XY Hamiltonian in a 1D Heisenberg Chain

Question

I've been trying to implement the 1D Heisenberg chain (i.e. the XXZ model) on Qiskit but have been having trouble. To recap, the Heisenberg hamiltonian is as follows: $$H_{XXZ} = \sum^{N}_{i = 1} [J(S^{x}_{i}S^{x}_{i+1} + S^{y}_{i}S^{y}_{i+1} + \Delta S^{z}_{i}S^{z}_{i+1})] $$ and we can take the XY hamiltonian to be $$H_{XY} = \sum^{N}_{i = 1} [J(S^{x}_{i}S^{x}_{i+1} + S^{y}_{i}S^{y}_{i+1})]$$ as I understand. I know that the matrix representation of this hamiltonian's time evolution takes the form $$XY(\theta) = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & \cos[{\theta}/2] & i\sin[{\theta}/2] & 0\\ 0 & i\sin[{\theta}/2] & \cos[{\theta}/2] & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$ however, I'm not exactly sure how to implement it on Qiskit with the available Quantum logic gates. I do know that a special case to this problem is the iSwapGate, where it is equal to $XY(\theta = \pi)$, but is there a way to implement $XY(\theta)$ for arbitrary angles?

Answer 1

Here is an implementation

from qiskit.circuit import QuantumCircuit, Parameter

theta = Parameter('θ')

qc = QuantumCircuit(2)
qc.cx(0, 1)
qc.crx(-1 * theta, 1, 0)
qc.cx(0, 1)

print(qc)
---
          ┌────────┐     
q_0: ──■──┤ RX(-θ) ├──■──
     ┌─┴─┐└───┬────┘┌─┴─┐
q_1: ┤ X ├────■─────┤ X ├
     └───┘          └───┘

and to evaluate that it works:

from qiskit.quantum_info import Operator

def XY(theta):
    c = np.cos(theta / 2)
    s = 1j * np.sin(theta / 2)
    
    return np.array([[1, 0, 0, 0], 
                     [0, c, s, 0], 
                     [0, s, c, 0], 
                     [0, 0, 0, 1]])

val = pi / 14
circ = qc.bind_parameters({theta: val})
np.allclose(Operator(circ).data, 
            XY(val))
---
True