Quantum gate teleportation is the act of being able to apply a quantum gate on the unknown state while it is being teleported. This is one of the ways in which measurement-based computation can be described using graph states.

Usually, teleportation works by having an unknown quantum state $|\psi\rangle$ held by Alice, and two qubits in the Bell state $|\Psi\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$ shared between Alice and Bob. Alice performs a Bell state measurement, getting one of 4 possible answers and Bob holds on his qubit, depending on the measurement result of Alice, one of the 4 states $|\psi\rangle,X|\psi\rangle,Z|\psi\rangle,ZX|\psi\rangle.$ So, once Bob learns what result Alice got, he can compensate by applying the appropriate Paulis.

Let $U$ be a 1-qubit unitary. Assume Alice and Bob share $(\mathbb{I}\otimes U)|\Psi\rangle$ instead of $|\Psi\rangle$. If they repeat the teleportation protocol, Bob now has one of $U|\psi\rangle,UX|\psi\rangle,UZ|\psi\rangle,UZX|\psi\rangle$, which we can rewrite as $U|\psi\rangle,(UXU^\dagger)U|\psi\rangle,(UZU^\dagger)U|\psi\rangle,(UZXU^\dagger)U|\psi\rangle.$ The compensations that Bob has to make for a given measurement result are given by the bracketed terms. Often, these are no worse than the compensations you would have to make for normal teleportation (i.e. just the Pauli rotations). For example, if $U$ is the Hadamard rotation, then the corrections are just $(\mathbb{I},Z,X,XZ)$ respectively. So, you can apply the Hadamard during teleportation just be changing the state that you teleport through (There is a strong connection here to the Choi-Jamiołkowski isomorphism). You can do the same for Pauli gates, and the phase gate $\sqrt{Z}=S$. Moreover, if you repeat this protocol to build up a more complicated computation, it is often sufficient to keep a record of what these corrections are, and to apply them later.

Even if you don't only need the Pauli gates (as is the case for $T=\sqrt{S}$), the compensations may be easier than implementing the gate directly. This is the basis of the construction of the fault-tolerant T gate.

In fact, you can do something similar to apply a controlled-NOT between a pair of qubits as well. This time, the state you need is $|\Psi\rangle_{A_1B_1}|\Psi\rangle_{A_1B_1}$, and a controlled-NOT applied between $B_1$ and $B_2$. This time, there are 16 possible compensating rotations, but all of them are just about how Pauli operations propagate through the action of a controlled-NOT and, again, that just gives Pauli operations out.

2I wish I could favourite answers, i created an account just to let you know how good this answer is – John – 2018-04-17T11:23:19.103

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– user – 2018-04-17T13:27:58.503"which we can rewrite as $|\psi\rangle$ ..." should be "$U |\psi\rangle$". – qbt937 – 2018-11-07T23:09:54.397

@qbt937 well spotted! Thanks. – DaftWullie – 2018-11-08T06:08:02.107