3
$\begingroup$

I am reading Litinski's paper, in particular the figure 4 shown below:

enter image description here

To implement a computation with surface code, we need to "commute" all the Clifford toward the end of the circuit, then we absorb these in the final Pauli measurements.

I am confused by the property in the red box: for me it is only true if we don't care about the quantum state in the computation (we only care about the classical outcome of the measurement). However, to perform many circuit simplifications it seems we also need to understand how the quantum state evolves through consecutive measurements.


To be specific, the upper equation in the red box, says that if $P P'=P' P$ ($P$ and $P'$ are Pauli), calling $M_{P'}$ a measurement of Pauli $P'$, then:

$$ M_{P'} e^{-i \pi/4 P}=M_{P'} $$

This is not true in general: take $|\psi\rangle = |00\rangle + |11\rangle$, $P'=Z_1 Z_2$, $P=Z_1$.

  • Doing the $Z_1 Z_2$ measurement will give eigenvalue $+1$ and post-measurement state $|\psi \rangle$.
  • Doing $e^{-i \pi/4 Z_1}$ rotation first will give the state $|\psi\rangle \to |00\rangle + i |11\rangle$. If I do afterward the $Z_1 Z_2$ measurement, I will also find the eigenvalue $+1$, but the post-measurement state will be $|00\rangle + i |11\rangle \neq |\psi \rangle$.

Hence the post-measurement state is different in both cases, hence the equality is not satisfied in general.

Is it that the equality should only be understood in terms of classical measurement outcomes? If so, how are we sure it doesn't cause any problem in the circuits simplification? For instance, when the circuit goes from (b) to (c) (see blue arrow), a sequence of Pauli measurements appears. It could be that not acknowledging the state evolution through the measurements implies that the circuit simplification is incorrect. How is this issue avoided?

[edit]

A simple example shows that while the equalities in the red box are true if one discards the measurements, they are not sufficient to perform the simplification that Litinski performs in his paper. This is because during such simplifications, consecutive Pauli measurements occur (and neglecting the post-measurement "Clifford correction" would lead to a wrong following measurement).

To be specific, in the example below, I need to "commute" $C$ and $C'$ after the first Pauli measurement as it modifies the second Pauli measurement to be performed. Of course we don't care commuting the $C$ and $C'$ after the last measurement.

enter image description here

$\endgroup$

1 Answer 1

3
$\begingroup$

The equality you've highlighted discards the qubits after the measurement. Their final value is irrelevant to the observational-equivalence of the two sides.

Note that the equality directly to the left of the one you highlighted is exactly the same concept, except it doesn't discard the qubits. As a result it needs an additional unitary operation on the right hand side to get the output state right to ensure observational equivalence.

enter image description here

The equality that discards the qubits is simply omitting that final unitary operation. Because applying a unitary operation to a set of qubits that's then immediately discarded can never change the observable function of a circuit.

$\endgroup$
6
  • 1
    $\begingroup$ Thanks for the answer and the useful comment in your second paragraph. I agree that we don't care about applying a unitary after a measurement if the qubits are then discarded. The thing that confuses me is that in principle the qubits are not discarded in his circuit simplifications. Look at the circuit at the end of my blue arrow: the output of the first measurement is an input of the second one (hence the quantum state could in principle matter). $\endgroup$ Aug 5, 2023 at 16:54
  • 1
    $\begingroup$ Ok I think I understood my issue: if we simplify from right to left starting from the circuit at the beginning of my blue arrow, the Clifford corrections are always applied after the very last measurement of the circuit (it can be seen recursively) so that we never care about the post-measurement correction in this case. $\endgroup$ Aug 5, 2023 at 17:00
  • $\begingroup$ I think I was wrong in my previous comment. I added an edit at the end of my message. If you have time to tell me if you agree with such edit it would be great. Basically, in his paper I think he must acknowledge the Clifford correction after the measurement as long as the measurement is not the last one in the computation. Hence his formula in the red box is not sufficient to understand how he simplifies his circuits. (I am asking as, perhaps, we don't need to take this Clifford correction because of some simplification I don't see). $\endgroup$ Aug 6, 2023 at 14:32
  • $\begingroup$ @MarcoFellous-Asiani That is being indicated by the fact that the qubit lines are being terminated within the diagram, instead of allowing them to continue to the right. Personally I'd put in an explicit note of "discard" on each qubit line, but there's certainly no need to keep showing the unitary. $\endgroup$ Aug 6, 2023 at 18:47
  • $\begingroup$ I agree with you, but what confused me is that from the way his paper is written, it looked like he never needed to know how a Clifford got transformed after a measurement, because he gave all the laws used in his simplification, apart from this one. However he conceptually needs this law to perform his simplifications. This is what I found confusing: I was afraid I missed a technical subtlety to understand his simplifications. In practice, I managed to reproduce his calculation (acknowledging the necessary Clifford correction after measurement), so everything is fine =). Thx for your time!! $\endgroup$ Aug 6, 2023 at 20:17

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.