13
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ I'm looking for more information on this theorem. Where did you fount it on the literature? $\endgroup$
    – Yair
    Nov 20, 2013 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). $\endgroup$ Dec 16, 2015 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, 2015 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. $\endgroup$ Dec 18, 2015 at 15:04
  • $\begingroup$ 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] $\endgroup$
    – Arkya
    Jun 7, 2020 at 23:24

1 Answer 1

9
$\begingroup$

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

$$(\vec{\epsilon}_V\times\vec\epsilon_1).\vec\epsilon_2=(\vec\epsilon_2\times\vec\epsilon_V).\vec\epsilon_1$$

which cancels the second term.

[there goes my bounty]

$\endgroup$
1
  • 1
    $\begingroup$ Congrats on figuring it out =) $\endgroup$ Oct 1, 2012 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.