# Why single photon decay into two photon in vacuum is not possible?

Suppose a photon with wavevector $$\vec{k}$$ and frequency $$\omega$$ decays into two photons with wavevector $$\vec{k}_1$$ and $$\vec{k}_2$$, and frequency $$\omega_1$$ and $$\omega_2$$. The energy and momentum conservation required:

$$\hbar\omega =\hbar \omega_1+\hbar\omega_2$$ $$\hbar \vec{k}=\hbar \vec{k}_1+\hbar \vec{k}_2\rightarrow \vec{k}=\vec{k}_1+\vec{k}_2$$ Further the dispersion relation $$\omega=kc$$ gives $$k=k_1+k_2$$ We know that these two conditions can be satisfied only if the vectors are along the same line. Does that mean that a single photon can decay into two photons? But this isn't the case, I think. Why is it so?

• Besides the fact that the relation seem to be applied to wave rather than photons, doesn't the second relation say the opposite? Sep 25 '21 at 6:35
• I don't understand, What do you mean? Sep 25 '21 at 6:56
• I mean that from the conditions you have listed no splitting follows even if the answers below wouldn't apply. Forget quantum but vect k1 + k2 would be like to split a rope tension in two components having the same direction. And still, that formalism seems (to me) for a number of photons. Anyway, Stephen G answer fix it beside my comment (not a specialist here). Sep 26 '21 at 6:26
• Charge conjugation? (Furry's theorem.) Oct 16 '21 at 0:39

We know that these two conditions can be satisfied only if the vectors are along the same line. Does that mean that a single photon can decay into two photons? But this isn't the case, I think. Why is it so?

Photons have spin 1. A photon decaying into two photons does not conservse spin, which must be conserved.

If you split a photon into two photons you end up with one of these combinations of spin :

• spin -1 and spin -1 - does not total spin 1
• spin -1 and spin +1 - does not total spin 1
• spin +1 and spin -1 - does not total spin 1
• spin +1 and spin +1 - does not total spin 1

None of them conservse spin so none of them are possible.

• Neat and to the point! :-) Sep 25 '21 at 7:46

Experimental observations are that the frequency of light does not change spontaneously, it has to go through a medium. If a photon could decay into two collinear photons there would be no constancy of colors as we know it.

When discussing photons one is in the quantum mechanical regime. The standard modelof particle physics does not have a three photon vertex, which would be necessary for the decay you envisage, there is not even a two photon vertex, the interaction has to go through virtual charged particles. That makes it very improbable. A three photon production through virtual exchanges would be even more improbable.

So it is no-go within the standard model, agreeing with the observation above.

Nevertheless there out of the mainstream theories that admit a mass for the photon, which changes things:

The idea that photons have a finite lifespan, and therefore mass, is difficult to imagine. Indeed, astronomers looking at distant cosmic objects regularly detect photons that are billions of years old. But some theories suggest that photons could have a non-zero rest mass, albeit a small one – the upper limit for the mass of the photon is constrained to $$10^{-18}$$ eV or $$10^{–54}$$ kg thanks to experiments with electric and magnetic fields.

• Is there a way to mathematically argue that, these vector can't be in the same direction? Sep 25 '21 at 6:56
• No, if I were ambitious I would try to prove the k and -k cannot be a zero momentum vector in the following problem physics.mcgill.ca/~andrzej/Peskin/Chap10.pdf . it concludes " Thus, the n-point function for odd n vanishes" and for a photon to two photons n=3 Sep 25 '21 at 8:23
• In Stephen's answer, conservation of angular momentum does not allow it , so it is still a law that comes from observations. Possibly the odd n point functions end up violating angular momentum Sep 25 '21 at 10:00