Checking my understanding about how massive particle states transform

[kent davidge]

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]

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]

I don't understand the question. For a complete treatment of the unitary representations of the Poincare group see Appendix B of

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf