In this paper (nature version), the authors state

We group the Pauli operators into tensor product basis sets that require the same post-rotations.

As a result, they have the table S2 in the suppl. I don’t understand this statement, could anyone please explain it, say for H2?


1 Answer 1


Let me show an example for grouping for this Hamiltonian:

$$H = 5 \cdot XI + 3 \cdot XZ - 2 \cdot YI + 1.5 \cdot IY$$

The expectation value:

$$\langle H \rangle = 5 \cdot \langle XI \rangle + 3 \cdot \langle XZ \rangle - 2 \cdot \langle YI \rangle + 1.5 \cdot \langle IY \rangle$$

Here I will group them in this way: the first group $XI$ and $XZ$, the second group $YI$ and $IY$. Note that (it is important) the members of the same group should commute with each other. Also, I should mention that this is not the only way of grouping.

enter image description here

For the first circuit:

\begin{align} &\langle X I \rangle = p(\text{00 or 01 measurements}) - p(\text{10 or 11 measurement}) \\ &\langle X Z \rangle = p(\text{00 or 11 measurements}) - p(\text{10 or 01 measurement}) \end{align}

For the second circuit:

\begin{align} &\langle Y I \rangle = p(\text{00 or 01 measurements}) - p(\text{10 or 11 measurement}) \\ &\langle I Y \rangle = p(\text{00 or 10 measurements}) - p(\text{01 or 11 measurement}) \end{align}

where $p$ denotes a probability of a measurement outcome described in parenthesis. The main idea here is: For a given Pauli term $P$ the expectation value is equal to:

$$\langle P \rangle = p_+ - p_-$$

where $p_+$ ($p_-$) is the probability of having an eigenstate that has eigenvalue $+1$ ($-1$). More details about this can be found in this answer about expectation value estimation. About why $HS^{\dagger}$ is applied in the second circuit can be understood from this answer.

  • $\begingroup$ Thank you Davit. That is a very clear explanation! Essentially we group a set of operators together, if there is one unitary operation to rotation all of them to be diagonal (consists of only Z and I). I have a second question, is that a way, either on qiskit, or any other manually written algorithm, that we can group a given arbitrary set of operators? Many thanks! $\endgroup$
    – fagd
    Sep 26, 2020 at 21:19
  • $\begingroup$ @fagd, you are welcome :). I don't know about the existence of software solutions for this problem. $\endgroup$ Sep 26, 2020 at 21:30

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.