Definition#

A general Poincaré transformation acts on spacetime coordinates as:

$x^\mu \to x'^\mu = \Lambda^\mu{}_\nu\, x^\nu + a^\mu$

where $\Lambda$ is a Lorentz transformation and $a^\mu$ is a constant translation four-vector. The Poincaré group is the set of all such transformations, and it’s a 10-parameter group: six parameters for $\Lambda$ (three rotations, three boosts) plus four parameters for $a^\mu$ (one time translation, three spatial translations).

The Poincaré group is a semi-direct product of the Lorentz group and the translation group:

$\text{Poincaré} = \mathbb{R}^{3,1} \rtimes SO^+(3,1)$

The semi-direct product (rather than a direct product) means that Lorentz transformations and translations don’t simply commute — a Lorentz transformation acts on translations by rotating/boosting the translation vector. This will show up in the algebra.

Generators#

The Poincaré group has 10 generators:

• $J^{\mu\nu}$ — the six Lorentz generators (three rotations $J^i$, three boosts $K^i$), as before
• $P^\mu$ — four translation generators (energy $P^0 = H$ and three-momentum $P^i$)

The Poincaré Algebra#

The commutation relations are:

$[J^{\mu\nu}, J^{\rho\sigma}] = i\left(\eta^{\nu\rho}J^{\mu\sigma} - \eta^{\mu\rho}J^{\nu\sigma} - \eta^{\nu\sigma}J^{\mu\rho} + \eta^{\mu\sigma}J^{\nu\rho}\right)$

$[J^{\mu\nu}, P^\rho] = i\left(\eta^{\nu\rho}P^\mu - \eta^{\mu\rho}P^\nu\right)$

$[P^\mu, P^\nu] = 0$

Let’s unpack what these say:

The first relation is the Lorentz algebra from Part II, now written in fully covariant tensor notation. It encodes $[J^i, J^j] = i\epsilon^{ijk}J^k$, $[J^i, K^j] = i\epsilon^{ijk}K^k$, and $[K^i, K^j] = -i\epsilon^{ijk}J^k$ all in one compact formula.

The second relation says that $P^\rho$ transforms as a 4-vector under Lorentz transformations. This is physically obvious: if you rotate your coordinate system, the components of momentum rotate accordingly. In terms of the rotation and boost generators:

$[J^i, P^j] = i\epsilon^{ijk}P^k \qquad \text{(momentum is a 3-vector under rotations)}$

$[J^i, P^0] = 0 \qquad \text{(energy is a scalar under rotations)}$

$[K^i, P^j] = -iP^0\delta^{ij} \qquad \text{(boosts mix energy and momentum)}$

$[K^i, P^0] = -iP^i \qquad \text{(boosts mix energy and momentum)}$

The last two are exactly what you’d expect from special relativity: a boost in the $x$-direction mixes $P^0$ (energy) with $P^x$ (momentum along $x$).

The third relation says that translations commute. This is the statement that spacetime is flat — there’s no curvature. In general relativity, spacetime is curved and translations along different directions don’t commute; the Poincaré group would be replaced by something more complicated. But for the flat spacetime of particle physics, $[P^\mu, P^\nu] = 0$.

First Casimir: $P^2$#

$C_1 = P^\mu P_\mu = P^2$

This is the squared four-momentum — the mass squared:

$P^2 = (P^0)^2 - \mathbf{P}^2 = m^2$

You can verify that $[P^2, J^{\mu\nu}] = 0$ (since $P^\mu$ transforms as a vector, $P^2$ is a Lorentz scalar) and $[P^2, P^\mu] = 0$ (since translations commute). So $P^2$ commutes with all ten generators.

On each irreducible representation, $P^2$ takes a definite value $m^2$. This is the mass of the particle. The fact that mass is a Casimir — a label of the representation, not a dynamical variable — is why all electrons have exactly the same mass: they all live in the same irreducible representation.

Second Casimir: $W^2$#

The second Casimir is constructed from the Pauli-Lubanski pseudovector:

$W^\mu = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}J_{\nu\rho}P_\sigma$

where $\epsilon^{\mu\nu\rho\sigma}$ is the totally antisymmetric Levi-Civita tensor with $\epsilon^{0123} = +1$.

The Pauli-Lubanski vector has several important properties:

• $W^\mu P_\mu = 0$ — it’s orthogonal to the four-momentum. This follows from the antisymmetry of $\epsilon^{\mu\nu\rho\sigma}$ contracted with the symmetric product $P_\sigma P_\mu$.

