# Can you split a photon?

I was wondering if a photon is divisible.

If you look at a photon as a particle, then you may be able to split it (in theory).

Is it possible and how do you split it?

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Actually, there once was a theory with composite photons, which I just found out about today. Each photon was actually supposed to be a neutrino-antineutrino pair. Although quite intriquing, it didn't work very well, since you couldn't reproduce the polarisation properties of light. Also, neutrinos turn out to have mass, which photons don't. See: en.wikipedia.org/wiki/Neutrino_theory_of_light – jdm Aug 22 '11 at 21:49

The photon cannot be split as one can split a nucleus. As it has zero mass it cannot decay. But it can interact with another particle lose part of its energy and thus change wavelength. It can be transmuted.

Have a look at the compton scattering entry in wikipedia.

Edit: Intrigued by the other answers I searched and found that within special crystals "splits" can happen, if one defines as a split that there can come out two photons whose energy adds up to the original energy of the photon. So in a collective crystal photon interaction there exists such a probability.

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+1 for special crystals – voix Aug 23 '11 at 16:06
Photon splitting cannot occur in vacuum, but it is a consequence not merely of the zero mass, but instead a combination of Lorentz invariance and satisfying QED Ward identities. One can then show that all possible amplitudes vanish kinematically. – Columbia Aug 26 '11 at 17:56

The answer somewhat depends on how you define "splitting" of a particle with a zero rest mass. Since it can never be stopped, there is always kinetic energy in a photon which can be converted into other particles. If this counts as "splitting" then the photon can be split e.g., into more photons in a parametric down-conversion. or even into pairs of massive particles and antiparticles if there is enough energy. However, if such excess energy conversion into matter does not count as splitting, then the answer is no. In this (rather loose) sense once can say that a photon is "pure energy".

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Just to add to this comment, PDC is a non-resonant phenomenon with constraints being energy and momentum conservation. The non-resonant aspect is experimentally critical because with resonant phenomena you have absorption that drastically reduces the efficiency of the conversion process. – Antillar Maximus May 17 '12 at 0:25

Photon is a stable particle. It has zero mass, so there is no lighter elementary particle for it to decay into.

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-1: it will not decay spontaneously, but if there is another body to accept momentum, splitting a photon becomes possible, see, e.g., parametric down-conversion. – Slaviks Aug 22 '11 at 18:57
@Slaviks: there is nothing wrong with my answer, it all depends on what Stephan meant by "splitting". I'm aware of the fact you've mentioned. – Physicsworks Aug 23 '11 at 6:23
You are right, sorry. It indeed depends on what splitting means, as I tried to explain in my answer. To moderators: I can not remove my -1 vote, can I? – Slaviks Aug 23 '11 at 6:49
@Slaviks you can, just toggle the downvote button again. – PPG Dec 13 '13 at 0:53
@PPG 7nfortunatwlly, I keep getting a message "You vote is locked untill the answer is edited" – Slaviks Dec 13 '13 at 21:55

there is a vertex gamma-> 3 gamma. So the photon can turn into 3 phtons, with smaller energy. To make this possible, they all have to be collinear, orthewise the kinematics forbids it.

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This vertex is already taken into account by renormalization: a single real photon is a perfect eigenstate of time evolution and does decay spontaneously into real particles, as already mentioned in other answers. – Slaviks Aug 23 '11 at 6:13

## protected by Qmechanic♦Dec 12 '13 at 23:35

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