# Checking my understanding about how massive particle states transform

I'd like to see whether or not I understood correctly how massive particle states will transform under a homogeneous Lorentz transformation, in terms of the standard four-momentum ##k = (0,0,0,M)##. I suppose we can write $$U(\Lambda) \Psi \propto D^{(j)} (W(\Lambda)) \Psi$$ where ##U(\Lambda)## is a unitary operator, ##\Psi## is a so called eigenstate (eigenvector) of the four-momentum operator ##P##, ##W(\Lambda)## is an ##SO(3)## element and ##D^{(j)} (W)## its jth-(reducible)-representation on the Hilbert Space.

I used the "proportional to" symbol instead of the "equal to" symbol because I left some terms out, which I guess, are not relevant to my specific question.

So the point of my question is whether or not we can write the matrices ##D^{(j)}## as $$D^{(j)} (R_z) = \begin{pmatrix} e^{-ij\varphi} & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & e^{ij\varphi} \end{pmatrix}$$
when ##R_z## is a rotation by ##\varphi## about the z-axis.

vanhees71
Gold Member
Yes, that's fine. The ##D^{(j)}## are the usual rotation matrix in the (2j+1)-dimensional representation of SU(2) with spin j.

kent davidge
An apparent problem is that one could instinctively look at the product ##U(R(\theta)) U(R(\bar \theta)) = U(R(\theta + \bar \theta))## and imply from this that ##U(R(\theta))## gives the irreducible representation ##\exp(i J \theta)##, with ##J## a generator, for finite ##\theta##, instead of the irreducible representations I wrote down in the matrix in the opening post.

Am I missing something?

vanhees71