One-bit teleportation

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.