# Photon emission from simple harmonic oscillator

1. Aug 27, 2010

### snoopies622

I've looked at a few introductory treatments of the quantum harmonic oscillator and they all show how one arrives at the discrete energy values

$$E_n = ( \frac {1}{2} + n ) hf \hspace {10 mm} n=0,1,2...$$

usually by setting up and then solving the Schrodinger equation for the system.

But so far I haven't found an answer to this question: how much time will pass before the oscillator will spontaneously jump from one energy state to a lower one, emitting a photon in the process?

This does happen, doesn't it?

2. Aug 27, 2010

### The_Duck

Griffiths does this calculation in chapter 9.

3. Aug 27, 2010

### snoopies622

Unfortunately, Griffiths' Introduction to Quantum Mechanics isn't available to me. According to their on-line catalog, the physics library at my local university has a copy, but someone has it out and it was due back more than four months ago, so it may never be returned.

4. Aug 28, 2010

### sweet springs

Hi. I assume from HUP it is approximately inverse of the difference of energy levels multiplied by Planck constant. Is it OK?
Regards.

Last edited: Aug 28, 2010
5. Aug 28, 2010

### Staff: Mentor

Surely other QM textbooks also cover this topic. Visit your library and browse its collection.

6. Aug 28, 2010

### dx

Don't forget that the probability of emission of a photon into a particular state depends on the number of photons already in it.

7. Aug 29, 2010

### snoopies622

8. Aug 29, 2010

### dx

If there are n photons in some state, then the probability of emission of a photon into that state is increased by a factor of n + 1:

$$\langle n + 1 | n \rangle = \sqrt{n+1} a$$

where a is the amplitude for spontaneous transition. This is a consequence of Bose statistics.

9. Aug 29, 2010

### Phrak

A sufficiently isolated system does not emit, in my understanding. In the case of electromagnetic radiation, if there is no null world line intersecting a future receiver, an atom in a raised energy state will remain in a raised state indefinitely.

I'm informed there is experimental evidence in support.

Last edited: Aug 29, 2010
10. Aug 29, 2010

### snoopies622

As this applies to the OP, is the number of photons in a state another way of describing the (discrete) energy level of the state?

11. Aug 29, 2010

### dx

Yes, we can either speak about n non-interacting bosons in some particular state or the n'th energy level of a quantum harmonic oscillator.

12. Aug 30, 2010

### snoopies622

Well that's interesting. Thanks, dx.

Since the formula for the Einstein-Bose distribution doesn't mention time, is there another one I should consider for this question?

13. Aug 30, 2010

### dx

Last edited by a moderator: Apr 25, 2017
14. Aug 30, 2010

### Iforgot

Hey Snoopies,

I wish I could remember my source for this:

there's no such thing as spontaneous emission. All emissions are stimulated emissions.

To calculate the "spontaneous" transition rates, I would construct a time dependent perturbation potential from the photon population. The photon population can be calculated using bose-einsten statistics. Then use time dependent perturbation theory to calculate the transition rates.

Last edited: Aug 30, 2010
15. Aug 30, 2010

### Cthugha

This is somewhat correct, but can be very misleading when quoted out of context. You can model spontaneous emission as the stimulated emission caused by the vacuum field. However, it has become the usual terminology to call such vacuum-assisted processes spontaneous.

The photon number population does not necessarily follow Bose-Einstein statistics. In most cases where stimulated emission is important (this means: lasers), it will be Poissonian. Usually spontaneous decay rates are treated perturbatively in terms of Fermi's golden rule. The rates depend on the transition of interest, mainly its dipole moment, the energy of the photon to be emitted and the density of states of the electromagnetic field the photon can be emitted to.
If not just situated in empty space, the surrounding geometry can have an influence on the rates as for example a microcavity can alter the density of states of the electromagnetic field.

Last edited: Aug 30, 2010
16. Aug 31, 2010

### snoopies622

By the way, are we talking about an oscillator that's electrically charged or neutral? I assume that the first one should emit radiation "spontaneously" - at least in the macroscopic limit - but I'm not sure about the second one.

17. Sep 3, 2010

### snoopies622

I've been wondering about this answer. It makes sense to me if a spontaneous photon emission can be thought of as being the same as an energy measurement. Either way, the system is in an energy eigenstate immediately afterwards, so maybe it can be.

18. Sep 3, 2010

### Cthugha

You mean the transition rate?

Fermi's golden rule gives a spontaneous emission rate of
$$\omega_{i\rightarrow f}=\frac{\omega^3_{if} d_{if}}{3 \pi \epsilon_0 \hbar c^3},$$
where $$d_{if}$$ is the transition matrix element and accounts for the dipole moment and the index of refraction. The $$\omega^3$$-dependency stems from the term corresponding to the photonic density of states and assuming a blackbody radiation spectrum. However, this calculation is only correct for free space or large cavities where the density of photonic states is not altered by geometry.

It is unfortunately not that easy. If you can know for sure that a spontaneous emission event has happened, an energy measurement has taken place. However, in many situations you end up with a superposition of the states

atom excited + no photon present and
atom in ground state + one photon present

for some amount of time until this superposition decoheres.

19. Sep 6, 2010

### snoopies622

Does Fermi's Golden Rule assume that the oscillators are electrically charged? I was wondering since the harmonic oscillator assumes the potential energy to be a function of $x^2$ instead of $1/x$, so I always assumed that it was uncharged.

20. Sep 6, 2010

### Cthugha

Fermi's golden rule is a rather general tool that can describe a lot of different situations. You just need to change the initial and sfinal states and the interaction matrix element accordingly.

The above estimate assumes a dipole transition of a neutral atom. If you want to describe a different scenario you can do a multipole expansion of the field-atom interaction energy. Then you get a monopole moment (only nonzero if the oscillator is charged), a dipole moment (only nonzero if the oscillator is a net dipole or an external light field induces a dipole, this is the case discussed above), a quadrupole moment (necessary for "dark" optical transitions which are not dipole-allowed) and so on.

So if you want to calculate a charged oscillator, the monopole term might already be sufficient.