In linear response theory, the Kubo formula describes the change in $\langle A \rangle$ for some observable $A$, due to applying a time-dependent perturbation $V(t)$ to the Hamiltonian $H_0$. It is given by:
\begin{align*} \langle A\rangle (t) = \langle A \rangle_0 - i \int_{-\infty}^{t} dt' \langle [\bar{A}(t), \bar{V}(t')] \rangle_0 \end{align*} where $\bar{A}(t) = \exp(iH_0t)A\exp(-iH_0t)$.
The expectation of the commutator has the subscript $0$, indicating that it is carried out over the equilibrium density matrix, typically $\rho_0 \sim \exp(-\beta H_0)$.
My question concerns the case(s) in which either $A$ or $V$ commute with $H_0$, and the response term vanishes. To illustrate, assume $A$ commutes with $H_0$. Then the expectation of the commutator can be written as:
$$\langle [\bar{A}(t), \bar{V}(t')] \rangle_0 = \mathrm{Tr}[\rho_0\bar{A}(t)\bar{V}(t') - \rho_0\bar{V}(t')\bar{A}(t)]$$ $$= \mathrm{Tr}[\rho_0 A\bar{V}(t') - \rho_0 \bar{V}(t')A]$$ $$= \mathrm{Tr}[\rho_0 A\bar{V}(t') - A\rho_0 \bar{V}(t')] \qquad(\text{Cyclic trace rule})$$ $$= \mathrm{Tr}[\rho_0 A\bar{V}(t') - \rho_0 A\bar{V}(t')] = 0 \qquad (A\text{ commutes with } H_0)$$
This result of zero confuses me, because I can think of a few experimental probes matching the above description that ordinarily show a finite response. For example, take $H_0$ to be an Ising Hamiltonian $-J\sum_{\langle ij \rangle}S^z_i S^z_j$ and $V(t)$ to be a constant field $-h\sum_i S^z_i$. Here, choosing $A$ to be the magnetization $\sum_i S^z_i$ is supposed to give a response that indicates the magnetic susceptibility.
The Ising model has non-zero susceptibility, yet the above calculation yields zero owing to $A$ commuting with $H_0$.
What's the resolution to this apparent discrepancy? My guess is that the corrections show up only at higher orders in this case, but I can't quite see it.