How Fourier components of vector potential becomes operators

[goodphy]

Hello.

I'm studying quantization of electromagnetic field (to see photon!) and on the way to reach harmonic oscillator Hamiltonian as a final stage, sudden transition that the Fourier components of vector potential A become quantum operators is observed. (See https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field)

I guess Wikipedia doesn't describes much of the reason for this jump. It only states that the operator is for boson. I know photon is boson but how it can explain Fourier components of A should become quantum operators? And besides, I originally expected to see the prove that photon is boson in this study.

Could anybody help me to understand this?

 

[stevendaryl]

It helps to see the analogy with single-particle quantum mechanics, in the case of the harmonic oscillator. You have, classically:

E = \frac{m}{2} \dot{x}^2 + \frac{K}{2} x^2

Then you can turn this classical equation into a quantum equation by letting p = m \dot{x} and rewriting it as:

E = \frac{p^2}{2m} + \frac{K}{2} x^2

where p and x obey the commutation relation [p,x] = -i \hbar

The next move is to rewrite it in terms of "raising and lowering" operators:

a = \frac{m \omega}{2 \hbar} x + i \frac{1}{2m \omega \hbar}

a^\dagger = \frac{m \omega}{2 \hbar} x - i \frac{1}{2m \omega \hbar}

Now, turning to quantum fields, instead of single particles. For a massless quantum field \Phi, there is energy associated with every Fourier mode \vec{k}. This energy is given by: (something like...)

E(\vec{k}) = C ( |\tilde{\dot{\Phi}}|^2 + |\vec{k} \tilde{\Phi}|^2)

where \tilde{\Phi} is the Fourier transform of \Phi, and \tilde{\dot{\Phi}} is similarly the Fourier transform of \dot{\Phi}

To quantize this, you similarly replace \dot{\Phi} by the canonical momentum \Pi to get:

E(\vec{k}) = C ( |\tilde{\Pi}|^2 + |\vec{k} \tilde{\Phi}|^2) = C ( |\tilde{\Pi}|^2 + \omega^2 |\tilde{\Phi}|^2) (where \omega = |\vec{k}|)

For a fixed mode \vec{k}, this looks like the hamiltonian of the harmonic oscillator. So you can similarly solve it by writing:

a(\vec{k}) = \omega \tilde{\Phi} + i \tilde{\Pi}
a^\dagger(\vec{k}) = \omega \tilde{\Phi} - i \tilde{\Pi}

What I've described here is a scalar field. The case of the electromagnetic field is more complicated, but only because there are more components of the field.

The point is that in QFT, you are treating \Pi and \Phi as operators, in the same way that you treat p and x as operators in the single-particle case. The point about Fourier transforms is that the hamiltonian is much simpler after the transform, and the raising and lowering operators are related to the Fourier transforms of \Phi and \Pi