Written in May 2026, backdated to when the work happened. This post is a reflection, not a contemporaneous journal entry.

Classical Field Theory: A Comprehensive Reference

Scalar fields, gauge theory, spinors, and the Standard Model Lagrangian; everything you need before QFT.

The Lagrangian mechanics document introduced the basic framework of classical field theory. The special relativity and tensors document built the mathematical language. This document combines them into the real subject: a complete classical treatment of the field theories that underlie modern physics.

By the end, you should understand:

• Why spontaneous symmetry breaking leads to Goldstone bosons (and to the Higgs mechanism)
• Why the photon exists at all; not as a postulate but as a consequence of local $U(1)$ symmetry
• How Yang-Mills theory generalizes QED to describe the strong and weak forces
• Why the Dirac equation automatically predicts antimatter and spin
• What the Standard Model Lagrangian actually says, term by term

That’s the bridge into quantum field theory proper.

Table of Contents

1. The Setup: What We’re Building
2. Real Scalar Fields
3. Complex Scalar Fields and Global U(1)
4. Spontaneous Symmetry Breaking and Goldstone Bosons
5. Gauge Symmetry: Why Local Matters
6. Scalar Electrodynamics and Minimal Coupling
7. Non-Abelian Gauge Theory (Yang-Mills)
8. Spinor Representations of the Lorentz Group
9. The Dirac Equation
10. Quantum Electrodynamics as a Classical Field Theory
11. The Higgs Mechanism
12. The Standard Model Lagrangian
13. Appendix: Formulas and Conventions

1. The Setup: What We’re Building

Every fundamental theory of physics is specified by a Lagrangian (density). From this single object; a Lorentz-invariant combination of fields and their derivatives; everything else follows: equations of motion (via Euler-Lagrange), conservation laws (via Noether), interactions, and ultimately (after quantization) scattering amplitudes and particle physics.

The rules for constructing fundamental Lagrangians are surprisingly constrained:

1. Lorentz invariance; the Lagrangian must be a Lorentz scalar
2. Locality; it’s a function of fields and their derivatives at a single spacetime point (no action-at-a-distance)
3. Hermiticity / reality; $\mathcal{L}$ is real (so the action is real)
4. Gauge symmetries; if present, the Lagrangian respects them
5. Renormalizability; for quantum field theory, typically only operators up to mass dimension 4 in 4D spacetime (this is a quantum criterion, but it massively constrains classical writing)
6. Discrete symmetries; depending on the theory, may respect P, C, T, or specific combinations

These rules plus particle content typically narrow the Lagrangian to a handful of possibilities, often just one. The Standard Model Lagrangian is essentially the Lagrangian you get from requiring Lorentz invariance, gauge invariance under $SU(3) \times SU(2) \times U(1)$, and the observed particle content.

With those rules in mind, let’s build.

Natural Units Reminder

Throughout: $\hbar = c = 1$, metric $\eta_{\mu\nu} = \text{diag}(+,-,-,-)$, Einstein summation implicit.

2. Real Scalar Fields

The simplest possible field: a single real number at every spacetime point. No direction, no internal structure, just a value.

$\phi(x) \in \mathbb{R}$

The Free Lagrangian

$\mathcal{L} = \tfrac{1}{2}(\partial_\mu \phi)(\partial^\mu \phi) - \tfrac{1}{2} m^2 \phi^2$

Each term is a Lorentz scalar. The first is a “kinetic” term (involves derivatives); the second is a “mass” term. The factors of $1/2$ are conventional, chosen to make the equations of motion look clean.

Euler-Lagrange

From the general field equation $\partial_\mu(\partial \mathcal{L}/\partial(\partial_\mu \phi)) - \partial \mathcal{L}/\partial \phi = 0$:

$\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} = \partial^\mu \phi, \qquad \frac{\partial \mathcal{L}}{\partial \phi} = -m^2 \phi$

So:

$\partial_\mu \partial^\mu \phi + m^2 \phi = 0$

$\boxed{(\Box + m^2)\phi = 0}$

This is the Klein-Gordon equation. It’s the relativistic wave equation for a spin-0 particle of mass $m$.

Plane-Wave Solutions

Try $\phi(x) = e^{-ik \cdot x} = e^{-i(k^0 t - \vec k \cdot \vec x)}$:

$\Box \phi = -k^\mu k_\mu \phi = -(k^{0 2} - |\vec k|^2)\phi$

So Klein-Gordon requires:

$k^\mu k_\mu = m^2 \quad \Longleftrightarrow \quad k^{0 2} = |\vec k|^2 + m^2$

This is exactly the relativistic energy-momentum relation if we identify $k^0 = E$ and $\vec k = \vec p$. Klein-Gordon plane waves describe particles of mass $m$.

General solution: a superposition of plane waves with both positive and negative $k^0$:

$\phi(x) = \int \frac{d^3 k}{(2\pi)^3 \, 2\omega_k}\left[a(\vec k)\, e^{-i k \cdot x} + a^*(\vec k)\, e^{+i k \cdot x}\right]$

where $\omega_k = +\sqrt{|\vec k|^2 + m^2}$ is the positive energy. In classical field theory this is a real-valued solution; in the quantum version, $a$ and $a^*$ get promoted to annihilation/creation operators.

The Canonical Momentum

$\pi(x) = \frac{\partial \mathcal{L}}{\partial \dot \phi} = \dot\phi = \partial^0 \phi$

This is what gets quantized into the commutator $[\hat\phi(\vec x, t), \hat\pi(\vec y, t)] = i\delta^3(\vec x - \vec y)$.

Adding Interactions

The simplest interacting scalar theory adds a $\phi^4$ term:

$\mathcal{L} = \tfrac{1}{2}(\partial_\mu \phi)(\partial^\mu \phi) - \tfrac{1}{2} m^2 \phi^2 - \frac{\lambda}{4!}\phi^4$

The $\phi^4$ term is the simplest Lorentz-invariant interaction for a real scalar that’s renormalizable (mass dimension 4). The equation of motion becomes:

$(\Box + m^2)\phi = -\frac{\lambda}{6}\phi^3$

Nonlinear. This theory; called ”$\phi^4$ theory”; is the simplest interacting field theory and the standard pedagogical example in QFT. It’s also a prototype for the Higgs sector (just with a different potential, as we’ll see).

