91影视

The transition rate for stimulated emission is given by:

$R_{a鈫�b}=\frac{\overline{)蟺}{\overline{)\mathfrak{H}}}^2}{3{慰}_0{魔}^2}p\left({蝇}_0\right)$ 鈥︹�︹��.. (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 $P\left({蝇}_0\right)$ as:

$P\left({蝇}_0\right)=\frac{1}{e^{h{蝇}_{0}I{K}_{a}T}-1}\frac{魔{蝇}_0^3}{{蟺}^2{C}^3}$ 鈥︹��.. (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:

$d_x=\frac{k^{2}v}{{蟺}^2}dk$

Using k=w/c , so we get:

$d_k=\frac{{蝇}^{2}V}{{蟺}^2{c}^2}d蝇$

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

$N_{蝇}=D_k=\frac{{蝇}^{2}V}{{蟺}^2{C}^2}d蝇$

Instead of:

$N_{蝇}=\frac{d_k}{e^{魔蝇}{B}^T-1}$

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

$P_0\left(蝇\right)=\frac{魔{蝇}^3}{{蟺}^2{C}^3}$

(b)

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

$$\begin{gathered} R_{b鈫�a}\overline{)\frac{蟺}{3o_0{h}^2}}{\overline{)\mathfrak{H}}}^2\frac{褣{蝇}^3}{{蟺}^2{c}^3} \\ {R}_{b鈫�a}=\frac{\overline{){蝇}^3}{\overline{)\mathfrak{H}}}^2}{3蟺o_0魔c^3} \end{gathered}$$

Hence, the solutions are,

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

(b) $R_{b鈫�a}=\frac{\overline{){蝇}^3}{\overline{)\mathfrak{H}}}^2}{3蟺o_{0}h{c}^3}$