# How Fourier components of vector potential becomes operators

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1. Sep 30, 2015

### 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?

2. Sep 30, 2015

### stevendaryl

Staff Emeritus
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 $\Pi$