• $W^\mu$ transforms as a four-vector under Lorentz transformations (it has one free index $\mu$) and commutes with translations ($[W^\mu, P^\nu] = 0$).

• The Casimir is $C_2 = W^\mu W_\mu = W^2$, which commutes with all ten generators.

The physical meaning of $W^2$ depends on whether the particle is massive or massless. We’ll see this shortly.

Why Unitary?#

In quantum mechanics, symmetry transformations must be represented by unitary operators (or anti-unitary, for time reversal). This is Wigner’s theorem: any symmetry of a quantum system — any map that preserves transition probabilities — must be implemented by a unitary or anti-unitary operator on the Hilbert space.

For the Lorentz group alone, we saw in Part III that finite-dimensional representations are not unitary (the boost generators are anti-Hermitian). This isn’t a contradiction: the finite-dimensional representations describe how field components transform, not how quantum states transform. Quantum states live in an infinite-dimensional Hilbert space, where unitary representations do exist.

The Poincaré group representations we’re looking for are infinite-dimensional (they act on the space of one-particle states of definite momentum), and they are unitary.

The Method of Induced Representations#

Wigner’s strategy is elegant:

Step 1: Since $P^2 = m^2$ is a Casimir, fix $m^2$ and consider the set of all four-momenta satisfying $p^2 = m^2$. This is a hyperboloid in momentum space (for $m^2 > 0$) or the light cone (for $m^2 = 0$).

Step 2: Pick a standard (reference) momentum $k^\mu$ on that surface. Any other momentum $p^\mu$ on the same surface can be reached from $k^\mu$ by a Lorentz transformation: $p^\mu = L(p)^\mu{}_\nu\, k^\nu$ for some “standard boost” $L(p)$.

Step 3: Find the little group — the subgroup of Lorentz transformations that leave $k^\mu$ invariant:

$\Lambda^\mu{}_\nu\, k^\nu = k^\mu$

The little group depends on the choice of $k^\mu$, but different choices on the same mass shell give isomorphic little groups.

Step 4: Classify the irreducible representations of the little group. These determine the “internal” degrees of freedom of the particle — spin, polarization, helicity.

Step 5: Build the full Poincaré representation by “boosting” the little group representation to all other momenta. This is the induced representation.

The result: every irreducible unitary representation of the Poincaré group is determined by the mass $m$ and an irreducible representation of the little group.

Let’s now carry this out for the two physical cases.

Standard Momentum#

For a massive particle, the natural choice of standard momentum is the rest frame:

$k^\mu = (m, 0, 0, 0)$

This is the momentum of a massive particle sitting at rest. Every other momentum $p^\mu = (E, \mathbf{p})$ on the mass shell $p^2 = m^2$ can be reached by boosting from the rest frame.

Little Group: $SO(3)$#

Which Lorentz transformations leave $k^\mu = (m, 0, 0, 0)$ invariant? The boosts certainly don’t — boosting a particle at rest gives it momentum. But rotations leave the rest-frame momentum unchanged: rotating a particle at rest still leaves it at rest, with the same energy $m$ and zero three-momentum.

Therefore, the little group for massive particles is the rotation group $SO(3)$, or equivalently its double cover $SU(2)$.

Spin#

The irreducible representations of $SU(2)$ are labeled by the spin $j = 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots$, and have dimension $2j + 1$.

In the rest frame, the Pauli-Lubanski vector simplifies dramatically. Since $k^\mu = (m, \mathbf{0})$:

$W^0 = \frac{1}{2}\epsilon^{0ijk}J_{ij}P_k = 0 \quad \text{(in the rest frame, } P_k = 0\text{)}$

$W^i = \frac{1}{2}\epsilon^{i0jk}J_{jk}P_0 = m\,\frac{1}{2}\epsilon^{ijk}J_{jk} = m\,J^i$

So in the rest frame, $W^\mu = (0, m\,\mathbf{J})$ — the Pauli-Lubanski vector is simply mass times the angular momentum. The Casimir becomes:

$W^2 = W^\mu W_\mu = -\mathbf{W}^2 = -m^2\,\mathbf{J}^2 = -m^2\,j(j+1)$

This is the physical content of the second Casimir: $W^2$ measures the spin. A massive particle is characterized by:

$\boxed{P^2 = m^2, \qquad W^2 = -m^2\,j(j+1)}$

A massive spin-$j$ particle has $2j + 1$ independent polarization states (the $2j + 1$ values of $J^z$ in the rest frame: $m_j = -j, -j+1, \ldots, j-1, j$). For example:

