# A condition on commutator implies quantization of field

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?

• 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). Commented Oct 18, 2016 at 11:05
• 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). Commented Oct 18, 2016 at 12:28

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.