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It is often written in books, the quantization conditions for classical field theory leading to Lagrangian of a real scalar field and thus to Klein Gordon equation. And these are introduced by introducing the conjugate momentum. I honestly don't get why we introduce conjugate momentum to set the quantization conditions? Can someone please explain this point to me?

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  • $\begingroup$ The book was by Mandl, Shaw on Quantum Field Theory p. 28 (for reference) $\endgroup$ – PhilosophicalPhysics Oct 31 '14 at 13:00
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Quantization is often treated as a mystery, yet it can be seen to arise naturally from classical Hamiltonian mechanics.1 This yields the quantization prescription

$$ \{\dot{},\dot{}\} \mapsto \frac{1}{{\mathrm{i}\hbar}}[\dot{},\dot{}]$$

for all classical observables on the phase space that are to be turned into quantum operators.2 Therefore, if we want to obtain our QFT through the process of canonical quantization, we have to first switch to the Hamiltonian formalism in order to carry out this prescription - and to get there from a Lagrangian/action formalism, you just Legendre transform, introducing the canonical momenta.

Note that this is not the only way to quantize a theory, one may also use the path integral formalism, which works with the original Lagrangian/action.


1For this formulation known as geometric quantization, see for example this excellent post.

2Caveat: This approach actually leads to deformation quantization, meaning that, sometimes, the above prescription only holds to first order in $\hbar$

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Quantization, in the Heisenberg picture is very much related to establishing quantum non-trivial commutators between the main observables of the theory. Originally this is linked to the classical Poisson bracket, which in the Hamiltonian formalism includes the cannonical momentum. Hope I gave some insight, as to why we are led to define the conjugade momentum. I have to add, that all this refers to "cannonical quantization" of fields. This is not the only approah towards quantization, since alternative ways including Feynmann's path integrals do exist.

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  • $\begingroup$ Thank you @Cala for your reply. But other than this analogy - what information does it give us? I mean if one is asked what are the quantization conditions, he goes as $$[\pi (x,t),\phi (x,t)] = $$ to some dirac formulation. So what? $\endgroup$ – PhilosophicalPhysics Oct 31 '14 at 13:18
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You need to introduce Momentum because you are using hamiltonian aprroach to quantize transforming the poisson brackets to conmutation relation. That process is trivial when you have theory without constrains, whit constrains you can't use lagrangian approach.

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