Re: Why is photon emitted during stimulated emission in phase with incoming photon?

Hi Abdullah,

That's a very interesting question, and probably the hardest I've been asked to answer until now. Let's start with the easiest question: Stimulated emission can be explained theoretically. In fact, Einstein theoretically predicted stimulated emission in 1917¹, while it was experimentally verified in 1928 by Ladenburg and Kopferman². The theoretical explanation by Einstein was made without the detailed knowledge of quantum mechanics we have nowadays, which is needed for a formal description.

So, let's take a brief look at the theoretical explanation, and see if we can understand why the photons must be in phase. First, we'll take a look at absorption. When a photon, or rather, its electric field, interacts with the potential that exists in an atom, there is a chance the energy in the photon is absorbed by the atom, and used to put an electron in a higher orbit. This chance is strongly dependent on how well the energy in the photon "matches" an existing orbit. If there is a poor match, the chance is negligible, but if the match is good, the chance is much higher. Anyway, the photon/atom interaction might end in a photon and a atom, or in an excited atom, both with a finite chance.

Now, let's hit this system with another photon that matches the transition we have just excited. We have the same photon electric field, and the same transition, only this time, it goes from up to down. The equations describing this problem are symmetrical with respect to the direction of the transition³, and we again might end up with an the old photon, an extra photon and an atom, or the old photon and the excited atom. The local electric field induced by the new photon should, because of the symmetry in the problem, match the field of the old photon. Furthermore, the energies of the photons should both match the energy of the transition in the atom. Because both the field and the total photon energy should match locally, the photons have to be in phase.

I hope this simplified and qualitative answer satisfies your curiosity. A more thorough explanation would involve some serious mathematics, in particular perturbation theory. There is a good description in chapter 9 of reference 3. Alternatively, you might look up reference 4 online.

Regards,

Bart Broks

[1] A. Einstein, Phys. Z. 18, 121 (1917)
[2] Kopferman and Ladenburg, Zeits. f. Physik 48, 26 (1928)
[3] David J. Griffiths, Intoduction to Quantum Mechanics, Prentice-Hall 1995
[4] http://farside.ph.utexas.edu/teaching/qm/perturbation/node17.html