Quantum teleportation

Quantum teleportation transfers an unknown qubit using one shared Bell pair, a Bell-basis measurement by the sender, two classical bits, and a Pauli correction by the receiver.

Two initial maximally entangled pairs $A{B}_1$ and $B_{2}C$ can be turned into an entangled pair $AC$ by a generalized Bell basis measurement on $B_1{B}_2$, followed by a local correction at an endpoint using its classical label. This is qudit teleportation of $B_1$, including its correlations with $A$, to $C$. The initial two pairs are consumed. Without the classical label, averaging the conditional endpoint states need not leave entanglement.

A sender and receiver share $n^{-1/2}{\sum }_j|jj⟩$. The sender measures the input and their resource half in the generalized Bell basis. Outcome $(r,s)$ gives the receiver $X^s{Z}^{-r}|\chi ⟩/n$, with probability $1/n^2$. After classical communication of the outcome, the correction $Z^r{X}^{-s}$ restores the unknown quantum state. The same calculation preserves correlations with a reference, as in teleportation as an identity channel on a reference.

Ideal corrected quantum teleportation transfers an input system to the receiver while preserving its entire joint density operator with an arbitrary reference. Expand a joint pure state as ${\sum }_j|j⟩|r_j⟩$. Every teleportation branch acts with the same known correction on each system basis vector and does nothing to the reference vectors; undoing that correction restores the complete sum. Linearity extends the result to mixed states. The sender's measured carrier no longer retains the original reference correlations.

Apply a Controlled-Z gate to input $a|0⟩+b|1⟩$ and a fresh $|+⟩$ quantum ancilla. Measuring the input in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s$, leaves the unnormalized outputEach branch has probability $1/2$, independently of the input. The logical output is $J(\alpha )|\psi ⟩$ with known Pauli frame $X^s$; the input qubit has been measured, so this does not clone it.

A two-vertex graph state starts with two $|+⟩$ states joined by a Controlled-Z gate. Measuring the first vertex in the $X$ basis leaves the second in $X^{s}H|+⟩=X^s|0⟩$, where $s$ is the measurement outcome. Thus a known zero input can be supplied to a measurement wire using only a graph-state resource and a known Pauli frame. This preparation link precedes the links implementing the desired logical J gate; later measurement angles and output-bit corrections absorb its byproduct.

Two angle-zero one-bit teleportations with outcomes $p,q$ implement $X^{q}H{X}^{p}H=X^q{Z}^p$ on an arbitrary input, up to global phase. A Controlled-Z gate with a fixed $|1⟩$ quantum ancilla implements $Z$, so $Z^p$ removes the known $Z$ factor. The result is $X^q$ up to phase. With probability one half the desired Pauli X gate is applied; otherwise the input is unchanged. Repeating until $q=1$ gives an exact heralded correction with two attempts on average and almost-sure termination. It has no finite worst-case measurement count. A Pauli frame avoids this repeat-until-success procedure when only logical action or classical output statistics are required.

For the printed equatorial-basis convention,It is the Hadamard gate after a phase gate. The identities $J(\alpha )Z=XJ(\alpha )$ and $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$ follow by multiplying their matrices. For $\alpha =0,\pi /2$ these are Clifford gates. In particular $J(-\pi /2)=XJ(\pi /2)$, permitting a sign change of the measurement angle to be absorbed into a Pauli frame.

The J gate is $J(\alpha )=HP(\alpha )$, where $P(\alpha )=\mathrm{diag}(1,e^{i\alpha })$: the phase gate acts first and the Hadamard gate acts second. Unitary invariance of the operator norm reduces its angle error to the single nonzero diagonal difference of the phase gate. Its norm is $|e^{i\delta }-1|=2|\sin (\delta /2)|\le |\delta |$. The quantum circuit gate-error telescoping bound then makes an angle tolerance of $ϵ/k$ sufficient for $k$ such imperfect gates and exact other gates.

For $|{\beta }_{xz}⟩=(Z^z{X}^x\otimes I)|{\beta }_{00}⟩$, an arbitrary qubit obeysA Bell-basis measurement reveals $(x,z)$, after which the receiver applies the inverse Pauli correction.

A deterministic finite-dimensional universal programmable quantum gate cannot implement every unitary exactly: programs for physically distinct unitaries would have to be mutually orthogonal, but a finite-dimensional program register contains only finitely many mutually orthogonal states.

When the shared resource is $|{\Psi }^{+}⟩$, Bell outcomes ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$ leave the receiver with $X|\alpha ⟩,XZ|\alpha ⟩,|\alpha ⟩,Z|\alpha ⟩$, respectively, up to global phases.