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We know that two fields commute - by locality and causality - iff there is spacelike separation

$\left[\phi_l^k(x) , \phi_m^{k'}(y)\right] = 0$ for $(x-y)^2<0$

In the canonical quantization of the Dirac field, if $b_\alpha(k)$ is the annihilation operator and $b^\dagger_\alpha(k)$ is the creation operator for a particle of 4-momentum $k$ with

$\left[b_\alpha(k), b^\dagger_\beta(q)\right] = (2\pi)^3\frac{\omega_\mathbf k}{m} \delta^{(3)}(\mathbf{k}-\mathbf{q})\delta_{\alpha\beta}$

and $\psi^{(+)}(x) = e ^{-ikx}u(k) $ is a solution with positive energy while $\psi^{(-)}$ is negative, when we use commutators all the way, the following

$\left[\psi_\xi(x) , \overline\psi_\eta(y)\right] = (i\not\partial_x+m)\int{\frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_\mathbf k}\left[e^{-ik(x-y)} + e^{+ik(x-y)}\right]|_{k=(\omega_k,\mathbf k )}}$

does not vanish for spacelike separations and results in a violation of causality. How is this problem overcome or why isn't it a problem?

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I understand that we have to use canonical anticommutation relations to satisfy micro-causality in this case. For example from Tong's lectures about "Quantizing the Dirac Field" we see that in case of fermions

$\left\{ \psi_\alpha(x) , \psi_\beta(y) \right\} = 0$ for $(x-y)^2<0$

The theory remains causal as long as fermionic operators are not observable. If you think this is a little weak, remember that no one has ever seen a physical measuring apparatus come back to minus itself when you rotate by 360 degrees!

Thus this is also related to another answer which explains that

Only products that are Grassmann-even – contain an even number of fermionic factors – are measurable due to the existence of superselection sectors.

So the "problem" is "solved" by noting that - as the wiki page puts it:

Since all reasonable observables (such as energy, charge, particle number, etc.) are built out of an even number of fermion fields, the commutation relation vanishes between any two observables at spacetime points outside the light cone

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