Relativity, Lorentz symmetry, and spin repair

Relativistic calculations become much easier when three questions stay separate: Which spacetime relationships are invariant? Which finite matrix acts on a field’s components? Which unitary representation acts on physical one-particle states? A Lorentz transformation appears in all three questions, but the carrier space and the physical meaning change.

This focused review connects invariant intervals and causal cones to the proper orthochronous Lorentz group, its spin cover, Weyl and Dirac fields, and the little-group labels of particles. Its central lesson is that a field’s finite-dimensional Lorentz label does not by itself determine the particle content of a theory.

Required background. You should be able to multiply small matrices, distinguish vectors from covectors, and test whether a bilinear form is preserved. Review Linear and tensor methods if those steps are uncertain. This page uses the site’s spacetime and spinor conventions.

Intervals, cones, and Lorentz components

For a displacement $x^\mu$ in Minkowski spacetime, the inherited metric gives

$$x^2 = \eta_{\mu\nu}x^\mu x^\nu = (x^0)^2-\lvert\mathbf x\rvert^2.$$

A real linear transformation is Lorentz when

$$\Lambda^{\mathsf T}\eta\Lambda=\eta.$$

Consequently $(\Lambda x)^2=x^2$. The sign of the interval classifies a nonzero displacement:

• $x^2>0$ is timelike;
• $x^2=0$ is null; and
• $x^2<0$ is spacelike.

This classification is frame independent. Time order is also invariant for timelike- or null-related events once a time orientation has been chosen. For spacelike-separated events, different inertial frames can assign opposite time order without reversing any causal influence, because neither event is inside the other’s light cone.

Interval preservation alone does not select the physically connected group. The Lorentz group has disconnected components. The proper orthochronous component is

$$SO^+(1,3) = \left\{ \Lambda: \Lambda^{\mathsf T}\eta\Lambda=\eta, \ \det\Lambda=1, \ \Lambda^0{}_0\geq1 \right\}.$$

“Proper” preserves spacetime orientation; “orthochronous” preserves the future cone. Parity and time reversal lie outside this connected component and require additional representation data.

For example, a boost of rapidity $\chi$ along $x^1$ acts by

$$\begin{pmatrix}x'^0\\x'^1\end{pmatrix} = \begin{pmatrix} \cosh\chi&-\sinh\chi\\ -\sinh\chi&\cosh\chi \end{pmatrix} \begin{pmatrix}x^0\\x^1\end{pmatrix},$$

while $x^2$ and $x^3$ are unchanged. The identity $\cosh^2\chi-\sinh^2\chi=1$ verifies interval preservation directly. For a future massive momentum $p^\mu=(E,p,0,0)$, the rest-frame condition is

$$p'^1=-E\sinh\chi+p\cosh\chi=0, \qquad \tanh\chi=\frac pE.$$

Because $E^2-p^2=m^2$ and $E>0$, this boost gives $p'^0=m>0$. The invariant mass shell and the positive-energy branch are two distinct pieces of the statement.

The spin cover and finite field representations

Spinors do not carry ordinary representations of $SO^+(1,3)$. They carry representations of its double and universal cover,

$$\operatorname{Spin}^+(1,3) \cong SL(2,\mathbb C) \longrightarrow SO^+(1,3),$$

whose kernel is $\{\mathbf1,-\mathbf1\}$. The cleanest four-dimensional construction turns a real vector into a Hermitian matrix. With the Pauli matrices $\boldsymbol\sigma$,

$$X(x) = x^0\mathbf1+x^i\sigma_i = \begin{pmatrix} x^0+x^3&x^1-ix^2\\ x^1+ix^2&x^0-x^3 \end{pmatrix}.$$

Its determinant is the Minkowski norm:

$$\det X(x) = (x^0)^2-\lvert\mathbf x\rvert^2 =x^2.$$

For $A\in SL(2,\mathbb C)$, define

$$X(x')=A X(x)A^\dagger.$$

Hermiticity is preserved, and

$$\det X(x') = \det A\,\det X(x)\,\det A^\dagger = \det X(x).$$

Thus $A$ induces a proper orthochronous Lorentz transformation. Both $A$ and $-A$ induce the same vector transformation, because the two signs cancel in $AXA^\dagger$. A Weyl spinor, however, can transform as $\xi' = A\xi$ and therefore distinguishes the two lifts. In particular, the lift of a $2\pi$ spatial rotation is $-\mathbf1$: it is invisible on vectors but multiplies a Weyl spinor by $-1$. A $4\pi$ rotation acts trivially on both. This is why the cover is physical representation data rather than decorative notation. The general descent test is developed in Groups, actions, quotients, and covers.

