We're familliar with talking about stimulated emission using energy and time domains (e.g. Wikipedia's Stimulated emission) but what about spatially?

My naive guess is that since the stimulating electric field of an incident plane wave is zero in the incident direction, the stimulated transition in the quantum system (e.g. an atom) will likewise produce zero electric field in that direction, so the radiated power at large distance will drop to zero along the plane perpendicular to it.

Does that turn out to be basically true for at least simple transitions (e.g. a hydrogen atom or a free exciton)?


2 Answers 2


Stimulated emission is in the same direction and has the same phase as the stimulating radiation. i.e. It has the same angular distribution as the incident radiation.

As the wikipedia page on stimulated emission correctly says

A transition from the higher to a lower energy state produces an additional photon with the same phase and direction as the incident photon; this is the process of stimulated emission.

Related question: Scattering vs Stimulated Emission

As for a deeper explanation of why this is the case: Bosons "want" to be in the same quantum state. Why is the photon emitted in the same direction as incoming radiation in Laser?

Some interesting discussion of time-reversal symmetry arguments are given here.

  • 1
    $\begingroup$ > "It has the same angular distribution as the incident radiation." -- only if many emitters are involved. Radiation of single emitter cannot be a plane wave. $\endgroup$ Commented Nov 1, 2020 at 12:08
  • $\begingroup$ @JánLalinský oh, so if there is a population of excited atoms in a volume then even if weakly stimulated (one photon at a time) there can be directionality in the radiated photons, yes this is helpful. $\endgroup$
    – uhoh
    Commented Nov 1, 2020 at 12:46
  • $\begingroup$ Thank you for the update and links! I think this will take some time for me to dig in as far as I can. Of course "They're bosons, so..." is anticlimactic to someone who keeps trying to understand rather than accept QM, but that seems to be (for me, at the moment) where this is going to end up. $\endgroup$
    – uhoh
    Commented Nov 1, 2020 at 13:46

the stimulated transition in the quantum system (e.g. an atom) will likewise produce zero electric field in that direction, so the radiated power at large distance will drop to zero along the plane perpendicular to it.

EDIT: I was wrong about the angular pattern of the stimulated emission, now I think only total radiation has this pattern. See below.

Yes, in case of dipole radiation, electric field component in the direction of wave propagation is zero in the radiation zone.

The simplest model of radiation from classical theory is that of oscillating charged particle (or oscillating dipole). The radiation goes in all directions from which the oscillation can be seen, the greater the projection of the acceleration vector seen, the greater the intensity of radiation. Mathematically, field strength of radiation varies as $\sin \theta$ in polar coordinates. This angular distribution is that of dipole radiator.

In quantum theory the radiation pattern depends on which transitions are involved in the interaction with EM field. The simplest cases are where the dominant contribution is that of "transition dipole moments" $\boldsymbol{\mu}_{ik} = \langle i|\sum_k q_k\mathbf r_k|k\rangle$. If only one such moment is involved (possible if the incident radiation is resonant with only one transition), the emitted radiation has the same dipole pattern as in classical theory, and has intensity given by the formula for spontaneous emission, independent of the incident radiation...

...except for intensity in direction of the incident wave, which does depend on intensity of incident radiation. Total intensity in the original direction is that of the spontaneous emission times a factor of $n+1$, where $n$ is number of photons in EM mode for this direction [1].

So total radiation emitted from the atom/molecule has dipole-like angular distribution, with a spike in the front. If we talk only about the stimulated emission part of that, this exists only in the original direction and is responsible for that spike.

[1] D. P. Craig, T. Thirunamachandran: Molecular Quantum Electrodynamics, formula 4.12.4., Academic Press (1984)

  • $\begingroup$ No, this isn't right , it describes resonant scattering. $\endgroup$
    – ProfRob
    Commented Nov 1, 2020 at 13:28
  • $\begingroup$ @RobJeffries what I describe is dipole transition theory, this applies to both absorption and emission. I think the dipole pattern may not be exactly right for the original direction of the incident field, as there is increased probability to emit in the original direction proportional to radiation intensity, but overall the dipole pattern should be correct for isolated system in incident radiation field. $\endgroup$ Commented Nov 1, 2020 at 21:49
  • $\begingroup$ It's not even remotely arguable. Stimulated emission photons are in phase and in the same direction as the original photon. Your answer also differs significantly from your answer to physics.stackexchange.com/questions/563487/… $\endgroup$
    – ProfRob
    Commented Nov 1, 2020 at 22:02
  • $\begingroup$ @RobJeffries perhaps you are using different definition of "stimulated emission". Some sources seem to define it only in the context of photon description as single mode process where one photon is added to the original mode. If you are using that definition, then you are trivially correct. I am interested in the actual physical process, as the OP is - does single excited atom stimulated by incident radiation radiate to different directions? I believe it does, the transition probability seems to have dipolar character, with additional enhancement in the original direction. $\endgroup$ Commented Nov 1, 2020 at 22:12
  • $\begingroup$ @RobJeffries my other answer addressed different question. I believe it is consistent with this one. $\endgroup$ Commented Nov 1, 2020 at 22:13

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