= Solution
The <canonical anticommutation relations> are
$$
\{a_{\mathbf p}^s,a_{\mathbf q}^{r\dagger}\}
=\{b_{\mathbf p}^s,b_{\mathbf q}^{r\dagger}\}
=(2\pi)^3\delta^{sr}\delta^{(3)}(\mathbf p-\mathbf q),
$$
with the mixed and equal-type <anticommutators> zero. Insert the <mode expansion of a Dirac field>. The phases cancel at equal time, and <2/a> gives
$$
\begin{aligned}
\{\psi_\alpha(t,\mathbf y),\psi_\beta^\dagger(t,\mathbf x)\}
&=\int\frac{d^3p}{(2\pi)^3}\,e^{i\mathbf p\cdot(\mathbf y-\mathbf x)}
\frac{\sum_s[u^s_\alpha(\mathbf p)u^{s*}_\beta(\mathbf p)
+v^s_\alpha(-\mathbf p)v^{s*}_\beta(-\mathbf p)]}{2E}\\
&=\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).
\end{aligned}
$$
Since $\bar\psi\gamma^0=\psi^\dagger$, the requested componentwise answer is
$$
\boxed{\{(i\bar\psi\gamma^0)_\alpha(t,\mathbf x),\psi_\beta(t,\mathbf y)\}
=i\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).}
$$
This is the <Equal-time canonical anticommutator of a Dirac field>: the <canonical momentum> conjugate to $\psi_\alpha$ is $\pi_\alpha=i\psi_\alpha^\dagger$. It is the fermionic counterpart of the canonical field <canonical commutation relation>, ensuring the correct local <canonical quantization>. For this first-order <Lagrangian density>, the conjugate variables are constrained; the displayed relation is the standard result after handling that constraint. It does not assert a nonzero <anticommutator> between the fields at separated spatial points.
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