I'm reading QFT: Vol 1 by Weinberg and I have a (perhaps trivial) question about a statement he makes on page 63. I can follow him to his derivation of equation (2.5.2): \begin{equation} P^\mu U(\Lambda) |p,\sigma \rangle = \Lambda^\mu{}_\rho p^\rho U(\Lambda) |p,\sigma \rangle \end{equation} where the label $\sigma$ denotes all other degrees of freedom in addition to the four-momentum. Now, I can see that, in light of equation (2.5.1), the above equation implies: \begin{equation} U(\Lambda) |p,\sigma \rangle \propto |\Lambda p,\sigma \rangle \end{equation} and so I would write: \begin{equation} U(\Lambda) |p,\sigma \rangle = C |\Lambda p,\sigma \rangle \end{equation} where $C$ is normalization constant to be determined. However, according to Weinberg, equation (2.5.2) implies: \begin{equation} U(\Lambda)|p,\sigma\rangle = \sum_{\sigma'} C_{\sigma' \sigma}(\Lambda,p)|\Lambda p, \sigma'\rangle \end{equation} Now, I do not understand what the above equation exactly means. What does $\sigma'$ represent and why are we summing over it?

Thanks in advance.


1 Answer 1


It is incorrect to say that $$ U(\Lambda) \left|p,\sigma\right> \propto |\Lambda p, \sigma\rangle~~~~~~ \text{WRONG!!} $$ Here is the correct logic. Consider the state $U(\Lambda) |p,\sigma\rangle$. We have just shown that (in eq. 2.5.2) that this state has a momentum eigenvalue $\Lambda p$. Now, there are a whole bunch of states with momentum $\Lambda p$, namely $|\Lambda p, \sigma'\rangle$ for all $\sigma'$. Thus, the only conclusion we can make is that $U(\Lambda) |p,\sigma\rangle$ is some linear combination of the states $|\Lambda p, \sigma'\rangle$. Mathematically, $$ U(\Lambda) |p,\sigma\rangle = \sum_{\sigma'} C_{\sigma\sigma'}(\Lambda,p)|\Lambda p, \sigma'\rangle $$ for some matrix $C$ which can possibly depend on $\Lambda$ or $p$ (or both)

  • $\begingroup$ thanks for your reply. When you say "there are a whole bunch of states with momentum $\Lambda p$", do you mean a whole bunch of state with that exact same momentum, but with different degrees of freedom $\sigma'$ (for instance spin)? $\endgroup$
    – Hunter
    Nov 21, 2013 at 17:27
  • 1
    $\begingroup$ Yes. That's what I mean $\endgroup$
    – Prahar
    Nov 21, 2013 at 17:28

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