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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've noted in comments that:

  • If you write fermions $f_1,\,f_2$ as $f_i=\eta_ic_i$ with $\{\eta_i,\,\eta_j\}=[\eta_i,\,c_j]=0$, $\{f_1,\,f_2\}=\eta_1\eta_2[c_1,\,c_2]$;
  • This works whether these objects are operators or not, but if they are operators, all matrix elements will be complex;
  • The operators in question will have uncountable dimension.

Let's also discuss what happens when you make a boson out of two fermions (if you use a larger even number of them, all but one of them makes a fermion, so this is the only case we need to consider). In particular write$$b_1=f_3f_4,\,b_2=f_5f_6\implies[b_1,\,b_2]=f_3\{f_4,\,f_5\}f_6-\{f_3,\,f_5\}f_4f_6+f_5f_3\{f_4,\,f_6\}-f_5\{f_3,\,f_6\}f_4$$in analogy with 9 here. If the anticommutators vanish at spacelike separations, so does $[b_1,\,b_2]$.

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  • $\begingroup$ Could you please point me to a link where the $\{\eta_i,\,\eta_j\}=[\eta_i,\,c_j]=0$ is derived? $\endgroup$ – Xlsx2020 Nov 23 '19 at 1:50
  • $\begingroup$ Is it still true that the $\left[\psi_\xi(x) , \overline\psi_\eta(y)\right]$ in the question does not vanish for spacelike separations and so we need to postulate superselection rules? $\endgroup$ – Xlsx2020 Nov 23 '19 at 2:11
  • $\begingroup$ Ok, I think I've found a couple of references: Fermionic Algebra and Fock Space and Supersymmetry, see par 4.3.2 etc.... $\endgroup$ – Xlsx2020 Nov 23 '19 at 12:09
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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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    $\begingroup$ If you write fermions $f_1,\,f_2$ as $f_i=\eta_ic_i$ for Grassmann numbers $\eta_i$ commuting with the $c_j$, $\{f_1,\,f_2\}=\eta_1\eta_2[c_1,\,c_2]$. $\endgroup$ – J.G. Nov 21 '19 at 8:09
  • $\begingroup$ @J.G. In my understanding you refer to the creation operators, about which I've found a related answer, but I've also noticed this comment there: "still Grassmann variables remain Grassmann variables, which are not any kind of operators and can't be reduced to a set of real numbers!" $\endgroup$ – Xlsx2020 Nov 22 '19 at 20:58
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    $\begingroup$ When we move from the classical theory in which bosons commute and fermions anticommute to a quantized one in which CCRs and CARs are nonzero, everything becomes an operator with complex matrix elements, including the $\eta_i$. (They'll be infinite-dimensional because of the $x$-dependence, contra the example to which the linked comment alluded.) But they'll still commute with the $c_j$. The calculation is then as before. $\endgroup$ – J.G. Nov 22 '19 at 21:05
  • $\begingroup$ @J.G. it would be great if you could write a bit more detailed answer (more expanded than these interesting but very succinct comments) that I'd be happy to accept (and study more in depth) $\endgroup$ – Xlsx2020 Nov 22 '19 at 21:28

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