= Solution
An always-on <Ising coupling> produces evolution during every nominally local pulse, and generally does not commute with the applied $x,y$ <Quantum Hamiltonians>. A naive product of single-<qubit> rotations therefore accumulates unwanted conditional phases and can entangle the qubits. Idling is also no longer an identity operation.
For a pulse of duration $\tau$, neglecting the coupling requires \b[$|J_{12}|\tau\ll1$] in units $\hbar=1$, or $|J_{12}|\tau/\hbar\ll1$ in ordinary units. For order-one rotation angles, this is the strong-local-drive regime $|B|\gg|J_{12}|$. Large drive amplitude alone does not justify neglect over a long total sequence: the accumulated interaction time or an appropriate error bound must also be small.
If it cannot be neglected, include it in the <quantum optimal control> model, or refocus it. A $\pi$ pulse $P=R_x^{(1)}(\pi)$ obeys $P(Z\otimes Z)P^\dagger=-Z\otimes Z$, giving the ideal <Ising spin echo>
$$
e^{-iH_I\tau/2}P e^{-iH_I\tau/2}P^\dagger=I.
$$
This cancellation assumes the refocusing pulses are instantaneous relative to $1/|J_{12}|$, or that the coupling during finite pulses is compensated. It explains how fast local controls can suppress the fixed coupling without mistaking it for a controllable zero interaction.
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