Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 35 (Quantum Information Theory)

The Pauli group includes scalar phases $\mathcal{Z}=\{\pm I,\pm iI\}$. For a nonzero stabilizer code, its Abelian subgroup must satisfy $-I\notin S$. Otherwise a common positive eigenvector would satisfy $-|\psi ⟩=|\psi ⟩$ and vanish; for example $S=\{I,-I\}$ is Abelian but yields no nonzero stabilizer code. We impose this standard admissibility condition.

Every element of such a stabilizer group is Hermitian and squares to $I$: an imaginary-phase Pauli with square $-I$ is excluded. The associated stabilizer code and its orthogonal projector areThe group multiplication proves $P^2=P$, Hermiticity gives $P^{†}=P$, and its range is exactly the common positive eigenspace. If $S$ has $r$ independent generators, $|S|=2^r$. Every nonidentity element has trace zero, while $\mathrm{tr}I=2^n$, soThus it encodes $k=n-r$ logical qubits.

For the symplectic representation of a stabilizer code, label $i^t{X}^a{Z}^b$ by $(a,b)\in {\mathbb{F}}_2^{2n}$, ignoring scalar phase. Products add labels, andThis is the nondegenerate alternating form $\omega ((a,b),(a^{\prime },b^{\prime }))=a\cdot b^{\prime }+b\cdot a^{\prime }$ on the binary symplectic space of Pauli labels. The image $L$ of $S$ is an r-dimensional isotropic subspace of a binary symplectic space, since the generators commute and no nontrivial scalar lies in $S$. Consequently $L\subseteq L^{\perp }$ and $r\le n$. A binary generator matrix $(A\mid B)$ satisfies $A{B}^T+B{A}^T=0$. Conversely an isotropic subspace can be lifted by choosing commuting Hermitian generators with independent labels; their products cannot equal $-I$, and the orthogonal projector formula supplies the stabilizer code. The labels of the centralizer of a stabilizer group are $L^{\perp }$, with nontrivial logical actions represented by $L^{\perp }\setminus L$.

For an example, the three-qubit bit-flip repetition code has $S=⟨Z_1{Z}_2,Z_2{Z}_3⟩$ and ${\mathcal{C}}_S=\mathrm{span}\{|000⟩,|111⟩\}$. Its two binary rows have zero X part and Z parts $110,011$. The errors $I,X_1,X_2,X_3$ have respective error syndromes $00,10,11,01$, so all single bit flips can be corrected. This is not a stabilizer code correcting every single-qubit error: $Z_1$ acts as a nontrivial logical phase error.

Let $C(S)$ denote the Pauli centralizer. The exact Pauli error criterion for a stabilizer code isEquivalently, no difference of error labels lies in $L^{\perp }\setminus L$. The scalar phases in $\mathcal{Z}S$ are essential: an error product $-I$ is harmless even though it is not in $S$.

We prove the criterion. For $T=E_a^{†}{E}_b$ outside $C(S)$, some $s\in S$ anticommutes with $T$. Since $sP=Ps=P$, we have $PTP=PsTsP=-PTP$, hence $PTP=0$. If $T=\lambda s\in \mathcal{Z}S$, then $PTP=\lambda P$. If $T\in C(S)\setminus \mathcal{Z}S$, it restricts to a unitary operator on the stabilizer code, butno $sT$ is scalar. A scalar restriction would have scalar coefficient zero by this trace equation, contradicting unitarity. Thus this third case is not scalar.

To justify why scalar compressions are exactly what correction requires, suppose one recovery corrects every $E_a$. Dilate that recovery to an isometry $W$ including its auxiliary system. Correct recovery of every pure stabilizer code state gives $W{E}_a|\psi ⟩=|\psi ⟩\otimes |{\eta }_a⟩$. The auxiliary state is independent of $\psi$: linearity on basis states and their superpositions forces their auxiliary vectors to coincide. Preservation of inner products then giveswhich is the Knill--Laflamme condition $P{E}_a^{†}{E}_{b}P=c_{ab}P$. The three cases above prove necessity of the boxed criterion.

