Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 34 (Quantum Information Theory)

Use base-two logarithms, so Von Neumann entropy is measured in bits, and set $0\log 0=0$. Write $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ for the quantum relative entropy, the quantity denoted $S(\rho \Vert \sigma )$ in the paper. It is $+\infty$ unless the support of a positive operator $\rho$ lies within that of $\sigma$.

We first establish nonnegativity of quantum relative entropy, including its equality condition. Let $\rho ={\sum }_{i:r_i>0}{r}_i|u_i⟩⟨u_i|$ and $\sigma ={\sum }_j{s}_j|v_j⟩⟨v_j|$, and suppose the support condition holds. Put $q_i=⟨u_i|\sigma |u_i⟩>0$. Concavity of the scalar logarithm givesTerms with $s_j=0$ have zero weight here. Working first with natural logarithms, the inequality $-\ln x\ge 1-x$ yieldssince the $q_i$ sum to at most $\mathrm{Tr}\sigma =1$. Equality forces $q_i=r_i$ for every positive-weight eigenvector, no remaining weight of $\sigma$ outside their span, and equality in the logarithm's strict concavity. The latter makes each $u_i$ an eigenvector of $\sigma$ with eigenvalue $q_i$. Thus $D(\rho \Vert \sigma )=0$ exactly when $\rho =\sigma$.

For the bipartite density operator,Indeed, a projector onto the kernel of ${\rho }_A$, tensored with $I_B$, has zero expectation in ${\rho }_{AB}$. Positivity implies that ${\rho }_{AB}$ annihilates its range: $\mathrm{Tr}({\rho }_{AB}P)=\Vert {\rho }_{AB}^{1/2}P{\Vert }_{\mathrm{HS}}^2=0$. The same argument applies to $B$. Hence the quantum relative entropy below is finite. On this support,Using the partial trace to evaluate the two local terms givesThis proves Subadditivity of Von Neumann entropy:The proof also identifies equality exactly with the product state ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

Introduce an orthonormal classical-label basis $|i⟩$ and the classical-quantum stateIts reduced density operators are ${\Gamma }_X={\sum }_i{p}_i|i⟩⟨i|$ and ${\Gamma }_A=\overline{\rho }={\sum }_i{p}_i{\rho }_i$. If ${\lambda }_{ij}$ are the eigenvalues of ${\rho }_i$, the eigenvalues of ${\Gamma }_{XA}$ are $p_i{\lambda }_{ij}$. Thus the entropy of an orthogonal quantum mixture iswhere $H(p)=-{\sum }_i{p}_i{\log }_2{p}_i$ is the Shannon entropy. Apply the proved Subadditivity of Von Neumann entropy to $X$ and $A$ and cancel $H(p)$. This gives Concavity of Von Neumann entropy:Zero-probability terms may simply be omitted. The equality condition from the preceding solution implies that equality holds precisely when all the positive-probability states are the same: the block equations $p_i{\rho }_i=p_i\overline{\rho }$ follow from ${\Gamma }_{XA}={\Gamma }_X\otimes {\Gamma }_A$.

Omit indices of probability zero. The support of a positive operator of every remaining ${\rho }_i$ lies within that of $\overline{\rho }$, because $\overline{\rho }$ is a positive sum containing $p_i{\rho }_i$. Thus $D({\rho }_i\Vert \overline{\rho })$ is finite.

First suppose $\mathrm{supp}\overline{\rho }\subseteq \mathrm{supp}\sigma$, so all quantities are finite. Expand the trace definitions of quantum relative entropy:Therefore Donald's identity isIf the support condition on $\sigma$ fails, at least one positive-weight ${\rho }_i$ also fails it. The left side and the final term on the right are then $+\infty$, while the first right-hand sum remains finite. The equality is valid in this extended sense, without taking any undefined difference of infinities.

A compression scheme consists of quantum channelswhere each is a CPTP map. The Hilbert space ${\mathcal{K}}_n$ is the compressed register. Its block rate in qubits per source qubit is $R_n=n^{-1}{\log }_2\dim {\mathcal{K}}_n$, with asymptotic rate $R={\mathrm{limsup}}_n{R}_n$ (or the limit when it exists).

Use the squared quantum fidelity convention for the ensemble overlap:This definition tests preservation of the emitted vectors themselves, including nonorthogonal ones; preserving only their classical labels would be a different task. The scheme is reliable when $F_n\to 1$ as $n\to \infty$. This is the ensemble-average criterion for reliable quantum source compression. A square root is not taken in this definition, even though an unsquared convention is also used for quantum fidelity.

