How to get explicit value of Wigner angle for massless one-particle state transformation?

The one-particle massless state $|\mathbf p , \sigma\rangle$ is transformed under the Lorentz group $U(\Lambda) \equiv U(\Lambda , 0)$ as $$U(\Lambda)|\mathbf p, \sigma \rangle = \sqrt{\frac{(\Lambda p)^{0}}{p^{0}}}e^{i\sigma\theta (\Lambda , p)}|\vec{(\Lambda p)}, \sigma\rangle ,$$ where $\theta (\Lambda , p)$ is the Wigner angle. How to extract it for the given $\Lambda$ and $p$?

• Weinberg has very explicit formulae for this. Apr 18, 2016 at 15:12
• @Prahar : unfortunately, I don't understand how to extract the explicit value of $\theta$ from $(2.5.43)$ (for given $\Lambda$, $\mathbf p$). Apr 18, 2016 at 15:16

By Wigner's general procedure of representing the little group, the $\theta(\Lambda,p)$ is the angle of rotation associated to the massless little group element $L(\Lambda,p)\in\mathrm{SE}(2) = \mathrm{SO}(2)\ltimes\mathbb{R}^2$ fulfilling $$L(\Lambda,p) = l^{-1}(\Lambda k)\Lambda l(k)$$ where $l(k)$ is the Lorentz transformation carrying the null vector $k$ to $(1,0,0,1)$. We can use the decomposition $$\Lambda = R_2 B_z R_1$$ where $R_i$ are pure rotations and $B_z$ is a boost in the $z$-direction. Now, to both rotations, we associate an angle $\omega_i$ as follows:
Denote the rotation that carries the z-axis to $v$ by $R(v)$, and denote by $\vec p$ the spatial part of $p$. Then, $R_i R(\vec p)R^{-1}(R_i \vec p)$ is a rotation around $\Lambda \vec p$, and we denote the angle of this rotation by $\omega_i$. It turns out that $\theta(\Lambda,p) = \omega_1+\omega_2$.