Starting from the free lagrangian
$$\mathscr L = \bar\Psi(i\displaystyle{\not}\partial - m)\Psi$$
I compute the canonical momenta
$$\Pi =\frac{\partial \mathscr L}{\partial\dot{\Psi}}=i\Psi^\dagger \qquad \bar\Pi =\frac{\partial \mathscr L}{\partial\dot{\bar\Psi}}= 0$$
and then I can perform canonincal quantization imposing the canonical anticommutation rules (equal time anticommutator)
$$\left\{\Psi_\alpha(x^0,\vec{x}),\Pi_\beta(x^0,\vec{y})\right\} = i\left\{\Psi_\alpha(x^0,\vec{x}),\Psi_\beta^\dagger(x^0,\vec{y})\right\} \equiv i\delta_{\alpha\beta}\delta^3\left(\vec{x}-\vec{y}\right) \tag1$$ $\alpha,\beta = 1,2,3,4$ are spinor indices.
I know the covariant anticommutator ($x$ and $y$ are 4-vectors)
$$\left\{\Psi_\alpha(x),\bar\Psi_\beta(y)\right\} = iS_{\alpha\beta}(x-y)=i(\displaystyle{\not}\partial+m)_{\alpha\beta}\Delta(x-y) \tag2$$
A possible definition of the $\Delta$ function is
$$\Delta(x) = -\frac{i}{2}\int\frac{d^3\vec{k}}{(2\pi)^3}\frac{e^{-ikx}-e^{ikx}}{\omega_k}= -\int\frac{d^3\vec{k}}{(2\pi)^3}\frac{\sin(kx)}{\omega_k}\qquad \text{where} \quad \omega_k = \sqrt{\left|\vec{k}\right|^2+m^2}$$
Now I can rewrite (2) as
$$\left\{\Psi_\alpha(x),\Psi^\dagger_\sigma(y)(\gamma_0)^\sigma_{\ \ \beta}\right\} = \left\{\Psi_\alpha(x),\Psi^\dagger_\sigma(y)\right\}(\gamma_0)^\sigma_{\ \ \beta} \tag3$$
If I take equal time anticommutator in (2), I get
$$\left\{\Psi_\alpha(x^0,\vec{x}),\bar\Psi_\beta(x^0,\vec{y})\right\} = i(\displaystyle{\not}\partial+m)_{\alpha\beta}\Delta(0,\vec{x}-\vec{y}) = 0$$
because
$$\Delta(0,\vec{x}) = -\int\frac{d^3\vec{k}}{(2\pi)^3}\frac{\sin(-\vec{k}\cdot\vec{x})}{\omega_k} = 0 \quad \text{integral of an odd function}$$
But looking at (3) one might conclude
$$0 = \left\{\Psi_\alpha(x^0,\vec{x}),\bar\Psi_\beta(x^0,\vec{y})\right\}=\left\{\Psi_\alpha(x^0,\vec{x}),\Psi^\dagger_\sigma(x^0,\vec{y})\right\}(\gamma_0)^\sigma_{\ \ \beta} \tag4$$
Which contradict the canonical anticommutation rules (1). For sure I made a mistake, but I can find it!
