Converting sigma model Hamiltonian into a quantum circuit

I would like to ask some specific questions about a paper I'm reading. In 1903.06577, authors are playing with the sigma model hamiltonian, and they are showing how they constructed the time evolution circuit with respect to this hamiltonian. They use four-dimensional Hilbert space, so each state is defined via two-qubit. I was wondering how I can compute the expectation value of this hamiltonian on a quantum circuit instead of time evolution.

So for the interaction terms (please correct me if I'm wrong), they are giving the $$j_{1,2,3}$$ definition (eq. 8) so I believe it is correct to use that definition to form the hamiltonian i.e. $$j_1 = \frac{\mathbf{1}\otimes\sigma_2}{\sqrt{3}} =$$ qml.Hamiltonian([1/np.sqrt(3.)], [qml.Identity(0) @ qml.PauliY(1)]). However, I'm not sure about the kinetic term, they are converting everything into this new $$T$$ basis they are using but without the conversion, (i.e. 1812.00944 eq 2), the kinetic term for the sigma model is $$\sum_k J^2_k$$. So would it be correct to write it as qml.Hamiltonian([(hbar/2)**2], [qml.PauliX(0) @ qml.PauliX(1)]) for $$k=0$$? So if I write a simple circuit to calculate the expectation value for only the terms $$k=0$$ and first interaction term

import pennylane as qml

dev = qml.device("default.qubit", wires = 2)

@qml.qnode(dev)
def circuit():
# prepare an initial state (not important)
for wire in range(2):
qml.RY(np.pi/2., wire)

return qml.expval(qml.Hamiltonian([(1/2)**2, 1/np.sqrt(3.)],
[qml.PauliX(0) @ qml.PauliX(1), qml.Identity(0) @ qml.PauliY(1)]))


Would this be correct? Any insight and/or reference is appreciated.

Thanks

• Welcome to QCSE. Please edit your question to directly link to the abstract of the paper to which you refer. This may help people get motivated to answer your question. May 4 at 15:07