Evolving Under a Single Pauli

The time-evolution operator for a Hamiltonian $H$ is the unitary $U(t) = e^{-iHt}$ . When $H$ is a single Pauli operator or a tensor product of Paulis (a Pauli string), the exponential has a closed form that maps directly to standard rotation gates.

Single-qubit Pauli rotations

For each single-qubit Pauli $P \in \{X, Y, Z\}$ we use the fact that $P^2 = I$ and expand the matrix exponential as a Taylor series:

e^{-i\theta P} = \cos\theta \cdot I - i\sin\theta \cdot P.

Comparing with the standard rotation gate definitions

R_x(\lambda) = e^{-i\lambda X/2}, \quad R_y(\lambda) = e^{-i\lambda Y/2}, \quad R_z(\lambda) = e^{-i\lambda Z/2},

we see that

e^{-i\theta X} = R_x(2\theta), \quad e^{-i\theta Y} = R_y(2\theta), \quad e^{-i\theta Z} = R_z(2\theta).

The angle passed to the rotation gate is twice the physical evolution parameter $\theta$ .

Two-qubit Pauli strings

For a two-qubit Pauli string such as $Z \otimes Z$ (often written $ZZ$ ), the eigenvalues of the operator determine the phase each basis state acquires:

(Z \otimes Z)|00\rangle = +|00\rangle, \quad (Z \otimes Z)|01\rangle = -|01\rangle, \quad (Z \otimes Z)|10\rangle = -|10\rangle, \quad (Z \otimes Z)|11\rangle = +|11\rangle.

Because the $ZZ$ eigenvalue of $|xy\rangle$ is $(-1)^{x \oplus y}$ , we can reduce the two-qubit problem to a single-qubit one: apply a CNOT from qubit 0 to qubit 1, which maps $|xy\rangle \to |x,\, x \oplus y\rangle$ . Now qubit 1 alone carries the parity $x \oplus y$ , so an $R_z$ on qubit 1 applies the correct phase:

e^{-i\theta\, Z \otimes Z} = \mathrm{CNOT}_{0\to1} \cdot \bigl(I \otimes R_z(2\theta)\bigr) \cdot \mathrm{CNOT}_{0\to1}.

The circuit has just three gates:

cx(0, 1) — parity-encode qubit 0 into qubit 1
rz(2 * theta, 1) — rotate by the interaction angle
cx(0, 1) — undo the encoding

This pattern generalises: for a $k$ -qubit Pauli string, use a ladder of CNOTs to collect the parity onto one qubit, apply $R_z$ , then reverse the ladder.

Try it

Implement $e^{-i(\pi/4)\, ZZ}$ on two qubits. The grader checks the full unitary matrix, so an empty circuit will not pass.

Run your code to see the quantum state.