Finite-dimensional complex field modules are conventionally labeled by $(j_L,j_R)$:

Field component module	Complex dimension	Content under spatial rotations
Scalar $(0,0)$	$1$	$0$
Weyl $(\tfrac12,0)$	$2$	$\tfrac12$
Conjugate Weyl $(0,\tfrac12)$	$2$	$\tfrac12$
Four-vector $(\tfrac12,\tfrac12)$	$4$	$0\oplus1$
Dirac $(\tfrac12,0)\oplus(0,\tfrac12)$	$4$	$\tfrac12\oplus\tfrac12$

Which Weyl module is called left- or right-handed depends on the declared boost and chirality conventions; the two modules are conjugate and parity exchanges them. A Dirac field contains both. These finite matrices are generally not unitary in a positive-definite component-space inner product, which is compatible with the noncompact Lorentz group.

For a Dirac field, no explicit gamma-matrix basis is required. If $S(\Lambda)$ is its spinor matrix, covariance is encoded by

$$S(\Lambda)^{-1}\gamma^\mu S(\Lambda) = \Lambda^\mu{}_{\nu}\gamma^\nu,$$

together with

$$\psi'(x')=S(\Lambda)\psi(x), \qquad \overline\psi'(x')=\overline\psi(x)S(\Lambda)^{-1}.$$

It follows immediately that

$$\overline\psi'\psi'=\overline\psi\psi, \qquad \overline\psi'\gamma^\mu\psi' = \Lambda^\mu{}_{\nu}\overline\psi\gamma^\nu\psi.$$

The first bilinear is a scalar and the second a vector. This is an intertwiner check, not a consequence of calling $\psi$ a spinor. The dimension-, signature-, and conjugation-dependent construction is treated in Spinors, conjugations, bilinears, chirality, and Fierz identities and Dreiner, Haber, and Martin 2010, §§2–3.

Worked bridge: one boost on vectors and Weyl spinors

Lift the $x^1$ boost above to

$$A(\chi) = \exp\!\left(-\frac\chi2\sigma_1\right) = \cosh\!\frac\chi2\,\mathbf1 - \sinh\!\frac\chi2\,\sigma_1.$$

The eigenvalues are $e^{-\chi/2}$ and $e^{+\chi/2}$, so $\det A(\chi)=1$. Because $A$ is Hermitian, direct multiplication gives

$$\begin{aligned} A\mathbf1 A^\dagger &=\cosh\chi\,\mathbf1-\sinh\chi\,\sigma_1,\\ A\sigma_1 A^\dagger &=-\sinh\chi\,\mathbf1+\cosh\chi\,\sigma_1,\\ A\sigma_2 A^\dagger&=\sigma_2, & A\sigma_3 A^\dagger&=\sigma_3. \end{aligned}$$

Substitution into $X(x')=AX(x)A^\dagger$ reproduces the boost matrix in the first section. The same group element acts on a Weyl field by

$$\xi'(x')=A(\chi)\xi(x).$$

This is the bridge: $X$ and $\xi$ transform under different representations of the same covering-group element. Replacing $A$ by $-A$ leaves the vector unchanged and reverses the spinor sign.

Now add a massive on-shell momentum. Choose a standard boost $L(p)$ that takes $k=(m,\mathbf0)$ to $p$. For a general Lorentz transformation, the combination

$$W(\Lambda,p) = L(\Lambda p)^{-1}\Lambda L(p)$$

fixes $k$ and therefore belongs to the massive little group. Plane-wave spinors provide an intertwiner between the field matrix and the little-group matrix. With covariantly normalized spinors and compatible phase and spin-basis conventions,

$$S(\Lambda)u_s(p) = \sum_r u_r(\Lambda p) D_{rs}\!\left(W(\Lambda,p)\right).$$

On the left, $S(\Lambda)$ mixes finite spinor components. On the right, $D(W)$ mixes the physical spin label attached to a one-particle momentum. The coefficient functions $u_s(p)$ connect the two representations; they do not identify them. For a collinear boost in a canonical spin basis the Wigner rotation is trivial, but for general non-collinear transformations it need not be.

