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.


1 Answer 1


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.

  • $\begingroup$ How can it be proved that the field operators obey the same equations as the classical equations of motion? I assume one starts with the commutation relation but where does one go from here? $\endgroup$
    – Whelp
    Oct 25, 2011 at 20:50
  • $\begingroup$ You commute the field with the field hamiltonian. But the answer is guaranteed by ehrenfest's theorem. $\endgroup$
    – Ron Maimon
    Oct 25, 2011 at 21:04
  • $\begingroup$ You should also do all the manipulations covariantly, using the covariant commutation relations on the relativistic creation and annihilation operators. $\endgroup$
    – Ron Maimon
    Oct 25, 2011 at 23:18
  • $\begingroup$ Is this the whole point of the Schwinger–Dyson equation? $\endgroup$
    – Whelp
    Oct 27, 2011 at 22:49
  • $\begingroup$ The Schwinger Dyson equation is just the equation of motion for the quantum fields. Historically, it was used as a surrogate for the path integral for deriving the perturbation expansion--- there are many equivalent ways. The equation of motion path to covariant perturbation theory dates back to Stueckelberg from the 1930s, but the Schwinger/Dyson credit stuck (probably to the chagrin of Schwinger and Dyson). The equations of motion for free fields are not usually called Schwinger-Dyson equation because they are too simple--- they are linear. There is no expansion necessary. $\endgroup$
    – Ron Maimon
    Oct 27, 2011 at 23:56

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