Write the spectral decomposition of the one-qubit density operator as $\pi ={\sum }_{j=0}^1{\lambda }_j|e_j⟩⟨e_j|$. For a word $\mathbf{j}=(j_1,\dots ,j_n)$, the tensor product vectoris an eigenvector of ${\pi }^{\otimes n}$, with eigenvalueThese product vectors form an orthonormal eigenbasis, allowing arbitrary orthonormal choices within degeneracies. Using ${\sum }_j{\lambda }_j=1$ and $0\log 0=0$,Thus the Von Neumann entropy isThis calculation uses the spectral ensemble of the memoryless quantum information source; the originally emitted vectors need not equal these eigenvectors.

For fixed $\epsilon >0$, let the quantum typical subspace ${\mathcal{T}}_{\epsilon }^{(n)}$ be spanned by the product eigenvectors with nonzero eigenvalues satisfyingEquivalently their spectral information $-n^{-1}{\log }_2{\lambda }_{\mathbf{j}}$ lies within $\epsilon$ of $S(\pi )$. Denote its orthogonal projection by $P_{\epsilon }^{(n)}$.

The typical subspace theorem states that for every $\delta >0$ there is $n_0$ such that for $n\ge n_0$,Also the operator bounds on its range areTo see the concentration statement, sample $j_s$ independently with probabilities ${\lambda }_j$. The mean of $-{\log }_2{\lambda }_{j_s}$ is $S(\pi )$, so the weak law of large numbers gives typical probability tending to one. Zero-eigenvalue letters have probability zero and are excluded. The dimension bounds follow by summing the upper and lower typical eigenvalue bounds and using the typical mass between $1-\delta$ and $1$.

For a block with nonzero typical dimension, write $P=P_{\epsilon }^{(n)}$ and $Q=I-P$. Let the compressed register be a copy of the typical space plus a one-dimensional orthogonal failure flag. Choose an isometry $V:P{\mathcal{H}}^{\otimes n}\to {\mathcal{K}}_n$ onto the successful code sector, extended by zero on $Q$, and a normalized flag $|f⟩$ orthogonal to its range. Choose any normalized vector $|\phi ⟩$ in the source Hilbert space. DefineThese are CPTP maps: the first performs the projective test and either encodes or prepares a flag; the second embeds the successful sector and prepares a fixed state on failure. Their trace identities follow from $V^{†}V=P$ and $V{V}^{†}+|f⟩⟨f|=I_{{\mathcal{K}}_n}$. Thus typical-subspace compression with a failure flag implementsFor a source vector $|{\Psi }_k⟩$, let $a_k=⟨{\Psi }_k|P|{\Psi }_k⟩={\alpha }_k^2$. Its overlap with the decoded state iswhere the final step is $(1-a_k{)}^2\ge 0$. Averaging proves average pure-source fidelity after typical projection:No orthogonality of the emitted vectors was used. The code dimension is $\dim {\mathcal{T}}_{\epsilon }^{(n)}+1$; the extra flag has vanishing asymptotic rate cost. Exceptional small block lengths with empty typical space can use any channel, since they do not affect reliability.

Choose $0<\epsilon <R-S(\pi )$. The preceding construction and the typical subspace theorem give, for all sufficiently large $n$,Embed this code into $m_n=\lceil nR\rceil$ qubits. The unused subspace can be decoded to a fixed source state, making the extended decoder trace preserving; encoded states never enter it. Hence its rate is $m_n/n\to R$.

For every $\delta >0$, sufficiently large $n$ has $\mathrm{Tr}({\pi }^{\otimes n}{P}_{\epsilon }^{(n)})\ge 1-\delta$. The proved fidelity bound yieldsSince $\delta$ is arbitrary, $F_n\to 1$. ThereforeThis establishes the achievability part of Schumacher compression, including the failure flag and without presuming the source ensemble is orthogonal. If $R$ exceeds one, the same rate-budget construction simply uses spare code qubits; no claim of reducing the physical register size is then needed.

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.

Fix paired orthonormal bases in the two $d$-dimensional Hilbert spaces. A maximally entangled state isIt is normalized, and the partial traces of its rank-one density operator are ${\rho }_A={\rho }_B=I/d$. Thus all its Schmidt coefficients are $1/\sqrt{d}$ and its entanglement entropy is ${\log }_{2}d$. No pure bipartite state of these dimensions can have larger entanglement entropy: nonnegativity of quantum relative entropy applied to a reduced state and $I/d$ gives $0\le D({\rho }_A\Vert I/d)={\log }_{2}d-S({\rho }_A)$. Hence it is maximally entangled.

Write $|{\varphi }_A⟩={\sum }_j{c}_j|j{⟩}_A$ and choose the conjugate index vectorWith $|{\tilde{\Psi }}_{AB}⟩=\sqrt{d}|{\Psi }_{AB}⟩={\sum }_j|j{⟩}_A|j{⟩}_B$, the partial inner product givesThis is the relative state construction. The coefficient conjugation is defined relative to the fixed paired bases; it is essential, because the bra conjugates those coefficients again. Using the normalized entangled vector instead would give $|{\varphi }_A⟩/\sqrt{d}$.

