= Solution
Define the constant <Dirac gamma matrix> combinations $\alpha^k=\gamma^0\gamma^k$ and $\beta=\gamma^0$. The <Hamiltonian operator> is
$$
H=\int d^3y\,\psi_b^\dagger(\mathbf y)
(-i\alpha^k_{bc}\partial_{y^k}+m\beta_{bc})\psi_c(\mathbf y).
$$
Evaluate its <commutator> with $n(\mathbf x)$ using the bilocal result in <3/a>. The <mass term> cancels immediately when $\mathbf z=\mathbf y$, because its two <Dirac delta distributions> coincide. For the derivative term first keep the two points distinct:
$$
C(\mathbf x)=-i\int d^3y\,\alpha^k_{bc}\left.
\partial_{z^k}\left\{[\delta^{(3)}(\mathbf x-\mathbf y)-\delta^{(3)}(\mathbf x-\mathbf z)]
\psi_b^\dagger(\mathbf y)\psi_c(\mathbf z)\right\}\right|_{\mathbf z=\mathbf y}.
$$
The terms in which the derivative acts on $\psi_c$ cancel against each other; the derivative of the second <Dirac delta distribution> remains. Thus
$$
C(\mathbf x)=i\int d^3y\,\psi^\dagger(\mathbf y)\alpha^k\psi(\mathbf y)
\partial_{y^k}\delta^{(3)}(\mathbf x-\mathbf y)
=-i\partial_{x^k}[\psi^\dagger(\mathbf x)\alpha^k\psi(\mathbf x)].
$$
The last equality is <integration by parts>, or the defining derivative rule for a <distribution>. Consequently the literal <conserved current> whose <divergence> equals $C$ is
$$
\boxed{C=\boldsymbol\nabla\cdot\mathbf J,\qquad
J^k=-i\psi^\dagger\alpha^k\psi=-i\bar\psi\gamma^k\psi.}
$$
Because a <commutator> of two <Hermitian operators> is anti-Hermitian, this $\mathbf J$ includes $-i$. The physical Hermitian <Dirac current> is instead $j^k=\bar\psi\gamma^k\psi$, so $C=-i\boldsymbol\nabla\cdot\mathbf j$. The <Heisenberg equation of motion> now gives
$$
\boxed{\partial_t n=i[H,n]=-iC=-\boldsymbol\nabla\cdot\mathbf j,\qquad
j^\mu=\bar\psi\gamma^\mu\psi.}
$$
This establishes the <continuity equation> directly from the required <commutator>, without assuming the <Dirac equation> or invoking the <Noether theorem> as a substitute for the computation.
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