Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 37 (Quantum Information Theory)

Use base-two logarithms for Von Neumann entropy, so the units are bits, and adopt $0{\log }_{2}0=0$. All systems here may be taken finite-dimensional; the entropy arguments also apply when the displayed Von Neumann entropies are finite. The density operator of the joint pure state is ${\rho }_{AB}=|{\Psi }_{AB}⟩⟨{\Psi }_{AB}|$. It is a rank-one orthogonal projector, whose only nonzero eigenvalue is one. ConsequentlyThe reduced density matrices are ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. The quantum conditional entropy is defined by the entropy differenceFor the pure state under consideration this becomes $S(B|A)=-S({\rho }_A)$. To determine when it is negative, write the Schmidt decompositionThe eigenvalues of ${\rho }_A$ are the ${\lambda }_j$, together with possible zeros. Each contribution $-{\lambda }_j{\log }_2{\lambda }_j$ is nonnegative, and their sum vanishes exactly when one eigenvalue equals one. Equivalently the Schmidt rank is one and $|{\Psi }_{AB}⟩$ is a product state. If the pure state is entangled, then $r\ge 2$, every nonzero eigenvalue is strictly below one, and the Von Neumann entropy is strictly positive. Thus the pure-state negative conditional entropy criterion givesThe purity assumption matters: for mixed density operators, negative quantum conditional entropy is not necessary for entanglement.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

Diagonalize the reduced density matrix on Alice's Hilbert space:For every positive eigenvalue, define a vector on Bob's Hilbert space byThe definition of the partial trace givesThus the $|b_j⟩$ are orthonormal. Extend the $|a_j⟩$ to an orthonormal basis of Alice's Hilbert space. If $|a⟩$ lies in the zero-eigenvalue subspace of ${\rho }_A$, thenThere is therefore no component of $|{\Psi }_{AB}⟩$ in those additional basis directions. Expansion in Alice's orthonormal basis yields the Schmidt decompositionIts positive Schmidt coefficients are $s_j=\sqrt{{\lambda }_j}$ and its Schmidt rank is $r$. Taking the two partial traces and using the orthonormality of both families givesHence the two reduced density matrices have exactly the same nonzero eigenvalues, including multiplicities. Their numbers of zero eigenvalues may differ when the local Hilbert spaces have different dimensions. In particular their Von Neumann entropies agree.

The coefficient matrix in the computational orthonormal bases, with Alice's index labelling rows, isSince the entries are real, the reduced density matrices areBoth have trace one and determinant $1/9$. Their characteristic polynomial is ${\lambda }^2-\lambda +1/9$, so the nonzero eigenvalues areBoth are positive. Thus the Schmidt rank, or the Schmidt number of this pure state, isIf “Schmidt numbers” denotes the individual positive Schmidt coefficients instead, those areTheir squares are ${\lambda }_{\pm }$; giving both the Schmidt rank and the Schmidt coefficients removes this terminology ambiguity. The pure state is entangled because its Schmidt rank exceeds one.

Factor the coefficients into a tensor product:This is already a Schmidt decomposition with one term. ConsequentlyEach reduced density matrix is the rank-one orthogonal projector onto $(|0⟩-|1⟩)/\sqrt{2}$, so its other eigenvalue is zero. The joint pure state is a product state, with no entanglement.

Yes: a deterministic LOCC protocol exists for every allowed value of the parameter. Here is an explicit implementation, so existence does not depend on merely quoting a pure-state conversion theorem.

First convert the input Bell state to $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$ by local unitary operations:Write $c=\cos \varphi$ and $s=\sin \varphi$. Alice performs the two-outcome measurement in quantum mechanics with Kraus operatorsThey define a physical measurement in quantum mechanics becauseActing on $|{\Phi }^{+}⟩$, the unnormalized branch vectors areBoth branches have probability $1/2$. Alice communicates the outcome to Bob. If it is one, both apply the Pauli X gate; this turns the normalized second branch into $c|00⟩+s|11⟩$. If it is zero, neither applies this correction. Finally Bob applies a Pauli X gate in either branch, obtainingEvery outcome produces the same target pure state, so this is deterministic local operations and classical communication, not a successful branch selected by postselection. This two-outcome LOCC dilution of a Bell pair remains valid at $\varphi =0$, where the output is a product state, and at $\varphi =\pi /4$, where it is another Bell state.

