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Derivation of second quantization (advanced)
This section describes the mathematics behind second quantization.

Purely classical part
Like any function, the electromagnetic vector potential $$\mathbf{A}(\mathbf{r}, t)$$ may be Fourier transformed in each of its arguments



\mathbf{\tilde{A}}(\mathbf{k}, \omega) \equiv \int dt \int d\mathbf{r} \ \mathbf{A}(\mathbf{r}, t) e^{i\omega t} e^{-i\mathbf{k} \cdot \mathbf{r}} $$

The corresponding inverse Fourier transform is



\mathbf{A}(\mathbf{r}, t) \equiv \int \frac{d\omega}{2\pi} \int \frac{d\mathbf{k}}{\left(2\pi\right)^{3}} \ \mathbf{A}(\mathbf{k}, \omega) e^{-i\omega t} e^{i\mathbf{k} \cdot \mathbf{r}} $$ which can be interpreted as a sum of waves of (possibly complex) amplitude $$\mathbf{\tilde{A}}(\mathbf{k}, \omega)$$ moving in direction $$\mathbf{k}$$ with phase velocity $$c = \omega / k$$ where $$k \equiv \left| \mathbf{k} \right|$$.

The commonly used Coulomb gauge condition can be written in real and Fourier space as

\boldsymbol\nabla \cdot \mathbf{A} = 0 = \mathbf{k} \cdot \mathbf{\tilde{A}} $$ This condition is also called the "transverse gauge", because $$\mathbf{\tilde{A}}$$ is perpendicular to the direction of propagation $$\mathbf{k}$$.

In empty space, we may set the scalar potential $$\phi = 0$$; hence, the electric and magnetic fields are described by



\mathbf{E} = -\frac{1}{c} \frac{\partial}{\partial t} \mathbf{A} = -\frac{i\omega}{c} \mathbf{\tilde{A}} $$



\mathbf{H} = \boldsymbol\nabla \times \mathbf{A} = i \mathbf{k} \times \mathbf{\tilde{A}} $$

Substituting these formulae into the Maxwell equation



\boldsymbol\nabla \times \mathbf{H} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} $$

yields the relativistic wave equation



\boldsymbol\nabla \times \boldsymbol\nabla \times \mathbf{A} \equiv -\nabla^{2} \mathbf{A} + \boldsymbol\nabla \left( \boldsymbol\nabla \cdot \mathbf{A} \right) = -\nabla^{2} \mathbf{A} = \frac{-1}{c^{2}} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}} $$

In Fourier space, this physical equation imposes the mathematical constraint $$\omega^{2} = c^{2} k^{2}$$



\frac{\omega^{2}}{c^{2}} \mathbf{\tilde{A}} = k^{2} \mathbf{\tilde{A}} $$

Using Parseval's theorem, the energy $$U_{tot}$$ of the electromagentic field can be expressed in both real and Fourier space



U_{tot} = \frac{1}{8\pi} \int d\mathbf{r} \left[ \mathbf{E}^{2} + \mathbf{H}^{2} \right] = \int \frac{d\mathbf{k}}{\left( 2\pi \right)^{3}} k^{2} \left| \mathbf{\tilde{A}} \right|^{2} $$

Define



\mathbf{Q}(\mathbf{k}) = \sqrt{\frac{V}{4\pi c^{2}}} \left[ \mathbf{\tilde{A}} + \mathbf{\tilde{A}}^{*} \right] $$



\mathbf{P}(\mathbf{k}) = -i\omega_{k} \sqrt{\frac{V}{4\pi c^{2}}} \left[ \mathbf{\tilde{A}} - \mathbf{\tilde{A}}^{*} \right] = \frac{\partial}{\partial t}\mathbf{Q} $$



H = \frac{1}{2} \int \frac{d\mathbf{k}}{\left( 2\pi \right)^{3}} \left[ \mathbf{P}^{2} + \omega_{k}^{2} \mathbf{Q}^{2} \right] $$

A is perp to propagation direction k, hence 2 polarizations