Fermi's Golden Rule and Selection Rules [Quantum Mechanics I Studied #40]

Previously we shot monochromatic light; now use weak, incoherent broadband light. Its random relative phases let us add the frequency contributions as probabilities after averaging. However, rather than building everything anew from scratch,

Monochromatic result: P_a→b(t) = P_b→a(t) = (|pE₀|²/ħ²) sin²((ω₀−ω)t/2)/(ω₀−ω)²

we’ll nicely Modify this thing we got from monochromatic light~~ Let’s get started~

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

Energy density: u = ½(ε₀E² + B²/μ₀), but E and B contribute equally to the energy, so u = ε₀E² = ε₀E₀² cos²ωt

For each ω, the average of u: ⟨u⟩ = ⟨ε₀E₀² cos²ωt⟩ = ε₀E₀²/2

(From Electromagnetic Waves (Electromagnetism I Studied #29))

I’m going to use this~~

The monochromatic result again: P_a→b(t) = P_b→a(t) = (|pE₀|²/ħ²) sin²((ω₀−ω)t/2)/(ω₀−ω)²

In this, we’re going to write the amplitude squared, $E_0^2$, using $\langle u\rangle$. Then

With E₀² = 2⟨u⟩/ε₀: P_a→b(t) = (2⟨u⟩/(ε₀ħ²))|p|² sin²((ω₀−ω)t/2)/(ω₀−ω)²

I have kept emphasizing “for each $\omega$” whenever I talk about $\langle u\rangle$. The continuum needs a spectral density: $\rho(\omega)\,d\omega$ is the energy per volume in a narrow frequency band. The picture’s ⟨u⟩ = ρ(ω) is shorthand; $\rho$ itself is energy density per unit angular frequency. Therefore

Therefore P_a→b(t) = (2ρ(ω)/(ε₀ħ²))|p|² sin²((ω₀−ω)t/2)/(ω₀−ω)²

Then P_a→b(t) turns out to mean ’the transition probability for one particular $\omega$’!!

But we shot every $\omega$??? That is, it seems like we really should be calling the sum of all these individual probabilities the Probability????!

We need to fix it like this. Take a time long enough to resolve the narrow resonance, with $\rho$ nearly constant across its width, but short enough that the transition probability remains small.

Adding up every frequency: P_a→b(t) = ∫P_a→b(t, ω)dω = (2|p|²/(ε₀ħ²))∫₀^∞ ρ(ω) sin²((ω₀−ω)t/2)/(ω₀−ω)² dω; as before, since it’ll absorb almost only ω ≅ ω₀ light, this is ≅ (2|p|²/(ε₀ħ²))ρ(ω₀)∫₀^∞ sin²((ω₀−ω)t/2)/(ω₀−ω)² dω

This is an integrable form. Substitution: x = (ω₀−ω)t/2, so −dω = (2/t)dx

With the substitution the integral is (t/2)∫ sin²x/x² dx over the whole x axis, and since ∫ sin²x/x² dx = π it equals πt/2. So P_a→b(t) ≅ π|p|²ρ(ω₀)t/(ε₀ħ²); differentiate once with respect to t and we can call it the Transition Rate, R_a→b ≅ π|p|²ρ(ω₀)/(ε₀ħ²)

The same step once more: P_a→b(t) ≅ (2|p|²/(ε₀ħ²))ρ(ω₀)∫₀^∞ sin²((ω₀−ω)t/2)/(ω₀−ω)² dω, since it’ll absorb almost only ω ≅ ω₀ light

This time, let’s say the light comes with its polarization vector pointing in every direction. And this light, holding polarization vectors in all directions, also comes in from all directions.

Sketch: an arrow p⃗, the atom’s dipole moment

On a particle with this kind of Polarization, I mean.

Sketch: wavy arrows of red, green and blue light coming in at the dipole p⃗ from every direction

It means we’re going to look at the situation where lights of all sorts of colors come in from every direction.

The same sketch with axes: p⃗ in the yz plane at angle θ from the z axis, and n̂ in the xy plane at angle φ from the x axis

When we calculate this, we’ll set up spherical coordinates like this and calculate. As for the calculation method: first, we’ll call the light’s polarization direction (the direction its electric field points) . Also,

Generalizing: p = q⟨ψ_a|z|ψ_b⟩ was what we had; p = q⟨ψ_a|r|ψ_b⟩ is the general form

and since the axis setting and variable setting have been done like that, we’re going to sum over all $\theta$ and all $\varphi$, that is, take the average. We’re going to find the average of |p⃗·n̂|² . Then let’s give it a try.

Components: p⃗ = p sinθ ĵ + p cosθ k̂ and n̂ = cosφ î + sinφ ĵ

(We said the vector $\vec{p}$ is stuck on the yz plane.)

Average over directions: p⃗·n̂ = p sinθ sinφ, and the average of |p⃗·n̂|² = (1/4π)∫p² sin²θ sin²φ (sinθ dθ dφ) = (|p|²/4π)∫sin³θ dθ ∫sin²φ dφ = |p|²/3

For light coming in from all directions,

For light from every direction: R_a→b ≅ (1/3)(π|p|²/(ε₀ħ²))ρ(ω₀)

We’ll stop here and draw conclusions.

1. Fermi’s Golden Rule

The conclusion we can draw by looking at ρ(ω₀) : only light with ω = ω₀ is selected in this long-time resonance approximation. At finite time, the peak has a finite width.

2. Selection Rule

The conclusion we can draw by looking at p = q⟨ψ_a|r|ψ_b⟩ : the transition or emission happens only when is non-Zero for an electric-dipole transition. These orbital $\Delta\ell$, $\Delta m$ rules apply to central-potential states; a dipole-forbidden line can still occur through weaker interactions, such as magnetic-dipole or electric-quadrupole coupling.

For p = q⟨ψ_a|r|ψ_b⟩ to be non-zero, Δℓ = ±1 and Δm = ±1, 0 must hold: the selection rule

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In the transition probability, the denominator is $(\omega_0 - \omega)^2$, not $(\omega_0 - \omega)$, in every picture that writes $\frac{\sin^2\left(\frac{\omega_0 - \omega}{2}t\right)}{\omega_0 - \omega}$ (the line after the substitution already uses the square, $\frac{4x^2}{t^2}$). And the energy density in a band $d\omega$ is $\langle u\rangle = \rho(\omega)\,d\omega$.
After the substitution $x = \frac{\omega_0 - \omega}{2}t$, the integral runs over $x$ from $-\infty$ to $\infty$ (for large $t$), and $\int \frac{\sin^2\left(\frac{\omega_0 - \omega}{2}t\right)}{(\omega_0 - \omega)^2}\,d\omega = \frac{\pi t}{2}$, not $\frac{\pi}{2}$; the final $P_{a\to b}(t) \simeq \frac{\pi|p|^2}{\varepsilon_0\hbar^2}\rho(\omega_0)\,t$ and the rate $R_{a\to b}$ are right.
In the average over polarization directions, the squares belong on $p$ and on the angular factors: $\frac{1}{4\pi}\int |p|^2\sin^2\theta\sin^2\phi\,\sin\theta\,d\theta\,d\phi = \frac{|p|^2}{3}$; the result $\frac{|p|^2}{3}$ is right.

1. Fermi’s Golden Rule

2. Selection Rule

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