such a case, we can choose $H_{\mathrm{easy}}$ to be $$H_{\mathrm{easy}}(t) = \sum_{k=1}^N m^{(k)}(t) \cdot \sigma^{(k)}.$$ Thus for any $\ket{\Psi} = \ket{\psi_1} \otimes \cdots \otimes \ket{\psi_N} \in
{\mathcal{A}}$, we have $$U(t,s) \ket{\Psi} =
  U^{(1)}(t,s) \ket{\psi_1} \otimes \cdots \otimes U^{(N)}(t,s) \ket{\psi_N}
  \in {\mathcal{A}},$$ where $$U^{(k)}(t,s) = \exp \left(
  -{\mathrm{i}}\left[ \int_s^t m^{(k)}(\tau) \,\mathrm{d}\tau \right] \cdot \sigma
\right),$$ where $\sigma = (\sigma_1, \sigma_2, \sigma_3)^{\top}$. The sample space $\Omega$ and the probability measure $\mu_{\omega}$ are given by $$\label{eq:Omega}
\Omega = \{(i,j,k) \mid i = 1,2,3, \, j,k = 1,\cdots,N, \, j < k \}, \qquad
{\mathbb{P}}(\omega = \omega_0) = 1 / |\Omega|, \quad \forall \omega_0 \in \Omega.$$ By defining $$\label{eq:A}
A(t,\omega) = |\Omega| \gamma_{jk}(t) \sigma_i^{(j)} \sigma_i^{(k)},$$ it is easy to find that ${\mathbb{E}}_{\omega} A(t,\omega) = H(t) -
H_{\mathrm{easy}}(t)$. Again, we choose $$\label{eq:lambda}
\lambda(t,\omega) = |\Omega| |\gamma_{jk}(t)|,$$ so that ${\widetilde{A}}(t,\omega)$ does not change the norm of the state.

In this method, the evolution equation for $\eta(t)$ defined in is $$\label{eq:eta1}
\frac{\mathrm{d}\eta}{\mathrm{d}t} =
  3 \sum_{1\le j < k \le N} |\gamma_{jk}(t)|.$$ In the method using $Z$-basis (see Section \[sec:Z-basis\]), we have $$\label{eq:eta2}
\frac{\mathrm{d}\eta}{\mathrm{d}t} =
  \sum_{k=1}^N \bigl\lvert m_1^{(k)} + {\mathrm{i}}m_2^{(k)} \bigr\rvert
  + 2 \sum_{1\le j < k \le N} |\gamma_{jk}(t)|.$$ Therefore, if the coupling intensity is significantly smaller 