Simulating the Heisenberg Model

The Heisenberg model is the canonical testbed for quantum simulation of many-body dynamics. Its spins interact through all three Pauli components, so the terms do not commute — exactly the regime where Trotterization is both necessary and interesting.

The Heisenberg Hamiltonian

For a chain of spin- $\tfrac12$ particles the (isotropic) Heisenberg Hamiltonian couples neighbours by

H = J \sum_{\langle i, j\rangle} \bigl(X_i X_j + Y_i Y_j + Z_i Z_j\bigr).

The $X X + Y Y$ part is a hopping (exchange) term that swaps spin excitations between sites; the $Z Z$ part is an Ising energy. Because $[X X, Y Y] \ne 0$ and $[X X, Z Z] \ne 0$ , we cannot exponentiate $H$ in closed form on hardware — we Trotterize each bond into three two-qubit rotations.

One bond, three rotations

Focus on a single bond ( $J = 1$ ). The exact one-step evolution factorises (to first order) into the three commuting-on-paper pieces:

e^{-iHt} \approx e^{-i t\, X X}\,e^{-i t\, Y Y}\,e^{-i t\, Z Z}.

Each factor is a two-qubit Pauli rotation $e^{-i\frac{\theta}{2} P\otimes P}$ with $\theta = 2 J t$ . The base case is the $ZZ$ rotation, which has a clean CNOT decomposition:

e^{-i\frac{\theta}{2} Z\otimes Z} = \text{CNOT}_{0\to1}\;\bigl(I\otimes R_Z(\theta)\bigr)\;\text{CNOT}_{0\to1}.

The CNOT maps the joint parity of the two qubits into qubit 1, an $R_Z(\theta)$ phases by that parity, and the second CNOT uncomputes it.

Changing basis for $XX$ and $YY$

$X X$ and $Y Y$ are just $Z Z$ in a rotated single-qubit basis:

$XX$ : since $H_{\text{gate}}\,Z\,H_{\text{gate}} = X$ , conjugating the $ZZ$ block by a Hadamard on each qubit gives the $XX$ rotation.
$YY$ : since $R_X(\tfrac{\pi}{2})\,Z\,R_X(-\tfrac{\pi}{2}) = -Y$ (the sign drops out of the symmetric $Y\otimes Y$ ), conjugating the $ZZ$ block by $R_X(\tfrac{\pi}{2})$ on each qubit gives the $YY$ rotation.

So all three Pauli rotations reuse the same CNOT– $R_Z$ –CNOT core, wrapped in a different single-qubit basis change.

Starting state: the Néel state

We initialise the two spins in the Néel (antiferromagnetic) state $|01\rangle$ — one spin up, one down — by flipping qubit 1 with an $X$ gate. Under the Heisenberg dynamics the $XX + YY$ exchange term makes the excitation hop, so amplitude leaks from $|01\rangle$ into $|10\rangle$ ; this spin transport is the physics the circuit is meant to capture.

Try it

Prepare $|01\rangle$ , then apply one first-order Trotter step of the 2-spin Heisenberg interaction at $J = 1$ , $t = 0.5$ (so $\theta = 2Jt = 1$ ). Build the $ZZ$ , $XX$ , and $YY$ rotation blocks as described. The grader checks the final state vector.

Run your code to see the quantum state.

The Heisenberg Hamiltonian

One bond, three rotations

Changing basis for XXXXXX and YYYYYY

Starting state: the Néel state

Try it

Changing basis for $XX$ and $YY$