Weinberg's approach to QFT starts with particles which are not necessarily more fundamental than fields but are known more for certain. A particle of a particular species can have different momenta $p$ in different frames but its internal degrees of freedom (like electron spin) is described by a discrete quantum number $\sigma$, and the one-particle state is denoted by $\Phi_{p,\sigma}$. A standard boost is defined to relate one-particle states with different momenta but the same quantum number of internal degrees of freedom $\Phi_{p,\sigma} = U(L(p))\Phi_{k,\sigma}$, where $k$ is a chosen standard momentum ($(m,0,0,0)$ for particles with mass $m$) and $L(p)$ is a standard boost that takes $k$ to $p$. With the non-compact boost components of the Lorentz group trivialized in this way, essentially particles are defined as finite-dimensional unitary representation of the little group, that is, the stabilizer of $k$.

My question is about the choice of the standard boost $L(p)$. Is there any physical consideration (it doesn't change the quantum number of internal degrees of freedom) that determines $L(p)$? Different choices of $L(p)$ differ by an element in the stabilizer of $p$ (which is $SO(3)$), the standard boost for one choice is not standard in another, hence mixed quantum numbers of internal degrees of freedom, does it matter for physics? Perhaps it's just redefinition of quantum number of internal degrees of freedom?

  • $\begingroup$ Is your question about why we can choose $U(L(p))\Phi_{k,\sigma}$ to be precisely $\Phi_{p,\sigma}$ and not some more general linear combination $\sum_{\sigma^\prime}C_{\sigma\sigma^\prime}(p)\Phi_{p,\sigma^\prime}$? $\endgroup$ Apr 4, 2021 at 6:46
  • $\begingroup$ Thank you for your interest. Sort of. A general Lorentz transformation will mix the quantum number of the internal degrees of freedom, hence the matrix $C_{\sigma \sigma^{'}}$, but in the book it defines a "standard boost" that doesn't change the quantum number of internal degrees of freedom, that is momentum transformed from $k$ to $p$ but $\sigma$ unchanged. $\endgroup$ Apr 5, 2021 at 1:47
  • $\begingroup$ Yeah, so what he's doing is not so much defining the operator $U$ as to not mix the other QN, but rather taking this equation to define the state $\Phi_{p,\sigma}$ for $p$ which are not the standard momentum. It's a slight change in point of view, but makes all the difference. You'll note this is allowed because he hasn't actually defined the state beyond saying it had to be an eigenvector of the 4-momentum to that point. $\endgroup$ Apr 5, 2021 at 4:37
  • $\begingroup$ I think your explanation makes sense, that is, it's like defining $\sigma$ for states with generic momentum. By the way, good luck at UCLA, where I used to be for a few years. $\endgroup$ Apr 5, 2021 at 5:25

1 Answer 1


I had some difficulty understanding a fair amount of chapter 2 because Weinberg’s book on QFOT (which you appear to be describing) sometimes skips some of the logic (like what are the “sigma’ degrees of freedom). I managed to find a set of videos that helped me a lot - much is filled in to Weinberg’s terse presentation. Here is the link:


  • $\begingroup$ Thank you very much! I will look at it. $\endgroup$ Apr 27, 2021 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.