Why the space is isotropic in the vector particle's decay?

[Ganesh Ujwal]

I come cross one proof the Landau-Yang Theorem, which states that a $J^P=1^+$ particle cannot decay into two photons, in this paper (page 4).

The basic idea is, the photon's wavefunction should be symmetric under exchange, however the spin part is anti-symmetric and the space part is symmetric and therefore forbidden.

I have trouble understanding the argument about the space part:

Since the photons conserve linear momentum in the particle rest frame and space is isotropic, they must be emitted in spherical waves.

Why the space is isotropic? Is isotropy an intrinsic property of original particle or just because the final particles are identical?

I guess the right answer is the latter one, because $\rho^+ \to \pi^+ \pi^0$ the final pions lie in $P$ wave and it doesn't bothered by the Bose-Einstein statistics. (Compare this with $\rho^0\to \pi^0 \pi^0$, which is forbidden.)

However, I still believe the isotropy is an intrinsic property of the original particle. I'm looking for an explanation more mathematically, or a definition of isotropy in the language of group theory.

Any suggestions?

 

[vanhees71]

You have conservation of angular momentum and parity in electromagnetic interactions. The vector boson must be massive to be able to decay into two photons due to energy-momentum conservation. So we can consider its decay in its rest frame. So if it decays in two photons their momenta must be back-to-back. We assume those vectors to be $\pm k \vec{e}_z$. The photons are massless and thus have only two polarization states $\sigma_z=\pm 1$. A possible basis for the final state thus is
$|\Psi_1 \rangle=|\vec{p},1 ;-\vec{p},1 \rangle + |-\vec{p},1;\vec{p},1 \rangle,$
$|\Psi_2 \rangle=|\vec{p},-1;-\vec{p},1 \rangle + |-\vec{p},1; \vec{p},-1 \rangle,$
$|\Psi_3 \rangle=|\vec{p},1;-\vec{p},-1 \rangle + |-\vec{p},-1; \vec{p},1 \rangle,$
$|\Psi_4 \rangle=|\vec{p},-1 ;-\vec{p},-1 \rangle + |-\vec{p},-1;\vec{p},-1 \rangle,$
Space reflection flips the momenta but keeps the $\sigma_z$. The vector boson and the photons have intrinsic parity -1. Thus

Thus the vector boson at rest, having no orbital angular momentum, must be a parity -1 state. The em. interaction conserves parity, and thus the 1st and the 4th two-photon states are ruled out. So $|\Psi_1 \rangle$ and $\Psi_2 \rangle$ are ruled out, because they have parity $+1$. Thus the photon state must be a linear combination
$$|\Psi \rangle=\alpha |\Psi_2 \rangle + \beta |\Psi_3 \rangle \stackrel{!}{=}-|\Psi \rangle \qquad (1).$$
Now $\hat{P} |\Psi_2 \rangle=|\Psi_3 \rangle$ and $\hat{P} |\Psi_3 \rangle=|\Psi_2 \rangle$. Thus from (1) we have
$$|\Psi \rangle=|\Psi_2 \rangle-|\Psi_3 \rangle.$$
Now we rewrite this state as a direct product of the orbital (momentum) and helicity part:
$$|\Psi \rangle = (|\vec{p},-\vec{p} \rangle -|-\vec{p},\vec{p} \rangle) \otimes (|-1,1 \rangle-|1,-1 \rangle).$$
Now consider a rotation by $\pi$ around the $y$ axis. Obviously this flips both $\vec{p}$ and $\sigma_z$ of the photons and thus $\hat{R}_{y}(\pi) |\Psi \rangle=|\Psi \rangle$.

On the other hand we have for $J=1$
$$\hat{R}_y(\pi)=\exp(\mathrm{i} \pi \hat{J}_y ) \mathrm{diag}(1,-1,1).$$
Since $|\Psi \rangle$ is a state of total $\sigma_z=0$ also the original vector-meson state must have $\sigma_z=0$ and thus flips sign under the rotation. Since total angular momentum is conserved this is in contradiction to our finding that $|\Psi \rangle$ is not changing under this rotation. Thus it cannot be a $J=1$, $\sigma_z=0$ state and thus it's not possible that your vector boson decays into two photons.

I don't know, whether this is the original proof by Landau and/or Yang.