3. Complex Scalar Fields and Global U(1)

A complex scalar has two real components:

$\phi(x) \in \mathbb{C}$

Equivalently $\phi = (\phi_1 + i\phi_2)/\sqrt{2}$ for two real fields $\phi_1, \phi_2$.

Lagrangian

$\mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^* \phi$

(Note: no factor of 1/2 in front. That’s because $\phi$ and $\phi^*$ are treated as independent fields in the variational principle, so there’s an implicit factor-of-2 doubling compared to the real case.)

Equations of Motion

Varying with respect to $\phi^*$ (treating $\phi$ as independent):

$(\Box + m^2)\phi = 0$

and similarly for $\phi^*$. Two complex Klein-Gordon equations; or equivalently, two real ones for $\phi_1$ and $\phi_2$.

The U(1) Symmetry

The Lagrangian is invariant under the transformation

$\phi \to e^{i\alpha} \phi, \qquad \phi^* \to e^{-i\alpha}\phi^*$

for any constant real $\alpha$. The two factors cancel in products like $\phi^*\phi$ and $(\partial \phi^*)(\partial \phi)$.

This is a global symmetry; $\alpha$ is the same at every spacetime point. It’s called $U(1)$ because the set of transformations $\{e^{i\alpha}\}$ forms the group $U(1)$ (unit circle in complex plane).

Noether Current

Apply Noether’s theorem. The infinitesimal variation is $\delta \phi = i\alpha \phi$, $\delta \phi^* = -i\alpha \phi^*$. The Noether current (derived in the Lagrangian doc):

$j^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta\phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi^*)}\delta\phi^*$

$= (\partial^\mu \phi^*)(i\alpha\phi) + (\partial^\mu \phi)(-i\alpha \phi^*) = i\alpha[\phi^*(\partial^\mu \phi) - \phi(\partial^\mu \phi^*)]$

Dropping the arbitrary $\alpha$ factor (conventional), and also dropping the $i$ to make the current real:

$j^\mu = i[\phi^*(\partial^\mu \phi) - \phi(\partial^\mu \phi^*)]$

You can verify directly: $\partial_\mu j^\mu = 0$ using the Klein-Gordon equation.

The conserved charge $Q = \int d^3 x \, j^0$ is what we call electric charge (up to overall factor). So electric charge conservation ultimately comes from a $U(1)$ symmetry of the Lagrangian; a beautiful instance of Noether’s theorem.

Interpretation

In quantum theory, the two complex components encode particles and antiparticles. A complex scalar describes a charged particle (like a $\pi^+$) together with its antiparticle ($\pi^-$). A real scalar (like a $\pi^0$, approximately) has no such internal structure; it’s its own antiparticle.

4. Spontaneous Symmetry Breaking and Goldstone Bosons

Now we modify the potential. Consider a complex scalar with Lagrangian:

$\mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - V(\phi^*\phi)$

with:

$V(\phi^*\phi) = -\mu^2 \phi^*\phi + \lambda (\phi^*\phi)^2$

Here $\mu^2 > 0$ and $\lambda > 0$. Note the wrong sign on the mass-like term: the coefficient of $\phi^*\phi$ is negative, whereas a normal mass term would be positive.

The “Mexican Hat” Potential

Plot $V$ as a function of $|\phi|$. At $|\phi| = 0$, we’re at a local maximum, not a minimum. The minimum is at:

$\frac{dV}{d(\phi^*\phi)} = -\mu^2 + 2\lambda \phi^*\phi = 0 \implies \phi^*\phi = \frac{\mu^2}{2\lambda} \equiv \frac{v^2}{2}$

So the minimum is a ring in the complex $\phi$ plane, at $|\phi| = v/\sqrt{2}$ where $v = \mu/\sqrt{\lambda}$ is the vacuum expectation value.

The potential is rotationally symmetric (invariant under $U(1)$), but the minimum is not a single point; it’s a circle, and the system has to pick one point on that circle.

Spontaneous Symmetry Breaking

The $U(1)$ symmetry of the Lagrangian is still exact. But the ground state (vacuum) is not symmetric; once the field picks a direction on the circle, you can’t transform to another without doing work.

This is spontaneous symmetry breaking: the laws are symmetric, the vacuum is not.

Classical analogy: a ball rolling on a rotationally symmetric Mexican-hat surface. The surface doesn’t favor any direction. But the ball eventually settles at one point, and from then on, the system as a whole (ball + surface) has broken the symmetry.

Expanding Around the Vacuum

Let’s choose the vacuum at $\phi = v/\sqrt{2}$ (real, no imaginary part). Write the field as small fluctuations:

$\phi(x) = \frac{1}{\sqrt{2}}[v + \eta(x) + i\xi(x)]$

where $\eta$ and $\xi$ are real fields. Substitute into the Lagrangian and expand. The algebra is tedious but instructive; the result is:

$\mathcal{L} = \tfrac{1}{2}(\partial \eta)^2 + \tfrac{1}{2}(\partial \xi)^2 - \mu^2 \eta^2 + (\text{cubic and quartic terms}) + \text{const}$

The key observations:

• The field $\eta$ (fluctuation along the radial direction; up the side of the hat) has a mass term: $m_\eta^2 = 2\mu^2$
• The field $\xi$ (fluctuation along the tangential direction; around the rim) has no mass term at all

Goldstone’s Theorem

The massless field $\xi$ is a Goldstone boson. Goldstone’s theorem says: for every spontaneously broken continuous symmetry, there is a massless scalar particle in the spectrum.

Why? Because fluctuations along the broken-symmetry direction cost no energy; you’re moving along the valley, where the potential is flat. “No energy cost” translates to “no mass term.”

More formally: broken symmetry generators produce massless particles. The number of Goldstone bosons equals the number of broken generators.

Generalization

For a broken symmetry group $G \to H$ (where $H$ is the unbroken subgroup), the number of Goldstone bosons is $\dim G - \dim H$. In our $U(1)$ example: $\dim U(1) - \dim\{\text{trivial}\} = 1 - 0 = 1$ Goldstone. Matches.

