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The canonical quantization procedure requires pairs of conjugate dynamical variables to be identified, which, after quantization, become operators whose commutator is $i\hbar$. How does the second quantization work? I mean just imposing a condition on the commutator leads to quantization of field, what is the underlying magic behind this prescription?

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    $\begingroup$ What do you mean by "the sceond quantization"? There's the "good" meaning as explained in this answer, and there's a "bad" meaning where people use second quantization to just mean the process of quantizing a classical field theory (as opposed to the classical particle theories for QM). $\endgroup$
    – ACuriousMind
    Commented Oct 18, 2016 at 11:05
  • $\begingroup$ First quantization is a mystery, but second quantization is a functor [cit. E. Nelson]. Apart from the suggestive quotes, all that you need is a real vector space with a non-degenerate antisymmetric bilinear form on it, and then you have a prescription on how to relate it to a C* algebra of quantum observables that encodes the canonical commutation relations (the so-called Weyl C* algebra). $\endgroup$
    – yuggib
    Commented Oct 18, 2016 at 12:28

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When you have a commutator of the form, $[A,B]=i\hbar$, you can relate this to $[x,p]= [-p,x]=i\hbar$. So, as $p=\frac{\hbar}{i}\frac{\partial}{\partial x}$ you can write $B = \frac{\hbar}{i}\frac{\partial}{\partial A}$ or $-A=\frac{\hbar}{i}\frac{\partial}{\partial B}$ and proceede with the calculation of states as in ordinary quantum mechanics using your energy expression as the Hamiltonian.

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  • $\begingroup$ Roughly speaking, this is only true in quantum mechanics, not in quantum field theory. $\endgroup$
    – yuggib
    Commented Oct 18, 2016 at 12:25
  • $\begingroup$ I have seen that comparision often given in text books. But it seems to me very casual. I believe there has to be some more fundamental reason. $\endgroup$
    – Seeker
    Commented Oct 18, 2016 at 12:26
  • $\begingroup$ This is the fundamental reasoning. Remember that quantum mechanics began with Heisenberg's matrices that obeyed this commutation relation. The calculation of the harmonic oscillator states and energies can be performed algebraically starting from the commutation relation. Pauli was able to calculate the hydrogen states as well. The instantiation of the commutation relation that is consistent with the Schrödinger equation is the one I described in my answer. More generally one might change p to p +f(x). $\endgroup$
    – Per Arve
    Commented Oct 18, 2016 at 19:27
  • $\begingroup$ I am not pleased to get my answer downvoted. I have used my time to help somebody by sharing my quite deep insights in this matter. Authors of QFT textbooks tend to not delve into this which might be a mistake. $\endgroup$
    – Per Arve
    Commented Oct 18, 2016 at 19:48
  • $\begingroup$ yuggib there is no mathematically satisfying formulation of QFT, so we only have recipes. However, the recipes give a consistent description of the physics. The recipes can be understood by making analogies with QM. C* algebras haven't really solved things. $\endgroup$
    – Per Arve
    Commented Oct 19, 2016 at 9:00

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