I Question about one-particle states

[hgandh]

From Weinberg's Quantum Theory of Fields vol 1. In Chapter 2.5, he lists the transformation rule of a one-particle state under a homogeneous Lorentz transformation:
\begin{equation}
U(\Lambda)\Psi_{p,\sigma} = \sum_{\sigma'}C_{\sigma'\sigma}(\Lambda,p)\Psi_{{\Lambda}p,\sigma'}
\end{equation}
Where sigma are discrete degrees of freedom (not yet shown to be spin). He then goes on to define a standard 4-momentum such that:
\begin{equation}
p^\mu = L^\mu_{v}(p)k^v
\end{equation}
Now he defines one-particle states of momentum p:

\begin{equation}
\Psi_{p,\sigma} {\equiv} N(p)U(L(p))\Psi_{k,\sigma}
\end{equation}

With N a constant.
Now, my question is why does this transformation not affect the sigma variable? According to the first equation, the result of the transformation should be a linear combination of the state vectors. Is it simply assumed that the Lorentz transformation which acts on the standard 4-momentum does not change the other degrees of freedom? Or is there another reason?

Thanks

 

[meopemuk]

In my understanding, the right hand side of your eq. (3) is the *definition* of the particle state with momentum p and spin projection σ. So, there is no need for a linear combination. One can choose different unitary mappings from the k-subspace to p-subspaces. They will correspond to different momentum-spin bases in the full Hilbert space and different definitions of the spin operator.

Eugene.