4.3 Emission and absorption of radiation

Consider an oscillating electric field $ℰ\epsilon cos\left(\omega t\right)$. (The unit vector $\epsilon$ indicates the direction of the field.) This corresponds to a perturbation

which has the form we considered when deriving Fermi’s golden rule.

We note that a field of this form is a long-wavelength-approximation to the electric field of an electromagnetic wave. Electric effects are much stronger than magnetic in this limit, so we are justified in ignoring the magnetic field in writing ${\hat{H}}^{\left(1\right)}$. This is called the electric dipole approximation because ${\hat{H}}^{\left(1\right)}\propto er\cdot \epsilon =d\cdot \epsilon$ where $d$ is the electric dipole moment (we don’t use $p$ for obvious reasons!). (Most textbooks more correctly start from the vector potential $A$ to derive the same perturbing Hamiltonian, see section 4.6. We follow Mandl here.)

To avoid unphysically sharp resonance as discussed above we actually need to consider a field with a range of frequencies, each being incoherent (so we add probabilities not amplitudes). The energy per unit volume in a frequency range $\omega \to \omega +\text{d}\omega$ is denoted $\rho \left(\omega \right)\text{d}\omega$, and $\frac{1}{2}{\epsilon }_0{ℰ}^2=\rho \left(\omega \right)$. (This expression allows for a factor of 2 from including the magnetic field energy, but also a factor of $\frac{1}{2}$ from the time average $⟨{cos}^2\omega t⟩$). Then we have

where the result applies equally to absorption (with $E_f>E_i$) or emission (with $E_f<E_i$). We note that as promised we now have a well-behaved result with no infinities around!

This is as far as we can go without being more specific about the system, except for one thing. The symmetry between emission and absorption is rather odd. Of course absorption needs a field to supply energy—or in quantum terms, to absorb a photon—a source of photons is needed. But why should the same be true for emission? What we have calculated here is stimulated emission, and it certainly does occur; it is behind the workings of a laser. Though our calculation is classical, stimulated emission can be thought of as due to the bosonic nature of photons - they “like to be in the same state”, so emission is more likely the more photons of the right energy are already present. But surely an excited atom in a vacuum at zero temperature can still decay! There must be spontaneous emission too, and the classical calculation can’t predict it.

There is however a very clever argument due to Einstein which connects the rates for stimulated and spontaneous emission, which says that if we regard the the energy density $\rho \left(\omega \right)$ as proportional to the number of photons in a mode of this frequency, $n\left(\omega \right)$, then to allow for spontaneous emission we simply need to replace $n\left(\omega \right)$ with $n\left(\omega \right)+1$. (Since classical fields correspond to huge numbers of photons, the difference between $n$ and $n+1$ is utterly negligible.)

• (Gasiorowicz ch 15.3)
• Mandl ch 9.5
• (Shankar ch 18.5)
• (Townsend ch 14.7)