The Complexification Trick#

The three commutation relations above mix $J$‘s and $K$‘s in an awkward way. The goal is to find linear combinations that decouple. Define:

$N_i^+ = \frac{1}{2}(J_i + iK_i), \qquad N_i^- = \frac{1}{2}(J_i - iK_i)$

Now compute the commutators. Using the Lorentz algebra relations (this is a straightforward but instructive exercise — work it out at least once):

$[N_i^+, N_j^+] = i\epsilon_{ijk}N_k^+$

$[N_i^-, N_j^-] = i\epsilon_{ijk}N_k^-$

$[N_i^+, N_j^-] = 0$

This is remarkable. The $N^+$ generators form an $\mathfrak{su}(2)$ algebra by themselves. The $N^-$ generators form a separate $\mathfrak{su}(2)$ algebra. And the two copies don’t talk to each other at all. The Lorentz algebra has decomposed:

$\mathfrak{so}(3,1)_\mathbb{C} \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$

A crucial subtlety: this decomposition works for the complexified Lorentz algebra — we had to multiply $K_i$ by $i$ to define $N_i^\pm$, so we’ve extended the algebra over the complex numbers. The real Lorentz algebra $\mathfrak{so}(3,1)$ is not the same as $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$ as a real Lie algebra. It is isomorphic to $\mathfrak{sl}(2,\mathbb{C})$, which is the complexification of $\mathfrak{su}(2)$. The distinction matters when we discuss unitarity — but for the purpose of classifying representations, the complexified version is exactly what we need.

Classification by $(j_+, j_-)$#

Since we have two independent $\mathfrak{su}(2)$ algebras, every finite-dimensional representation of the Lorentz algebra is labeled by two half-integers:

$(j_+, j_-)$

where $j_+$ labels the representation of $N^+$ and $j_-$ labels the representation of $N^-$. Each $j_\pm = 0, \frac{1}{2}, 1, \frac{3}{2}, \ldots$ follows the same rules as angular momentum in quantum mechanics. The dimension of the representation is:

$\dim = (2j_+ + 1)(2j_- + 1)$

This is the payoff of the decomposition. Instead of wrestling with the full Lorentz algebra (where boosts make everything non-compact and painful), we’ve reduced the classification problem to something we already know: two copies of angular momentum.

From the definitions $N_i^\pm = \frac{1}{2}(J_i \pm iK_i)$, we can recover the physical generators:

$J_i = N_i^+ + N_i^-, \qquad K_i = -i(N_i^+ - N_i^-)$

The Casimir operators for the two $\mathfrak{su}(2)$‘s are $\mathbf{N}^{+\,2}$ and $\mathbf{N}^{-\,2}$, with eigenvalues $j_+(j_+ + 1)$ and $j_-(j_- + 1)$ respectively.

$(0, 0)$ — Scalar#

Dimension: $1 \times 1 = 1$. Both $N^+ = 0$ and $N^- = 0$, so $J_i = 0$ and $K_i = 0$. Nothing transforms. This is the trivial representation, which we already met in Part II.

$(\frac{1}{2}, 0)$ — Left-Handed Weyl Spinor#

Dimension: $2 \times 1 = 2$. The $N^+$ generators act as the spin-$\frac{1}{2}$ representation (i.e., the Pauli matrices divided by 2), while $N^- = 0$:

$N_i^+ = \frac{\sigma_i}{2}, \qquad N_i^- = 0$

Recovering the physical generators:

$J_i = N_i^+ + N_i^-, \qquad K_i = -i(N_i^+ - N_i^-) = -i\frac{\sigma_i}{2}$

The rotation generators $J_i = \frac{\sigma_i}{2}$ are Hermitian — this is expected, since rotations are generated by Hermitian operators in quantum mechanics.

The boost generators $K_i = -\frac{i}{2}\sigma_i$ are anti-Hermitian. This is the fingerprint of non-compactness: the Lorentz group is not compact, so its finite-dimensional representations cannot be unitary. Concretely, boosting a left-handed Weyl spinor does not preserve its norm. This is physically sensible — Lorentz boosts are not symmetries of any positive-definite inner product on spinor space.

An object transforming in the $(\frac{1}{2}, 0)$ representation is called a left-handed Weyl spinor, denoted $\psi_L$ or $\chi_\alpha$ (with a two-component undotted index $\alpha = 1, 2$). Under a Lorentz transformation:

$\psi_L \to \exp\left(-i\boldsymbol{\theta}\cdot\frac{\boldsymbol{\sigma}}{2} - \boldsymbol{\eta}\cdot\frac{\boldsymbol{\sigma}}{2}\right)\psi_L$

Note the relative sign: the rotation and boost terms enter with opposite signs (one has $-i$, the other has $-1$).

