Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 51 / 1

Quantum teleportation transfers an arbitrary unknown qubit state to a separated receiver while using local operations and classical communication, rather than sending that input qubit. Alice and Bob first share one maximally entangled state of two qubits, such as a Bell pair; a shared spin singlet state is equivalent after a known local unitary operation.

Alice performs a joint Bell-basis measurement on the input qubit and her half of the pair. There are four possible outcomes. Bob's conditional qubit is the input state transformed by a known Pauli operator determined by that outcome. Alice sends the outcome as two classical bits, and Bob applies the inverse Pauli operator. Thus one shared Bell pair and two classical bits, together with local measurement and correction, transmit one unknown qubit exactly. The entangled resource is consumed.

The procedure does not produce a classical description of the unknown amplitudes and does not leave a second copy at Alice, so it respects the no-cloning theorem. Before receiving the classical outcome, Bob's reduced density matrix is $I/2$, independent of the input; this respects the no-communication theorem. The next part proves the full identity-channel action, including inputs entangled with another system.

Label an orthonormal basis of the $n$-dimensional Hilbert space by $j=0,\dots ,n-1$, with indices modulo $n$. LetThese are the shift and phase Heisenberg-Weyl operators. Alice holds the input $S$ and one half $A$ of the shared maximally entangled stateShe measures $SA$ in the generalized Bell basisIts inner products areThere are $n^2$ orthonormal vectors in an $n^2$-dimensional space, so this is a complete projective measurement.

For an unknown normalized state $|u⟩={\sum }_j{u}_j|j⟩$, its contraction with measurement outcome $(p,q)$ gives Bob's unnormalized stateEach outcome has probability $1/n^2$, independently of the input. Alice communicates the two $n$-valued labels, and Bob applies the ordered inverseThe order matters because shift and phase operators do not generally commute. Normalizing the branch recovers the unknown state exactly, and all outcomes are corrected, so success is deterministic. The quantum resource is one maximally entangled pair of local dimension $n$. The classical resource is an $n^2$-valued message, with information content $2{\log }_{2}n$ bits; $\lceil {\log }_2(n^2)\rceil$ ordinary bits suffice in fixed-length binary encoding.

This is faithful qudit teleportation even for an input entangled with a reference $R$. Expand a joint pure state as ${\sum }_j|j{⟩}_S|r_j{⟩}_R$, without requiring the reference vectors to be orthogonal. The same contraction and correction act on each system basis vector, producing $n^{-1}{\sum }_j|j{⟩}_B|r_j{⟩}_R$. Thus the whole joint state, not just a reduced state, is preserved. Linearity extends the conclusion to mixed inputs. This is teleportation as an identity channel on a reference.

Yes: deterministic entanglement swapping gives the required singlet. Denote Bob's two local qubits by $B_1,B_2$. Known local unitary operations first convert the two given spin singlet states into $|{\Phi }^{+}{⟩}_{A{B}_1}$ and $|{\Phi }^{+}{⟩}_{B_{2}C}$. For example applying $ZX$ to the second qubit of a singlet gives $|{\Phi }^{+}⟩$, where $X,Z$ are Pauli operators.

Use the binary-labelled Bell statesThey are ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ for labels $(0,0),(1,0),(0,1),(1,1)$. Reordering registers to group Bob's qubits, direct expansion givesAll four basis vectors are real in this convention; otherwise the paired expansion must account for complex conjugation. Bob performs the Bell-basis measurement on $B_1{B}_2$. This joint operation is local to him, so it is allowed by LOCC. Each outcome occurs with probability $1/4$ and leaves $AC$ in the corresponding Bell state.

Bob sends $(p,q)$ to Charlie. Charlie applies $Z^p{X}^{-q}$ to obtain $|{\Phi }^{+}{⟩}_{AC}$, and then $XZ$ to obtainThe correction works for every outcome, so no postselection or successful-outcome assumption is needed. This is entanglement swapping, equivalently teleportation of $B_1$ to $C$ while preserving its entanglement with $A$. If Bob's measurement result is ignored, the average $AC$ state is $I_4/4$; the protocol therefore does not supply entanglement usable without the classical outcome or violate the no-communication theorem.