The target Schmidt coefficients are $c,s$, while those of the input are both $1/\sqrt{2}$. For an interior value $0<\varphi <\pi /4$, local unitary operations alone could not change these Schmidt coefficients; the local measurement in quantum mechanics is the step that reduces the entanglement.

Expand an initial bipartite pure state in local orthonormal bases asBy the Schmidt decomposition, or equivalently the singular value decomposition of $C$, its Schmidt rank is $\mathrm{rank}C$. In a specified local measurement branch, the product Kraus operator $A\otimes B$ changes the coefficient matrix toIndeed the coefficient of $|k⟩|l⟩$ is ${\sum }_{i,j}{A}_{ki}{C}_{ij}{B}_{lj}$. The elementary matrix rank inequality givesNormalizing a nonzero branch multiplies $C^{\prime }$ by a scalar and therefore leaves its matrix rank unchanged.

Now refine a complete LOCC transcript to record every individual Kraus operator. Although later choices depend on the earlier classical outcomes, fixing a transcript fixes all those choices. Composing the local operations therefore gives a single product Kraus operator $A_{\tau }\otimes B_{\tau }$ in that branch. Each nonzero branch obeys the same Schmidt-rank contraction under product operators. Local auxiliary systems prepared independently are included by local isometries and do not add entanglement. Discarding a local auxiliary system is included by resolving its partial trace in an orthonormal basis; this again yields refined product Kraus operators.

If the protocol produces a specified pure state deterministically, its output density operator is a sum of positive rank-one branch density operators. Every nonzero branch vector must be proportional to that output vector: its squared overlap with any vector orthogonal to the output sums to zero, so each such overlap is separately zero. Thus the output Schmidt rank equals that of each surviving branch, andThis proves monotonicity of Schmidt rank under LOCC, even for a nonzero branch selected by postselection. If the classical transcript is instead discarded and the result is mixed, the refined branch vectors supply an ensemble with Schmidt ranks at most $r_{\mathrm{in}}$; by definition its Schmidt number is at most $r_{\mathrm{in}}$. This is the correct mixed-state extension, rather than a bound on the ranks of its reduced density matrices. In finite dimensions, any limiting exact protocol obeys the same pure-state bound because the set of coefficient matrices of rank at most $r_{\mathrm{in}}$ is closed, being defined by vanishing minors.

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.

Under the intended assumption that one use of the quantum channel transmits one qubit without noise, Alice can communicate two classical bits with certainty if she and Bob already share a Bell pair. She uses superdense coding. The prior distribution of the Bell pair must be a resource available before the permitted channel use; distributing it through that same channel would consume another use.

Take the shared Bell state to be $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$, with Alice holding the first qubit. For a two-bit message $(a,b)\in \{0,1{\}}^2$, Alice applies the local unitary operation $U_{ab}=X^a{Z}^b$ to her qubit. The four encoded pure states are, up to irrelevant global phases,They are orthogonal: for any matrix $V$ on the first qubit, $⟨{\Phi }^{+}|(V\otimes I)|{\Phi }^{+}⟩=\mathrm{Tr}V/2$, and the four Pauli operators are orthogonal in the trace inner product. Explicitly,Alice sends her qubit through the quantum channel once. Bob now holds both qubits and performs a Bell-basis measurement. Its four orthogonal projectors identify $(a,b)$ with certainty, givingBefore the transmission, Bob's reduced density matrix is $I/2$ for every message, so shared entanglement alone does not convey the message.

Without shared entanglement, a noiseless single-qubit transmission cannot perfectly distinguish four messages: perfect discrimination requires orthogonal state supports, and its two-dimensional Hilbert space has room for at most two nonzero mutually orthogonal supports. The Holevo bound also limits its accessible classical information to one bit. For the usual protocol starting with a shared two-qubit pure state and using local unitary encodings, exact two-bit transmission requires a maximally mixed Bob marginal. Indeed all encoded joint states are pure and have the same ${\rho }_B$, while Subadditivity of Von Neumann entropy gives their average entropy at most $1+S({\rho }_B)$. Four equally likely perfectly distinguishable messages have average Von Neumann entropy two, forcing $S({\rho }_B)=1$ and hence a maximally entangled state.

The specification “quantum channel” alone does not determine its dimension or noise, so the one-qubit, noiseless interpretation is needed for this particular conclusion. A noiseless four-dimensional channel could send four orthogonal states without any shared entanglement; a completely depolarizing qubit channel cannot transmit the message even with a shared Bell pair, since its output is independent of Alice's encoding. Shared entanglement is therefore the extra resource for the intended single-qubit channel, not a guarantee for every unspecified quantum channel.

