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I have a question regarding the dependence on energy density of the probability rate of emission/absorption while studying lasers in particular. It is given that,the probability rate of absorption as well as stimulated emission depends on the energy density as well as the no. of atoms on either levels but the probability rate of spontaneous emission depend only on the no of atoms and not on the energy density.Why is it so?

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Absorption and emission (of any kind) of light by an "atom" (I'll use this word for all discrete absorbers, whether elemental atoms or molecules) is a manifestation of one phenomenon alone: the coupling of the atom's electrons with the second quantized electromagnetic field. The EM field is thought of as being made up of modes and each of these modes is a harmonic oscillator which is replaced by the simple quantum harmonic oscillator in the simplest, nonrelativistic quantum model of the EM field.

So if an atom is in an excited state, its spontaneous emission is set by the linear coupling co-efficients with each of the electromagnetic field mode oscillators, and these coupling co-efficients are the same no matter how many photons each mode oscillator has in it. From the outside, the quantum harmonic oscillator looks the same, whether it is absorbing a photon to rise from the ground state to the one photon state, or whether it is absorbing a photon to switch from the 100 to 101 photon state.

It might be, however, that high levels of light in the cavity will excite the atom again very soon after it emits a photon.

Now for (stimulated) absorption. The probability of this absorption is actually proportional to the population density of ground state atoms. So a gain (or fluorescent) medium can become saturated: if the atoms are truly two-level atoms, then they cannot absorb a further photon whilst not in their ground state. Each photon propagates according to Maxwell's equations, so it is equally likely to be absorbed on any contour of equal intensity of a laser cavity mode, for example. So, if photon propagation is confined to a region of $A\mathrm{m^2}$ and the intensity of the propagating mode is roughly constant over this cross section, the probability of absorption for each photon is roughly the proportion of area "shrouded" by the absorbing atoms. So, if a slice of our cavity is $\delta z$ long, there are $\rho$ atoms per cubic metre and each "atom" has a cross-section $\sigma\mathrm{\,m^2}$ (effective target size) then the photon's probability of absorption is simply:

$$\frac{\mathrm{Number\,of\,absorbers\,in\,cross\,section} \times \sigma}{A} = \frac{\rho\,A\,\sigma\,(1-p)\,\delta z}{A} = \rho\,\sigma\,(1-p)\,\delta z$$

where $p$ is the proportion of atoms that are already excited. Since this is the probability per photon, the overall rate of absorption is:

$$\frac{I\,A}{h\,\nu}\,\rho\,\sigma\,(1-p)\,\delta z$$

since the photon flux per unit time through the area is $\frac{I\,A}{h\,\nu}$.

This explains the dependence of absorption as being proportional to the intensity of throughgoing light.

The reasoning for stimulated emission is like absorption: there is an effective cross section $\sigma^\prime$ for stimulated emission too, and it equals $\sigma$, by the Einstein relations (that the Einstein co-efficients $B_{1,2}$ and $B_{2,1}$ are equal).

However, only atoms in their raised state can undergo stimulated emission, so we replace $1-p$ by $p$ in the above expressions; the rate of stimulated emission is then:

$$\frac{I\,A}{h\,\nu}\,\rho\,\sigma\,p\,\delta z$$

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  • $\begingroup$ Fantastic answer! Maybe you have a reference of equal clarity to a book or a review paper? $\endgroup$ – ZeroTheHero Dec 19 '19 at 14:00

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