The one particle states are defined as the states of an irreducible unitary representation of the Poincare Group. If this was not true, there would be states of a reducible representation that would not be connected by a Poincare transformation. These states are rather different particles. 

The Poincare Group has two Casimir operators, $P_\mu P^\mu$ and $W_\mu W^\mu$, where $P^\mu$ is the momentum generator and
$$W^\mu=-\frac{1}{2}\epsilon^{\mu\nu\sigma\rho}J_{\nu\sigma}P_\rho,$$
is the Pauli-Lubanski vector. The $J^{\mu\nu}$ are the Lorentz Group generator. The eigenvalues of these Casimirs label the irreducible representations.

We write the one particle states as
$$|p,\sigma\rangle,$$
where $p$ is the four momentum and $\sigma$ is the other label to be determined. The eigenvalues of $P_\mu P^\mu$ are $m^2$, the square mass of the particle. This gives rise to an infinite dimensional representation whose states are labeled by four momentum $p$. So we are left to find the irreducible representations of the homogenous Lorentz Group. However we have to consider the massive and massless cases separately.

The Little Group
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Let us first pick up a particular four momentum $k$. We write a general Lorentz group transformation as
$$\Lambda=L(\Lambda p)W(\Lambda,p)L^1(p),$$
where $L(p)$ is the boost relating $k$ and $p$,
$$L(p)k=p,$$
and
$$W(\Lambda,p)\equiv L^1(\Lambda p)\Lambda L(p),$$
is the so-called Wigner rotation. These elements form the so-called  Little Group which leaves the rest frame momentum $k$ invariant,
$$W(\Lambda,p)k=k.$$
Acting with $\Lambda$ on a state $|p,\sigma\rangle$,
$$\Lambda |p,\sigma\rangle=L(\Lambda p)W(\Lambda,p)|k,\sigma\rangle,$$ 
and noticing the resulting state must have four momentum $\Lambda p$ and be in a linear combination of states with the unknown label $\sigma$ we conclude that the $W(\Lambda,p)$ act on the unknown label $\sigma$. Therefore knowing the irreducible representation of the Little Group is what we need to know the irreducible representations of the Poincare Group.


Massive Particles
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In this case we can go to the rest frame, $p^\mu=(m,0,0,0)\equiv k^\mu$. We see that the Little Group leaving $k^\mu=(m,0,0,0)$ can be the rotation group in three dimensions, $SO(3)$, or even the more general $SU(2)$ which is a double cover of $SO(3)$. For the later case we know (standard Quantum Mechanics) that their irreducible representations are labeled by the spin $j=0,1/2,1,3/2,...$ and the total number of states for a given spin is $2j+1$.

Massless Particles
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There is no rest frame so we choose $P^\mu=(k,0,0,k)$. The Little Group leaving $k$ invariant is the Euclidean group in two dimensions $ISO(2)$ which consists of two translations and rotations in the $x^1x^2$ plane. The two translation generators give rise to another continuous eingenvalue $\theta$ but it is an experimental fact that there is no particle with $\theta\neq 0$. So we only need to consider the plane rotations. These rotations (about the $x^3$ axis) form the Abelian group $SO(2)$ whose elements are $e^{i\phi \vec J\cdot\vec e_3}$. Each representation of this group **has only one state**, and they are labeled by integers
$$h\equiv \vec J\cdot\vec e_3,$$
which we will call helicity. A massless particle in principle has one possible value of the helicity $h$ but from its definition the helicity is a pseudo-scalar. For a massless particle interacting through a parity conserving interaction we have to assign the two representations $h$ and $-h$ to represent the particle. That is why the phtoton has helicity $+1$ and $-1$ and the graviton has helicity $+2$ and $-2$.