You could derive the spontaneous emission rate (Equation 11.63) without the detour through Einstein’s A and B coefficients if you knew the ground state energy density of the electromagnetic field ${\mathit{P}}_{\mathbf{0}}\left(\omega \right)$for then it would simply be a case of stimulated emission (Equation 11.54). To do this honestly would require quantum electrodynamics, but if you are prepared to believe that the ground state consists of one photon in each classical mode, then the derivation is very simple:

(a) Replace Equation $\mathbf{5}\mathbf{.}\mathbf{11}1$by localid="1658381580036" ${\mathit{N}}_{\mathbf{0}}\mathbf{=}{\mathit{d}}_{\mathbf{k}}$and deduce ${\mathit{P}}_{\mathbf{0}}\left(\omega \right)$ (Presumably this formula breaks down at high frequency, else the total "vacuum energy" would be infinite ... but that's a story for a different day.)

(b) Use your result, together with Equation 9.47, to obtain the spontaneous emission rate. Compare Equation 9.56.

The transition rate for stimulated emission is given by:

$R_{a\to b}=\frac{\overline{)\pi }{\overline{)\mathfrak{H}}}^2}{3o_0{ħ}^2}p\left({\omega }_0\right)$ ……….. (1)

Where, $\mathfrak{H}$ is the dipole moment averaged between the two states a and bwhich is given by:

If the stimulating radiation occurs due to the thermal processes, we can write $P\left({\omega }_0\right)$ as:

$P\left({\omega }_0\right)=\frac{1}{e^{h{\omega }_{0}I{K}_{a}T}-1}\frac{ħ{\omega }_0^3}{{\pi }^2{C}^3}$ …….. (2)

(a)

But for the zero-point field this formula doesn't hold, so we need another formula. Assume that the number of photons per available state is 1, then the number of states in the shell from wave number k to k+dk is:

$d_x=\frac{k^{2}v}{{\pi }^2}dk$

Using k=w/c , so we get:

$d_k=\frac{{\omega }^{2}V}{{\pi }^2{c}^2}d\omega$

Here, V is the volume of the box in which we put the photon, then the number of photons is:

$N_{\omega }=D_k=\frac{{\omega }^{2}V}{{\pi }^2{C}^2}d\omega$

Instead of:

$N_{\omega }=\frac{d_k}{e^{ħ\omega }{B}^T-1}$

So simply remove the factor $\left(e^{ħ\omega I{k}_{a}T}-1\right)$ in the denominator of equation (2), so we get:

$P_0\left(\omega \right)=\frac{ħ{\omega }^3}{{\pi }^2{C}^3}$

(b)

Substitute into (1) with $P_0\left(\omega \right)$so we get:

$$\begin{gathered} R_{b\to a}\overline{)\frac{\pi }{3o_0{h}^2}}{\overline{)\mathfrak{H}}}^2\frac{ђ{\omega }^3}{{\pi }^2{c}^3} \\ {R}_{b\to a}=\frac{\overline{){\omega }^3}{\overline{)\mathfrak{H}}}^2}{3\pi o_0ħc^3} \end{gathered}$$

Hence, the solutions are,

(a) $P_0\left(\omega \right)=\frac{h{\omega }^2}{{\pi }^2{c}^3}$

(b) $R_{b\to a}=\frac{\overline{){\omega }^3}{\overline{)\mathfrak{H}}}^2}{3\pi o_{0}h{c}^3}$