I'm watching Christian Schaffner's talk on quantum position-based cryptography (link here) and have a question about a particular application of teleportation.

At about the 16:40 mark, he seems to say something along these lines (notation is mine)

Alice receives a quantum state $|\psi\rangle$, which is an element of some basis $\mathcal{B}$. Alice teleports it to Bob, which gives him the state $|\psi^\prime\rangle = X^{M_1}Z^{M_2}|\psi\rangle$, for some values $M_1,M_2$ which Alice learns as a result of teleportation. What's next is the part that confuses me. He says that Bob can measure $|\psi^\prime\rangle$ in the $\mathcal{B}$ basis, and then when he receives the classical values $M_1,M_2$, he can apply the $Z^{M_2}X^{M_1}$ to recover $|\psi\rangle$.

My question is: Since $|\psi^\prime\rangle = X^{M_1}Z^{M_2}|\psi\rangle$, how can we be sure that Bob's measurement will always return the same value? Certainly if he measured $|\psi\rangle$ in the $\mathcal{B}$ basis, he would get result $|\psi\rangle$ with 100% certainty. But if $|\psi^\prime\rangle$ is some superposition of basis states, then Bob's answer will not always be the same.

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    $\begingroup$ I haven't watched the talk, but does he perhaps restrict the set of possible bases? If the basis B, when represented as an operator $\vec{n}\cdot\vec{\sigma}$ either commutes or anticommutes with both $X$ and $Z$, then you can certainly still proceed. For example, if B were a choice between $X$, $Y$ or $Z$, you'd be fine. Basically because you'd be guaranteed that the error $X^{M_1}Z^{M_2}$ maps basis states to basis states. $\endgroup$
    – DaftWullie
    Commented Mar 27, 2019 at 8:56

1 Answer 1


Alice receives a quantum state $|\psi\rangle$, which is an element of some basis $\mathcal{B}$, though she does not know what $\mathcal B$ is. She then teleports this to Bob, who is told by someone else what $\mathcal B$ is. Furthermore, it seems that specifically either $\mathcal{B} = \{ \lvert 0 \rangle, \lvert 1 \rangle \}$ or $\mathcal{B} = \{ \lvert + \rangle, \lvert - \rangle \}$.

The last detail is important here. It means that the basis is either the eigenbasis of the $Z$ operator, or the eigenbasis of the $X$ operator — which means that when Bob wants to do the corrections, one of the two corrections which he will perform will in principle have no effect, while the other will simply interchange the two possible basis elements.

  • In a normal teleportation procedure, just after Alice's measurement, the state on Bob's side looks perfectly random to him. If Bob didn't know whether Alice would be sending one of the states $\{ \lvert 0 \rangle, \lvert 1 \rangle \}$ or one of the states $\{ \lvert + \rangle, \lvert - \rangle \}$, there is very little that he can do with this qubit which would be of any use: he would have to wait to obtain Alice's measurement outcomes. If he did wait for Alice's measurement outcomes, he could perform the necessary $X$ and/or $Z$ correction to obtain the state.

  • If someone (Alice or someone else) then came along and told him what basis $\mathcal B$ his state was in, he could perform a (non-demolition) measurement in the basis $\mathcal B$ to find out which state it is without disturbing it. But the outcome of this measurement is not affected by one of the corrections he did from Alice's measurement outcomes. If $\mathcal B = \{ \lvert 0 \rangle, \lvert 1 \rangle \}$, then the $Z$ correction from Alice's teleportation results didn't have any effect on the teleported state; and if $\mathcal B = \{ \lvert + \rangle, \lvert - \rangle \}$, then the $X$ correction from Alice's teleportation results didn't have any effect. The effect of the other correction, on the other hand, would be just to interchange the two possible states of the $\mathcal B$ measurement, which could have been done before or after the measurement.

  • It follows that if someone told Bob what basis $\mathcal B$ Alice's state was going to be in, he could perform the $\mathcal B$ measurement first, and then wait for Alice's measurement results. The measurement will disturb the state, but it will only collapse it to a state of the same basis $\mathcal B$ that Alice is sending; then Alice's teleportation results will give Bob enough information to correct the state.

Now, if Bob performs the measurement before Alice sends him the correction information, the result of his measurement will be random. But because the two bases are either the $X$ eigenbasis or the $Z$ eigenbases, the outcome of his measurement will either be the state he is meant to obtain or the orthogonal state; he will always know that the state he obtains is one of those two states, and what those two states are; and it will always be correctable using the information that Alice sends. (The same is also true if there is a third possibility, that the state is in the $Y$ eigenbasis, for essentially the same reason using the fact that $Y \propto XZ$.)

  • $\begingroup$ So, in short, Bob can perform non-demolition measurement on teleported qubit without Alice's info (2 bits) because he got info from verifier2 (that hides the correct basis). And after receiving Alice's 2 bits (that finalize teleportation) Bob can just classically correct the result. Right? $\endgroup$
    – Danylo Y
    Commented Mar 27, 2019 at 16:46
  • $\begingroup$ Basically, yes. (If he's acting on the post-measurement state of his measurement, I wouldn't call it "classically correcting" the result.) $\endgroup$ Commented Mar 27, 2019 at 16:56
  • $\begingroup$ Thanks for a really great answer. I've opened a bounty on this question and will award it to you once I'm able, as recognition for such a helpful post. $\endgroup$
    – NNN
    Commented Mar 30, 2019 at 2:11
  • $\begingroup$ This is a widespread technique in quantum error correction. As long as the future lightcone of a Pauli correction only contains stabilizer gates, which is always the case if you're doing non-Clifford operations via magic state injection, you can apply the corrections by flipping appropriate measurement results instead of by doing anything to the qubits. This can remove the need to wait for the corrections before continuing to do additional work on the qubits, so instead of having a hard real time feedback problem you can take it slow and still get the same answer in the end. $\endgroup$ Commented Dec 1, 2021 at 4:44

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