In essence, Maxwell's equations explain how electric and magnetic fields interact, propagate and are affected by objects.

$$ \begin{equation} \label{maxeq01} \nabla \cdot B = 0 , \end{equation} $$

$$ \begin{equation} \label{maxeq02} \nabla \times E = -\dfrac{1}{c}\dfrac{\partial B}{\partial t}, \end{equation} $$

$$ \begin{equation} \label{maxeq03} \nabla \cdot E = 4\pi\rho, \end{equation} $$

$$ \begin{equation} \label{maxeq04} \nabla \times B = \dfrac{4\pi}{c}J+\dfrac{1}{c}\dfrac{\partial D}{\partial t}. \end{equation} $$

In the equations, $B$ and $E$ indicate the magnetic and electric fields respectively. Here eq. (\ref{maxeq01}) (Magnetic law) states that there can not be magnetic monopoles, eq. (\ref{maxeq02}) (Faraday's law) states that a magnetic field that is changing in time will give rise to a circulating electric field, eq. (\ref{maxeq03}) (Gauss's law) states that the electric flux out of any closed surface is proportional to the total charge enclosed within the surface and eq. (\ref{maxeq04}) (Ampere’s law) states that the magnetic field created by an electric current is proportional to the size of that electric current and permeability. Here $\rho$ and $J$ are the charge density and current density respectively. If we consider a source-free state ($\rho = 0$ and $J = 0$), we can obtain the following equations (consider $B = \mu H$ and $D = \epsilon E$).

$$ \begin{equation} \label{maxeq1} \nabla \cdot B = 0 , \end{equation} $$

$$ \begin{equation} \label{maxeq2} \nabla \times E = -\dfrac{\partial B}{\partial t}, \end{equation} $$

$$ \begin{equation} \label{maxeq3} \nabla \cdot D = 0, \end{equation} $$

$$ \begin{equation} \label{maxeq4} \nabla \times H = -\dfrac{\partial D}{\partial t}. \end{equation} $$

Maxwell's equations are gauge invariant when no sources are present. The convenient choice of gauge in quantum optics is the Coulomb gauge. In the Coulomb gauge, it sets the divergence of the vector potential to zero, which eliminates eq. (\ref{maxeq03}). This simplifies the equations of motion and makes it easier to solve for the electric and magnetic fields. In the Coulomb gauge, both $B$ and $E$, are determined from a vector potential $A$.

$$ \begin{equation} \label{maxeq5} B = \nabla \times A , \end{equation} $$

$$ \begin{equation} \label{maxeq6} E = -\dfrac{\partial A}{\partial t}. \end{equation} $$

We consider $\nabla \cdot A = 0$ with the coulomb condition. Then substituting eq. (\ref{maxeq5}) into eq. (\ref{maxeq4}) we obtain the following equation.

$$ \begin{equation} \label{maxeq7} \nabla \times \frac{B}{\mu_0} = \dfrac{\partial D}{\partial t}, \end{equation} $$

$$ \begin{equation} \label{maxeq8} \nabla \times \left(\nabla \times \frac{A}{\mu_0}\right) = \dfrac{\partial D}{\partial t}, \end{equation} $$

$$ \begin{equation} \label{maxeq9} \nabla \left(\nabla\cdot\dfrac{A}{\mu_0}\right) - \nabla^2 \left(\frac{A}{\mu_0}\right) = \dfrac{\partial D}{\partial t}, \end{equation} $$

because $\nabla \cdot A = 0$,

$$ \begin{equation} \label{maxeq10} \nabla^2 \frac{A}{\mu_0} = \dfrac{\partial D}{\partial t}, \end{equation} $$

$$ \begin{equation} \label{maxeq11} \nabla^2 A = \mu_0 \dfrac{\partial D}{\partial t}, \end{equation} $$

$$ \begin{equation} \label{maxeq12} \nabla^2 A = \mu_0 \epsilon_0 \dfrac{\partial E}{\partial t}, \end{equation} $$

As $\mu_0\epsilon_0 = c^{-2}$,

$$ \begin{equation} \label{maxeq13} \nabla^2 A = \dfrac{1}{c^2} \dfrac{\partial E}{\partial t}, \end{equation} $$

$$ \begin{equation} \label{maxeq14} \nabla^2 A = \dfrac{1}{c^2} \dfrac{\partial^2 A}{\partial t^2}. \end{equation} $$

In eq. (\ref{maxeq14}) we have obtained the wave equation $\left(\frac{\partial^2z}{\partial t^2} = c^2\frac{\partial^2z}{\partial x^2}\right)$. Next, consider separating the vector potential into two complex terms.

$$ \begin{equation} \label{maxeq15} A(r,t) = A^{+}(r,t) + A^{-}(r,t). \end{equation} $$

We can update eq. (\ref{maxeq15}) considering a complex travelling wave which explains the vector potential operator of a photon in quantum field theory as follows.

$$ \begin{equation} \label{maxeq16} \hat{A}(r,t) = \hat{A}_0 e^{i(kr-\omega t)} \hat{a} + \hat{A}_0^{*} e^{-i(kr-\omega t)} \hat{a}^{\dagger}. \end{equation} $$

Here annihilation and creation operators are given as $\hat{a}$ and $\hat{a}^{\dagger}$ respectively. Term $\hat{A}_0$ indicates the amplitudes of the vector potential. $k$ and $\omega$ imply the wave number vector and the angular frequency of photons respectively. With eq. (\ref{maxeq16}) we can update eq. (\ref{maxeq5}) and eq. (\ref{maxeq6}) to keep consistency as follows.

