The procedure to quantize free field theories is imposing a commutation/anticommutation relation with the field and its conjugate momentum, as $$\mathcal L = i\bar\psi\gamma^\mu\partial_\mu\psi\rightarrow\left\{ \psi(t,\vec x),i\psi^\dagger(t,\vec y)\right\}=i\delta(\vec x-\vec y)$$ But how would this procedure be applied to an interacting field theory? In example, $$\mathcal L = -\frac14F_{\mu\nu}F^{\mu\nu}+i\bar\psi\gamma^\mu\partial_\mu\psi-eA_\mu\bar\psi\gamma^\mu\psi$$ How should $A^\mu$ and $\psi$ behave under the quantizing? Should they commute/anticommute with each other? $$[ A^\mu ,\psi]=0$$ Or something more complicated than that?
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$\begingroup$ You can impose the same/usual commutation or anticommutation relations for the individual particles. E.g., $\{\psi_x,\pi^{\psi}_y\}=i\delta_{xy}$ and $[A_x, \pi^{A}_y]=i\delta_{xy}$. For the relationship between different particle fields I think you are good with always allowing them to commute $[\psi_x, A_y]=0$. Although, I'm not positive that you couldn't choose the anticommutator, since in QED the $\psi$ field is always coupled to the $A$ field as a pair like $\psi_x\psi^\dagger_x A_x$... So that would commute regardless... So, I won't post an answer, just a comment. $\endgroup$– hftCommented Jan 19 at 22:04
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1$\begingroup$ Why does the field theory being interacting matter? We don't change the canonical commutation relations $[x,p] = \mathrm{i}\hbar$ in ordinary QM when we add non-free terms, either... $\endgroup$– ACuriousMind ♦Commented Jan 19 at 22:16
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1$\begingroup$ Well, the matter of interaction is to mix two different kinds of fields, with different statistics. In level of QM, the way I see why the commutation relation does not change, at least in the scenarios I've encontered, is that almost always the interactions are in respect to the same particle, like $x^4$ in H.O., but, as we put two different particles to interact between each other, shouldn't this imply that the measure of one interfere in the other? An this isn't explicitly the statment of a non-zero commutator? $\endgroup$– vfigueiraCommented Jan 19 at 22:32
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$\begingroup$ When you have operators $x_i,p_i$ where $i$ labels different particles in QM, they still commute for $i\neq j$ even when the particles interact, no? My point is: Whatever the confusion here is, it doesn't appear to have much to do with field theory. $\endgroup$– ACuriousMind ♦Commented Jan 19 at 22:38
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$\begingroup$ @ACuriousMind I think the OP's confusion/question is with respect to fermionic fields vis-a-vis bosonic fields, which is only a problem/question that would arise in an interacting theory. In particular, the question seems to be whether or not a fermionic field commutes or anti-commutes with a bosonic field. For example, should one write $[\psi_x, A_y]=0$ or should one write $\{\psi_x, A_y\}=0$? $\endgroup$– hftCommented Jan 20 at 19:37
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Yes, one should expect that $[A_{\mu}, \psi] = 0$. But how to do it is quite complicate. I describe a way to do it below (not the only way).
First of all, in old fashion operator formalism, since QED is a gauge theory, $A_{\mu}$ don't have standard commutator relation and it depends on the gauge you chose. An old standard way to construct the commutator is called "Dirac Procedure". In section 7.6 of Weinberg's "The quantum theory of field I", he introduces this methods. (Another way I've seen is proposed by Jackiw in his paper :(Constrained) Quantization Without Tears.)
Secondly, since the commutator is modified (by the method of Dirac's), $A_{\mu}$ and $\psi$ commutator could be non trivial. This is described again in Weinberg's the same book in section 8.3, p349. He computes the modified commutator between $A_{\mu}$ and other matter field (it could be Dirac field $\psi$) in the Coulomb gauge, which is not trivial. However, one expects that this commutator should be trivial. Thus, another modification is needed. In the same section in Weinberg's book, you can see how he solves this problem.
Finally, with these modified brackets, one can construct the Hamiltonian.
Conclusion : Defining the commutators in a gauge(constraint) theory or interaction theory are not easy usually. Although one may anticipate $[A_{\mu},\psi] = 0$, it is hard to construct such theory.
There are few QFT text book talks about the quantization in this way since it is too complicated and may not applicable. Instead, most of them just sloppy argue commutator in gauge theory (like free gauge boson $A_{\mu}$). As for interaction theory, they turn into path integral to calculate the correlation function, e.g. $\langle A_{\mu}A_{\nu} \rangle_{int}$ perturbatively and then by LSZ formula to construct the S-matrix.
If you are interested in old fashion operator formalism, you can see some reference like "Quantization of Gauge Systems" by Marc Henneaux. Also, some old papers like "Feynman Integral for Singular Lagrangian" by L.D. Faddeev or "Path Integral Quantization of Field Theories with Second-Class Constraints" by P.Senjanovic may help. However, these methods are too old, although they contain valuable concepts of constructing quantum theory, few people nowadays follow it.
