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.

  • $\begingroup$ Why do you say that the light cannot promote the system to a higher energy level? $\endgroup$
    – garyp
    Commented Sep 25, 2016 at 11:58
  • $\begingroup$ @garyp I think op means that if there is no state at 2 \hbar \omega, it will not populate a secondary excited state. $\endgroup$
    – anon01
    Commented Sep 25, 2016 at 12:17
  • $\begingroup$ Photons can couple states of +/- \hbar \omega; I don't know of a concise way to explain this better... $\endgroup$
    – anon01
    Commented Sep 25, 2016 at 12:21
  • $\begingroup$ This question is ambiguous, so no one will know how to answer. Please clarify what you mean by "Even if the photon cannot excite the electron to a higher orbit." $\endgroup$
    – garyp
    Commented Sep 25, 2016 at 16:59
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    $\begingroup$ Don't think of this in terms of particles. The photons doing the stimulation are an oscillating electric field that makes the electron want to oscillate with the wave and this causes the in-phase collapse (stimulated emission). $\endgroup$ Commented Sep 28, 2016 at 7:52

3 Answers 3


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.


You are essentially asking about the model/mathematics of this:

Stimulated Emission: If an electron is already in an excited state (an upper energy level, in contrast to its lowest possible level or "ground state"), then an incoming photon for which the quantum energy is equal to the energy difference between its present level and a lower level can "stimulate" a transition to that lower level, producing a second photon of the same energy.


The answer to a related question by Motl here may help, although it is one step further in the query, "why do many photons come out coherently out of an inverted population", because it discusses the second quantization frame.

In this frame there exists the photon field on which creation and annihilation operators are working on the ground state wave function of the photon.

Quantum mechanical states are described by wavefunctions. These are complex functions able to predict the probability density distribution for a particle to exist at (x,y,z,t) and equivalently with a given energy momentum vector.

I will handwave what I believe is happening at the individual atomic level, I hope somebody good at QED will correct me or better, give a satisfactory answer:

An atom, having a single electron on the excited state E2 has a probability of falling from E2 to E1 by emitting a photon of energy hnu=E_2-E_1. It is modeled by a complex wavefunction.

A photon has a wavefunction (formula 11 in the link) that contains the electric and magnetic fields of the classical wave which will emerge from a large number of such photons, as well as the frequency nu which characterizes the photon.

The correct wavefunction for the atom will have the electric field information that characterizes the atom and the future photon to be emitted with some probability.

The wavefunction of a photon overlapping in space time with the wavefunction of the atom are in a superposition, mathematically.

Here we have to be clear on superposition. Superposition of two wavefunctions is not interaction. A good example is the interference pattern observed from two laser beams overlapping in space. Even though photon photon interactions are very small interference patterns appear in laser beams.

This is due to the difference between quantum mechanical superposition and interaction. Superposition adds the complex wavefunctions of the two independent waves. Observation/interaction squares( Psi*Psi) the superposed wavefunction and interference appears due to the mathematics of complex numbers, in the probability density of observing the photons in the laser beam.

The excited atom has its wavefunction which contains the pattern of the specific frequency photon. In a first approximation, i.e. ignoring photon-atom scattering off the fringe field of the atom, the wavefunctions are superposed.

In superposition with an atom where the electron is in the excited state, the wavefunction of the photon and the wavefunction of the atom add linearly.

To find the probability of the electron falling to E1, the square (Psi*Psi) of the combined wavefunction has to be taken. This will give a very high probability for the electron to fall to the lower energy level if the incoming photon has the exact energy difference, and two photons leave the interaction area.

I expect if the mathematics is done correctly one will see that the interference appearing because of the squaring of the wavefunction will increase the probability for the atom to end in the lower E1 state. In a sense one is manipulating probabilities in the laser set up.


Let's assume that the system is lazing. So it is producing a massive number of photons that are in the same quantum state. If a photon interacts with a molecule in the excited state, a stimulated emission may occur and another photon will be release in the same quantum state. This leaves the molecule in the non-excited state. Now if a photon interacts with a molecule in the non-excited state it may be absorbed and the molecule would return to the higher state. The probability of both of these is the same. The goal in a laser it to make sure that there are more molecules in the excited state than there are in the non-excited state. In general this does not happen. But in a laser there is more going on. We want to create an inversion where there are more molecules in the excited state than there are in the non-excited state. There is some additional mechanism (perhaps an electric charge) that will raise molecules in the ground state to the higher state. Once you have more in the excited state then in the ground state there is more chance for a photon to interact with an excited state molecule. The trick is to keep the excitation methodology running to keep the inversion as lazing happens.

NOTE: This is simplified explanation in that lasers have several state for the molecules and the excitation method actually excites to an even higher state where there is a good probability that the spontaneous decay will get the molecule into the excited state for lazing.

EDIT: I did not fully answer the questions so I am adding more. For stimulated emission to happen the molecule must go from the excited energy state to non-excited state. Because of the structure of the energy states, this is a very specific energy. Additionally stimulated emission produces a photon in the exact same quantum state as the incident photon. So when the excited state molecule interacts with the incident photon, it goes to the non-excited state and releases the photon.

As to why this happens you need to look at the eigenstates of the electrons for the molecule. When the molecule interacts with the incident photon in either the excited or non-excited state, the Hamiltonian for the system changes. This causes a change in the eigenstates for the electrons and causes either the photon to be absorbed in the case of a non-excited state or a photon to be emitted if it is in the excited state.

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    $\begingroup$ This is a decent explanation of how a laser works, but I don't see how it addresses the question. $\endgroup$
    – garyp
    Commented Sep 25, 2016 at 13:50
  • $\begingroup$ @garyp You are correct. I have added to the answer. Thank you $\endgroup$
    – Michael
    Commented Sep 25, 2016 at 16:51
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    $\begingroup$ Version2 still does not answer the question and actually says some stuff that's incorrect. For example, "When the molecule interacts with the incident photon in either the excited or non-excited state, the Hamiltonian for the system changes" <-- That is not true. The Hamiltonian is the same regardless of the state of the system. $\endgroup$
    – DanielSank
    Commented Sep 25, 2016 at 17:00
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    $\begingroup$ The system itself is different. There are 2 Hamiltonians. The first without the incident photon. The second with the incident photon. The E&M fields of the incident photon causes a perturbation of the Hamiltonian and causes a shift in the solutions for the eigenstates. This can be solved either with perturbation theory or as a different Hamiltonian. Either way the eigenstates changes and thus the shift between the two state is allowed. $\endgroup$
    – Michael
    Commented Sep 26, 2016 at 15:10

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