Stimulated Emission Rates
Most recent answer: 07/28/2012
Q:
Concerning how a laser works, why does a photon with just the right energy increase the likelihood that an excited electron will drop down to a less excited level? I understand an electron in a lower state absorbing a photon with just the right energy to kick it up to another level. That makes sense to me, but the stimulated emission of radiation does not. I'm guessing its in the details of quantum mechanics. Any direction would be appreciated. thanks!
- shaun kelly (age 46)
plymouth,ma,usa
- shaun kelly (age 46)
plymouth,ma,usa
A:
This is one of those great questions that make us go back to think out the argument for something that we've gotten to take too much for granted. I'll try to make a simple argument along what I think are the lines that first led Einstein to his "A and B coefficients", the initial discovery of this effect.
Let's try to pick a simple model: an atom that can be in low-energy state S0 or high-energy state S1, bathed in a sea of just one type of photon, whose energy just matches the energy difference between the atom states. We can specify the state by (M, N), where M= 0 or 1 tells us what state the atom is in and N is a non-negative integer saying how many photons are around. The states (0, N+1) and the state (1, N) have exactly the same energy. So far just notation.
Let's let there be some very weak interaction with a thermal bath at temperature T, so that gradually we can get the probabilities of everything from thermal physics.
Now we get to the key part. There is exactly one state of each type, because the photons are indistinguishable particles. In thermal equilibrium the probabilities of those states (0, N+1) and (1, N) with equal energies must be equal. The rates of converting from one to the other must be equal. (Here I'm using a principle called detailed balance.)
So if you accept that as N goes up the rate of getting excitations (0, N+1) -> (1, N) goes up, then you get that the rate of getting emission (1, N) -> (0, N+1) must also go up as N goes up.
Mike W.
Let's try to pick a simple model: an atom that can be in low-energy state S0 or high-energy state S1, bathed in a sea of just one type of photon, whose energy just matches the energy difference between the atom states. We can specify the state by (M, N), where M= 0 or 1 tells us what state the atom is in and N is a non-negative integer saying how many photons are around. The states (0, N+1) and the state (1, N) have exactly the same energy. So far just notation.
Let's let there be some very weak interaction with a thermal bath at temperature T, so that gradually we can get the probabilities of everything from thermal physics.
Now we get to the key part. There is exactly one state of each type, because the photons are indistinguishable particles. In thermal equilibrium the probabilities of those states (0, N+1) and (1, N) with equal energies must be equal. The rates of converting from one to the other must be equal. (Here I'm using a principle called detailed balance.)
So if you accept that as N goes up the rate of getting excitations (0, N+1) -> (1, N) goes up, then you get that the rate of getting emission (1, N) -> (0, N+1) must also go up as N goes up.
Mike W.
(published on 07/28/2012)