho }_x$ and $|v_x⟩$ of ${\sigma }_x$ in a common reference space. By Uhlmann's theorem, the second can be chosen, including its overall phase, so thatConstruct the two flagged purifications of a quantum ensembleTheir reduced states are the respective mixtures, and orthogonality of the flags gives $⟨U|V⟩={\sum }_x{p}_{x}F({\rho }_x,{\sigma }_x)$. A particular purification overlap cannot exceed the maximizing overlap in Uhlmann's theorem. HenceThis proves joint concavity of quantum fidelity. Choosing each overlap nonnegative prevents cancellation of different phases; the same probability weights in the two mixtures yield $p_x$ rather than distinct square-root weights.

Every conditional output distribution is a permutation of $q$. Since Shannon entropy is invariant under a permutation, its value is the same for every input symbol. Thus, for any input distribution,All Shannon entropies in this solution are in bits. A different fixed logarithm base changes every entropy and capacity by the same constant factor.

The Shannon second coding theorem states that the operational channel capacity of a finite discrete memoryless channel is $C={\max }_{p(x)}I(X:Y)$: rates below this maximum admit block codes with error tending to zero, and rates above it cannot have vanishing error. For this channel the preceding calculation givesusing maximum entropy on a finite alphabet. The transition matrix is a doubly stochastic matrix. Therefore the uniform input has uniform output: $p(y)=\frac{1}{3}{\sum }_{x}p(y\mid x)=1/3$. It achieves the entropy upper bound, and henceEquivalently this is the weakly symmetric channel capacity theorem: permutations of a common row and equal column sums make the uniform input optimal. If $q$ is uniform, the output contains no information about the input and the formula gives zero.

The error indicator is a Bernoulli random variable with probabilities $P(Z=1)=p_e$ and $P(Z=0)=1-p_e$. Its Shannon entropy is therefore the binary entropyUse $0{\log }_{2}0=0$, so this expression also covers error probabilities zero and one.

Apply the definitions of conditional entropy and insert $H(X,Y)$:Inserting $H(Z,Y)$ instead gives the alternative chain rule for conditional entropyBoth are expansions of the same joint conditional uncertainty, with the variables exposed in opposite orders.

Because $Z$ is determined by $X,Y$, its conditional entropy satisfies $H(Z\mid X,Y)=0$. Equate the two forms of the chain rule for conditional entropy to obtainConditioning cannot increase classical Shannon entropy: $H(Z)-H(Z\mid Y)=I(Z:Y)\ge 0$ by nonnegativity of mutual information. Thus $H(Z\mid Y)\le H(Z)=h(p_e)$. When $Z=0$, the value of $X$ is exactly $f(Y)$ and $H(X\mid Y,Z=0)=0$. When $Z=1$ and $Y=y$, the value $f(y)$ is excluded, leaving at most $|J_X|-1$ possibilities. By maximum entropy on a finite alphabet,Averaging the two conditional cases now yieldsand consequentlyThis is Fano's inequality via an error indicator. Events of zero probability contribute zero to the average and need no conditional distribution. For $|J_X|=1$, the inference is automatically correct and the entropy is zero; the displayed logarithmic form is intended for $|J_X|\ge 2$. Optimality of the guess is not needed for the inequality: it holds for every deterministic $f$.

Fix an orthonormal basis $\{|i⟩{\}}_{i=1}^d$ of the input Hilbert space, where $d=\dim \mathcal{H}$, and a reference copy $R\simeq \mathcal{H}$. Use the normalized maximally entangled vector $|{\Omega }_d⟩=d^{-1/2}{\sum }_i|i{⟩}_R|i{⟩}_{\mathcal{H}}$. The normalized Choi–Jamiołkowski state isComplete positivity makes $J\ge 0$, and trace preservation gives ${\mathrm{Tr}}_{\mathcal{K}}J=I_R/d$, hence $\mathrm{Tr}J=1$. It is therefore a genuine density operator. In the unnormalized Choi matrix convention the factor $1/d$ is omitted and the trace is $d$; specifying normalization distinguishes a Choi state from that matrix convention. The reference basis also fixes the transpose in the Choi reconstruction formula, $\Lambda (X)=d{\mathrm{Tr}}_R[(X^T\otimes I)J]$.

