Consider any two internal states, s₁ and s₂, of an atom. Let s₂ be the higher-energy state, so that $E\left(s_2\right)-E\left(s_1\right)=ϵ$ for some positive constant. If the atom is currently in state s₂, then there is a certain probability per unit time for it to spontaneously decay down to state s₁, emitting a photon with energy e. This probability per unit time is called the Einstein A coefficient:

A = probability of spontaneous decay per unit time.

On the other hand, if the atom is currently in state s1 and we shine light on it with frequency $f=ϵ/h$, then there is a chance that it will absorb photon, jumping into state s₂. The probability for this to occur is proportional not only to the amount of time elapsed but also to the intensity of the light, or more precisely, the energy density of the light per unit frequency, u(f). (This is the function which, when integrated over any frequency interval, gives the energy per unit volume within that frequency interval. For our atomic transition, all that matters is the value of $u\left(f\right)\text{at}f=ϵ/h$) The probability of absorbing a photon, per unit time per unit intensity, is called the Einstein B coefficient:

$B=\frac{\text{probability of absorption per unit time}}{u\left(f\right)}$

Finally, it is also possible for the atom to make a stimulated transition from s₂down to s₁, again with a probability that is proportional to the intensity of light at frequency f. (Stimulated emission is the fundamental mechanism of the laser: Light Amplification by Stimulated Emission of Radiation.) Thus we define a third coefficient, B, that is analogous to B:

$B^{\text{'}}=\frac{\text{probability of stimulated emission per unit time}}{u\left(f\right)}$

As Einstein showed in 1917, knowing any one of these three coefficients is as good as knowing them all.

(a) Imagine a collection of many of these atoms, such that N₁ of them are in state s₁ and N₂ are in state s₂. Write down a formula for $d{N}_1/dt$ in terms of A, B, B', N₁, N₂, and u(f).

(b) Einstein's trick is to imagine that these atoms are bathed in thermal radiation, so that u(f) is the Planck spectral function. At equilibrium, N₁and N₂ should be constant in time, with their ratio given by a simple Boltzmann factor. Show, then, that the coefficients must be related by

$B^{\text{'}}=B\text{and}\frac{A}{B}=\frac{8\pi h{f}^3}{c^3}$

Consider any two internal states, s₁ and s₂, of an atom. Let s₂ be the higher-energy state, so that $E\left(s_2\right)-E\left(s_1\right)=ϵ$for some positive constant. If the atom is currently in state s₂, then there is a certain probability per unit time for it to spontaneously decay down to state s₁, emitting a photon with energy e. This probability per unit time is called the Einstein A coefficient:

(a) Consider N₁ and N₂ atoms in state s₁ and s₂, respectively. The probability of atoms decaying to state s1 equals the coefficient A, and the rate of atoms per unit time equals the number of atoms in this state N₂ multiplied by the constant A, resulting in:

$A{N}_2$

Second, when we shine a light with energy e, there is a chance that the atoms in state s₁ will absorb the photons and jump to state s₂. The probability of this action occurring equals the constant B multiplied by the energy density of the beam u(f), so the rate per unit time of the atoms jumping to the second state equals the negative sign of the probability multiplied by the number of the first state, that is:

$-N_{1}Bu\left(f\right)$

Finally, we have stimulated emission from state s₂ to state s₁, which has a probability of the constant B' multiplied by the energy density of the beam u(f), so the rate per unit time of atoms jumped to the second state equals the probability multiplied by the number of atoms in the first state, i.e.

$N_2{B}^{\text{'}}u\left(f\right)$

The total rate $d{N}_1/dt$equals to the sum of these rates:

role="math" localid="1647754521419" $\frac{d{N}_1}{dt}=A{N}_2+N_2{B}^{\text{'}}u\left(f\right)-N_{1}Bu\left(f\right)\left(1\right)$

(b) Because the number of atoms N1 and N2 in the equilibrium is independent of time, therefore

$$\begin{gathered} \frac{d{N}_1}{dt}=0 \\ A{N}_2+N_2{B}^{\text{'}}u\left(f\right)-N_{1}Bu\left(f\right)=0 \\ A{N}_2+\left(N_2{B}^{\text{'}}-N_{1}B\right)u\left(f\right)=0\left(2\right) \end{gathered}$$

the ratio of the number of atoms can be written in terms of Boltzmann factor, as:

$$\begin{gathered} \frac{N_2}{N_1}=\frac{e^{-E\left(s_2\right)/kT}}{e^{-E\left(s_1\right)/kT}} \\ \frac{N_2}{N_1}=e^{-\left(E\left(s_2\right)-E\left(s_1\right)\right)/kT} \end{gathered}$$

But the energy difference between the levels is $ϵ=hf$, therefore

$\frac{N_2}{N_1}=e^{-hf/kT}\left(3\right)$

Equation 7.83 is :

$\frac{U}{V}=\frac{8\pi }{(hc{)}^3}{\int }_0^{\infty }\frac{{ϵ}^3}{e^{ϵ/kT}-1}dϵ$

Let $ϵ=hf\to dϵ=hdf$, then

$$\begin{gathered} \frac{U}{V}=\frac{8\pi }{(hc{)}^3}{\int }_0^{\infty }\frac{(hf{)}^3}{e^{hf/kT}-1}hdf \\ \frac{U}{V}=\frac{8\pi h}{c^3}{\int }_0^{\infty }\frac{f^3}{e^{hf/kT}-1}df \\ u\left(f\right)=\frac{8\pi h}{c^3}\frac{f^3}{e^{hf/kT}-1} \end{gathered}$$

Substitute this into (2)

$A{N}_2+\left(N_2{B}^{\text{'}}-N_{1}B\right)\frac{8\pi h}{c^3}\frac{f^3}{e^{hf/kT}-1}=0$

Divide both sides by N₁and substitute from (3)

$A{e}^{-hf/kT}+\left(B^{\text{'}}{e}^{-hf/kT}-B\right)\frac{8\pi h}{c^3}\frac{f^3}{e^{hf/kT}-1}=0$

Multiply both sides by $e^{hf/kT}$

$$\begin{gathered} A+\left(B^{\text{'}}-B{e}^{hf/kT}\right)\frac{8\pi h}{c^3}\frac{f^3}{e^{hf/kT}-1}=0 \\ A=\left(B{e}^{hf/kT}-B^{\text{'}}\right)\frac{8\pi h}{c^3}\frac{f^3}{e^{hf/kT}-1} \end{gathered}$$

The constant A is temperature independent, as are the constants B and B'. This indicates that the RHS must be temperature independent, thus the temperature portion must cancel out. This only happens when B = B', so we get:

$$\begin{gathered} A=B\left(e^{hf/kT}-1\right)\frac{8\pi h}{c^3}\frac{f^3}{e^{hf/kT}-1} \\ A=B\frac{8\pi h{f}^3}{c^3} \end{gathered}$$