Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 66 / 2

Quantum teleportation transfers the quantum state of an unknown qubit from a sender's carrier to a receiver's carrier. It also transfers correlations with any external reference. It does not determine classical amplitudes or send the original particle.

The standard deterministic protocol consumes one shared maximally entangled Bell pair and two classical bits from Alice to Bob. Alice performs a local Bell-basis measurement on the input and her half of the pair; Bob applies the corresponding local Pauli gate. The classical message is essential: before receiving it, Bob has the outcome-averaged density operator $I/2$ and cannot recover or detect the unknown state. The original input is consumed by the measurement, so quantum teleportation is compatible with the no-cloning theorem.

Write the arbitrary joint pure state of the input $A$ and an external reference $R$ asThe reference vectors need not be normalized or orthogonal. Share $|{\Phi }^{+}{⟩}_{aB}$. Label Alice's Bell states byProjecting onto this Bell basis gives the unnormalized stateEach outcome has probability $1/4$. Alice sends $(r,s)$ and Bob applies $Z_B^r{X}_B^s$, reversing the two Pauli gates. The final state is exactly $|\Omega {⟩}_{BR}$ for every outcome. In particular all input entanglement and other correlations with $R$ now belong to $B$, while the reference marginal is unchanged. This is teleportation as an identity channel on a reference.

Conditioned on the record, Alice's two measured carriers are in $|{\beta }_{rs}⟩$ and factor from $BR$. Thus $A$ no longer retains its original correlations with $R$. There is no second copy, even when the teleported qubit was initially entangled with a larger system.

The root $\omega$ must be a primitive $n$th root of unity, for example $e^{2\pi i/n}$. The printed statement merely says $n$th root; that is insufficient. With $n=2$ and $\omega =1$, the states with $r=0$ and $r=1$ coincide. More generally, a root of order $d<n$ repeats the $r$ labels modulo $d$. For $n=1$ the construction is trivial.

For a primitive root of unity, the generalized Bell basis obeysThe last equality follows from the finite geometric series: a nonzero exponent difference modulo $n$ has sum zero, while zero difference has sum $n$. There are $n^2$ orthonormal vectors in an $n^2$-dimensional Hilbert space, so they form a complete basis.

For qudit teleportation, Alice's input is $|\chi {⟩}_U={\sum }_j{c}_j|j⟩$. Alice and Bob share $|{\psi }_{00}{⟩}_{aB}=n^{-1/2}{\sum }_k|kk⟩$. Define the qudit shift and phase operators by $X|j⟩=|j+1⟩$ and $Z|j⟩={\omega }^j|j⟩$. Alice measures $Ua$ in the generalized Bell basis. On outcome $(r,s)$, Bob's unnormalized state isAll outcomes have probability $1/n^2$. Alice communicates $(r,s)$ and Bob applieswhich restores $|\chi ⟩$ exactly. The protocol consumes one maximally entangled pair of $n$-level systems, a local $n^2$-outcome measurement, and a classical message with $n^2$ possibilities. For a fixed-length binary encoding, $\lceil {\log }_2{n}^2\rceil$ bits suffice. No measurement depends on the unknown amplitudes.

Bob uses qudit teleportation to teleport his carrier $B_1$ to Clare through the pair $B_{2}C$. This transfers Bob's entanglement with Alice to Clare. Explicitly, Bob performs a generalized Bell basis measurement on $B_1{B}_2$. Contraction of the two initial pairs givesThe probability is $1/n^2$ for each outcome. Bob sends $(r,s)$ to Clare, who applies $Z_C^r{X}_C^{-s}$. Alice and Clare then share $|{\psi }_{00}{⟩}_{AC}$ with certainty. This is entanglement swapping; neither an additional entangled pair nor a quantum transmission during the protocol is required.

The two initial pairs are consumed, and Bob's measured carriers cease to be entangled with $AC$. Without Bob's classical record, averaging the possible generalized Bell states leaves $AC$ maximally mixed. Thus the conditional entanglement swapping does not supply faster-than-light communication.