= Solution
For three odd <fermionic operators>, the identity
$$
[A,BC]=\{A,B\}C-B\{A,C\}
$$
converts the <commutator> with a quadratic <Hamiltonian> into the given <canonical anticommutation relations>. Writing a mode label as $i=(\mathbf p,s)$, this yields
$$
[a_i,a_j^\dagger a_j]=\delta_{ij}a_j,\qquad
[b_i^\dagger,b_j^\dagger b_j]=-\delta_{ij}b_j^\dagger.
$$
The opposite-species number terms commute with each <fermionic creation operator> or <fermionic annihilation operator>, because two anticommutations give a positive sign. Restoring the continuous normalization, the <Dirac delta distribution> in the mode brackets cancels $d^3\mathbf q/(2\pi)^3$. Consequently
$$
\boxed{[a^s_{\mathbf p},H]=E_{\mathbf p}a^s_{\mathbf p},\qquad[b^{s\dagger}_{\mathbf p},H]=-E_{\mathbf p}b^{s\dagger}_{\mathbf p}.}
$$
Substitute these into the <mode expansion of a Dirac field>. The result is
$$
[\psi(x),H]=\int\frac{d^3\mathbf p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}\sum_s E_{\mathbf p}
\left[e^{-ip\cdot x}a^s_{\mathbf p}u^s(\mathbf p)-e^{ip\cdot x}b^{s\dagger}_{\mathbf p}v^s(\mathbf p)\right].
$$
Since $i\partial_t e^{-ip\cdot x}=E_{\mathbf p}e^{-ip\cdot x}$ and $i\partial_t e^{ip\cdot x}=-E_{\mathbf p}e^{ip\cdot x}$, this is exactly
$$
\boxed{i\partial_t\psi(t,\mathbf x)=[\psi(t,\mathbf x),H].}
$$
The <Heisenberg equation of motion> uses an ordinary <commutator> with the even <Hamiltonian>, despite the odd <fermionic operators> obeying <anticommutators> among themselves.
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