$(0, \frac{1}{2})$ — Right-Handed Weyl Spinor#

Dimension: $1 \times 2 = 2$. Now $N^+ = 0$ and $N^-$ acts as spin-$\frac{1}{2}$:

$N_i^+ = 0, \qquad N_i^- = \frac{\sigma_i}{2}$

The physical generators:

$J_i = \frac{\sigma_i}{2}, \qquad K_i = -i(0 - \frac{\sigma_i}{2}) = +i\frac{\sigma_i}{2}$

The rotation generators are the same as for the left-handed spinor — both transform as spin-$\frac{1}{2}$ under rotations. But the boost generators have the opposite sign. This is the entire distinction between left-handed and right-handed: they rotate the same way but boost differently.

A right-handed Weyl spinor is denoted $\psi_R$ or $\bar{\chi}^{\dot{\alpha}}$ (with a dotted index). Under a Lorentz transformation:

$\psi_R \to \exp\left(-i\boldsymbol{\theta}\cdot\frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\eta}\cdot\frac{\boldsymbol{\sigma}}{2}\right)\psi_R$

Comparing with the left-handed case: the rotation piece $-i\boldsymbol{\theta}\cdot\frac{\boldsymbol{\sigma}}{2}$ is the same, but the boost piece flips sign. A parity transformation ($\mathbf{x} \to -\mathbf{x}$) reverses the boost direction but not the rotation — so parity exchanges $(\frac{1}{2}, 0) \leftrightarrow (0, \frac{1}{2})$, left-handed $\leftrightarrow$ right-handed.

$(\frac{1}{2}, \frac{1}{2})$ — Vector#

Dimension: $2 \times 2 = 4$. This is a 4-dimensional representation — and it is, in fact, equivalent to the vector representation we already constructed with explicit $4\times4$ matrices in Part II. The four-vector $V^\mu$ transforms in the $(\frac{1}{2}, \frac{1}{2})$ representation.

The connection can be made explicit via the map $V^\mu \to V_{\alpha\dot{\alpha}} = V^\mu (\sigma_\mu)_{\alpha\dot{\alpha}}$, where $\sigma_\mu = (\mathbf{1}, \sigma_1, \sigma_2, \sigma_3)$. Under a Lorentz transformation, the undotted index $\alpha$ transforms under $(\frac{1}{2}, 0)$ and the dotted index $\dot{\alpha}$ transforms under $(0, \frac{1}{2})$. The product gives $(\frac{1}{2}, 0) \otimes (0, \frac{1}{2}) = (\frac{1}{2}, \frac{1}{2})$.

Higher Representations#

The pattern continues. The $(1, 0)$ and $(0, 1)$ representations are 3-dimensional each and correspond to self-dual and anti-self-dual antisymmetric tensors. The $(1, 1)$ representation is 9-dimensional and corresponds to symmetric traceless tensors. The representation $(j_+, j_-)$ for general $j_\pm$ describes higher-spin objects. In practice, most of particle physics lives in the representations listed above.

Spinors in Non-Relativistic QM#

In non-relativistic quantum mechanics, you learn that spin-$\frac{1}{2}$ particles are described by two-component objects $\chi = \begin{pmatrix} \chi_1 \\ \chi_2 \end{pmatrix}$ that transform under rotations as:

$\transform{\chi} \to e^{-i\boldsymbol{\theta}\cdot\boldsymbol{\sigma}/2}\,\chi$

This is a representation of the rotation group $SU(2)$. There’s no mention of boosts because non-relativistic QM doesn’t have Lorentz symmetry — only rotational symmetry. The two-component spinor is a representation of $SU(2)$ and nothing more.

Spinors in Relativistic QM#

When we upgrade to special relativity, we need representations of the full Lorentz group, not just the rotation subgroup. A single two-component spinor is no longer enough to describe a massive particle — we need to specify how it boosts, not just how it rotates. This is where the $(j_+, j_-)$ classification becomes essential.

A left-handed Weyl spinor $\psi_L$ and a right-handed Weyl spinor $\psi_R$ both transform as spin-$\frac{1}{2}$ under rotations, but they transform differently under boosts. Non-relativistic QM doesn’t see the difference because there are no boosts. Relativistic QM must, and this is why we need the full Lorentz representation theory.

Scalar Fields — $(0, 0)$#

A scalar field $\phi(x)$ has no indices and transforms as:

$\phi(x) \to \phi'(x) = \phi(\Lambda^{-1}x)$

The field value doesn’t change — it just gets “moved” to the new location. The Higgs field in the Standard Model is a (complex) scalar field. The Klein-Gordon equation $(\partial^2 + m^2)\phi = 0$ describes a free scalar field.

Weyl Fields — $(\frac{1}{2}, 0)$ and $(0, \frac{1}{2})$#

A left-handed Weyl field $\psi_L(x)$ has two components and transforms as:

$\psi_L(x) \to \exp\left(-i\boldsymbol{\theta}\cdot\frac{\boldsymbol{\sigma}}{2} - \boldsymbol{\eta}\cdot\frac{\boldsymbol{\sigma}}{2}\right)\psi_L(\Lambda^{-1}x)$

A right-handed Weyl field $\psi_R(x)$ similarly transforms with the opposite boost sign.

