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I am reading Sec. 2.5 of Weinberg's Quantum Theory of Fields, Volume I. There he talks about the classification of relativistic one-particle states according to their transformation under the Poincare group.

In Eq. 2.5.2, Weinberg derives that the action of an arbitrary proper, orthochronous Lorentz transformation, $\Lambda$, on a single-particle state $\Psi_{p,\sigma}$, is a linear combination of state vectors $\Psi_{\Lambda p,\sigma\prime}$: $$U(\Lambda)\Psi_{p,\sigma}=\sum_{\sigma^\prime}C_{\sigma^\prime\sigma}(\Lambda,p)\Psi_{\Lambda p,\sigma^\prime}.$$ I completely understand how he arrives at this result.

I have the following doubt about this result. If $\Psi_{p,\sigma}$ is a member of an irreducible representation i.e., if $\Psi_{p,\sigma}$ is a basis vector of the invariant subspace of fixed $p^2$ and fixed ${\rm sgn}(p^0)$, shouldn't the RHS involve a linear combination of $\Psi_{p^\prime,\sigma^\prime}$ with all $p^\prime$ subject to the restriction fixed $p^2$ and fixed ${\rm sgn}(p^0)$)? However, since for a given $\Lambda$, the RHS contains only one momentum in the orbit, namely $\Lambda p$, and leaves out all other momenta in the orbit of fixed $p^2$ and fixed ${\rm sgn}p^0$, can we say that $\Psi_{p,\sigma}$ is a basis for an irrep?

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  • $\begingroup$ Do you mean you think you can transform out of the subspace and therefore it is no irrep? $\endgroup$ Mar 9 at 8:22
  • $\begingroup$ I didn't get you. The action of an arbitrary group element on any member of an irrep, in general, should mix all components of that irrep. For example, the action of the rotation operator on the subspace of fixed $j$, in general, mixes all $m$ values i.e. $U(R)|j,m\rangle=\sum_{m^\prime=-j}^{j}D^{(j)}_{m^\prime m}(R)|j,m^\prime\rangle$. $\endgroup$ Mar 9 at 8:52
  • $\begingroup$ There are rotations that do not mix all of them (around the $z$-axis for instance). But indeed with all available rotations you should be able to mix them all, or else it is reducable, so no irrep. My question was about your question: do you mean Weinberg doesn't mix them all, or do you mean he mixes them in a way that ends up outside the original subspace? (Those are the only two ways it could go wrong, I would say...) $\endgroup$ Mar 9 at 9:18
  • $\begingroup$ Now, I completely get the point. You are correct. Momentarily, I was confused. Because under a most general rotation $R(\hat{n},\theta)$, applied to an irrep to fixed $j$, in general, mixes all $m$ as I've written in my previous comment. However, a most general Lorentz transformation does not mix all $p$ at the same time. Of course, I must look at it in the way you mention. $\endgroup$ Mar 9 at 9:23

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Weinberg proves directly before the equations you mention that $$P^\mu U(\Lambda)\Psi_{p, \sigma} = \Lambda^\mu{_\rho}p^\rho U(\Lambda)\Psi_{p, \sigma}.$$ In particular, $U(\Lambda)\Psi_{p, \sigma}$ is precisely a momentum eigenstate with eigenvalue $\Lambda p$. The most general state with eigenvalue $\Lambda p$ is precisely $$\sum_{\sigma'}C_{\sigma, \sigma'}(\Lambda,p)\Psi_{\Lambda p, \sigma'}.$$

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    $\begingroup$ This is not what I am asking. I understood how he proves this equation. $\endgroup$ Mar 9 at 9:01
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Here is my answer. Feel free to correct me.

Under an arbitrary proper, orthochronous Lorentz transformation, $\Lambda$, takes us from $p\to p^\prime=\Lambda p$. However, we have to consider the set of all proper, orthochronous Lorentz transformations belonging to ${\rm SO}(3,1)^{+}$. Then, starting from any particular $p$, all other $p^\prime$ belonging to an orbit (i.e., fixed $p^2$ and fixed ${\rm sgn}~p^0$) can be reached by applying suitable $\Lambda\in SO(3,1)^+$ to that $p$. Therefore, as far as I understand, an orbit is an invariant subspace and the set of all $p$ in an orbit belongs to an irrep.

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    $\begingroup$ This is correct. This is by definition of an invariant subspace. Emphasis being on the fact you consider the action of the entire represented group, not just one $U(\Lambda)$. $\endgroup$ Mar 9 at 9:13
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    $\begingroup$ This mathematics makes intuitive physical sense because it defines a particle as a collection of states related to each other by Lorentz transformations: I consider both an object at rest and that object moving with constant velocity to be the same object. $\endgroup$ Mar 9 at 9:21

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