How does stimulated emission work, exactly?

I am currently in my first year of college, and I'm having a hard time understanding how stimulated emission works. I was told by my professor that an excited electron jumps back to the ground state upon interaction with a photon, in other words, the photon stimulated the electron's de-excitation. Wikipedia gives the same thing.

I just don't understand one thing, if photons carry energy, shouldn't the electron be forced to higher excited states? What makes it drop even lower? Its almost like a "I pass a heat ray through water, and the water freezes" kind of analogy. Can anyone point out how stimulated emission happens, and why an electron chooses to drop rather than raise to higher levels upon photon absorption.

• Before the photon comes along, the atom is in a metastable excited state - which means that it's not in the ground state, but it's also not going to immediately decay. The incident photon "knocks" the electron down to the ground state, which releases an additional photon. This is an extremely rough analogy, but at this level it's hard to be more precise than that. Aug 28, 2017 at 8:28

If you consider an electron in an isolated atom then we get a set of states for that electron, these states being the eigenstates of the Hamiltonian. These eigenstates are time independent so excited states are stable and never decay, which seems odd at first glance because of course excited states do decay.

The reason for this is that an EM wave has an oscillating electric field and you have to include the electric field in the Hamiltonian. When you do this the eigenstates of the isolated atom are no longer eigenstates of the atom + EM wave so the system will evolve with time. So if we start with a hydrogen atom in the ground $1s$ state and shine light with the right energy on it the original $1s$ state evolves with time into a superposition of $1s$ and $2p$ states. This superposition may collapse into a $2p$ state, i.e. the atom absorbed the photon and moved to an excited state, or it may collapse to a $1s$ state, i.e. the photon wasn't absorbed and nothing changed.

Now suppose we start in the excited $2p$ state and shine a light on it. Exactly the same argument applies. The light changes the Hamiltonian and the original $2p$ state starts evolving with time into a superposition, and that superposition ultimately collapses into one of various possible final states, one of which is the ground state i.e. we get stimulated emission. The other possibilities are that nothing happens or that light excites the atom to an even higher energy state (for the specific case of the hydrogen $1s\rightarrow 2p$ transition the $10.2$eV light would ionise the $2p$ state).

The probabilities of the various outcomes can be calculated using perturbation theory and specifically Fermi's golden rule.

• I'm sorry, but I don't understand what Hamiltonian and Eigenstates mean. Aug 28, 2017 at 7:39
• @PrittBalagopal: maybe we should discuss this in the chat room Aug 28, 2017 at 7:41
• Very specific, yet, I think it misses the point of the question. Jul 10, 2020 at 6:50

I just don't understand one thing, if photons carry energy, shouldn't the electron be forced to higher excited states?

This typically does not happen because the photon usually considered when discussing spontaneous emission is not on resonance with any of the higher excited states.

But you are correct that what you describe can happen.

Consider the following diagram:

A photon can make the state transition from $$|1\rangle \rightarrow |2\rangle$$, if the photon is the right energy and the atom is in state $$|1\rangle$$. But they key is that the photon needs to match the energy level difference between 2 and 1. Likewise, the same thing can happen if the atom is in $$|2\rangle$$ initially and the photon has an energy level difference equal to the energy level difference of the excited states 2 and 3.

So, we can get what you described if we sent a photon with energy $$\omega = \omega_3 - \omega_2$$ when the state is $$|2\rangle$$, then it will transition to $$|3\rangle$$.

But what if we sent a photon with energy $$\omega = \omega_2 - \omega_1$$ when the state is $$|2\rangle$$? Now it can't get excited again to the excited state $$|3\rangle$$ because its too far away in frequency. This is the case that causes stimulated emission. When light is sent on-resonance (same energy) with a transition, it generally causes the atom to flip-flop between excited states. So light will cause the atom to transition from $$|1\rangle \rightarrow |2\rangle \rightarrow |1\rangle \rightarrow |2\rangle \rightarrow ...$$. When you consider just a single photon, then essentially only a single step of this process can happen. In $$|1\rangle \rightarrow |2\rangle$$ a photon is absorbed, in $$|2\rangle \rightarrow |1\rangle$$ an extra photon is created.

• Answered quite some time ago, but I think OP was asking about what creates the actual flip-flopping between excited states and why is the resulting photon the same frequency, phase, and polarization as the original.
– TTa
Nov 23, 2023 at 0:48