• Spin-0 (Higgs boson): 1 state
• Spin-$\frac{1}{2}$ (electron): 2 states (spin up, spin down)
• Spin-1 ($W$ boson): 3 states (three polarizations)
• Spin-2 (hypothetical massive graviton): 5 states

Standard Momentum#

For a massless particle, there is no rest frame — massless particles travel at the speed of light in every frame. The standard momentum is conventionally chosen along the $z$-axis:

$k^\mu = (\omega, 0, 0, \omega)$

for some energy $\omega > 0$. This satisfies $k^2 = \omega^2 - \omega^2 = 0$.

Little Group: $ISO(2)$#

Which Lorentz transformations leave $k^\mu = (\omega, 0, 0, \omega)$ invariant?

Rotations about the $z$-axis (the direction of motion) clearly do — they rotate the transverse $xy$-plane while leaving $k^\mu$ unchanged. This gives a $U(1)$ subgroup generated by $J^z$.

But there are also less obvious transformations. It turns out there are two additional generators that leave $k^\mu$ invariant. These are combinations of rotations and boosts in the transverse directions:

$\Pi_1 = K^x + J^y, \qquad \Pi_2 = K^y - J^x$

You can verify directly that $[\Pi_1, P^\mu]k_\mu = 0$ and $[\Pi_2, P^\mu]k_\mu = 0$ — these generators leave the standard momentum invariant. Together with $J^z$, they satisfy:

$[J^z, \Pi_1] = i\Pi_2, \qquad [J^z, \Pi_2] = -i\Pi_1, \qquad [\Pi_1, \Pi_2] = 0$

This is the algebra of $ISO(2)$ — the Euclidean group in two dimensions, consisting of rotations and translations of a 2D plane.

Compare this with the massive case:

The little group has changed topology — from the compact group $SO(3)$ to the non-compact group $ISO(2)$. This is not a smooth deformation. You cannot continuously go from one to the other, and this will have physical consequences.

Dealing with $ISO(2)$: The Translation Generators#

The representations of $ISO(2)$ come in two types:

Faithful representations: The “translation” generators $\Pi_1, \Pi_2$ have nonzero eigenvalues. These give the continuous spin representations, where the spin is not discrete but takes continuous values. These representations describe particles with an infinite number of polarization states for each momentum — exotic objects that have never been observed in nature. They are discarded on physical grounds (they lead to pathological behavior: infinite degeneracy, problems with localizability, and conflicts with observed particle spectra).

Unfaithful representations: We set $\Pi_1 = \Pi_2 = 0$ and keep only the $J^z$ generator. The little group effectively reduces to $SO(2) \cong U(1)$, the group of rotations in the transverse plane.

Every physical massless particle corresponds to the second case.

Helicity#

The irreducible representations of $U(1)$ are one-dimensional, labeled by a single number — the helicity:

$J^z \,|k, \lambda\rangle = \lambda\, |k, \lambda\rangle$

where $\lambda$ is the component of angular momentum along the direction of motion. Since $J^z$ generates rotations by angle $\theta$ via $e^{-i\theta J^z}$, and a $2\pi$ rotation must give $\pm 1$ (for bosons/fermions respectively), $\lambda$ must be an integer or half-integer: $\lambda = 0, \pm\frac{1}{2}, \pm 1, \pm\frac{3}{2}, \pm 2, \ldots$

A massless particle with helicity $\lambda$ has only one polarization state for each $\lambda$. But CPT symmetry (required in any local relativistic quantum field theory) demands that if a state with helicity $+\lambda$ exists, so must a state with helicity $-\lambda$. So physical massless particles come in pairs of helicity states: $(+\lambda, -\lambda)$.

The contrast with massive particles is stark:

For spin-$\frac{1}{2}$, the count happens to agree — both massive and massless have 2 states. But for higher spins, the massless case has fewer states. A massive spin-1 particle (like the $W$ boson) has 3 polarizations; a massless spin-1 particle (the photon) has only 2. The “missing” longitudinal polarization of the photon is a direct consequence of the little group being $ISO(2)$ rather than $SO(3)$.

This is why the Higgs mechanism is necessary: when a gauge boson acquires mass (via the Higgs field), it must pick up a third polarization — the longitudinal mode — which it “eats” from the Goldstone boson. The transition from 2 polarizations to 3 is the transition from $ISO(2)$ to $SO(3)$ as the little group.