Canonical quantization of quantum field

The canonical quantization of a quantum field prescribes that given a lagrangian, one can quantize the theory by imposing the commutation relations between the field operators and their conjugated momenta.The field operators are then obtained by solving the equation of motion for plane waves.

However, why is this correct? The equation of motions are derived in the classical theory from the requirement that the action be stationary. This requirement does not apply in quantum theory where all possible paths contribute to the action.

Answer

The reason this works is because the field operators obey linear equations of motion. The Heisenberg equations of motion for the canonically quantized field are the same linear equations as the classical equations of motion, and they are translationally invariant. So you take the Fourier transform of the field, and the equation of motion guarantees that you only get a superposition of those plane waves which solve the linear equation. The Fourier transform coefficients of these plane waves are operators which manifestly have a definite frequency, and therefore are creation/annihilation operators.

It is a consequence of the commutation relation with the Hamiltonian that a definite frequency operator acting on an energy state produces a state with energy incremented by the frequency of the operator.

Digression on Normalization

When doing the field expansion, field theory books usually don't put things in manifestly covariant form, so the formulas are very ugly. The proper relativistic conventions are as follows:

The k integration measure on a relativistic mass shell is given by a delta function concentrated on the mass shell.

$$ \int {d^4k\over (2\pi)^4} 2\pi \delta(k^2-m^2) = \int {d^3k\over (2\pi)^3 2\omega_k}$$

All k delta-functions carry $2\pi$ factors, all integrations over dk have a $2\pi$ in the denominator, this is the physicist conventions for Fourier transforms. These factors should be implicit when you write down the things, so the above should be written:

$$ \int d^4 k \delta(k^2 - m^2) = \int {d^3k\over 2\omega_k} = \int d\vec{k}$$ Where the last equality is a definition of notation. The k-states of a particle should be normalized so that their inner product is orthogonal when integrating with the measure above (the $2\pi$ factors in the delta function are supressed, as immediately above)

$$ \langle k|k'\rangle = 2\omega_k \delta^3(k-k') $$

This means that the state $|k\rangle$ is really $\sqrt{2\omega_k}$ bigger than the nonrelativistically normalized state, where the right hand side of the above is just a delta-function. It is also $(2\pi)^{3/2}$ times bigger than the state normalized with a non-2pi-absorbing delta function on the right hand side.

The relativistic creation operators need to create these bigger-normalized states, so they are bigger than the naive creation operators

$$ \alpha^\dagger(k) = (2\pi)^{3\over 2} \sqrt{2\omega_k} a(k) $$

And likewise for the annihilation operators. With these conventions, the Fourier expansion for the scalar field $\phi$ reads

$$\phi(x) = \int \alpha(k) e^{i(k\cdot x + \omega_k t)} + \alpha^\dagger(k) e^{-i (k\cdot x + \omega_k t)} d\vec{k} $$

and the naturalness of the expansion is manifest.