Spontaneous Emission and Einstein Coefficients

Remember how I said spontaneous emission happens because of energy that’s hanging around even at zero temperature?

So — who’s the scientist who actually pinned it all down with equations?

Ugh…

…Einstein. Of course.

OK, here we go.

Einstein A & B coefficients

Picture a box stuffed with atoms, each one a two-level system.

Inside the box, call the number of atoms in state $\psi_a$ → $N_a$, and the number in state $\psi_b$ → $N_b$.

Now about the labels in that figure, including $A$:

$$A \;:\; \text{spontaneous emission rate}\\ B_{ab}\rho(w_0) \;:\; \text{absorption rate}\\ B_{ba}\rho(w_0) \;:\; \text{stimulated emission rate}$$

Obviously — absorption and stimulated emission should scale with the energy density of the incoming light. And specifically the energy density right at $w_0$. So yeah. Proportional. Makes sense.

Write out the rate of change of $N_b$:

$$\frac{dN_b}{dt} \;=\; -N_b A \;-\; N_b B_{ba}\rho(w_0) \;+\; N_a B_{ab}\rho(w_0)$$

Now here’s the thing — we don’t actually care about the instantaneous rate of change. Can’t measure it, and honestly, what would it even mean?

What we care about is equilibrium, as $t \to \infty$. At that point,

the rate of change is $0$!!!!

Because that’s literally what equilibrium means!!!!!

Which is to say $N_b$ is sitting constant. (We’ve hit thermal equilibrium at temperature $T$.)

So we set it to zero and solve:

$$\frac{dN_b}{dt} = 0 \;\Rightarrow\; \rho(w_0) \;=\; \frac{A}{(N_a/N_b) B_{ab} \;-\; B_{ba}}$$

And — inside our box, every atom’s energy is either $E_a$ or $E_b$, right????

In a system at temperature $T$, the populations at $E_a$ and $E_b$ follow the Boltzmann distribution!!!!

(Check my old thermo / stat-mech notes if this is fuzzy.)

So:

$$N_a \;=\; (\text{total number}) \times e^{-E_a / k_B T}\\ N_b \;=\; (\text{total number}) \times e^{-E_b / k_B T}$$$$\begin{aligned} \frac{N_a}{N_b} &\;=\; \frac{e^{-E_a / k_B T}}{e^{-E_b / k_B T}} \\ &\;=\; e^{\frac{1}{k_B T}(E_b - E_a)} \\ &\;=\; e^{\frac{\hbar}{k_B T} \cdot \frac{E_b - E_a}{\hbar}} \\ &\;=\; e^{\frac{\hbar w_0}{k_B T}} \end{aligned}$$

Plug that into

$$\rho(w_0) \;=\; \frac{A}{(N_a/N_b) B_{ab} \;-\; B_{ba}}$$

and we get

$$\rho(w_0) \;=\; \frac{A}{\left(e^{\hbar w_0 / k_B T}\right) B_{ab} \;-\; B_{ba}}$$

And now — the move. We connect this to Planck’s blackbody radiation formula.

Quick refresher on blackbody radiation. (Also see my old post: http://gdpresent.blog.me/220460671337.)

My Modern Physics Studies #7. Rayleigh-Jeans Formula.

Short version: heat something up, light comes out, right????

Study that light, and:

…that’s what it looks like.

Before Planck, Rayleigh-Jeans had pretty much nailed the pre-visible (long-wavelength) region, but it totally choked on the visible part. Then big-brother Planck swoops in and modifies the formula:

$$\rho(w) \;=\; \frac{\hbar w^3}{\pi^2 c^3 \left(e^{\hbar w / k_B T} \;-\; 1\right)}$$

(And thanks to this little move, the curtain rose on quantum mechanics. No big deal.)

So — we can compare! A box of two-level atoms at thermal equilibrium is emitting light, and the energy density at $w_0$ is what we just wrote using the Einstein coefficients. Planck tells us what that energy density actually is, from experiment. Set them equal!!!!

$$\frac{\hbar w^3}{\pi^2 c^3 \left(e^{\hbar w / k_B T} \;-\; 1\right)} \;=\; \frac{A}{\left(e^{\hbar w / k_B T}\right) B_{ab} \;-\; B_{ba}}$$

Match it up, and —

$$B_{ab} \;=\; B_{ba}$$

Which is just

$$P_{a \to b} \;=\; P_{b \to a}$$

…which we already proved!!! OK yeah, maybe obvious in retrospect, heh.

And for $A$:

$$A \;=\; \frac{\hbar w^3}{\pi^2 c^3} B_{ab}$$

We also derived earlier that

$$P_{b \to a} \;=\; \frac{\pi}{3 \epsilon_0 \hbar^2} |p|^2 \rho(w_0)$$

right?? So

$$B_{ba} \;=\; \frac{\pi |p|^2}{3 \epsilon_0 \hbar^2}$$

and therefore

$$A \;=\; \frac{\hbar w^3}{\pi^2 c^3} \cdot \frac{\pi |p|^2}{3 \epsilon_0 \hbar^2} \;=\; \frac{w^3}{3 \pi \epsilon_0 \hbar c^3} |p|^2$$

(What is this, a full cameo roll call of every universal physics constant lol)

And that’s it — we can now actually compute the spontaneous emission rate $A$.

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Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.