# Proof of Yang's theorem

Yang's theorem states that a massive spin-1 particle cannot decay into a pair of identical massless spin-1 particles. The proof starts by going to the rest frame of the decaying particle, and relies on process of elimination of possible amplitude structures.

Let $\vec\epsilon_V$ be the spin vector of the decaying particle in its rest frame, and let $\vec\epsilon_1$ and $\vec\epsilon_2$ be the polarization 3-vector of the massless particles with 3-momenta $\vec{k}$ and $-\vec{k}$ respectively.

In the literature, I've seen arguments saying that

$\mathcal{M_1}\sim(\vec\epsilon_1\times\vec\epsilon_2).\vec\epsilon_V$, and $\mathcal{M_2}\sim(\vec\epsilon_1.\vec\epsilon_2)(\vec\epsilon_V.\vec{k})$ don't work because they don't respect Bose symmetry of the final state spin-1 particles.

But, why is $\mathcal{M_3}\sim(\vec\epsilon_V\times\vec\epsilon_1).\epsilon_2+(\vec\epsilon_V\times\vec\epsilon_2).\epsilon_1$ excluded? Sure, it's parity violating (if parent particle is parity even), but that's not usually a problem

Thanks

• I'm looking for more information on this theorem. Where did you fount it on the literature? – Yair Nov 20 '13 at 5:09
• Your amplitude is also impossible because it's bilinear in epsilon-V - but quantum mechanics has to produce evolution amplitudes that are linear in the initial state (the evolution operator is a linear one). – Luboš Motl Dec 16 '15 at 14:53
• @LubošMotl that's true. But I must be blind; which amplitude do you refer to that is bilinear in epsilon-V? – QuantumDot Dec 16 '15 at 15:14
• I must have mislooked, sorry. I would swear that I saw $(\epsilon_V\times \epsilon_1) \times (\epsilon_V\times \epsilon_2)$ or something like that. – Luboš Motl Dec 18 '15 at 15:04
• Regarding your comment about parity violation: as far as I understand, Yang's paper mentions that the selection rules are derived with the assumption of rotation and (parity) inversion symmetries. So you should probably not even consider parity-violating channels if trying to prove Yang's theorem. [doi:10.1103/PhysRev.77.242] – Arkya Jun 7 '20 at 23:24

Because $\mathcal{M}_3$ as written above actually vanishes by a simple vector identity. On the first term, write
$$(\vec{\epsilon}_V\times\vec\epsilon_1).\vec\epsilon_2=(\vec\epsilon_2\times\vec\epsilon_V).\vec\epsilon_1$$