Quantization of an electromagnetic field using plane waves

Also common to find in the literature is the following version of the quantization of the electromagnetic field.

We rejoin the previous discussion, interjecting now the more complicated attempted solution.

$${\bf A}({\bf r},t)=\sum_m\sqrt{\frac{\hbar}{2\omega_m \epsilon_o}}\left(a_m(t){\bf u}_m({\bf r}) +a^{\dagger}_m(t){\bf u}^*_m({\bf r})\right)$$ (4.1)

As before, it follows that,

$$\nabla^2u_m(r) + \frac{\omega^2}{c^2}u_m(r)=0$$ (4.2)

$$\frac{\partial^2a_m}{\partial t^2} + \omega^2a_m = 0$$ (4.3)

It follows, using plane waves,

$$a_m(t)=a_me^{-i\omega_m t}$$ (4.4)

$$a_m^{\dagger}(t) = a^{\dagger}_m e^{i\omega_m t}$$ (4.5)

$$u_m(r)=\frac{e_m}{\sqrt{v}} e^{ik_m \cdot r} \ \ @ \ \ k^2_m = \frac{\omega_m^2}{c^2}$$ (4.6)

Here, following the notation of Orszag, $e_m$ represents the appropriate unit vector. Requiring only periodic boundary conditions: $A({\bf r} + L {\bf e_m}) = A({\bf r}) \ \ni \ {\bf k_m} = 2\pi(m_1{\bf i} + m_2{\bf j} + m_3{\bf k})/L$. We thus have,

$${\bf A}({\bf r},t)=\sum_m \sqrt{\frac{\hbar}{2 \omega_m \epsilon_... ...(k_m\cdot r - \omega_mt)} + a^{\dagger}_me^{-i(k_m\cdot r - \omega_m t)}\right)$$ (4.7)

Again,

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

By sight we see that the result of this integral will come down to:

$$\left(-iae^{\alpha} + ia^{\dagger}e^{-\alpha}\right)\cdot \left(-... ...agger}e^{-\alpha}\right)\cdot\left(iae^{\alpha}-ia^{\dagger}e^{-\alpha}\right)$$

Thus we are really concerned with:

$$2\left(-iae^{\alpha} + ia^{\dagger}e^{-\alpha}\right)\cdot \left(... ...}-a^{\dagger}a^{\dagger}e^{-2\alpha} +2\left(aa^{\dagger} + a^{\dagger}a\right)$$

Only the latter term survives through the integration (due to the periodic boundary conditions), and we again conclude that:

$$H = \frac{1}{2}\sum_l \hbar \omega_l \left(a_l^{\dagger}a_l + a_la_l^{\dagger}\right)$$ (4.8)