As described e.g. in this post, quantum gate teleportation can be framed as a variation of quantum state teleportation where a gate is applied beforehand on the receiver, and this results in the final state being teleported being slightly different.

More precisely, if $|\psi\rangle$ is the state to send, $|\Psi\rangle\equiv\sum_i |i,i\rangle$ the shared maximally entangled state, then the standard state teleportation protocol can be seen as consequence of the fact that we can write (ignoring normalisation constants): $$|\psi\rangle|\Psi\rangle = \sum_a |\Psi_a\rangle\otimes (U_a^\dagger|\psi\rangle), \qquad |\Psi_a\rangle\equiv (U_a\otimes I)|\Psi\rangle,$$ where is a(ny) basis of unitary matrices, so that $\operatorname{Tr}(U_a^\dagger U_b)=d\delta_{ab}$, with $d$ dimension of the underlying space. This formulation is nicely described e.g. in this answer.

Similarly, for gate teleportation, we can write $$|\psi\rangle(I\otimes U)|\Psi\rangle = \sum_a |\Psi_a\rangle\otimes (UU_a^\dagger|\psi\rangle),$$ and we thus see how the "gate to teleport" $U$ is applied to the (corrected) teleported state. One can now go ahead and correct the teleported state as usual. Observing that $UU_a^\dagger|\psi\rangle=(UU_a^\dagger U^\dagger) U|\psi\rangle$, we see that to obtain $U|\psi\rangle$ we need to implement a correction of the form $UU_a^\dagger U^\dagger$ for some $a$.

The question is: if we assume to be able to implement local operations on the register containing $|\psi\rangle$, why not just apply directly $U$, either before or after performing standard state teleportation? After all, we are assuming to be able to apply $U$ to (a part of) $|\Psi\rangle$, so why not apply it to $|\psi\rangle$ itself? What's the context where this makes sense?


1 Answer 1


The interesting thing is where the set of possible corrections required is somehow easier to implement than the unitary itself.

For example, if you were trying to implement a $T$ gate ($\pi/8$ phase gate), the corrections that you'd need would be $I, Z$ and $(X\pm Y)/\sqrt{2}$. All of these corrections are in the Clifford group. So, this is where something like magic state distillation comes in. Imagine you have an error correcting code (such as the Steane [[7,1,3]] code) where you can perform any Clifford gate transversally. To get universality, you need to add a $T$ gate. This cannot be performed transversally (by Eastin-Knill theorem). So, put lots of effort into preparing a state capable of teleporting the gate $T$. Any corrections you need to perform are in the set of available operations.

(Viewing things in this way leads to a classification of how "easy" different gates are to implement. IIRC this is the Gottesman-Chuang hierarchy)

  • $\begingroup$ Thank you for your answer, but I still have a lot of doubts, just like Quantum gate teleportation T gate link, Although we can understand the whole process, I can not find there is any difference between making a bell state that goes through a T gate and doing a T gate on a target state. $\endgroup$
    – Curry
    Feb 18 at 9:13
  • $\begingroup$ @Curry Mathematically, there's no difference. Practically, there are scenarios where one sequence of events may be easier to accomplish than the other. $\endgroup$
    – DaftWullie
    Feb 19 at 9:00

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