- #1

hgandh

- 27

- 2

*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