Let $D={\mathcal{C}}_{\mathrm{ev}}$, and write ${\mathcal{C}}_{\mathrm{odd}}=t+D$ for any odd-weight word $t\in {\mathcal{C}}_H$. The parity functional on the four-dimensional binary linear code ${\mathcal{C}}_H$ is nonzero, so its kernel $D$ has dimension three and its other fibre is the coset $t+D$. For example $t=(1,\dots ,1)$ belongs to the standard length-seven Hamming code and has odd Hamming weight. Denote the two logical basis states by $|{\psi }_0⟩$ and $|{\psi }_1⟩$, corresponding to these two cosets. The eight computational orthonormal basis vectors in each sum are distinct, so both logical states have norm one; their disjoint supports make them orthogonal.

For a binary vector $w\in {\mathbb{F}}_2^7$, define the product of Pauli Z gateswhere the dot product is computed modulo two. The key diagonal overlap, with $t_0=0$ and $t_1=t$, isIf $w\in D^{\perp }$, every summand is one. If $w\notin D^{\perp }$, choose $x_0\in D$ with $w\cdot x_0=1$. Translation $x\mapsto x+x_0$ permutes $D$ and negates every summand. The sum must then be zero. ThusOff-diagonal logical overlaps vanish for every $w$:because a diagonal Pauli operator cannot move a computational basis word from one coset to the disjoint other coset.

The possible errors are $E_0=I$ and $E_j=Z(e_j)$ for $j=1,\dots ,7$. Set $e_0=0$. Each phase flip is its own adjoint and inverse, soIf $i\ne j$, the vector $e_i+e_j$ is nonzero and has Hamming weight one or two. By the given dual code identity $D^{\perp }={\mathcal{C}}_H$ and the minimum Hamming distance three of the Hamming code, it cannot belong to $D^{\perp }$. Hence all its diagonal logical overlaps vanish. If $i=j$, the product is the identity and its logical overlaps are ${\delta }_{ab}$. Combining both cases givesThis proves the Knill--Laflamme condition with an identity error-overlap matrix, not merely a possibly degenerate scalar matrix. In particular the eight two-dimensional error subspaces $E_i{\mathcal{X}}_{\mathrm{Steane}}$ are mutually orthogonal. This is precisely nondegenerate phase-flip correction in the Steane code.

To exhibit recovery rather than only the criterion, let $P$ be the orthogonal projector onto the code and $P_i=E_{i}P{E}_i^{†}$. The projective measurement consisting of these eight mutually orthogonal projectors, completed by $I-{\sum }_i{P}_i$, determines which error occurred. For an arbitrary encoded pure state $|\chi ⟩=\alpha |{\psi }_0⟩+\beta |{\psi }_1⟩$ affected by $E_j$, outcome $j$ occurs with certainty and leaves $E_j|\chi ⟩$ unchanged. Applying $E_j$ again recovers $|\chi ⟩$. The outcome gives the error syndrome and does not reveal $\alpha$ or $\beta$, so recovery preserves logical quantum superpositions.

An equivalent explicit error syndrome uses three commuting X-type stabilizer generators. Choose a parity-check matrix for ${\mathcal{C}}_H$ with the seven different nonzero binary columns:Its rows span ${\mathcal{C}}_H^{\perp }=D$. For a row $h_l$, the operator $X(h_l)={⨂}_j{X}_j^{(h_l{)}_j}$ permutes each of $D$ and $t+D$, and therefore has eigenvalue $+1$ on both logical states. A phase flip at position $j$ changes its eigenvalue to $(-1{)}^{H_{lj}}$, since $X(h_l)Z_j=(-1{)}^{H_{lj}}{Z}_{j}X(h_l)$. Measuring the three commuting stabilizer generators gives the $j$th column of $H$ as a binary syndrome; no error gives the zero column. All eight outcomes are different, and applying the identified $Z_j$ restores the code state. Measuring these stabilizer generators never distinguishes the two logical states.

Finally, if a noise Kraus operator is a coherent linear combination $F={\sum }_j{c}_j{E}_j$, the same syndrome measurement separates its mutually orthogonal components $c_j{E}_j|\chi ⟩$. Conditional recovery returns $|\chi ⟩$ in every nonzero branch. Linearity then also corrects any quantum channel whose Kraus operators lie in this span. The argument therefore corrects arbitrary unknown single-site phase flips nondegenerately, including coherent phase-error amplitudes, rather than only a known error on a basis codeword.