Fermi's Golden Rule and Selection Rules

Last time we shot monochromatic light. Now — polychromatic. All the colors at once.

But we’re not starting from scratch. That would be painful. Instead, we’re going to take the thing we already got for monochromatic light —

$$P_{a\to b}(t) = P_{b\to a}(t) = \frac{|pE|^2}{\hbar^2}\frac{\sin^2\!\left(\frac{\omega_0-\omega}{2}\cdot t\right)}{\omega_0-\omega}$$

— and just modify it. Let’s go.

For each $\omega$, the energy density $u$ of an EM wave is

$$u = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right)$$

The $E$ and $B$ pieces contribute the same amount, so

$$u = \epsilon_0 E^2 = \epsilon_0 E_0^2\cos^2\omega t$$

Average it over a cycle:

$$\langle u \rangle = \langle\epsilon_0 E_0^2\cos^2\omega t\rangle = \frac{\epsilon_0 E_0^2}{2}$$

(I worked through all this back in the EM series — see EM #29 if you want the long version.)

OK, taking this result and jamming it in:

$$P_{a\to b}(t) = P_{b\to a}(t) = \frac{|pE|^2}{\hbar^2}\frac{\sin^2\!\left(\frac{\omega_0-\omega}{2}\cdot t\right)}{\omega_0-\omega}$$

Express $E^2$ in terms of $\langle u\rangle$:

$$= \frac{2\langle u\rangle}{\epsilon_0\hbar^2}|p|^2\frac{\sin^2\!\left(\frac{\omega_0-\omega}{2}\cdot t\right)}{\omega_0-\omega}$$

Notice I keep saying “for each $\omega$” whenever I write $\langle u\rangle$. That wasn’t just filler — it was me saying $\langle u\rangle$ is actually a function of $\omega$. So let me rename it:

$$\langle u \rangle = \rho(\omega)$$

Then

$$P_{a\to b}(t) = \frac{2\rho(\omega)}{\epsilon_0\hbar^2}|p|^2\frac{\sin^2\!\left(\frac{\omega_0-\omega}{2}\cdot t\right)}{\omega_0-\omega}$$

But wait. This $P_{a\to b}(t)$? That’s the transition probability for one specific $\omega$. And we just fired every $\omega$ at the atom. So the actual probability is the sum — er, integral — of all of these little per-$\omega$ probabilities. Right??

So we fix it:

$$P_{a\to b}(t) = \int P_{a\to b}(t,\,\omega)\,d\omega = \int_0^\infty \frac{2\rho(\omega)}{\epsilon_0\hbar^2}|p|^2\frac{\sin^2\!\left(\frac{\omega_0-\omega}{2}\cdot t\right)}{\omega_0-\omega}d\omega = \frac{2|p|^2}{\epsilon_0\hbar^2}\int_0^\infty \rho(\omega)\frac{\sin^2\!\left(\frac{\omega_0-\omega}{2}\cdot t\right)}{\omega_0-\omega}d\omega$$

Same trick as before — basically all the absorption happens around $\omega \cong \omega_0$, so pull $\rho(\omega_0)$ out:

$$\cong \frac{2|p|^2}{\epsilon_0\hbar^2}\rho(\omega_0)\int_0^\infty \frac{\sin^2\!\left(\frac{\omega_0-\omega}{2}\cdot t\right)}{\omega_0-\omega}d\omega$$

And this — this we can actually integrate.

Substitution:

$$\frac{\omega_0-\omega}{2}t = x \qquad -d\omega = \frac{2}{t}dx$$$$P_{a\to b}(t) \cong \frac{2|p|^2}{\epsilon_0\hbar^2}\rho(\omega_0)\int_0^\infty \frac{\sin^2 x}{\frac{4}{t^2}x^2}\left(-\frac{2}{t}dx\right)$$

And since $\displaystyle\int_{-\infty}^{\infty}\frac{\sin^2 x}{x^2}\,dx = \pi$, the integral we need comes out to $\dfrac{\pi}{2}$.

Which means:

$$P_{a\to b}(t) \cong \frac{\pi|p|^2}{\epsilon_0\hbar^2}\rho(\omega_0)\,t$$

Now look at that — probability grows linearly in $t$. So the rate of change is just a constant. Differentiate once with respect to $t$ and boom — that’s the Transition Rate:

$$R_{a\to b}(t) \cong \frac{\pi|p|^2}{\epsilon_0\hbar^2}\rho(\omega_0)$$

Light from every direction, every polarization

OK so far we’ve been a little cheap. Let’s make the light really realistic: polarization vectors pointing every which way, and the light coming in from every direction too.

Hitting a particle with some polarization of its own, like this:

So basically: every color, from every direction, polarized every way. The whole buffet.

Set up spherical coordinates and go.

Let the direction of travel of the light be $\hat{n}$.

Also, we had

$$p = q\langle\psi_a|z|\psi_b\rangle$$

but now that we’re not locking things to the $z$-axis, let’s generalize:

$$p = q\langle\psi_a|r|\psi_b\rangle$$

With the axes and variables set up this way, we’re going to integrate over all $\theta$ and all $\varphi$ — i.e. take the average of

$$|\vec{p}\cdot\hat{n}|^2$$

over all directions. Let’s do it.

$$\vec{p} = p\sin\theta\,\hat{j} + p\cos\theta\,\hat{k}$$$$\hat{n} = \cos\varphi\,\hat{i} + \sin\varphi\,\hat{j}$$

(Remember, $\vec p$ sits in the $yz$-plane.)

$$\vec{p}\cdot\hat{n} = p\sin\theta\sin\varphi$$$$\langle|\vec{p}\cdot\hat{n}|^2\rangle = \frac{1}{4\pi}\int p^2\sin^2\theta\sin^2\varphi\cdot(\sin\theta\,d\theta\,d\varphi) = \frac{|p|^2}{4\pi}\int\sin^3\theta\,d\theta\int\sin^2\varphi\,d\varphi = \frac{|p|^2}{3}$$

So for light coming from every direction, stick that $1/3$ in:

$$R_{a\to b}(t) \cong \frac{1}{3}\frac{\pi|p|^2}{\epsilon_0\hbar^2}\rho(\omega_0)$$

Two conclusions

1. Fermi’s Golden Rule.

Look at the $\rho(\omega_0)$ sitting in there. What it’s telling us: only the frequency component at exactly $\omega = \omega_0$ gets absorbed. Transitions (and emissions) happen only at that frequency. Nothing else gets a word in.

2. Selection Rule.

Now look at

$$p = q\langle\psi_a|r|\psi_b\rangle$$

If $p = 0$, the transition rate is zero. No transition, no emission, nothing. So transitions only happen when that matrix element is non-zero.

And for $\langle\psi_a|r|\psi_b\rangle$ to be non-zero, you need

$$\Delta\ell = \pm 1 \quad \text{and} \quad \Delta m = \pm 1,\, 0$$

That’s the Selection Rule.

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