# Can UV photons spontaneously convert into IR photons, and vice versa?

My general question is: can high energy photons convert into many lower energy photons? Could the reverse reaction occur spontaneously?
Let's say we have a single photon that was emitted from a distant supernova. We detect it here on Earth. The photon hasn't converted into multiple lower energy photons during the path from the supernova to the Earth. It just gets red-shifted as space expands.

So, to start we have: Energy = 12 eV, Spin = 1, Momentum = 12 eV/c, Charge = 0

If the photon could split into 3 lower energy photons of Spin (+1,-1,+1) all in the same direction, we would have:

Energy = 4+4+4=12 eV, Spin = 1-1+1=1, Momentum = 4+4+4=12 eV/c, Charge = 0+0+0=0

Since bosons are allowed to be in the same energy state, we could have all 3 new photons be exactly the same energy. Though, the energy values could have be any number of different combinations.

What law of physics prevents this splitting from happening? And vice versa, what prevents the 3 photons from converting into 1 higher energy photon?
Thanks,
Eddie

mathman
There is no law of physics to describe what you are looking for (either way).

DrChinese
Gold Member
My general question is: can high energy photons convert into many lower energy photons? Could the reverse reaction occur spontaneously?
On their own, I don't believe so. (Or the likelihood is too small to be of any consideration.)

There is a process called Spontaneous Down Conversion (and its reverse Up Conversion) in which 1 photon becomes 2 and vice versa. However, a special crystal is required to make that happen.

http://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion

DrDu
I think this should be possible. You can write down a Feynman diagram with an electron loop with four vertices.
On one vertex, the uv photon is absorbed, on the other three, an ir photon is being emitted.
Probably only very improbable.

Ah, see here:
http://en.wikipedia.org/wiki/Delbruck_scattering

DrDu
fzero
Homework Helper
Gold Member
What law of physics prevents this splitting from happening? And vice versa, what prevents the 3 photons from converting into 1 higher energy photon?
Thanks,
Eddie
In an external electromagnetic field, it is possible to for a single photon to split into two or more lower energy photons. The process is related to the "scattering of light by light" diagram in DrDu's link. In the absence of an external field, the splitting of a single photon into other photons is impossible by a kinematical argument, discussed in this thread. The details of the kinematic argument are summarized in this post. I didn't discuss the collinearity constraint in detail there. That is the point that, since the initial photon has no center of mass frame, all of the photons in the process must have their momenta on the same line.

DrDu
Very interesting. It should be possible to describe this in terms of a classical picture drawing cute little rotating arrows, don't you think so?

fzero
Homework Helper
Gold Member
Very interesting. It should be possible to describe this in terms of a classical picture drawing cute little rotating arrows, don't you think so?
This would be a one-loop effect, so I don't think there's a purely classical version of the argument. It is true that the Euler–Heisenberg Lagrangian vanishes for a sum of collinear EM waves, since ##\mathbf{E}\cdot \mathbf{B} = \mathbf{E}^2 - \mathbf{B}^2 =0##.

Bill_K
This would be a one-loop effect, so I don't think there's a purely classical version of the argument.
There's a classical limit in which photon-photon scattering is described by adding quartic terms to the usual Lagrangian, such as FαβFβγFγδFδα.

DrDu
Delbruck scattering is something slightly different - photon scattering in a Coulomb field.
Of course, but the relevant Feynman diagram is drawn in the article.