Field representations are not particle spins

A particle classification uses the covering Poincaré group, including translations. It acts unitarily on a one-particle Hilbert space. Momentum ranges over an orbit, so even a fixed-spin irreducible representation is normally infinite-dimensional. Under the usual irreducibility and positive-energy assumptions:

• for $m>0$, the standard momentum is $(m,\mathbf0)$ and the little group is $SU(2)$, giving spin $s=0,\tfrac12,1,\ldots$;
• for $m=0$, a null standard momentum has little group given by the cover of $ISO(2)$. In ordinary finite-helicity sectors its translation-like part acts trivially, leaving a helicity label.

These are the ordinary massive and finite-helicity branches of the unitary Poincaré classification Wigner 1939, §§4–7.

The contrast is therefore structural:

Question	Field representation	One-particle representation
Acting group	Spin cover of the Lorentz group on component indices, plus a transformed spacetime argument	Covering Poincaré group, including translations
Carrier	Finite-dimensional component space	Positive-definite Hilbert space over a momentum orbit
Typical transformation	Generally nonunitary finite matrix	Unitary, normally infinite-dimensional operator
Main labels	$(j_L,j_R)$, direct sums, chirality or reality data	Mass shell plus unitary little-group representation

One counterexample is enough to break the shortcut “four-vector field means spin-one particle.” The derivative $\partial^\mu\phi$ transforms as a four-vector but can create a scalar particle. A Proca field, after its field equation removes the nondynamical component, describes a massive spin-one sector. A Maxwell potential uses the same Lorentz component module but has gauge redundancy and, in the ordinary physical sector, creates massless helicity $+1$ and $-1$ states. Dynamics, constraints, gauge equivalence, the state space, and the mass shell perform the conversion from field components to particle content. A systematic account appears in Lorentz field representations and Poincaré particle representations and Weinberg 1995, §§2.5, 5.1, and 5.5.

Limits of this repair

Keep the following qualifications visible:

• The discussion uses four-dimensional flat spacetime and the connected proper orthochronous group. Parity, time reversal, curved spacetime, and other dimensions require additional structures.
• Chirality is a finite-dimensional field-representation property in even dimensions. Helicity labels massless one-particle states. They are related by a field equation and energy branch in familiar free theories, but they are not synonyms. For a massive state, helicity can change under a boost that reverses the momentum, and spin is organized by the $SU(2)$ little group, while chirality continues to act on field components.
• A Lorentz-covariant field transformation does not by itself establish locality, positivity, a spectrum condition, an equation of motion, or the existence of particles.
• The massless little group also admits continuous-spin representations when its translation-like part acts nontrivially. Those sectors are outside this entry lesson, so “massless means helicity” is being used only for ordinary finite-helicity sectors.
• Spin–statistics is a theorem with locality, positivity, and other hypotheses. It does not follow from the double cover alone.

Continue to Fermions, spin, and anticommutation for a free Dirac field, or to Vector fields, constraints, and gauge redundancy for the spin-one branch.

Exercises

1. Rest frame, future cone, and spacelike time order

Let $v^\mu=(5,3,0,0)$ and choose the boost satisfying $\tanh\chi=3/5$. Find $v'^\mu$ and verify its norm and time orientation. Then take $s^\mu=(1,2,0,0)$ and find a proper orthochronous boost for which $s'^0<0$. Explain why the second result does not violate causality.

Solution

For $\tanh\chi=3/5$,

$$\cosh\chi=\frac54, \qquad \sinh\chi=\frac34.$$

Therefore

$$\begin{aligned} v'^0&=\frac54(5)-\frac34(3)=4,\\ v'^1&=-\frac34(5)+\frac54(3)=0. \end{aligned}$$

The norm is preserved:

$$v^2=25-9=16=(v'^0)^2-(v'^1)^2.$$

The original vector is future timelike and the transformed time component is positive, as required for an orthochronous transformation.