Physical Examples

• Pions in QCD are (approximately) Goldstone bosons of broken chiral symmetry. They’re not exactly massless (the symmetry is only approximate), so pions are pseudo-Goldstones.
• Phonons in solids are Goldstone bosons of broken translational symmetry.
• Magnons in ferromagnets are Goldstones of broken rotational symmetry.

Goldstone’s theorem is a deep statement about broken continuous symmetries, with applications well beyond particle physics.

Foreshadowing

In a moment, we’ll make the $U(1)$ symmetry local (gauge). When a gauge symmetry is spontaneously broken, Goldstone’s theorem gets modified: instead of a massless Goldstone boson, the would-be Goldstone gets “eaten” by the gauge boson, giving it a mass. This is the Higgs mechanism. We’ll work it through in section 11.

5. Gauge Symmetry: Why Local Matters

Global vs. Local Symmetry

A global symmetry transforms the field the same way everywhere: $\phi \to e^{i\alpha}\phi$ with $\alpha$ a constant.

A local (or gauge) symmetry allows $\alpha$ to depend on spacetime: $\phi(x) \to e^{i\alpha(x)}\phi(x)$.

Global symmetries are relatively tame; they yield conservation laws via Noether. Local symmetries are much stronger: they constrain the form of the Lagrangian itself, forcing new fields into existence.

The Problem

Start with the complex scalar Lagrangian $\mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^*\phi$ and try to make it locally invariant. Under $\phi \to e^{i\alpha(x)}\phi$:

$\partial_\mu \phi \to e^{i\alpha} \partial_\mu \phi + i(\partial_\mu \alpha) e^{i\alpha}\phi$

The extra term $i(\partial_\mu \alpha)e^{i\alpha}\phi$ breaks the invariance of the kinetic term. Under local $U(1)$, the Lagrangian is not invariant.

The Fix: the Covariant Derivative

Introduce a new four-vector field $A_\mu(x)$, transforming as

$A_\mu \to A_\mu + \frac{1}{e}\partial_\mu \alpha$

and define the covariant derivative:

$D_\mu \phi \equiv (\partial_\mu + ie A_\mu)\phi$

Check: under the combined local $U(1)$ transformation (on $\phi$) and the above shift (on $A_\mu$):

$D_\mu \phi \to (\partial_\mu + ieA_\mu + i\partial_\mu \alpha)(e^{i\alpha}\phi)$

$= e^{i\alpha}\partial_\mu \phi + i(\partial_\mu \alpha)e^{i\alpha}\phi + ieA_\mu e^{i\alpha}\phi + i(\partial_\mu\alpha)e^{i\alpha}\phi \cdot (-1)$

Wait; let me redo this carefully. Under $\phi \to \phi' = e^{i\alpha}\phi$ and $A_\mu \to A'_\mu = A_\mu + (1/e)\partial_\mu \alpha$:

$D'_\mu \phi' = (\partial_\mu + ieA'_\mu)(e^{i\alpha}\phi)$

$= \partial_\mu(e^{i\alpha}\phi) + ie[A_\mu + (1/e)\partial_\mu \alpha] e^{i\alpha}\phi$

$= e^{i\alpha}\partial_\mu \phi + i(\partial_\mu \alpha)e^{i\alpha}\phi + ieA_\mu e^{i\alpha}\phi + i(\partial_\mu \alpha)e^{i\alpha}\phi$

Hmm, the $i(\partial_\mu \alpha) e^{i\alpha}\phi$ terms add, not cancel. Let me reconsider the sign in the $A_\mu$ transformation. The right choice is:

$A_\mu \to A_\mu - \frac{1}{e}\partial_\mu \alpha$

Then:

$D'_\mu \phi' = e^{i\alpha}\partial_\mu \phi + i(\partial_\mu \alpha)e^{i\alpha}\phi + ieA_\mu e^{i\alpha}\phi - i(\partial_\mu \alpha)e^{i\alpha}\phi$

$= e^{i\alpha}[\partial_\mu \phi + ieA_\mu \phi] = e^{i\alpha} D_\mu \phi$

The covariant derivative transforms covariantly; just like $\phi$ itself, with the same phase. This is the whole point.

Summary: the Gauge Structure

Under a local $U(1)$ transformation with parameter $\alpha(x)$:

$\phi \to e^{i\alpha}\phi$

$A_\mu \to A_\mu - \frac{1}{e}\partial_\mu \alpha$

$D_\mu \phi \to e^{i\alpha}(D_\mu \phi)$

The covariant derivative transforms like the field. Therefore $(D_\mu \phi)^* (D^\mu \phi)$ is gauge invariant, and so is $\phi^*\phi$. We can write a gauge-invariant Lagrangian:

$\mathcal{L}_{\text{matter}} = (D_\mu \phi)^*(D^\mu \phi) - m^2 \phi^*\phi - \lambda (\phi^*\phi)^2$

What About $A_\mu$ Itself?

The gauge field $A_\mu$ needs its own kinetic term. From the tensor doc, we know the gauge-invariant combination is:

$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$

Under the gauge transformation $A_\mu \to A_\mu - (1/e)\partial_\mu \alpha$:

$F_{\mu\nu} \to F_{\mu\nu} - (1/e)(\partial_\mu \partial_\nu \alpha - \partial_\nu \partial_\mu \alpha) = F_{\mu\nu}$

(Partial derivatives commute.) So $F_{\mu\nu}$ is gauge invariant, and $-\tfrac{1}{4}F_{\mu\nu}F^{\mu\nu}$ is the correct kinetic term.

Can $A_\mu$ Have a Mass?

A mass term for $A_\mu$ would be $\tfrac{1}{2} m_A^2 A_\mu A^\mu$. But under the gauge transformation, $A_\mu A^\mu$ is not invariant. A naive mass term breaks gauge symmetry.

So gauge symmetry forces gauge bosons to be massless. The photon is massless because electromagnetism has gauge symmetry.

(Exception: if the gauge symmetry is spontaneously broken; the Higgs mechanism; mass can sneak in without explicitly breaking the symmetry. Section 11.)

The Full Classical Theory

Putting it all together:

$\mathcal{L} = (D_\mu \phi)^*(D^\mu \phi) - m^2 \phi^*\phi - \lambda(\phi^*\phi)^2 - \tfrac{1}{4} F_{\mu\nu}F^{\mu\nu}$

This is a complete, gauge-invariant, Lorentz-invariant, renormalizable theory describing a complex scalar field coupled to an Abelian gauge field. It’s called scalar electrodynamics.

