# Quantization and Commutation Relations

Why do we use commutation relations when quantizing any system? In the case of developing quantum mechanics from classical mechanics, we write the hamiltonian and then quantize it by having the conjugate variable/observables obey the commutation relation. And this process is valid for any quantum system.

The same is the case when we are trying to quantize fields, we write down the hamiltonian and quantize the conjugate fields by the commutation relations.

So, why does adding the additional condition of commutation on conjugate variables (after promoting them to operators of course) lead to a quantum theory of the same system? Is it just a postulate or is there some reasoning behind the same?

PS: My original idea was that it's required for uncertainty principle but that's just a circular argument.

• Actually the procedure you outline is known to fail since van Hove. see physics.stackexchange.com/q/573908/36194 or even here physics.stackexchange.com/q/345859/36194. Oct 12, 2020 at 18:15
• @ZeroTheHero The answers that you have referred to were really interesting, but even then it does not explain why promoting the observable to operators and using commutation relations lead to a quantum hamiltonian that is equal to (upto ordering of p,q) the classical hamiltonian. Oct 12, 2020 at 18:30
• "Why" is an impossibly broad question. It just happened that Heisenberg discovered non commuting matrices that did the trick (described quantum radiation) and his boss, Born, understood these obeyed commutation relations. In another universe/planet, they could have deformed classical variables by pseudo differential operators, to describe the same systems. In yet another planet, Dirac/Feynman could have hit upon path integrals to do the same job. What is "why"? Oct 12, 2020 at 20:02
• @CosmasZachos, could you please provide some source for the other two methods that you mentioned i.e the pseudo differential operators and the path integrals. BTW, by path integrals, do you mean the ones we use in QFT? Cause, in that also, we quantize the hamiltonian first, r8? Oct 14, 2020 at 6:45
• You should first define what you mean of the "same system". The statement is extremely imprecise. The fact is we know some procedures for quantization, one of which is the one you describe (canonical quantization), but there are more. Now the point is that they are able to describe phenomena that we cannot by means of classical theory. So if you quantize a harmonic oscillator you get the QUANTUM harmonic oscillator which doesn't describe a (classical) harmonic oscillator directly but perhaps in some limit. They are not the same system.... P.D. Did you just use "BTW" and "r8"? Oct 14, 2020 at 13:29

## 1 Answer

In response to your reference request, phase space formulation; path integrals. Perhaps this should be reassigned to community status resource recommendation. If full books were required, in your shoes I might well opt for CTQMPS, and Feynman & Hibbs respectively.

No, the Hamiltonian in a path integral is a c-number! Admittedly, the recipe through which you actually evaluate infinitesimal shifts and averages does pick ordering prescriptions (L. Cohen 1970), and thus some non-commutativity prescription in the conceptual background.

(Analogously, the star-products of deformation quantization determine telltale star-commutation relations. Quantization is a mystery.) So it's fair to think of quantization and the commutation relation joined at the hip, but the narrative of the 1920s you are considering fits in a much-much broader landscape.

My original comment eliciting the resource request was

"Why" is an impossibly broad question. It just happened that Heisenberg discovered non commuting matrices that did the trick (described quantum radiation) and his boss, Born, understood these obeyed commutation relations. In another universe/planet, they could have deformed classical variables by pseudo-differential operators, to describe the same systems. In yet another planet, Dirac/Feynman could have hit upon path integrals to do the same job.