Massive and Massless Representations

A particle’s internal spin labels are controlled by the transformations that leave one reference momentum unchanged, its little group. A massive momentum can be brought to rest, where this group is $SU(2)$ on the Lorentz cover. A nonzero massless momentum has no rest frame and has a different stabilizer, related to plane rotations and translations. Its finite-helicity representations require the translation part to act trivially.

Required background. Poincaré Group fixes the invariants, SL(2, ℂ) and the Lorentz Group constructs the covering map, and Helicity and Chirality defines helicity.

A little group preserves a standard momentum

Use natural units $\hbar=c=1$ on the positive-energy shell. Choose a standard momentum $k$ and define

$$G_k=\{\Lambda:\Lambda k=k\}.$$

Any momentum on the same orbit is $p=L(p)k$ for a suitable Lorentz transformation $L(p)$. Stabilizers at different momenta are related by conjugation, so choosing $k$ simplifies the calculation without changing the representation type.

The little group acts on the internal labels at fixed momentum. It is distinct from the finite-dimensional Lorentz action on all components of a covariant wave equation.

Massive particles have a rest-frame rotation group

For $m>0$, take $k=(m,0,0,0)$. Its Hermitian matrix is $K=mI$. A lift $A\in SL(2,\mathbb C)$ preserves it exactly when

$$AKA^\dagger=K \quad\Longleftrightarrow\quad AA^\dagger=I.$$

Thus the stabilizer in the cover is $SU(2)$, mapping to ordinary rest-frame rotations. Its irreducible unitary representations have spin $s=0,1/2,1,\ldots$ and dimension $2s+1$.

In the Pauli–Lubanski convention used here, $W^0=0$ and $\mathbf W=-m\mathbf J$ at rest. The invariants are

$$P^2=m^2,\qquad W^2=-m^2s(s+1).$$

These labels describe a positive-energy one-particle representation. A second charge species, an antiparticle, or an internal flavor label is extra information, not a factor already included in $2s+1$.

The massless stabilizer contains null rotations

For a nonzero massless momentum, take $k=(\kappa,0,0,\kappa)$ with $\kappa>0$, so $K=\operatorname{diag}(2\kappa,0)$. Writing a general determinant-one matrix and imposing $AKA^\dagger=K$ gives

$$A= \begin{pmatrix} e^{-i\theta/2}&z\\ 0&e^{+i\theta/2} \end{pmatrix}, \qquad z\in\mathbb C.$$

The first column must be $(e^{-i\theta/2},0)^T$; the determinant fixes the lower diagonal entry. The angle supplies a rotation about the momentum direction, while the two real components of $z$ supply null rotations. The group is the double cover of the orientation-preserving Euclidean group of the plane, $ISO(2)$.

“Translations” here means the two-parameter normal subgroup of this little group. They are Lorentz null rotations in spacetime, not the spacetime translations generated by $P^\mu$.

Using the dimensionless generator convention of Weyl Spinors, one may choose

$$N_1=K_x-J_y,\qquad N_2=K_y+J_x.$$

They leave the reference null momentum fixed and obey

$$[N_1,N_2]=0,\qquad [J_z,N_1]=iN_2,\qquad [J_z,N_2]=-iN_1.$$

This is the plane-motion algebra, unlike the three rotation generators of a massive rest frame.

Finite helicity and the excluded continuous-spin branch

For the usual finite-helicity particle representations, the null-rotation subgroup acts trivially on the physical one-particle states. The remaining rotation acts as

$$D(\theta)=e^{-ih\theta},$$

where $h$ is helicity. Since the rotation parameter has period $4\pi$ on the cover, $2h$ is an integer. Each fixed-helicity representation has one internal state per momentum. Multiple helicities form a direct sum.

With $\epsilon^{0123}=+1$ and $W^\mu=\frac12\epsilon^{\mu\nu\rho\sigma}P_\nu M_{\rho\sigma}$, the present signs give

$$W^\mu=-hP^\mu,\qquad P^2=W^2=0.$$

The minus sign can be checked from $W^0=-\mathbf P\cdot\mathbf J$. Sources defining the Pauli–Lubanski vector with the opposite overall sign instead write $W^\mu=+hP^\mu$. Neither convention changes the Casimirs.

Allowing a nontrivial unitary action of the null-rotation subgroup leads to the continuous-spin branch with infinitely many internal states and nonzero negative $W^2$. It is excluded by the finite-helicity assumption, not by the massless dispersion relation alone. The two zero Casimirs also fail to distinguish different finite helicities.

Counting states and taking a massless limit

Representation or field content	Physical internal states per momentum
Massive spin $s$ particle species	$2s+1$
One massless finite-helicity irrep	One
Photon with both helicities	Two, $h=+1$ and $h=-1$
Massive Dirac particle, charge fixed	Two spin states
Massless left-chiral field	One particle helicity and the opposite antiparticle helicity

The four components of a vector potential or Dirac spinor must not be read directly as four physical polarizations. Equations, constraints, gauge equivalence, frequency sectors, and internal charges all enter the count.

For example, a massive spin-one species has three spin states, whereas a Maxwell photon has two helicities. Simply setting $m=0$ in a count $2s+1$ does not implement the gauge constraints or decide how a longitudinal sector behaves. A limiting theory can retain an additional decoupled sector or a nontrivial coupling, depending on its dynamics. The representation-theoretic change of stabilizer must be combined with a specified dynamical limit.

The little-group analysis determines kinematic labels. It does not derive interactions, enforce particle statistics by itself, or establish that every mathematical representation is realized by an observed particle.

Exercises

1. Why does the massive little group not contain a nontrivial boost that leaves $k=(m,0,0,0)$ fixed?

Solution

A boost changes the rest momentum to one with a nonzero spatial part. In the Hermitian construction, the stabilizer condition forces $A$ to be unitary, excluding a nontrivial positive Hermitian boost lift.

2. Verify $[N_1,N_2]=0$ from the Lorentz algebra.

Solution

The commutator is $[K_x,K_y]+[K_x,J_x]-[J_y,K_y]-[J_y,J_x]$. The middle terms vanish; the others are $-iJ_z$ and $+iJ_z$, which cancel.

3. Do $P^2=W^2=0$ distinguish a scalar from a photon?

Solution

No. They vanish for every finite-helicity massless representation. The little-group rotation phase supplies $h=0$ for a scalar and $h=\pm1$ for the photon sectors. Including both photon helicities is an additional direct-sum statement, not information contained in those two Casimirs.

References

• N. Straumann, “Unitary Representations of the inhomogeneous Lorentz Group and their Significance in Quantum Physics,” 2008, arXiv:0809.4942 — standard momenta and little-group representations.
• S. Weinberg, The Quantum Theory of Fields, Vol. I: Foundations, Cambridge University Press, 1995 — particle classification and helicity.
• E. P. Wigner, “On Unitary Representations of the Inhomogeneous Lorentz Group,” Annals of Mathematics 40, 149–204, 1939, doi:10.2307/1968551 — original classification.