For sufficiency, put errors in the same class when $E_a^{†}{E}_b\in \mathcal{Z}S$. They act identically on the stabilizer code up to scalar phase, so give the same error-image subspace. Under the criterion, different classes give mutually orthogonal subspaces $E_a{\mathcal{C}}_S$, because their cross-compressions vanish. Measure which of these subspaces is occupied and apply the inverse of a representative error. This returns every stabilizer code state unchanged; extend the recovery on the orthogonal complement by preparing any fixed stabilizer code state. Linearity also corrects any valid noise channel whose Kraus operators are linear combinations of the specified errors: within each syndrome all amplitudes multiply the same logical state, and trace preservation normalizes the recovered state. This proves sufficiency, including degenerate error classes.

For a classical error-correcting code of length $n$, alphabet size $q$, minimum Hamming distance $d$ and $M$ codewords, the Singleton bound isFor a linear $[n,k,d{]}_q$ code this is $n-k\ge d-1$. Deleting any $d-1$ coordinates is injective on the codewords, since two words agreeing on the remaining coordinates would differ in fewer than $d$ positions; this also explains the classical bound.

For an n-qubit quantum code carrying $k>0$ logical qubits and having quantum code distance $d$, the quantum Singleton bound isMore generally replace $k$ by ${\log }_{2}K$ for code dimension $K>1$. The proof applies to degenerate codes too.

Let $r=d-1$. Purify the maximally mixed encoded input by a reference $R$ of dimension $K$:Distance $d$ makes erasure of any subset of at most $r$ qubits correctable. The allowed quantum erasure correction condition says that a correctable erased set $A$ is decoupled from $R$:This is the matrix-element correction condition in another form: all operators on $A$ have constant diagonal matrix elements and zero off-diagonal matrix elements in the logical basis, so tracing the other physical qubits in $|\Omega ⟩⟨\Omega |$ yields the displayed product.

First justify the sizes needed for the partition. If $2r>n$, partition all physical qubits into $A,B$, both of size at most $r$. Both sets are correctable. Purity of $RAB$ and the Schmidt decomposition giveAdding forces $k=0$, a contradiction. Thus $2r\le n$, without assuming the bound we aim to prove.

Choose disjoint sets $A,B$ of size $r$, and let $C$ contain the remaining $n-2r$ qubits. The joint state $RABC$ is pure, so decoupling and Subadditivity of Von Neumann entropy giveAdding and cancelling gives $k\le S(C)$. The Von Neumann entropy of $D$-dimensional states is at most ${\log }_{2}D$: for positive eigenvalues $p_i$, concavity of the logarithm gives ${\sum }_i{p}_i{\log }_2(1/p_i)\le {\log }_2{\sum }_i{p}_i/p_i\le {\log }_{2}D$. Here $D=2^{n-2r}$, so $k\le n-2r$, proving the quantum bound. The code is assumed to encode information ($K>1$); a one-dimensional code requires a separate distance convention, since simply re-preparing a known state makes its erasure-correction condition vacuous.

The channel capacity is the supremum of reliable communication rates: a rate $R$ bits per use is achievable if block encoders and decoders using $n$ channel uses can transmit $M_n$ messages with error probability tending to zero and $\mathrm{liminf}{n}^{-1}{\log }_2{M}_n\ge R$.

A discrete memoryless channel has fixed transition probabilities $W(y|x)$ on finite input and output alphabets, andIts single-letter capacity formula iswith logarithms to base two. The Shannon second coding theorem identifies this maximum with the operational reliable rate.

For the binary erasure channel, use output symbols $0,1,e$, where $e$ is a recognizable erasure flag. Given input $x$, the output is $x$ with probability $1-p$ and $e$ with probability $p$. If $\mathrm{Pr}(X=1)=a$, the output probabilities are $(1-p)(1-a),(1-p)a,p$. Thus, writing $h_2$ for binary entropy,and $I(X;Y)=(1-p)h_2(a)$. Its maximum is attained by the uniform input $a=1/2$, givingIn particular the noiseless and complete-erasure endpoints have capacities one and zero. The erasure flag reveals which positions were lost; it does not reveal the erased input values.

For a density matrix $\rho$ on a finite-dimensional Hilbert space, its Von Neumann entropy in bits iswhere ${\lambda }_j$ are its eigenvalues and $0{\log }_{2}0=0$. It is the Shannon entropy of the spectrum, is zero for a pure state, and is invariant under unitary conjugation.

The concavity inequality is true, but its printed equality condition is false. At $b=1$ equality always holds, even for distinct states; for example take ${\rho }_1=|0⟩⟨0|$ and ${\rho }_2=|1⟩⟨1|$. The correct statement on $0\le b\le 1$ isFor $0<b<1$, entropy is therefore strictly concave.