For $s^\mu=(1,2,0,0)$, choose any boost with $1/2<\tanh\chi<1$; for example $\tanh\chi=3/4$. Then

$$s'^0 = \cosh\chi\left(1-2\tanh\chi\right) <0.$$

No causal order has been reversed because $s^2=1-4=-3<0$: the two events are spacelike separated. A timelike or null future-directed displacement could not acquire a negative time component under the same proper orthochronous boost.

2. Verify the double-cover boost

With $A(\chi)=c\mathbf1-s\sigma_1$, where $c=\cosh(\chi/2)$ and $s=\sinh(\chi/2)$, verify the four conjugation formulas in the worked bridge. Show explicitly why $A$ and $-A$ give the same vector transformation but opposite Weyl-spinor transformations.

Solution

Using $\sigma_1^2=\mathbf1$,

$$A^2 = (c^2+s^2)\mathbf1-2cs\,\sigma_1 = \cosh\chi\,\mathbf1-\sinh\chi\,\sigma_1.$$

Since $A$ commutes with $\sigma_1$,

$$A\sigma_1A = \sigma_1A^2 = -\sinh\chi\,\mathbf1+\cosh\chi\,\sigma_1.$$

The Pauli anticommutators give $\sigma_1\sigma_2+\sigma_2\sigma_1=0$ and $\sigma_1\sigma_2\sigma_1=-\sigma_2$, so

$$A\sigma_2A = (c^2-s^2)\sigma_2 =\sigma_2.$$

The same calculation gives $A\sigma_3A=\sigma_3$. Here $A^\dagger=A$. Substituting these results into $AX(x)A^\dagger$ yields precisely the $x^1$-boosted components.

For the other lift,

$$(-A)X(-A)^\dagger=AXA^\dagger,$$

so the induced vector transformation is unchanged. But $\xi' = A\xi$ becomes $\xi'=-A\xi$ for the same vector transformation, so the Weyl spinor detects the nontrivial element of the covering kernel.

3. One field module, three particle outcomes

All of $\partial^\mu\phi$, a Proca field $B^\mu$, and a Maxwell potential $A^\mu$ transform as four-vectors. State the ordinary one-particle content associated with each free theory and name the extra input that distinguishes the three answers.

Solution

The operator $\partial^\mu\phi$ is a vector-valued derivative of a scalar field. Acting on the vacuum, it can have nonzero overlap with the same massive or massless spin-zero state created by $\phi$; the vector index comes from the momentum factor, not from an independent spin-one little-group multiplet.

A massive Proca field obeys an equation that implies a transversality constraint. On the massive shell the remaining three polarizations transform as the spin-one representation of the $SU(2)$ little group.

A Maxwell potential has gauge redundancy. After imposing the constraint and quotienting null or gauge directions, the ordinary massless physical sector has helicities $+1$ and $-1$, not a three-state massive spin multiplet.

Thus the common four-vector component module is insufficient. The field equations, mass shell, constraints, gauge equivalence, state-space inner product, and little-group action determine the particle outcome.

Re-check and return

Repeat the Lorentz-and-spin part of the classical-field and relativity diagnostic with a new boost and a field different from the one used above. Your work is ready to carry forward when it includes:

• an interval and causal-class check;
• the Lorentz-group component and time orientation being used;
• the transformed spacetime argument and the finite field-component matrix;
• the cover when a spinor representation is involved;
• one scalar or covariant bilinear verified by an intertwiner identity; and
• a separate mass shell, little group, and spin or helicity label for a one-particle state.

If all six parts are explicit, return to Core QFT or to the specialist path that sent you here. If the component calculation works but the acting space remains unclear, write “spacetime vector,” “field component,” or “one-particle state” beside every index and repeat the worked bridge. If the norm or future-cone check fails, return to the first section and diagnose the boost before adding spinors.

References

• Herbi K. Dreiner, Howard E. Haber, and Stephen P. Martin, “Two-Component Spinor Techniques and Feynman Rules for Quantum Field Theory and Supersymmetry,” Physics Reports 494, nos. 1–2 (2010): 1–196, doi:10.1016/j.physrep.2010.05.002, Open PDF.
• Steven Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press, 1995, doi:10.1017/CBO9781139644167.
• Eugene P. Wigner, “On Unitary Representations of the Inhomogeneous Lorentz Group,” Annals of Mathematics 40, no. 1 (1939): 149–204, doi:10.2307/1968551.