The linear operator $M_A$ acts on the first factor, while the partial inner product acts on the second. They therefore commute in the contraction:Thus the operated vector is exactly the relative state obtained from the operated bipartite vector using the same conjugate index vector. For a physical normalized pure state, divide by $\Vert M_A{\varphi }_A\Vert$ when that norm is nonzero. If $M_A{\varphi }_A=0$, the vector identity still holds, but there is no normalized output state for that zero-probability branch. An arbitrary operator is not assumed unitary or norm preserving.

Use the unnormalized Choi matrixIt is positive because ${\Phi }_A$ is a completely positive map. Take its spectral decomposition, absorbing square roots of nonzero eigenvalues into the vectors:Expand $|v_k⟩={\sum }_{i,j}{v}_{k,ij}|i{⟩}_A|j{⟩}_B$ and define the Kraus operator $A_k$ by $(A_k{)}_{ij}=v_{k,ij}$. Equivalently,The preceding relative state identity then gives $(I\otimes ⟨{\varphi }_B^{\ast }|)v_k=A_k{\varphi }_A$. Inserting the spectral decomposition into the stated reconstruction identity yieldsThe conjugated index vector is required here; it is present in the PDF but missing in the converted TeX version of this identity.

Rank-one projectors span all operators by polarization. Explicitly, putting $P(w)=|w⟩⟨w|$ givesThus linearity extends the equality to every $X\in \mathcal{B}({\mathcal{H}}_A)$:Finally trace preservation and cyclicity of the trace giveFor pure-projector inputs this says the Hermitian operator ${\sum }_k{A}_k^{†}{A}_k-I$ has zero expectation on every vector, so it vanishes. ThereforeThis is the spectral Kraus decomposition. Using a normalized Choi state $J_{\Phi }/d$ instead would require the extra factor $\sqrt{d}$ when reshaping its eigenvectors into Kraus operators.

The generalized measurement postulate specifies Kraus operators $M_{i\alpha }$, where $i$ is the recorded outcome and $\alpha$ permits unobserved alternatives for that outcome, satisfyingFor a density operator $\rho$, outcome $i$ has probability and conditional stateWith one operator per outcome, this is the usual formula $p_i=\mathrm{Tr}(M_i\rho M_i^{†})$, ${\rho }_i=M_i\rho M_i^{†}/p_i$. The positive operators $E_i={\sum }_{\alpha }{M}_{i\alpha }^{†}{M}_{i\alpha }$ form a POVM, while the state-update maps form a quantum instrument.

The full postulate reduces to the usual projective measurement with the Lüders update when there is one operator per outcome and $M_i=P_i$, with $P_i=P_i^{†}=P_i^2$, $P_i{P}_j=0$ for $i\ne j$, and ${\sum }_i{P}_i=I$. Then $p_i=\mathrm{Tr}(P_i\rho )$ and ${\rho }_i=P_i\rho P_i/p_i$. Projective effects $E_i=P_i$ alone ensure these probabilities but do not fix the state update: an additional outcome-dependent unitary could still be applied. The choice $M_i=P_i$ makes the required projective state-update condition explicit.

The unread measurement is the pinching map $\mathcal{P}(\rho )={\rho }^{\prime }={\sum }_i{P}_i\rho P_i$, a nonselective projective measurement. Its output is block diagonal and commutes with each $P_i$, so $\log {\rho }^{\prime }$ also commutes with the projectors on its support.

First the quantum relative entropy $D(\rho \Vert {\rho }^{\prime })$ is finite. If $v$ lies in the kernel of ${\rho }^{\prime }$, positivity givesEvery summand is zero; summing ${\rho }^{1/2}{P}_{i}v=0$ gives ${\rho }^{1/2}v=0$. Thus $\ker {\rho }^{\prime }\subseteq \ker \rho$, equivalently $\mathrm{supp}\rho \subseteq \mathrm{supp}{\rho }^{\prime }$.

The block-diagonal logarithm obeys ${\sum }_i{P}_i(\log {\rho }^{\prime })P_i=\log {\rho }^{\prime }$ on this support. Cyclicity of the trace therefore givesConsequently the relative entropy of a pinched state is exactly the entropy increase:By the proved nonnegativity of quantum relative entropy and its equality condition,Equivalently, equality holds exactly when all off-diagonal blocks $P_i\rho P_j$ for $i\ne j$ already vanish, or when $\rho$ commutes with every measurement projector. This argument proves the equality condition even when the projectors are degenerate and the states singular.

The orthogonal projection onto the $+1$ eigenspace of the Pauli matrix ${\sigma }_z$ is $P_{+}=(I+{\sigma }_z)/2$. By the Born rule and the Bloch vector representation,Since $s_z=1/5$,The given vector has length $19/30<1$, so it lies inside the Bloch ball and indeed defines a valid mixed qubit state. For the physical spin operator $S_z=(\hslash /2){\sigma }_z$, the corresponding outcome is $+\hslash /2$ with the same probability.