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Please allow me to set the context based on my understanding before I present the question.

In quantum field theory, one-particle states are the basis states of the infinite-dimensional unitary irreducible representation spaces of the Poincare group. Consider the massive case, we commonly denote such a one-particle state by $|\vec p, \lambda\rangle$, where $\vec p$ is the 3-momentum of the particle with a rest mass $M = \sqrt{-p^2}$, where $p$ is the $4$-momentum, and $\lambda$ is defined below. Such a state can be obtained by the method of induced representations from the rest-frame state $|\vec 0, \lambda\rangle$ of the particle by a proper orthochronous Lorentz transformation $L(p) = R(\phi,\theta,0) B_z(\beta)$, such that $L(p)|\vec 0,\lambda\rangle = |\vec p,\lambda\rangle$. Here $R(\phi,\theta,0)$ is a rotation in the Euler-angle parametrization, and $B_z(\beta)$ is the $z$-boost with rapidity $\beta$. The normalization of $|\vec p,\lambda\rangle$ is not worried here, as it's irrelevant to the question.

The one-particle states $|\vec p, \lambda\rangle$ are supposed to be called the helicity states because $\lambda$ is not the spin — that is, the eigenvalue of the rotation generator $J_3$ — but the helicity as the eigenvalue of the helicity operator $\vec J\cdot \vec P/|\vec p|$. The reason is that $[J_3,\vec P] \neq 0$ but $[\vec J\cdot P, \vec P] = 0$. Only in the rest frame or when $\vec p$ is set to be along the $z$-axis, $\lambda$ is the eigenvalue of $J_3$.

Now if we include the parity $\mathscr{P}$, since $[\mathscr{P},\vec J] = 0$ and $\{\mathscr{P},\vec P\} =0$, we would expect that $$ \mathscr{P} |\vec p,\lambda\rangle \propto |-\vec p, -\lambda\rangle, \tag1 $$ which can indeed be derived by noting that $[R_y(\pi)\mathscr{P}, B_z(\beta)]=0$. The derivation is then straightforward, as may be found in e.g. Pages 232-233 of Wu-Ki Tung's "Group theory in physics".

Nevertheless, in Weinberg's "Quantum field theory", although the way he obtained the one-particle states $|\vec p, \lambda\rangle$ in apparently the same way as by the method of induced representations, he seemed to assume that $\lambda$ is the eigenvalue of $J_3$, and thus $$ \mathscr{P} |\vec p, \lambda\rangle \propto |-\vec p, \lambda\rangle. \tag2 $$ The derivation is rather trivial: \begin{align*} \mathscr{P}|\vec p, \lambda\rangle&= \mathscr{P} L(p)|\vec 0, \lambda\rangle \\ &= R(\phi,\theta,0) \mathscr{P} B_z(\beta)|\vec 0, \lambda\rangle \\ &= R(\phi,\theta,0) \mathscr{P} B_z(\beta) \mathscr{P}^{-1} \mathscr{P}|\vec 0, \lambda\rangle\\ &= R(\phi,\theta,0)B_z(-\beta) \eta |\vec 0, \lambda\rangle \\ &= \eta L(\mathscr{P} p) |\vec 0, \lambda\rangle\\ &= \eta |\mathscr{P}\vec p, \lambda\rangle\\ &= \eta |-\vec p, \lambda\rangle, \tag3 \end{align*} where since when $\mathscr{P}$ is included, the little group of $\vec 0$ is $O(3)$, and $|\vec0, \lambda\rangle$ is an eigenstate of $\mathscr{P}$ with eigenvalue $\eta$ being the intrinsic helicity that depends on the species of the particle only.

I can't see any flaw neither in my derivation above nor in the derivation in Tung's book, so I'm confused whether $\mathscr{P}$ should flip $\lambda$ or not. I suppose perhaps the $\lambda$ (which is the $\sigma$) in Weinberg is different from the that in Tung; however, I can't see how because they both simply apply the method of induced representations.

Can anyone help? Thanks!

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  • $\begingroup$ You seem to be using the same symbol $\lambda$ for the pseudoscalar helicity $\hat\sigma\cdot\hat p$ and for the eigenvalues of the pseudovector $\vec\sigma$, which are different under parity. Any one source should be self-consistent, but both conventions are useful in different contexts. In my experience the notation $\left|\text{foo},\lambda\right>$ isn’t associated with a particular definition of $\lambda$ but is a shorthand for “we are really interested in ‘foo’ but we don’t want to forget that there are other variables which matter to our model.” Is that an answer, or can you clarify? $\endgroup$
    – rob
    Commented Jan 31, 2021 at 15:01
  • $\begingroup$ Right, I suppose the $\lambda$ differs from the $\sigma$ in Weinberg; however, I could not see how because the derivation in Weinberg seems implying that his $\sigma$ is the very helicity $\lambda$. The $\sigma$ was set to be foo in the first place but then was shown to be the internal index of the irreducible representation space of the little group. In the massive case, this $\sigma$ would be an index of the Wigner $D$-matrix. As such, I couldn't see why it is different from the $\lambda$. $\endgroup$
    – braids
    Commented Feb 1, 2021 at 0:36

2 Answers 2

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I finally realized that my confusion was due to a silly mistake. In Eq. (3) in the question, the second to the last step was wrong because $R(\phi,\theta,0)B_z(-\kappa) \neq L(\mathscr{P}p)$.

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I think your problem depends on whether the particle is massless or not, that is, the standard momenta associated to massless particle and massive particle are different. And $\sigma$ plays different roles in massless and massive particles.

For massive particle, it represents the spin; for massless particle, it represents the helicity in Weinberg’s book.

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  • $\begingroup$ No. See my own answer. $\endgroup$
    – braids
    Commented Feb 29 at 16:25
  • $\begingroup$ Hey @braids, could you explain why my points is wrong because I read through Weinberg's book, and he points out that $\sigma$ changes (doesn't change) sign of massless (massive) particle under parity. And I am confused about your question now. Could you explain it more precisely? Did you try to address the sign of helicity for massive and massless particle? Thanks! $\endgroup$ Commented Mar 1 at 14:35

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