Weyl fields describe massless fermions (in the standard treatment). The equation of motion for a left-handed Weyl field is:

$i\bar{\sigma}^\mu\partial_\mu\psi_L = 0$

where $\bar{\sigma}^\mu = (\mathbf{1}, -\sigma_1, -\sigma_2, -\sigma_3)$. This is a two-component equation for a two-component field — elegant and minimal.

Why are Weyl fields associated with massless particles? A mass term would look like $m\psi_L^\dagger\psi_R$, which requires both a left-handed and a right-handed field. A single Weyl field by itself cannot have a Lorentz-invariant mass term (a left-handed field alone can’t form a scalar bilinear with itself under the full Lorentz group — including boosts — because $\psi_L^\dagger\psi_L$ is not Lorentz invariant). So a theory with only $\psi_L$ and no $\psi_R$ describes a massless particle.

Before the discovery of neutrino oscillations (which imply neutrino masses), neutrinos were thought to be described by a single left-handed Weyl field. The situation is now more subtle, but Weyl fields remain the fundamental building blocks.

Dirac Fields — $(\frac{1}{2}, 0) \oplus (0, \frac{1}{2})$#

To describe a massive fermion, we need both chiralities. The Dirac field is a reducible representation of the Lorentz group:

$\Psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$

This is a four-component object built from a left-handed Weyl spinor (top two components) and a right-handed Weyl spinor (bottom two). It transforms as the direct sum $(\frac{1}{2}, 0) \oplus (0, \frac{1}{2})$ — reducible, because the two Weyl spinors don’t mix under Lorentz transformations.

What couples them is the mass term. The Dirac equation:

$\left(i\gamma^\mu\partial_\mu - m\right)\Psi = 0$

mixes $\psi_L$ and $\psi_R$ through the mass $m$. In the massless limit $m \to 0$, the equation decouples into two independent Weyl equations, and the left-handed and right-handed components propagate independently.

The gamma matrices $\gamma^\mu$ are $4\times4$ matrices that intertwine the two Weyl representations. In the chiral (Weyl) basis:

$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{pmatrix}$

where $\sigma^\mu = (\mathbf{1}, \boldsymbol{\sigma})$ and $\bar{\sigma}^\mu = (\mathbf{1}, -\boldsymbol{\sigma})$. The off-diagonal structure is precisely what couples $\psi_L$ to $\psi_R$.

The electron, muon, quarks — all the massive fermions in the Standard Model — are described by Dirac fields.

Majorana Fields#

A Majorana field is a Dirac field with an additional constraint: the particle is its own antiparticle. Formally, this is the condition:

$\Psi^c = \Psi$

where $\Psi^c = C\bar{\Psi}^T$ is the charge conjugate, and $C$ is the charge conjugation matrix (whose explicit form depends on your gamma matrix convention).

In terms of Weyl components, the Majorana condition sets $\psi_R = i\sigma_2\psi_L^*$. So a Majorana field has only two independent degrees of freedom, not four — the right-handed component is fully determined by the left-handed component (or vice versa).

The key difference from a Dirac field:

• A Dirac field has four independent components: $\psi_L$ and $\psi_R$ are unrelated.
• A Majorana field has two independent components: $\psi_R$ is the charge conjugate of $\psi_L$.

Majorana fields can have mass (unlike a single Weyl field), but the mass term looks different — it’s a Majorana mass $\frac{1}{2}m\psi_L^T(i\sigma_2)\psi_L + \text{h.c.}$, which violates lepton number by two units. This is why Majorana masses for neutrinos, if they exist, would have deep implications: they would imply lepton number violation and could be connected to the matter-antimatter asymmetry of the universe.

Whether neutrinos are Dirac or Majorana particles is one of the major open questions in particle physics. Neutrinoless double beta decay experiments are designed specifically to answer it.

Vector Fields — $(\frac{1}{2}, \frac{1}{2})$#

A vector field $A^\mu(x)$ has four components and transforms as:

$A^\mu(x) \to \Lambda^\mu{}_\nu\,A^\nu(\Lambda^{-1}x)$

This is the representation we studied in Part II, now promoted to a field. The photon field in electrodynamics and the $W^\pm, Z$ bosons of the weak interaction are vector fields.

A subtlety: a massive vector field has three physical degrees of freedom (the three polarizations), not four. The timelike component $A^0$ is not an independent propagating degree of freedom — it’s eliminated by the constraint equations (or, in the language of gauge theory, by gauge fixing). For a massless vector field like the photon, only two polarizations are physical (the two transverse modes), and gauge invariance is essential for consistency.