I know this is the basis of LASER action but why does an electron fall to a lower vacant orbit when exposed to a photon of energy equal to the difference between the initial and final orbit? The concerned incident photon lacks the energy needed to excite the electron to a higher orbit. In other words, given that the electron cannot be excited to a higher level, why would a photon induce it to fall to a lower level and emit another photon? Although, understanding the basics of the stimulation process, I lack some understanding of the underlying physics that causes this to happen.
Answer
It is a characteristic of bosons that they like to be in the same state. So when a photon encounters an atom where the electron is in an excited state with an energy difference equal to that of the photon, the probability for that electron to fall down and radiate a photon is increased.
One can see how this works by looking at the creation and annihilation operators for bosons. To create a photon from the vacuum state we have $$ a^{\dagger} |0\rangle = |1\rangle . $$ Here $|1\rangle$ is a one-particle state. However, if there is already a photon and we want to create another we get $$ a^{\dagger} |1\rangle = |2\rangle \sqrt{2} , $$ where $|2\rangle$ is a two-particle (Fock) state. So one can see the process is enhanced by a factor of $\sqrt{2}$. This comes out of the requirement for the normalization of the state. In general we have $$ a^{\dagger} |n\rangle = |n+1\rangle \sqrt{n+1} , $$ where $|n\rangle$ is an $n$-particle Fock state. So the more particles we start with the higher the probability to create another one.
In other words, it is not so much that the photon makes the electron fall down to the lower level. It's more a case that the photon wants that energy to create another photon and for the sake of energy conservation the electron must fall to the lower level.
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