Set $K_y=|y⟩⟨y|$. The map is in Kraus representation:For any ancillary Hilbert space $R$ and positive operator $M$ on $R\otimes \mathcal{H}$,This verifies complete positivity directly, not merely positivity on unextended states. Cyclicity of the trace gives $\mathrm{Tr}\Lambda (X)=\mathrm{Tr}[X{\sum }_y{K}_y^{†}{K}_y]=\mathrm{Tr}X$. Thus $\Lambda$ is a CPTP map, the completely dephasing channel in the given basis.

Let $q_y=⟨y|\rho |y⟩$. Positivity and normalization of the density operator imply $q_y\ge 0$ and ${\sum }_y{q}_y=1$. The completely dephasing channel produceswhich is already a spectral decomposition. Its Von Neumann entropy is thereforeThe quantum output entropy is exactly the Shannon entropy of the basis-measurement probabilities, including possible zero eigenvalues via $0{\log }_{2}0=0$.

Use Klein's inequality, or equivalently nonnegativity of quantum relative entropy. First check the support needed for the logarithm. If $q_y=0$, positivity gives $|{\rho }_{yz}{|}^2\le {\rho }_{yy}{\rho }_{zz}=0$ for every $z$; the corresponding row and column vanish. Thus support inclusion under rank-one dephasing gives $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$, and the logarithms may be evaluated on this support.

Since $\log \sigma$ is diagonal in the dephasing basis,The relative-entropy identity for rank-one dephasing follows:Klein's inequality gives $D(\rho \Vert \sigma )\ge (\mathrm{Tr}\rho -\mathrm{Tr}\sigma )/\ln 2=0$. ThereforeEquality holds precisely when $\rho =\sigma$, meaning that the input was already diagonal in the chosen basis. This quantifies why rank-one dephasing removes coherence without reducing the entropy.

Nielsen's pure-state conversion theorem gives the exact deterministic LOCC criterion. Let the Schmidt decompositions be $|\psi ⟩={\sum }_j\sqrt{{\lambda }_j}|a_j{b}_j⟩$ and $|\varphi ⟩={\sum }_j\sqrt{{\mu }_j}|a_j^{\prime }{b}_j^{\prime }⟩$. The vectors $\lambda ,\mu$ consist of squared Schmidt coefficients, equivalently the eigenvalues of either reduced density operator. Order each in decreasing order and pad with zeros to a common length $d$.

Then deterministic conversion is possible if and only ifThis is majorization, with the input vector majorized by the output vector. The direction matters: a maximally entangled state has a uniform vector, which is majorized by a product state's vector $(1,0,\dots )$, so entanglement can be discarded by LOCC. The criterion is for certain exact conversion, without catalysts or postselection on a successful branch.

Let $r$ be the input Schmidt rank, so ${\lambda }_j=0$ for $j>r$ and ${\sum }_{j=1}^r{\lambda }_j=1$. If $r=d$, the output rank is already at most $r$. Otherwise, Nielsen's pure-state conversion theorem and majorization at $k=r$ giveThus all output coefficients beyond $r$ vanish, provingThis is monotonicity of Schmidt rank under LOCC. It also holds separately in any nonzero postselected branch: represent the input amplitudes by a matrix $C$; a local branch maps it to $AC{B}^T$, whose rank cannot exceed the rank of $C$. The deterministic result requested here follows already from the majorization criterion.

