(adsbygoogle = window.adsbygoogle || []).push({}); How to understand the "standard" momentum introduced in Weinberg's QFT

I'm reading Weinberg's QFT Volume I. In page 63, you can find a formula (2.5.3) which states that the new state vector obtained by a Lorentz transformation is a linear combination of a whole bunch of other vectors.

[itex]U(\Lambda)\Psi_{p,\sigma}=\sum_{\sigma'} C_{\sigma'\sigma}(\Lambda,p)\Psi_{\Lambda p,\sigma'}[/itex]

But in page 64, he tells us that we can choose a so-called "standard" momentum [itex]k^\mu[/itex] and associated with it, a "standard" Lorentz transformation [itex]L^\mu_{\phantom{x}\nu}(p)[/itex] such that we can define

[itex]\Psi_{p,\sigma}=N(p)U(L(p))\Psi_{k,\sigma}[/itex]

with [itex]p^\mu=L^\mu_{\phantom{x}\nu}(p)k^\nu[/itex].

Then the state vector [itex]U(L(p))\Psi_{k,\sigma}[/itex] is no longer a linear combination of several different [itex]\Psi_{p,\sigma}[/itex]'s. Why can we do this?

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# How to understand the standard momentum introduced in Weinberg's QFT

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