The Moral

Gauge symmetry isn’t a convenience or a mathematical trick. It’s a principle that demands the existence of gauge fields. Start with a free complex scalar and insist on local $U(1)$ invariance; a new field $A_\mu$ must exist, with specific transformation law and specific coupling. The photon is not postulated; it is the unique consequence of demanding local $U(1)$.

This generalizes. Every force in the Standard Model comes from gauging a symmetry:

• $U(1)_{\text{EM}}$ → photon
• $SU(2)_L$ → W and Z bosons
• $SU(3)_C$ → gluons

Gauge symmetry is the principle that generates all known non-gravitational forces. General relativity, similarly, can be understood as gauging translations (though this is technically more subtle).

6. Scalar Electrodynamics and Minimal Coupling

Let’s unpack what we just wrote.

The Lagrangian Expanded

$\mathcal{L}_{\text{SQED}} = (D_\mu \phi)^*(D^\mu \phi) - V(\phi^*\phi) - \tfrac{1}{4} F_{\mu\nu}F^{\mu\nu}$

Expanding the covariant derivative:

$(D_\mu \phi)^*(D^\mu \phi) = (\partial_\mu \phi^* - ieA_\mu \phi^*)(\partial^\mu \phi + ieA^\mu \phi)$

$= (\partial_\mu \phi^*)(\partial^\mu \phi) + ieA^\mu[\phi^*(\partial_\mu \phi) - \phi(\partial_\mu \phi^*)] + e^2 A_\mu A^\mu \phi^*\phi$

Three terms:

• Free scalar kinetic term
• Current-gauge field coupling: $A^\mu \cdot j_\mu$ where $j_\mu = ie[\phi^*(\partial_\mu\phi) - \phi(\partial_\mu\phi^*)]$
• “Seagull” interaction $e^2 A^2 \phi^2$; unique to scalar QED (fermion QED has no such direct term)

Minimal Coupling

The prescription “replace $\partial_\mu$ with $D_\mu$” is called minimal coupling. It’s the simplest way to couple a charged field to a gauge field, and it’s minimal in the sense of using only the fewest derivatives and powers of fields. Standard Model couplings are all minimal.

Equations of Motion

Varying with respect to $A^\mu$:

$\partial_\mu F^{\mu\nu} = J^\nu$

where $J^\nu$ is the matter current; now including pieces from both the scalar field and its interaction with $A^\mu$. Maxwell’s equations, with a specific source.

Varying with respect to $\phi^*$:

$D_\mu D^\mu \phi + \frac{\partial V}{\partial \phi^*} = 0$

This is Klein-Gordon with covariant derivatives. In the presence of an electromagnetic field, a charged scalar satisfies this equation, not the ordinary Klein-Gordon.

Charge Quantization

In classical scalar QED, the charge $e$ is just a parameter; any value is allowed. Quantum mechanics, through Dirac’s argument about magnetic monopoles, suggests that charges would be quantized in units of $e$ if monopoles existed. This argument remains theoretical, since monopoles haven’t been observed; but in the Standard Model, charge quantization is built in via the gauge structure (hypercharge assignments).

7. Non-Abelian Gauge Theory (Yang-Mills)

Everything we just did for $U(1)$ generalizes to larger symmetry groups; but with crucial new features. This is where the strong and weak forces come from.

Non-Abelian Groups

$U(1)$ is abelian: any two elements commute ($e^{i\alpha} e^{i\beta} = e^{i\beta} e^{i\alpha}$). Physical gauge groups used in the Standard Model include:

• $SU(2)$: 2×2 unitary matrices with determinant 1. Three generators. Used for weak isospin.
• $SU(3)$: 3×3 unitary matrices with determinant 1. Eight generators. Used for color.

These groups are non-abelian: matrix multiplication doesn’t commute.

Matter Fields in Non-Abelian Gauge Theory

The matter field is now a multiplet; not a single complex scalar, but a vector of complex scalars transforming as a representation of the group. For $SU(N)$, the simplest is the fundamental representation: an $N$-component complex vector.

$\phi = \begin{pmatrix} \phi_1 \\ \vdots \\ \phi_N \end{pmatrix}$

Under a gauge transformation:

$\phi(x) \to U(x)\phi(x)$

where $U(x)$ is an $N\times N$ unitary matrix that can vary in spacetime.

The Lie Algebra

Near the identity, group elements can be written:

$U(x) = \exp\left(i g \alpha^a(x) T^a\right) \approx 1 + ig\alpha^a T^a + \cdots$

where $T^a$ ($a = 1, \ldots, N^2 - 1$ for $SU(N)$) are the generators of the group. They’re Hermitian matrices. For $SU(2)$, $T^a = \sigma^a/2$ (Pauli matrices over 2). For $SU(3)$, $T^a = \lambda^a/2$ (Gell-Mann matrices over 2).

Key property: generators don’t commute:

$[T^a, T^b] = i f^{abc} T^c$

The numbers $f^{abc}$ are the structure constants of the group. For $SU(2)$, $f^{abc} = \epsilon^{abc}$ (Levi-Civita). For $SU(3)$, $f^{abc}$ is more complicated but specific.

The Gauge Field

For non-abelian groups, you need one gauge field per generator:

$A_\mu^a(x), \quad a = 1, \ldots, N^2-1$

$SU(2)$ has three: $W^1_\mu, W^2_\mu, W^3_\mu$ (the weak gauge bosons, before mixing).

$SU(3)$ has eight: the eight gluons.

Often package into a matrix-valued field:

$A_\mu(x) \equiv A_\mu^a(x) T^a$

The Covariant Derivative

$D_\mu \phi = (\partial_\mu + ig A_\mu)\phi = (\partial_\mu + igA_\mu^a T^a)\phi$

where $g$ is the coupling constant (analog of $e$).

The Transformation Law of $A_\mu$

For the covariant derivative to transform simply ($D_\mu \phi \to U(x) D_\mu \phi$), the gauge field must transform as:

$A_\mu \to U A_\mu U^{-1} + \frac{i}{g}(\partial_\mu U) U^{-1}$

The first term is a matrix conjugation (reflecting that $A_\mu$ transforms in the adjoint rep of the group); the second is the analog of the inhomogeneous $\partial_\mu \alpha$ shift from $U(1)$.

