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

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

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

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 $-2$.

In Quantum Field Theory 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 Casimirs

If we have an irreducible representation of a group then Schur's Lemma says that an operator that commutes with all generators, a Casimir Operator, must be a multiple of the identity. Then applying this operator to any state of the representation gives the same eigenvalue (sometimes also called Casimir). We use the eigenvalues of different representations to label them. This is exaclty what we do in Quantum Mechanics when we use the Casimir $J^2$ and they eigenvalues $j$ to label irreducible representations of the angular momenum algebra.

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. We can assume therefore that we have two labels for the irreducible representations of the Poincare Group.

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

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 $p$, $$L(p)k=p,$$ $$W(\Lambda,p)\equiv L^{-1}(\Lambda p)\Lambda L(p),$$ is the so-called Wigner rotation and the $L^{-1}$ denote the inverse transformation. 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

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

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 $-2$.

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

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

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

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 $x^1-x^2$ 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 $-2$.

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

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

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

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 $-2$.