In the inhomogenous Lorentz group $ISO(1,3)$, you have the space-time translation group $\mathbb{R}^{1,3}$, and the Lorentz group $SO(1,3)$.
You begin to find a representation of the space-time translation group, by choosing a momentum $p$. So your representation must have a $p$ index,
$$\psi_p \, .$$
After this, you will have to get the full representation, by finding a representation of the Lorentz group compatible with the momentum $p$, this will add another index $\sigma$ which corresponds to the polarization, so you will have a representation,
$$\psi_{p, \sigma} \, ,$$
which is the representation of the inhomogenous Lorentz group.
This answer is based on this article by A. Ungar.
Ungar computed the Thomas rotation formula which is almost what you need. I'll describe the general procedure, and in some cases, I'll refer you to Ungar for the proof. I'll express (just like Ungar), the Boosts in terms of velocities rather than momenta. If you wish you can repeat the exercise with the momentum parametrization. From Wikipedia we have
$$B(\mathbf{v}) = \begin{bmatrix}
\gamma_{\mathbf{v}}& \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v}^t \\
\frac{\gamma_{\mathbf{v}}}{c}\mathbf{v} & 1_{(3\times3)}+
\frac{\gamma_{\mathbf{v}}^2}{1+\gamma_{\mathbf{v}}}
\frac{\mathbf{v}\mathbf{v}^t}{c^2}
\end{bmatrix}$$
The key point is finding the Wigner rotation is the observation that every Lorentz transformation can be decomposed (nonuniquely) as a product of a Boost and a rotation:
$$\Lambda = B(\mathbf{u}) R$$
Any choice of the decomposition method will do, but we need to work in a fixed method of decomposition. I'll describe to you a possible method at the end of the answer
Now, when we multiply two boosts, we get a boost with the relativistic addition velocity + a rotation (often referred to as the Thomas rotation):
$$ B(\mathbf{u}) B(\mathbf{v}) = B(\mathbf{u}\oplus \mathbf{v}) \mathrm{Tom}(\mathbf{u},\mathbf{v})$$
Where $\mathrm{Tom}$ is a rotation matrix
Ungar found the general solution for the Thomas rotation (Equation : (13) in the article)
$$\mathrm{Tom}(\mathbf{u},\mathbf{v}) = B(-\mathbf{u}\oplus \mathbf{v}) B(\mathbf{u}) B(\mathbf{v})$$
Now, we are in a position to solve the equation for the Wigner rotation. We need to solve:
$$ \Lambda B(\mathbf{v}) = B(\Lambda \mathbf{v}) W$$
for $W$. We parametrize $\Lambda$, we get for the left handside:
$$\begin{align*}
\Lambda B(\mathbf{v}) & = B(\mathbf{u}) R B(\mathbf{v}) \\ & = B(\mathbf{u}) R B(\mathbf{v}) R^{-1} R \\ & = B(\mathbf{u}) B(\mathbf{Rv}) R \\ & = B(\mathbf{u}\oplus R\mathbf{v}) \mathrm{Tom}(\mathbf{u},R\mathbf{v}) R
\end{align*} $$
and for the right hand side
$$ B(\Lambda \mathbf{v}) W = B(B(\mathbf{u}) R\mathbf{v}) W = B(\mathbf{u}\oplus R\mathbf{v}) W$$
Thus:
$$W = \mathrm{Tom}(\mathbf{u},R\mathbf{v}) R$$
What is left is to describe a specific parametrization of a general Lorentz matrix into a boost and a rotation:
We need to find $B(\mathbf{u})$ and $R$ such that:
$$\begin{bmatrix}
\lambda_0 &\mathbf{\xi}^t \\
\mathbf{\eta} & \Lambda_1
\end{bmatrix} = \begin{bmatrix}
\gamma_{\mathbf{v}}& \frac{\gamma_{\mathbf{v}}}{c}\mathbf{v}^t \\
\frac{\gamma_{\mathbf{v}}}{c}\mathbf{v} & 1_{(3\times3)}+
\frac{\gamma_{\mathbf{v}}^2}{1+\gamma_{\mathbf{v}}}
\frac{\mathbf{v}\mathbf{v}^t}{c^2}
\end{bmatrix} \begin{bmatrix}
1&0 \\
0 &R_1
\end{bmatrix}$$
We observe that $R_1$ needs to satisfy the following relation
$$\mathbf{\xi} = R_1 \mathbf{\eta}$$
Thus $R_1$ needs to rotate the 3-vector $\mathbf{\mathbf{\eta}}$ into the 3-vector $\mathbf{\xi}$
The solution (using a similar method as Ungar) can be written as:
$$R_1 = 1_{(3\times3)}+ \sin(\theta) \Omega + (\cos(\theta-1)) \Omega^2$$
Where $\theta$ is the angle between the vectors $\mathbf{\eta}$ and $\mathbf{\xi}$
$$\sin(\theta) = \frac{\mathbf{\eta} \times \mathbf{\xi}}{|\mathbf{\eta}||\mathbf{\mathbf{\xi}}|}$$
and
$$\Omega_{ij} =\frac{ \eta_i \xi_j - \xi_i \eta_j}{|\mathbf{\eta}||\mathbf{\xi}|}$$
Best Answer
This just says that you can decompose any unitary representation of the Poincare group (= inhomogeneous Lorentz group) into irreducible representations.
He suggests to identify the irreducible representations with elementary particles, as suggested by the analogy irreducible = no longer decomposable = elementary. He doesn't really explain why (but only asserts that) it is natural to do that - this is summing up the experience of several generations of particle physicists: A single particle can be moved, rotated, and boosted, hence (in flat space-time) its Hilbert space must carry a unitary rep of the Poincare group. The particle is treated as elementary if this rep is irreducible as it cannot be decomposed. What is considered elementary depends on the resolution: in relativistic chemistry, all nuclei are treated as elementary particles, as the Poincare group acts irreducibly on their Hilbert space. In nuclear physics, nuclei are modelled in more detail as composite particles with a much more complex Hilbert space and a reducible representation of the Poincare group on it. Thus effectively Weinberg defines the notion of an elementary particle (in mathematical models) as being an irreducible representation of the Poincare group.
In view of Wigner's classification of irreducible unitary representations of Poincare, rederived by Weinberg in Chapter 5, elementary particles are classified into particle types by their mass and spin. The Hilbert space of an elementary particle of mass $m>0$ and spin $s$ is the space of $2s+2$-component wave functions $\psi(p)$ with $p_0=\sqrt{\mathbf{p}^2+(mc)^2}$ (defining the mass shell of mass $m$), with the corresponding irreducible representation of the Poincare group. (For the massless case see point 4 below.) A unitary representation consists of a Hilbert space and operators on this space generating the group (or a homomorphic image of it). For details see chapters B1 and B2 of my FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html
The standard model refines this classification by also specifying the irreducible representation of the gauge group of internal symmetries, giving rise to further quantum numbers. Conserved quantum numbers are nothing else than labels that tell you which irreducible representations is associated with the particle labelled by these numbers.
I don't think anyone understand Weinberg at first reading; though it is the best QFT book around if you want to understand the deeper reasons for why relativistic QFT is the way it is. So you may need to take some things based on a preliminary understanding, as proper understanding of what it all means requires at least that you covered the first 6 chapters.