For an infinitesimal transformation $U = 1 + ig\alpha^a T^a$:

$\delta A_\mu^a = \partial_\mu \alpha^a + g f^{abc}\alpha^b A_\mu^c$

The second term; involving structure constants; has no $U(1)$ analog. It says the gauge field itself transforms non-trivially under the symmetry. This will have dramatic consequences.

The Field Strength

The obvious generalization $\partial_\mu A_\nu - \partial_\nu A_\mu$ doesn’t transform nicely by itself in the non-abelian case. The correct gauge-covariant generalization:

$F_{\mu\nu}^a = \partial_\mu A_\nu^a - \partial_\nu A_\mu^a + g f^{abc} A_\mu^b A_\nu^c$

Or in matrix form:

$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + ig[A_\mu, A_\nu]$

The extra term $ig[A_\mu, A_\nu]$ (involving the commutator) is the crucial new feature.

Gauge Boson Self-Interactions

Here’s where things get genuinely different from electromagnetism. Expand $-\tfrac{1}{4} F^a_{\mu\nu} F^{a\,\mu\nu}$ in the Lagrangian. Schematically:

$F^a_{\mu\nu} F^{a\,\mu\nu} \sim (\partial A)^2 + g \partial A \cdot A^2 + g^2 A^4$

The first term gives kinetic terms for $A^a$. But the other two terms describe gauge bosons interacting with each other.

• Three-gluon vertex (schematically $g \partial A \cdot A^2$)
• Four-gluon vertex ($g^2 A^4$)

Gluons; unlike photons; carry color charge, and hence interact with each other directly. This is the mathematical root of asymptotic freedom and confinement in QCD. It is also why non-abelian gauge theories are so much richer than QED.

The Yang-Mills Lagrangian

$\mathcal{L}_{\text{YM}} = -\tfrac{1}{4} F^a_{\mu\nu} F^{a\,\mu\nu}$

Despite looking similar to the Maxwell Lagrangian, this theory is radically different from QED:

• Nonlinear: the gauge fields interact with each other
• Has a dimensionless coupling $g$ that runs with energy
• Exhibits asymptotic freedom (coupling decreases at high energy)
• Confinement (coupling diverges at low energy, if there are no Higgs-like fields)
• Much harder to solve, both classically and in quantum theory

The Standard Model uses Yang-Mills with gauge groups $SU(3)_C \times SU(2)_L \times U(1)_Y$.

8. Spinor Representations of the Lorentz Group

This is the hardest section. Up to this point we’ve worked with scalar fields (Lorentz invariant) and vector fields (transforming with the matrix $\Lambda^\mu{}_\nu$). Fermions transform in a third way; as spinors; that has no analog in pre-relativistic physics.

Take this section slowly. The machinery here is genuinely new.

The Lorentz Algebra

The Lorentz group’s Lie algebra has six generators: three for rotations ($J^i$) and three for boosts ($K^i$). Commutation relations:

$[J^i, J^j] = i\epsilon^{ijk} J^k$

$[J^i, K^j] = i\epsilon^{ijk} K^k$

$[K^i, K^j] = -i\epsilon^{ijk} J^k$

The minus sign in the last one; boosts don’t close under commutation; their product is a rotation; is what makes the Lorentz group non-compact and distinguishes it from the rotation group.

The Trick: Complex Combinations

Define:

$A^i = \tfrac{1}{2}(J^i + iK^i), \qquad B^i = \tfrac{1}{2}(J^i - iK^i)$

These satisfy:

$[A^i, A^j] = i\epsilon^{ijk} A^k, \quad [B^i, B^j] = i\epsilon^{ijk} B^k, \quad [A^i, B^j] = 0$

Two independent copies of the rotation algebra. So the complexified Lorentz algebra factorizes into $SU(2) \times SU(2)$.

Representations of $SU(2)$ are labeled by a non-negative half-integer $j$ (spin). Representations of the Lorentz group are labeled by a pair $(j_A, j_B)$.

The Fundamental Representations

• $(0, 0)$: scalar. One component.
• $(\tfrac{1}{2}, 0)$: left-handed Weyl spinor. Two complex components.
• $(0, \tfrac{1}{2})$: right-handed Weyl spinor. Two complex components.
• $(\tfrac{1}{2}, \tfrac{1}{2})$: vector. Four components.
• $(\tfrac{1}{2}, \tfrac{1}{2})$ appears as the rank-2 antisymmetric tensor, spin-1 massive field, etc.

The two Weyl representations are the building blocks for all spinor physics. Left-handed and right-handed Weyl spinors transform differently under the Lorentz group. This is the mathematical fact behind the observation that the weak interaction treats left-handed and right-handed fermions asymmetrically.

Weyl Spinors Explicitly

A left-handed Weyl spinor $\psi_L$ is a two-component complex object. Under a Lorentz transformation with rotation parameters $\vec\theta$ and boost parameters $\vec\eta$:

$\psi_L \to e^{-i\vec\sigma \cdot \vec\theta/2 + \vec\sigma \cdot \vec\eta/2}\psi_L$

where $\vec\sigma = (\sigma^1, \sigma^2, \sigma^3)$ are the Pauli matrices. Note: the boost generator has no $i$; boosts act differently from rotations.

A right-handed Weyl spinor transforms with the opposite sign on the boost:

$\psi_R \to e^{-i\vec\sigma \cdot \vec\theta/2 - \vec\sigma \cdot \vec\eta/2}\psi_R$

Under rotations, both $\psi_L$ and $\psi_R$ transform the same way. Under boosts, they transform oppositely.

The Double Cover

A $2\pi$ rotation, which you’d think returns you to the same state, sends $\psi \to -\psi$ for a spinor. You need a $4\pi$ rotation to truly come back. This is because spinors are representations not of the Lorentz group itself, but of its double cover; which for the Lorentz group is $SL(2, \mathbb{C})$.

This ”$-$ sign under $2\pi$ rotation” is the experimental signature of spin-½ particles. It’s why Fermi-Dirac statistics work the way they do, and why electrons can’t all pile into the same ground state.

