Life with Physics

What is a particle? S. Weinberg described a particle simply as "a physical system that has no continuous degrees of freedom except for its total momentum".  Recall that the spacetime admits the so-called Poincare symmetry $$\begin{equation} {x'}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,x^{\nu} + a^{\mu}\,.\end{equation}$$ By Noether's theorem , each continuous symmetry indicates a conservation. Thus, the total momentum $p^{\mu}$ is a good quantum because of the spacetime translation invariance. As we shall see below, there are other discrete degrees of freedom, denoted by $\sigma$, that are associated with the Lorentz invariance. So the quantum state of a particle is described as $|p, \sigma\rangle$.  Under the Lorentz transformation $\Lambda$, the one-particle state $|p, \sigma\rangle$ changes to a new state $U(\Lambda)|p,\sigma\rangle$ by a unitary operator $U(\Lambda)$. Note that the Lorentz transformation only changes the reference frames, elementary particles should remain...

Problem:  Show that the Lorentz transformation $W$ that keeps $k^{\mu}\equiv[1, 0,  0,  1]^T$ unchanged forms a two-dimensional Euclidean group SE(2). Solution: As shown in this post , the general solution of the Lorentz transformation $W$ that satisfies $W^{\mu}_{\,\,\,\,\nu}\,k^{\nu}=k^{\mu}$ can be parametrized in three parameters $\alpha,\beta,\theta$ as $$\begin{equation}W(\alpha, \beta, \theta)=\left[\begin{matrix} 1+(\alpha^2+\beta^2)/2  & -\alpha & -\beta & -(\alpha^2+\beta^2)/2  \\  -\alpha & 1 & 0 & \alpha \\ -\beta & 0 &  1 & \beta \\ (\alpha^2+\beta^2)/2 & -\alpha & -\beta & 1-(\alpha^2+\beta^2)/2 \end{matrix}\right]\left[\begin{matrix} 1 &  &  & \\  & \cos\theta & -\sin\theta & \\ & \sin\theta & \cos\theta  &   \\ & &  & 1\end{matrix}\right]\,.\tag{1}\end{equation}$$ To work out the Lie algebra, we expand $W(\...