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


  • $\begingroup$ I'm looking for more information on this theorem. Where did you fount it on the literature? $\endgroup$ – Yair Nov 20 '13 at 5:09
  • $\begingroup$ 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). $\endgroup$ – Luboš Motl Dec 16 '15 at 14:53
  • $\begingroup$ @LubošMotl that's true. But I must be blind; which amplitude do you refer to that is bilinear in epsilon-V? $\endgroup$ – QuantumDot Dec 16 '15 at 15:14
  • $\begingroup$ 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. $\endgroup$ – Luboš Motl Dec 18 '15 at 15:04

Because $\mathcal{M}_3$ as written above actually vanishes by a simple vector identity. On the first term, write


which cancels the second term.

[there goes my bounty]

  • 1
    $\begingroup$ Congrats on figuring it out =) $\endgroup$ – Marton Trencseni Oct 1 '12 at 18:24

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