Dirac Spinors

A Dirac spinor $\psi$ combines a left-handed and a right-handed Weyl spinor into a four-component object:

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

This is the representation used in standard Dirac theory. It’s reducible; it’s the direct sum of two Weyl representations; but for a massive particle, the two halves are coupled by the mass term, making the Dirac description natural.

Gamma Matrices

To write Lorentz-invariant equations involving spinors, we need matrices that connect the different spinor components. The gamma matrices $\gamma^\mu$ (four of them, $\mu = 0, 1, 2, 3$) are defined by the anticommutation relations:

$\{\gamma^\mu, \gamma^\nu\} \equiv \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2\eta^{\mu\nu} \mathbb{1}$

This is the Clifford algebra of Minkowski space. Any set of 4×4 matrices satisfying these relations works; different choices are called “representations” or “bases.”

Chiral (Weyl) Basis

A convenient choice:

$\gamma^0 = \begin{pmatrix} 0 & \mathbb{1} \\ \mathbb{1} & 0 \end{pmatrix}, \qquad \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}$

(4×4 matrices built from 2×2 blocks, with $\sigma^i$ the Pauli matrices.) In this basis, the Dirac spinor naturally splits into its left- and right-handed components.

The Chirality Operator

Define:

$\gamma^5 \equiv i\gamma^0 \gamma^1 \gamma^2 \gamma^3$

Properties:

• $(\gamma^5)^2 = 1$
• $\{\gamma^5, \gamma^\mu\} = 0$
• Eigenvalues $\pm 1$

In the chiral basis:

$\gamma^5 = \begin{pmatrix} -\mathbb{1} & 0 \\ 0 & \mathbb{1} \end{pmatrix}$

So $\gamma^5$ distinguishes left-handed ($\gamma^5 = -1$) from right-handed ($\gamma^5 = +1$) components. Projection operators:

$P_L = \tfrac{1}{2}(1 - \gamma^5), \qquad P_R = \tfrac{1}{2}(1 + \gamma^5)$

extract the left- and right-handed parts: $\psi_L = P_L \psi$, $\psi_R = P_R \psi$.

Dirac Adjoint

For spinor Lagrangians to be Lorentz scalars, we need a specific combination of complex conjugation and multiplication by $\gamma^0$:

$\bar\psi \equiv \psi^\dagger \gamma^0$

Then $\bar\psi \psi$ (a product of a row and column of spinors) is a Lorentz scalar, and $\bar\psi \gamma^\mu \psi$ is a Lorentz four-vector. These are the invariants we can put in the Lagrangian.

9. The Dirac Equation

With spinors in hand, we can write the field equation for a free spin-½ particle.

The Derivation

Dirac’s challenge (1928): write a Lorentz-invariant wave equation that is first-order in derivatives, so that particle energies are not $\pm$ square roots (as in Klein-Gordon).

The guess: $(i\gamma^\mu \partial_\mu - m)\psi = 0$.

Check by squaring. Act on the equation with $(i\gamma^\nu \partial_\nu + m)$:

$(i\gamma^\nu \partial_\nu + m)(i\gamma^\mu \partial_\mu - m)\psi = -\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu \psi - m^2 \psi$

Using $\gamma^\nu \gamma^\mu = \tfrac{1}{2}(\{\gamma^\nu, \gamma^\mu\} + [\gamma^\nu, \gamma^\mu])$ and that $\partial_\mu \partial_\nu$ is symmetric (so the antisymmetric part drops):

$-\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu \psi = -\eta^{\mu\nu}\partial_\mu \partial_\nu \psi = -\Box \psi$

So $(i\gamma^\nu\partial_\nu + m)(i\gamma^\mu \partial_\mu - m)\psi = -(\Box + m^2)\psi$.

If $\psi$ satisfies the Dirac equation, it automatically satisfies Klein-Gordon. Good; the right relativistic dispersion $E^2 = p^2 + m^2$ is built in.

The Dirac Equation

$\boxed{(i\gamma^\mu \partial_\mu - m)\psi = 0}$

Or using the “Feynman slash” notation $\slashed{\partial} \equiv \gamma^\mu \partial_\mu$:

$(i\slashed{\partial} - m)\psi = 0$

The Dirac Lagrangian

$\mathcal{L}_{\text{Dirac}} = \bar\psi(i\slashed{\partial} - m)\psi$

Varying with respect to $\bar\psi$ gives the Dirac equation. Varying with respect to $\psi$ gives the conjugate equation:

$i\partial_\mu \bar\psi \gamma^\mu + m\bar\psi = 0$

The Spin Comes for Free

The Dirac equation doesn’t postulate spin. It falls out automatically because $\psi$ has four components and the Lorentz transformation law for spinors includes the Pauli matrices. Solutions turn out to come in spin-up and spin-down varieties; two spin states per particle.

Antimatter Falls Out Too

Plane-wave solutions to the Dirac equation come in two types, corresponding to positive and negative frequency. Dirac initially interpreted the negative-energy solutions as a “sea” of filled states, with antimatter being holes in the sea. The modern interpretation: negative-frequency solutions describe antiparticles.

A free electron has two spin states. An antiparticle (positron) also has two spin states. Total: four states per momentum, matching the four components of the Dirac spinor. The mathematical structure of the Dirac equation; dictated by Lorentz invariance and first-order derivatives; requires antiparticles. Dirac’s prediction of the positron is one of the most beautiful examples of mathematics anticipating experiment.

Plane-Wave Solutions

Try $\psi(x) = u(p) e^{-ip\cdot x}$ with $p^\mu = (E, \vec p)$:

$(\slashed{p} - m) u(p) = 0$

This is an algebraic equation for the four-component spinor $u(p)$. Solutions exist when $p^2 = m^2$ (on shell). For each such $p$, there are two linearly independent solutions corresponding to the two spin states.

Similarly, antiparticles: $\psi(x) = v(p) e^{+ip\cdot x}$ with $(\slashed{p} + m) v(p) = 0$, also two spin states.

Chirality and the Dirac Equation

Project onto left- and right-handed components:

$i\slashed{\partial} \psi_L = m \psi_R$

$i\slashed{\partial} \psi_R = m \psi_L$

The mass term couples $\psi_L$ and $\psi_R$! For a massless particle ($m = 0$), left- and right-handed components decouple, and you have two independent Weyl equations. This is exactly what’s needed for neutrinos in the original Standard Model (massless and left-handed only).

