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Say I have the following Hamiltonian (given in terms of Pauli operators):

\begin{equation} H=aX_1Z_2+bZ_1X_2. \end{equation}

Both Pauli terms commute with each other. I want to make a measurement of $\langle H\rangle$ but only want to make one measurement using the Pauli terms' common eigenbasis. How do I find the set of Clifford gates which allows me to do this?

What I've tried:

I've found a unitary $U$ such that $U.X_1Z_2.U^\dagger$ and $U.Z_1X_2.U^\dagger$ are diagonal but not the Clifford gates which create $U$.

If the Hamiltonian were $aX_1+bX_2$ then $U=H_1H_2$ would work.

If the Hamiltonian were $aZ_1+bZ_2$ then no $U$ would be required.

If the Hamiltonian were $aX_1+bZ_2$ then $U=H_1X_2$ would work.

If the Hamiltonian were $aZ_1+bX_2$ then $U=X_1H_2$ would work.

This seems really simple but is somehow eluding me. Any help would be very appreciated!

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  • $\begingroup$ Have you tried applying a controlled-Z gate? $\endgroup$
    – DaftWullie
    Nov 15, 2021 at 20:04

1 Answer 1

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If you're not worried about the depth of the circuit, or about chunking observables into commuting sets, then you can solve the observables one by one:

  1. Pick an observable that has a term $X_q$ or $Y_q$.
  2. For each other term $P_{q_2}$ in that observable, apply a controlled-$P$ operation controlled by $q$ targeting $q_2$. The observable you want to measure (in order to measure the original intended observable before the inserted operation) no longer has the $P_{q_2}$ term.
  3. The observable-to-measure is now a single $X_q$ or $Y_q$ term. Apply a single-qubit rotation to change that term into a $Z_q$ term. All other observables also no longer have an $X_q$ or $Y_q$ term (on $q$) since that would imply they anti-commuted with the observable you just isolated.
  4. Keep repeating steps 1-3 until there are no $X$ or $Y$ terms remaining. Make sure you're properly keeping track of the signs of the observables as you operate on them.
  5. All observables are now products of single-qubit $Z$ observables. Measure all qubits in the $Z$ basis. Multiply together appropriately to get the desired observables.
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  • $\begingroup$ @Craig_Gidney, I am looking to chunk them together though. As I said I want to make only 1 measurement. $\endgroup$ Nov 15, 2021 at 19:09
  • $\begingroup$ @KenRobbins By chunking I was referring to the case where you have commuting and anti-commuting measurements, and need to find a minimal number of groups to divide the measurement into such that within each group all measurements commute. You said all measurements commute so you're in the case with just 1 group, and don't need to solve the grouping part. $\endgroup$ Nov 15, 2021 at 20:05

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