We give the matrix inequality needed for the proof. Define the quantum relative entropy $D(\tau \Vert \sigma )=\mathrm{tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ if the support of $\tau$ is contained in that of $\sigma$, and $+\infty$ otherwise. Diagonalize the two states with eigenvalues $r_i,s_j$, and put $q_{ij}=|⟨i|j⟩{|}^2$. Each row and column of $q$ sums to one. The elementary scalar inequalityfollows from $x\ln x-x+1\ge 0$, with equality exactly at $x=1$; use limits at zero. ConsequentlyEquality forces $r_i=s_j$ whenever $q_{ij}>0$. Hence every eigenvector $|i⟩$ of $\tau$ is also a $\sigma$-eigenvector with eigenvalue $r_i$, so $\tau =\sigma$. Conversely identical states give zero relative entropy. This proves nonnegativity of quantum relative entropy and its equality condition, including states with zero eigenvalues under the support convention.

Now set $\rho =b{\rho }_1+(1-b){\rho }_2$. For $0<b<1$ its support contains both component supports. Direct expansion givesSince both coefficients are positive, equality holds exactly when both component states equal $\rho$, equivalently ${\rho }_1={\rho }_2$. At either endpoint the mixture has only one component and equality is automatic. This proves the corrected strict concavity of Von Neumann entropy.

False. Take the Bell stateThe joint state is pure, so $S(\rho )=0$. Taking either partial trace kills the off-diagonal terms and gives ${\rho }_1={\rho }_2=I_2/2$, whose two eigenvalues are $1/2$. Thus $S({\rho }_1)=S({\rho }_2)=1$, contradicting the proposed inequality $1\le 0$. Quantum entanglement allows a pure joint state to have mixed subsystem states.

True: quantum entropy is subadditive. Write $\sigma ={\rho }_1\otimes {\rho }_2$. Positivity ensures the support of $\rho$ lies in that of $\sigma$. Indeed, if $|v⟩\in \ker {\rho }_1$, then ${\sum }_j⟨v,j|\rho |v,j⟩=0$; all summands are nonnegative, so ${\rho }^{1/2}|v,j⟩=0$ for every $j$. Thus $\rho$ annihilates $\ker {\rho }_1\otimes {\mathcal{K}}_2$, and similarly the kernel from the other factor.

On the resulting support, ${\log }_2({\rho }_1\otimes {\rho }_2)={\log }_2{\rho }_1\otimes I+I\otimes {\log }_2{\rho }_2$. Using the defining property of the partial trace and nonnegativity of quantum relative entropy proved in part (i),HenceEquality holds exactly for the product state $\rho ={\rho }_1\otimes {\rho }_2$. This proves Subadditivity of Von Neumann entropy directly.

For a memoryless quantum channel $\mathcal{N}$, successive uses are described by ${\mathcal{N}}^{\otimes n}$. Its entanglement-assisted classical capacity is the supremum of classical bits per use achievable with error tending to zero when sender and receiver may use unlimited preshared quantum entanglement, independent of the message. Only the channel is used to transmit the message.

For input state ${\rho }_A$, choose a purification $|{\psi }_{\rho }{⟩}_{RA}$ and set ${\omega }_{RB}=({\mathrm{id}}_R\otimes \mathcal{N})(|{\psi }_{\rho }⟩⟨{\psi }_{\rho }|)$. The capacity formula isPurifications differ only by an isometry on the reference, so this expression depends only on the input density matrix. The quantity optimized is quantum mutual information.

The binary quantum erasure channel sends a qubit to a qubit subspace plus an orthogonal erasure flag:On the nonerased branch it transmits the entire state, including its coherences. Put $s=S(\rho )$. The entropy of an orthogonal quantum mixture givesThe reference-output state has two orthogonal branches:The first branch is pure; the second has entropy $S({\rho }_R)=s$. Thus $S({\omega }_{RB})=h_2(p)+ps$ andA qubit has entropy at most one bit, with equality at $\rho =I_2/2$. We concludeThe endpoint capacities are two and zero. As an operational check, superdense coding encodes four classical messages in one member of a shared Bell pair. A successful channel use lets the receiver identify the message; an erased qubit yields the flag. This induces a four-symbol classical erasure channel of capacity $(1-p){\log }_{2}4=2(1-p)$, agreeing with the entropy calculation.