For massive particles, $\psi_L$ and $\psi_R$ are coupled, and the relevant description is the full Dirac equation.

Helicity vs. Chirality

Two closely related but distinct concepts:

• Helicity: spin projected along momentum direction. Frame-dependent for massive particles.
• Chirality: the eigenvalue of $\gamma^5$. Frame-independent but not conserved unless $m = 0$.

For massless particles, chirality = helicity. For massive particles, they coincide only at very high energies.

10. Quantum Electrodynamics as a Classical Field Theory

Now we have all the pieces to write QED at the classical level.

Coupling Dirac Fermions to EM

The Dirac Lagrangian has a global $U(1)$ symmetry: $\psi \to e^{i\alpha}\psi$. Promote to local: $\psi \to e^{ieQ\alpha(x)}\psi$ where $Q$ is the electric charge in units of $e$ (for the electron, $Q = -1$).

Replace $\partial_\mu$ with the covariant derivative $D_\mu = \partial_\mu + ieQA_\mu$.

The QED Lagrangian

$\boxed{\mathcal{L}_{\text{QED}} = \bar\psi(i\slashed{D} - m)\psi - \tfrac{1}{4}F_{\mu\nu}F^{\mu\nu}}$

Expanding $\slashed{D} = \gamma^\mu(\partial_\mu + ieQ A_\mu)$:

$\mathcal{L}_{\text{QED}} = \bar\psi(i\slashed{\partial} - m)\psi - eQ\bar\psi \gamma^\mu \psi A_\mu - \tfrac{1}{4}F_{\mu\nu}F^{\mu\nu}$

Three terms:

• Free Dirac Lagrangian
• Interaction: $-eQA_\mu \bar\psi \gamma^\mu \psi$. The quantity $j^\mu = \bar\psi\gamma^\mu \psi$ is the electromagnetic current; the electron couples to the photon through this current.
• Free Maxwell Lagrangian

That’s QED at the classical level. Every precision prediction of QED; anomalous magnetic moments, Lamb shift, pair production; is derived by quantizing this Lagrangian.

The QED Vertex

The interaction term has exactly one photon and two fermion lines. In quantum theory, this becomes the QED vertex: an electron line emits or absorbs a photon, nothing else. Every QED Feynman diagram is built from this single vertex.

Equations of Motion

For $\psi$: $(i\slashed{D} - m)\psi = 0$ (covariant Dirac equation).

For $A_\mu$: $\partial_\mu F^{\mu\nu} = eQ \bar\psi \gamma^\nu \psi = j^\nu$ (Maxwell’s equations with Dirac current as source).

These are coupled, nonlinear, and in general impossible to solve exactly. QED calculations proceed perturbatively in $e$ (or equivalently $\alpha = e^2/4\pi$), generating the Feynman diagram expansion.

Gauge Invariance Check

$\bar\psi \gamma^\mu \psi$ is a Noether current; it’s automatically conserved ($\partial_\mu j^\mu = 0$ on the equations of motion). This is necessary for Maxwell’s equations to be consistent: $\partial_\mu \partial_\nu F^{\mu\nu} = 0$ identically by antisymmetry, so the source $j^\nu$ must be conserved.

Current Structure

The form $\bar\psi \gamma^\mu \psi$ is a vector current; it transforms as a Lorentz four-vector. Its time component is the charge density $\bar\psi\gamma^0\psi = \psi^\dagger \psi$.

QED’s interaction is purely vectorial; the same coupling for left- and right-handed electrons. Parity is conserved in QED.

Compare: the weak interaction uses $\bar\psi \gamma^\mu(1 - \gamma^5)\psi$; a V-A current (vector minus axial vector); which couples only to left-handed fermions. This is why the weak interaction violates parity.

11. The Higgs Mechanism

We return to spontaneous symmetry breaking, now with a gauge symmetry. The results are dramatically different.

The Setup: U(1) Gauge Theory with Scalar

Take scalar QED with a Mexican-hat potential:

$\mathcal{L} = (D_\mu \phi)^*(D^\mu \phi) - V(\phi^*\phi) - \tfrac{1}{4}F_{\mu\nu}F^{\mu\nu}$

with $V = -\mu^2 \phi^*\phi + \lambda(\phi^*\phi)^2$, so the minimum is at $|\phi| = v/\sqrt{2}$.

Choice of Vacuum

Choose (say) $\phi_0 = v/\sqrt{2}$, real. The $U(1)$ symmetry is broken.

Parametrize Fluctuations

A clever parametrization: write

$\phi(x) = \frac{1}{\sqrt{2}}[v + h(x)] e^{i\theta(x)/v}$

Here $h$ is a real field (fluctuation along the radial direction) and $\theta$ is a real field (fluctuation along the tangential/phase direction). The Goldstone field is $\theta$.

The Gauge Choice That Eliminates $\theta$

In a gauge theory, we’re allowed to make a gauge transformation. Choose the gauge $\alpha(x) = \theta(x)/(ev)$. Then:

$\phi(x) \to e^{-i\alpha(x)}\phi(x) = \frac{1}{\sqrt{2}}[v + h(x)]$

After this gauge transformation, $\phi$ is purely real. The Goldstone $\theta$ has vanished from $\phi$.

But: the gauge field transforms too. Working through:

$A_\mu \to A_\mu + \frac{1}{ev}\partial_\mu \theta$

So $\theta$ reappears, absorbed into $A_\mu$.

This gauge choice; where we “eat” the Goldstone; is called unitary gauge.

What Happens to the Lagrangian

Substitute $\phi = (v + h)/\sqrt{2}$ (real) into the Lagrangian and expand. The kinetic term $(D_\mu \phi)^*(D^\mu \phi)$ produces:

$(D_\mu \phi)^*(D^\mu \phi) = \tfrac{1}{2}(\partial_\mu h)(\partial^\mu h) + \tfrac{1}{2} e^2 (v + h)^2 A_\mu A^\mu$

Expand the second term:

$\tfrac{1}{2} e^2(v + h)^2 A_\mu A^\mu = \tfrac{1}{2}e^2 v^2 A_\mu A^\mu + e^2 v h A_\mu A^\mu + \tfrac{1}{2}e^2 h^2 A_\mu A^\mu$

The first term is a mass term for $A_\mu$!

