Measuring Many Pauli Terms

Every VQE energy estimate requires computing $\langle H \rangle$ for a molecular Hamiltonian that may contain dozens or even thousands of Pauli string terms. Measuring each term in its own dedicated circuit run is correct but wasteful. Commuting Pauli terms can be measured simultaneously, so grouping them reduces the total circuit count and shot budget.

Why commutativity enables simultaneous measurement

Two observables $A$ and $B$ can be measured simultaneously (in the same basis rotation circuit) if and only if they commute: $[A, B] = AB - BA = 0$. For two Pauli strings of the same length, commutativity has a simple combinatorial test: count the positions where the two strings differ and neither is $I$. If that count is even the strings commute; if it is odd they anti-commute.

For example, on two qubits:

• $ZI$ and $IZ$: they differ at every position, but at qubit 0 we have $Z$ vs $I$, and at qubit 1 we have $I$ vs $Z$ — every discrepancy involves an $I$, so the count of non-trivial differences is 0 (even). They commute.
• $ZI$ and $ZZ$: qubit 0 has $Z$ vs $Z$ (same), qubit 1 has $I$ vs $Z$ — again the discrepancy involves $I$. Count is 0 (even). They commute.
• $ZI$ and $XX$: qubit 0 has $Z$ vs $X$ (both non-identity, a genuine conflict), qubit 1 has $I$ vs $X$ (involves $I$, not counted). Count is 1 (odd). They anti-commute.

Because $ZI$, $IZ$, and $ZZ$ all mutually commute they form a commuting group that can be measured in a single circuit — no basis-rotation overhead beyond the standard computational-basis measurement.

Measurement basis rotations

Each commuting group defines a shared measurement basis. The standard procedure is:

1. Z terms require no rotation — measure directly in the computational basis.
2. X terms require a Hadamard on each affected qubit before measurement, rotating $|+\rangle \to |0\rangle$ and $|-\rangle \to |1\rangle$.
3. Y terms require an $S^\dagger H$ (also written $HS^\dagger$) rotation before measurement.

For the $ZZ$ term on a two-qubit state, after measuring bit string $xy$ the contribution is:

$\langle ZZ \rangle = \sum_{x,y \in \{0,1\}} (-1)^{x+y}\, P(xy).$

The sign rule comes directly from $Z|0\rangle = +|0\rangle$ and $Z|1\rangle = -|1\rangle$: each qubit contributes a factor of $(-1)^{\text{bit}}$ and the factors multiply.

From grouping to circuit count

Given a Hamiltonian with $T$ Pauli terms partitioned into $G$ commuting groups, you need exactly $G$ distinct measurement circuits per VQE iteration instead of $T$. Grouping reduces the shot budget by a factor of roughly $T/G$, which can be significant: a hydrogen molecule in the STO-3G basis has 15 Pauli terms but only 2 commuting groups, cutting the measurement circuits from 15 to 2.

The full VQE measurement loop becomes:

1. Prepare $|\psi(\boldsymbol{\theta})\rangle$ with the ansatz.
2. For each commuting group: apply the shared basis rotation, measure, collect statistics.
3. From each group's measurement results extract $\langle P_k \rangle$ for every term in the group.
4. Return $E(\boldsymbol{\theta}) = \sum_k c_k \langle P_k \rangle$ to the classical optimizer.

Try it

Consider the two-qubit Hamiltonian

$H = 0.5\,ZI + 0.3\,IZ + 0.2\,ZZ + 0.4\,XX.$

Use the commutativity rule to identify how many distinct measurement circuits are needed. Return that number.

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