, it is clear that $\lambda(t,\omega) = \alpha$ gives the optimal intensity function.

The above analysis also explains the reason for our choice . When $\gamma_{jk}(t)$ is either zero or a fixed constant $\gamma$ as stated in the begining of the previous paragraph, we can naturally remove the tuples $(i,j,k)$ with $\gamma_{jk}(t) \equiv 0$ from the sample space $\Omega$ defined in , and then $|\Omega| = N_c$ and becomes $\lambda(t,\omega) = N_c \gamma = \alpha$, which gives exactly the optimal intensity function.

Numerical results
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Examples with few spins: Validity check {#sec:ex1}
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To verify the algorithm, we first consider some simple cases with only a few spins. The magnetic field $(m_1(t), m_2(t))$ is chosen as $$m_1(t) = \cos(\omega t), \qquad m_2(t) = \sin(\omega t)$$ withe $\omega = -0.5$, and $m_3^{(k)} = 1$ is used for all the spins. Assuming that each spin interacts only with its adjacent spins, and all the interaction strengths are equal, we have that $$\gamma_{jk}(t) = \gamma_0 \delta_{1,|j-k|}.$$ Initially, we assume that all the spins have the same state $\ket{\uparrow}$: $$\label{eq:init}
\ket{\Psi(0)} = \ket{\uparrow}^{\otimes N},$$ and we are concerned about the evolution of the probability for the “all spin-down” state: $$p(t) = \expval{B}{\Psi(t)}, \qquad
  B 