$m_A^2 = e^2 v^2$

The gauge boson has acquired a mass; despite the fact that explicit mass terms for $A_\mu$ are forbidden by gauge invariance. The mass came from the gauge boson’s coupling to the Higgs VEV.

Counting Degrees of Freedom

Before symmetry breaking:

• Complex scalar $\phi$: 2 degrees of freedom
• Massless gauge field $A_\mu$: 2 degrees of freedom (transverse polarizations)
• Total: 4

After symmetry breaking (unitary gauge):

• Real Higgs field $h$: 1 degree of freedom
• Massive gauge field $A_\mu$: 3 degrees of freedom (transverse + longitudinal)
• Total: 4

The would-be Goldstone boson became the longitudinal polarization of the massive gauge boson. “The gauge boson ate the Goldstone.”

Physical Consequences

The photon remains massless because the electromagnetic $U(1)$ is not broken; only a different $U(1)$ is, which isn’t electromagnetism.

The W and Z get their masses from the Higgs VEV via the analogous mechanism in $SU(2)_L \times U(1)_Y$.

$M_W = \tfrac{1}{2} g v, \qquad M_Z = \tfrac{1}{2}\sqrt{g^2 + g'^2} v$

The physical Higgs $h$ is the remaining scalar. Its mass comes from the shape of the potential:

$m_h^2 = 2\mu^2 = 2\lambda v^2$

Measured at the LHC: $m_h \approx 125$ GeV, with $v \approx 246$ GeV.

Fermion masses come from Yukawa couplings to the Higgs:

$\mathcal{L}_{\text{Yukawa}} = -y_f \bar\psi \phi \psi$

When $\phi$ gets its VEV, this becomes $-y_f v/\sqrt{2} \cdot \bar\psi \psi$; a mass term $m_f = y_f v/\sqrt{2}$. Every fermion’s mass comes from its coupling to the Higgs field.

The Bigger Picture

The Higgs mechanism is the one known way to give gauge bosons masses without destroying gauge invariance. Without it, the Standard Model is mathematically sick above ~1 TeV (cross-sections violate unitarity). The 2012 Higgs discovery confirmed not just one more particle, but the consistency of the whole theoretical framework.

It’s worth pausing to appreciate what’s happening here. By positing one scalar field with the Mexican-hat potential and coupling it to the $SU(2) \times U(1)$ gauge fields with the right charges, we get:

• Photon: massless (good, reality)
• W, Z: massive with specific predicted masses
• Electron, quarks, etc.: masses proportional to their Yukawa couplings
• A new scalar boson (the Higgs) with a specific mass from $\lambda$ and $v$

Every one of these predictions has been experimentally verified. The Standard Model is astonishingly tight.

12. The Standard Model Lagrangian

Here it is, piece by piece.

Gauge Structure

$G_{\text{SM}} = SU(3)_C \times SU(2)_L \times U(1)_Y$

• $SU(3)_C$: color (QCD). 8 gluons $G_\mu^a$, coupling $g_s$.
• $SU(2)_L$: weak isospin, acts only on left-handed fermions. 3 bosons $W_\mu^a$, coupling $g$.
• $U(1)_Y$: weak hypercharge. 1 boson $B_\mu$, coupling $g'$.

After electroweak symmetry breaking, the physical gauge bosons are:

• Photon $A_\mu$ (massless)
• $W^\pm_\mu$ (mass $M_W$)
• $Z_\mu$ (mass $M_Z$)
• 8 gluons $G_\mu^a$ (massless)

Matter Fields

Three generations of quarks and leptons, each in specific representations:

Left-handed quarks in $(3, 2)$ of $SU(3) \times SU(2)$ with hypercharge $Y = +1/6$:

$Q_L = \begin{pmatrix} u_L \\ d_L \end{pmatrix}$

Right-handed up-type quarks in $(3, 1)$, $Y = +2/3$: $u_R$.

Right-handed down-type quarks in $(3, 1)$, $Y = -1/3$: $d_R$.

Left-handed leptons in $(1, 2)$, $Y = -1/2$:

$L_L = \begin{pmatrix} \nu_L \\ e_L \end{pmatrix}$

Right-handed charged leptons in $(1, 1)$, $Y = -1$: $e_R$.

(No right-handed neutrinos in the original Standard Model. They’re added in extensions to give neutrinos mass.)

Three copies of this pattern for the three generations.

Higgs Field

Complex scalar doublet under $SU(2)$, hypercharge $Y = +1/2$:

$\phi = \begin{pmatrix} \phi^+ \\ \phi^0 \end{pmatrix}$

The Lagrangian

$\mathcal{L}_{\text{SM}} = \mathcal{L}_{\text{gauge}} + \mathcal{L}_{\text{fermion}} + \mathcal{L}_{\text{Higgs}} + \mathcal{L}_{\text{Yukawa}}$

Gauge term:

$\mathcal{L}_{\text{gauge}} = -\tfrac{1}{4} G^a_{\mu\nu} G^{a\,\mu\nu} - \tfrac{1}{4} W^a_{\mu\nu} W^{a\,\mu\nu} - \tfrac{1}{4} B_{\mu\nu} B^{\mu\nu}$

Kinetic terms for the gauge fields (with self-interactions for the non-abelian ones).

Fermion kinetic + gauge coupling:

$\mathcal{L}_{\text{fermion}} = \sum_\psi \bar\psi i \slashed{D}\psi$

summed over $\psi \in \{Q_L, u_R, d_R, L_L, e_R\}$ for each generation. The covariant derivative $D_\mu$ contains all gauge couplings appropriate to each field’s representation. For the left-handed quark doublet:

$D_\mu Q_L = \left(\partial_\mu - i g_s G_\mu^a T_C^a - i g W_\mu^a T^a - i g' Y_{Q_L} B_\mu\right) Q_L$

Each fermion couples to gauge fields according to its charges under each group.

Higgs:

$\mathcal{L}_{\text{Higgs}} = (D_\mu \phi)^\dagger(D^\mu \phi) - V(\phi)$