$$ \begin{equation} \label{maxeq17} B = i\omega \hat{A}_0 e^{i(kr-\omega t)} \hat{a} - \hat{A}_0^{*} e^{-i(kr-\omega t)} \hat{a}^{\dagger}, \end{equation} $$

$$ \begin{equation} \label{maxeq18} E = ik\hat{A}_0 e^{i(kr-\omega t)} \hat{a} - \hat{A}_0^{*} e^{-i(kr-\omega t)} \hat{a}^{\dagger}. \end{equation} $$

For an electromagnetic field, the Hamiltonian is given as follows.

$$ \begin{equation} \label{maxeq19} H = \dfrac{1}{2}\int \left(\epsilon_0 \hat{E}^2 + \dfrac{1}{\mu_0}\hat{B}^2\right) dr . \end{equation} $$

We can update the above general equation using eq. (\ref{maxeq17}) and eq. (\ref{maxeq18}) as follows.

$$ \begin{equation} \label{maxeq20} H = \dfrac{1}{2}\int \left(\epsilon_0 \hat{E}(r,t)\cdot \hat{E}(r,t) + \dfrac{1}{\mu_0}\hat{B}(r,t)\cdot \hat{B}(r,t)\right) dr . \end{equation} $$

Taking the $\hat{E}(r,t)$ and $\hat{B}(r,t)$ terms in eq. (\ref{maxeq20}) individually and considering $e^{i(kr-\omega t)} = \Theta$,

$$ \begin{equation} \label{maxeq21} \hat{E}(r,t) = i\omega\left(\hat{A}_0 \Theta \hat{a} - \hat{A}_0^{*} \bar{\Theta} \hat{a}^{\dagger}\right) , \end{equation} $$

$$ \begin{equation} \label{maxeq22} \hat{B}(r,t) = ik\times\left(\hat{A}_0 \Theta \hat{a} - \hat{A}_0^{*} \bar{\Theta} \hat{a}^{\dagger}\right) , \end{equation} $$

is obtained. Consider the first term of eq. (\ref{maxeq20}),

$$ \begin{equation} \label{maxeq23} H_1 = \dfrac{1}{2}\int \epsilon_0 i^2\omega^2\left(\hat{A}_0 \Theta \hat{a} - \hat{A}_0^{*} \bar{\Theta} \hat{a}^{\dagger}\right)\left(\hat{A}_0 \Theta \hat{a} - \hat{A}_0^{*} \bar{\Theta} \hat{a}^{\dagger}\right)dr , \end{equation} $$

$$ \begin{equation} \label{maxeq24} H_1 = -\dfrac{1}{2}\int \epsilon_0 \omega^2\left(\hat{A}_0^2 \Theta\Theta \hat{a}\hat{a} - \hat{A}_0\hat{A}_0^{*} \Theta\bar{\Theta} \hat{a}\hat{a}^{\dagger} - \hat{A}_0^{*}\hat{A}_0 \bar{\Theta}\Theta \hat{a}^{\dagger}\hat{a} + \hat{A}_0^{*}\hat{A}_0^{*} \bar{\Theta}\bar{\Theta} \hat{a}^{\dagger}\hat{a}^{\dagger}\right)dr . \end{equation} $$

Because $\hat{a}\hat{a} = 0$ and $\hat{a}^{\dagger}\hat{a}^{\dagger} = 0$,

$$ \begin{equation} \label{maxeq25} H_1 = \dfrac{1}{2}\int \epsilon_0 \omega^2\left(|\hat{A}_0|^2 \Theta\bar{\Theta} \hat{a}\hat{a}^{\dagger} - |\hat{A}_0|^2 \bar{\Theta}\Theta \hat{a}^{\dagger}\hat{a}\right)dr . \end{equation} $$

Considering the second term of eq. (\ref{maxeq20}), we obtain the following equation.

$$ \begin{equation} \label{maxeq26} H_2 = \dfrac{1}{2\mu_0}\int k^2\left(|\hat{A}_0|^2 \Theta\bar{\Theta} \hat{a}\hat{a}^{\dagger} - |\hat{A}_0|^2 \bar{\Theta}\Theta \hat{a}^{\dagger}\hat{a}\right)dr . \end{equation} $$

Stating $H = H_1 + H_2$ and $e^0 = 1$,

$$ \begin{equation} \label{maxeq27} H = \left(\dfrac{\epsilon_0 \omega^2}{2}+\dfrac{k^2}{2\mu_0}\right)\int |\hat{A}_0|^2 dr \left(\hat{a}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a}\right) . \end{equation} $$

Considering $\int A_o^2 dr = \frac{\hbar}{2\epsilon_0 \omega}$,

$$ \begin{equation} \label{maxeq28} H = \left(\dfrac{\epsilon_0 \omega^2}{2}+\dfrac{k^2}{2\mu_0}\right)\dfrac{\hbar}{2\epsilon_0 \omega} \left(\hat{a}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a}\right) . \end{equation} $$

Using $\epsilon_0 = \frac{1}{c^2\mu_0}$ and $k = \frac{\omega}{c}$, the Hamiltonian can be stated as follows.

$$ \begin{equation} \label{maxeq29} H = \dfrac{\hbar\omega}{2} \left(\hat{a}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a}\right) , \end{equation} $$

$$ \begin{equation} \label{maxeq30} H = \dfrac{\hbar\omega}{2} \left(\hat{a}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a}\right) . \end{equation} $$

Considering $\hat{a}\hat{a}^{\dagger}-\hat{a}^{\dagger}\hat{a} = 1$,

$$ \begin{equation} \label{maxeq31} H = {\hbar\omega} \left(\hat{a}^{\dagger}\hat{a}+\dfrac{1}{2} \right) . \end{equation} $$

This eq. (\ref{maxeq30}) represents the sum of the number of photons in a mode multiplied by the energy of a photon in the mode and the addition of vacuum fluctuations in the mode.