Quantization of an electromagnetic field with standing waves

Maxwell's equations for a source free environment are

$${\bf\nabla \cdot D} = {\bf\nabla \cdot B } = 0$$ (3.1)

$${\bf\nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}$$ (3.2)

$${\bf\nabla} \times {\bf H} = \frac{\partial {\bf D}}{\partial t}$$ (3.3)

In this environment, $B =\mu_oH$ and $D = \epsilon_o E$

The simplicity of these four equations begs for even further simplification, whereby we introduce the vector potential,

$${\bf\nabla} \times {\bf A}{ \equiv {\bf B} \ \ \ \longrightarrow \ \ \ {\bf E}= -\frac{\partial {\bf A}}{\partial t} - \nabla V$$

There are many ways to fit these equations while maintaining the validity of the Maxwell equations. The coulomb potential suffices and is preferrable due to its simplicity: $\bf {\nabla}\cdot {\bf A} = 0$, $V = 0$. The identity $\nabla \times \nabla \times A = \nabla(\nabla \cdot A) + (\nabla \cdot \nabla) A$ coupled with our previous assertation and equation 3 above yield

$$\nabla^2A = \frac{1}{c^2}\frac{\partial^2 A}{\partial t^2}$$ (3.4)

We will quantize this centralized magnetic potential. To completely specify the field we would have to describe its values for all points in space; it is customary to develop the quantization in a theoretical cube, and then let the volume of the cube expand to infinity to accomplish a full discription.

Our derivation can consider standing or plane waves. The case of standing waves is quicker. We assume the magnetic potential has a solution of form

$${\bf A}({\bf r},t)=\frac{1}{\epsilon_o}\sum_lq_l{\bf u}_l({\bf r})$$ (3.5)

Under this seperation of variables,

$${q_l \nabla^2 u_l = \frac{u_l}{c^2}\frac{d^2 q_l}{dt^2} \ \ \ \lo... ...{u_l}\nabla^2 u_l = \frac{1}{q_l}\frac{d^2 q_l}{dt^2} \equiv -\omega_l^2 \ \ \$$ (3.6)

$$\ni \ \nabla^2 u_l + \frac{\omega_l^2}{c^2}u_l = 0, \ \ \frac{d^2q_l}{dt^2} + \omega_l^2q_l = 0$$ (3.7)

Being in the standing wave regime, there can be no currents on the boundary, implying that $u_l\vert _{tangential} = 0$ and $\nabla \times u_l\vert _{normal} = 0$. But of course, we have already doomed ourselves to the fact that $\nabla \cdot u_l = 0$. These consequences become important as follows.

The energy stored by our electromagnetic field is

$$H = \frac{1}{2}\int(\epsilon_oE^2 + \mu_oH^2)dv = \frac{1}{2}\int... ...\partial A}{\partial t}\right)^2 + \mu_o\left(\nabla \times A\right)^2\right)dv$$ (3.8)

We assume via our target solution that our spatial modes will have the orthogonality

$$\int u_l \cdot u_m dv = \delta_{l,m}$$ (3.9)

Using this orthogonality, we have

$$H = \frac{1}{2}\sum_l\left(\frac{dq_l}{dt}\right)^2\int u_l^2 dv + \frac{c^2}{2}\sum_l q_l^2\int \left(\nabla \times u_l\right)^2dv$$ (3.10)

The rightmost integral can be taken with an algebraic manipulation. We examine it as follows:

$$(\nabla \times u_l) \cdot (\nabla \times u_l) \equiv B \cdot (\nabla \times A)$$

A snap shot of the latter yields:

$$B_k(\nabla_i A_j) = \nabla_i (A_j B_k)$$

This suggest the generating equation

$$\nabla \cdot (A \times B) \ : \ \nabla_i A_jB_k - \nabla_i A_kB_j = B_k(\nabla _i A_j) - A_k(\nabla_i B_j)$$

Thus our vector identity is

$$\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A\cdot(\nabla \times B)$$

More appropriate for our needs, we rearrange the previous equation to find:

$$B \cdot (\nabla \times A ) = \nabla \cdot (A \times B) + A \cdot (\nabla \times B)$$

Reidentifying $A = u_l$ and $B = \nabla \times u_l$, we have

$$(\nabla \times u_l)\cdot (\nabla \times u_l) = \nabla \cdot (u_l \times \nabla \times u_l) + u_l \cdot (\nabla \times \nabla \times u_l)$$

Remembering that this is under a volume integral, we quickly see that the first term Stokes away as a surface integral which our previously established boundary conditions founded as zero. The second term is taken care of by the use of a previous vector identity. Catching up, we have

$$H = \frac{1}{2}\sum_l \left(\frac{dq_l}{dt}\right)^2 + \frac{c^2}{2}\sum_l q_l^2\int u_l\cdot(\nabla(\nabla \cdot u_l)-\nabla^2u_l)dv$$ (3.11)

With equation 7 and our boundary conditions, this becomes,

$$H = \frac{1}{2}\sum_l\left(\left(\frac{dq_l}{dt}\right)^2 + \omega_l^2q_l^2\right)\equiv \sum_l H_l$$ (3.12)

This form is fully equivalant to the harmonic oscillator, and with the following set:

$$p_l = \frac{dq_l}{dt}$$ (3.13)

$$q_l = \sqrt{\frac{\hbar}{2\omega_l}}(a^{\dagger}_l + a_l)$$ (3.14)

$$p_l=i\sqrt{\frac{\hbar \omega_l}{2}}(a_l^{\dagger} - a_l)$$ (3.15)

we complete the standing wave quantization by concluding that

$$H = \frac{1}{2}\sum_l \hbar \omega_l \left(a_l^{\dagger}a_l + a_la_l^{\dagger}\right)$$ (3.16)