# Atomic Physics: stimulated emission

I'm studying a chapter about atomic physics right now but there's thing I just don't seem to understand. When stimulated emission occurs, there's an incoming photon which stimulates the atom to go an energy level lower with the result that a new photon gets emitted.

Can someone explain the reason why the incoming photon stimulates the atom to emit a second one? To me it seemed more logic for the atom to absorb the photon and go an energy level up. There seems to be a phenomenon which I don't know about.

• You might want to have a look at my answer to this question as it covers related territory. – John Rennie Jan 27 '14 at 11:03

Lubos's answer is no doubt correct, but here's an 'intuitive' classical picture that I keep in mind: Energy levels in an atom are similar to those in a harmonic oscillator. So imagine a spring oscillating at its natural frequency with some amplitude. Then you come and excite it (push and pull it) at the same frequency. Will the spring always gain energy, i.e. absorb a 'photon' and thus increase in amplitude? Not always - it depends on the phase of the excitation. If you push and pull out of phase with the existing motion, then the oscillation is dampened and a 'quantum' of energy is removed. This is sort of like stimulated emission. You perturb an oscillator at its natural frequency, but in a way that extracts energy from it instead of adds energy to it.

I am not saying that this is a rigorous physics argument...just a tool to help give a classical 'feeling' of how something non-classical works.

• Thank you! This explanation gave me a clear view of what happens. I trust Lubos' answer but my physics knowledge is just not advanced enough to fully understand... I'm just a computer science student who chose this physics class because it interests me a lot :) – Confituur Jan 30 '14 at 0:02

The probability of the emission of a photon to a state where $N$ photons are already present is proportional to $(N+1)$. This fact follows from the symmetry of physics under the time reversal $(t\to -t)$. Why? Because the probability of absorption of a photon, i.e. process going from $N+1$ to $N$ photons in the given state (reduction), is clearly proportional to the initial number of photons which I called $N+1$. The more photons, the more likely the atom is to absorb. By time-reversal (or CPT) symmetry, the emission process has to exist as well and it has to depend on the existing number of photons that are already "out there" because the absorption process clearly depends on it, too.

The probability of emission is therefore proportional to $N+1$. The $N$ part is "stimulated" and the $1$ part is "spontaneous".

The time-reversal calculation above was already known to Einstein before quantum mechanics was found by Heisenberg et al. in 1925. A kosher quantum mechanical derivation of this factor $N+1$ in both probabilities comes from the squared matrix element in the harmonic oscillator $$|\langle n+1| a^\dagger|n\rangle|^2=n+1$$ or its Hermitian conjugate. It's because the operator responsible for the emission/absorption contains a creation/annihilation operator for the photon.

• I convinced myself of a similar argument, arguing from a symmetry point of view. What I'm not sure to get is why in a Laser emission dominates instead of absorption. – nabla Jan 17 '15 at 13:56
• Because the probability of emission is proportional to $N+1$ while the probability of absorption is $N$. Moreover, absorption needs the atoms to be in the unexcited state, for the energy $E_0+hf$ to be in the spectrum. So absorption appears when the atoms are unexcited, and emission when they're excited. Just to be sure, when the atoms are excited, $N+1$ goes to emission and $0$ goes to absorption - the difference is in no way small. – Luboš Motl Jan 18 '15 at 7:57
• @gerd: Maybe I am missing your point, but a laser requires an inversion of states, i.e. more atoms have to be in the upper than in the lower state. This is achieved by having a third state that is pumped and that populates the upper state from above, rather than from below. – CuriousOne Jun 21 '15 at 6:38
• Right, it's a technicality to achieve a given state - and the method isn't necessarily unique. The laser starts to operate when this state is achieved and the laser actually emits. – Luboš Motl Jun 21 '15 at 7:01

Another point to mention is that atoms typically are not harmonic oscillators (which have equally spaced energy levels.) For example, Hydrogen atoms' energy spacings go as $n^{-2}$, and other atoms are more complicated, but generally, the energy from the atomic level $|n\rangle$ to level $|n+1\rangle$ is different from the energy from level $|n+1\rangle$ to level $|n+2\rangle$. Given this, an excited atom can either emit a photon or do nothing.

Of course, the above comment about Einstein coefficients is totally correct regardless, but this is just to address your last point about why should an atom undergo stimulated emission rather than re-absorption.