The following figure displays a circuit implementing quantum teleportation. enter image description here

It uses Haramard transform (H), controlled not (+), measurements (meas), a Not gate (N) and a phase shift gate (P) to teleport quantum state q from Alice to Bob.

I have learned that all quantum computations should be reversible (see e.g. here), but this gate does not seem reversible, because it contains measurements.

Is quantum teleportation reversible? What does being reversible even mean in this context? Most quantum algorithms I know only measure in the end. I am assuming that reversibility in this case means that the gates are reversible before the measurement. But in quantum teleportation, this is not possible (because Alice must send the result of the measurement to Bob).

It is particularly confusing that at the red line below, we have a mixture of bits and quantum bits. How should I interpret this?

enter image description here

Possible answer

The deferred measurement principle states that I could move the measurements to the end and still get the same result. Should I interpret reversibility as "reversibility after moving the measurements to the end"?


1 Answer 1


Quantum teleportation is not reversible since it exhausts its resource - the entangled Bell state initially shared by Alice and Bob. At the end of the teleportation Alice and Bob don't share anymore an entangled state.

Any unitary operations (here H and CNOT) are reversible. The measurements, hovewer, are not. But the measurement is crutial for the teleportation to work, specifically - from the teleportation identity it is the measurement performed by Alice that actually kicks on the teleportation. This measurement uses the entanglement resource, since afterward Alice's and Bob's states are entangled no more.

After Alice's measurement the state is teleported but encoded. It is known that classical communication should be used to transfer to Bob the proper measurement basis (otherwise superluminal communication channel could be constructed from this procedure). Bob's measurement doesn't use the entanglement resource (it's just trying to measure a single state in proper basis), so it can be deffered.

  • $\begingroup$ What do you mean by Bob's measurement? In quantum teleportation, only Alice performs measurements, right? $\endgroup$
    – Peter
    Commented Apr 5, 2018 at 12:55
  • $\begingroup$ Bob's measurement is not necessary. The decoded state can be used for further quantum computation processes. $\endgroup$
    – Alexander
    Commented Apr 5, 2018 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.