For any tripartite density operator, Strong subadditivity of Von Neumann entropy isUse quantum relative entropy $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$, with the usual support condition. The equivalent relative-entropy comparison isIndeed the two sides expand respectively as $S(A)+S(BC)-S(ABC)$ and $S(A)+S(B)-S(AB)$. Subtracting cancels $S(A)$ and leaves exactly the strong-subadditivity gap. Marginal-product supports contain the support of the joint state, so these expressions are finite; singular marginals can also be handled by full-rank regularization and a limit.

Finally, tracing out $C$ sends the numerator and denominator of the first relative entropy to those of the second. The data-processing inequality for quantum relative entropy therefore proves the comparison. This is strong subadditivity from relative-entropy monotonicity, rather than an assumption that classical entropy proofs automatically apply to quantum states.

Take a Stinespring dilation $V:B⟶B^{\prime }E$ of the local quantum channel and define ${\tau }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. Tracing out $E$ gives the prescribed output, with $A^{\prime }=A$. An isometry preserves the nonzero eigenvalues of a density operator; therefore $S(B^{\prime }E{)}_{\tau }=S(B{)}_{\rho }$, $S(A{B}^{\prime }E{)}_{\tau }=S(AB{)}_{\rho }$, and $S(A{)}_{\tau }=S(A{)}_{\rho }$.

Consequently $I(A:B^{\prime }E{)}_{\tau }=I(A:B{)}_{\rho }$. The loss of quantum mutual information isThe last inequality is Strong subadditivity of Von Neumann entropy, or nonnegativity of quantum conditional mutual information. HenceThis mutual-information loss as conditional mutual information shows exactly which correlations are discarded into the environment. No purity assumption on the original state is needed.

Choose a Stinespring dilation $V:A⟶B_{1}E$ of the quantum channel. Applying it to the input purification gives the pure vectorIts $R{B}_1$ marginal is the output state in the definition of coherent information. Complementary subsystems of a pure state have the same nonzero eigenvalues, by the Schmidt decomposition. Thus $S(B_1)=S(RE)$ and $S(R{B}_1)=S(E)$. It follows thatHere $S(R\mid E)$ denotes quantum conditional entropy, evaluated on the complementary output $RE$. This is coherent information as an environment conditional entropy. Equivalently $I_c=-S(R\mid B_1)$; the environment expression has the positive sign, while the receiving-system expression has the negative sign.

Subadditivity of Von Neumann entropy applied to $RE$ gives $S(RE)\le S(R)+S(E)$. Together with the environment expression for coherent information,The channel does not act on $R$. Its reduced state has the same nonzero eigenvalues as the original input $\rho$, because $RA$ initially purifies that input. HenceMore precisely, $I_c=S(\rho )-I(R:E)$: the coherent information upper bound by input entropy is a consequence of nonnegative quantum mutual information with the environment. Equality holds when $R$ and $E$ are uncorrelated.

Dilate the second quantum channel by an isometry $W:B_1⟶B_2{E}_2$, and set ${\tau }_{R{B}_2{E}_2}=(I_R\otimes W){\sigma }_{R{B}_1}(I_R\otimes W^{†})$. Its $R{B}_2$ marginal is the final output. Isometry invariance of Von Neumann entropy gives $S(B_1{)}_{\sigma }=S(B_2{E}_2{)}_{\tau }$ and $S(R{B}_1{)}_{\sigma }=S(R{B}_2{E}_2{)}_{\tau }$. The difference between the two values of coherent information is thereforeNonnegativity follows from Strong subadditivity of Von Neumann entropy. This proves the data-processing inequality for coherent information:The dilated state here need not be pure. The proof uses only isometry invariance and strong subadditivity, so it applies even when the first channel has already entangled the input with its own discarded environment.

Both output states have the same reference marginal, and $S(R)=S(\rho )$. Expand quantum mutual information to obtain the mutual information and coherent information identityThe input-entropy term is identical in the two expressions. Subtract them and use the preceding data-processing inequality for coherent information:Thus the final quantum channel cannot increase the reference-output quantum mutual information, as required by data processing for quantum mutual information.