# Irreducible representations of the Poincaré Group: states obtained from standard momentum states

This question is an extension of Irreducible Representations Of Lorentz Group.

Following the exposition in Weinberg, let $k^\mu$ be a standard 4-momentum vector from which we obtain any other momentum $p^\mu$, $p^2=k^2$, by a Lorentz transformation $L_p k = p$. Let $|k,\sigma\rangle$ be an eigenstate of momenta $P^\mu$ and let operator $S$ be an operator, commutable with $P^\mu$, such that $$S |k, \sigma \rangle = \sigma |k, \sigma\rangle. \qquad (*)$$ From the answers and comments to the question linked to above, it seems to follow that it is not in general true that $$S U(L_p) |k, \sigma \rangle = \sigma |k, \sigma\rangle, \qquad (**)$$ where $U$ is a unitary irreducible representation of the Poincaré group.

However, in the book

Costa, Fogli -- Symmetries and Group Theory in Particle Physics: An Introduction to Space-Time and Internal Symmetries

on pages 71-72, the approach is somewhat different. They note that the $L_p$ for a given $k$ is not fixed: if $l$ is a Lorentz transformation such that $lk=k$, then $\Lambda_p k \equiv L_p l k = L_p k$. Then they make the following claim:

We can fix $\Lambda_p$ in such a way that $U(\Lambda_p)$ leaves the quantum number $\sigma$ unchanged

in the equation $$|p, \sigma \rangle = U(\Lambda_p) |k, \sigma\rangle.$$

My questions are:

1) Do they essentially mean the same thing as in Weinberg?

2) Where am I going wrong in the example below?

or

3) Do they imply that for a given $\sigma$, i.e. $S$, one can find such $l$ that equations $(*)$ and $(**)$ hold, with $L_p = \Lambda_p$

4) If 3) holds, could it also be true that given any $L_p$, one can find such an operator $S$ that equations $(*)$ and $(**)$ hold?

5) Could you point me to a proof of 3) or 4)?

An example of 2). Consider a particle of mass $m$ and choose the standard momentum $k^\mu = (m,0,0,0)$. Let $\sigma$ be the eigenvalue $w^3$ of the third component $W^3$of the Pauli-Lubanski vector $$W^\mu = \frac{1}{2} \varepsilon^{\mu\nu\rho\sigma} M_{\nu\rho} P_{\sigma}.$$ Since we can factor any Lorentz transformation into a product of a pure boost and a pure rotation, let $B_p$ and $R_p$ be such a boost and rotation that $\Lambda_p k \equiv B_p R_p k = B_p k = p$. In other words, $B_p$ is determined and $R_p$ plays the role of $l$ defined above. Requiring that $$W^3 U(\Lambda_p) |k,w^3\rangle = w^3 U(\Lambda_p) |k,w^3\rangle$$ and noting that $W^\mu$ transforms as a vector under Lorentz transformations we obtain $$U(\Lambda_p) W^3 |k,w^3\rangle = U(\Lambda_p) U^{-1}(\Lambda_p) W^3 U(\Lambda_p) |k, w^3\rangle = U(\Lambda_p) (\Lambda_p)^{3}_{\;\nu} W^\nu |k, w^3\rangle.$$ Since we have $W^\nu$ acting on a rest state, we know that they act as $W^i = -m J^i$ and $W^0=0$, $J^i$ being the angular momentum generators. Since $J^i$ are images of a Lie algebra (vector) homomorphism, they must be linearly independent and it follows that $$\delta^3_{i} = (\Lambda_p)^{3}_{\;i}= (B_p R_p)^3_{\;i}.$$ Since $B_p$ is known, the above equations constitute three equations in three (independent) parameters of the rotation $R_p$ and, hence, at least in principle, $R_p$ is uniquely determined. The above equations also imply that $\Lambda_p$ leaves $w^3$ invariant, which is what one would intuitively expect? And equations $(*)$ and $(**)$ hold for $S=W^3$ and $L_p = B_p R_p$.