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!
