Spin, and the price of relativistic quantum mechanics.

QFT document 2: fermions, antiparticles, and the spin-statistics theorem, falling out of the mathematics.

The scalar field document showed how to quantize a bosonic field. Now we do the same for the Dirac field; a relativistic spin-½ field describing electrons, quarks, and every other fundamental fermion. Same canonical procedure, but with one crucial twist: commutators get replaced by anticommutators. That single change generates every distinctive feature of fermionic physics: Pauli exclusion, antisymmetric wave functions, and the spin-statistics theorem itself.

Pre-requisites: the scalar field quantization document, plus the sections on spinors and the Dirac equation from the classical field theory reference. If gamma matrices, left/right-handed Weyl components, and $\slashed{\partial}\psi = m\psi$ feel familiar, you’re ready.

Conventions (Same as Document 1)

• Metric $\eta_{\mu\nu} = \text{diag}(+,-,-,-)$ (mostly-minus)
• $\hbar = c = 1$
• $\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}$
• $\bar\psi = \psi^\dagger \gamma^0$
• $\slashed{p} = \gamma^\mu p_\mu$

Table of Contents

1. Why Fermions Need Different Rules
2. The Classical Dirac Field in Review
3. Solutions to the Dirac Equation: u and v Spinors
4. The Canonical Quantization Attempt That Fails
5. Anticommutators and the Spin-Statistics Theorem
6. The Dirac Field as Creation and Annihilation Operators
7. The Hamiltonian and Vacuum
8. Fermionic Fock Space
9. The Fermion Propagator
10. Discrete Symmetries: C, P, T on the Dirac Field
11. Bilinears and Physical Currents
12. Physical Content and Preview of QED
13. Appendix: Formulas and Identities

1. Why Fermions Need Different Rules

The Spin-Statistics Puzzle

Experimentally, identical electrons (and all other spin-½ particles) obey the Pauli exclusion principle: no two can occupy the same quantum state. This is why atoms have shell structure, why metals conduct, why white dwarfs resist collapse. Equivalently: multi-fermion states are antisymmetric under particle exchange.

Spin-0 particles, by contrast, can pile into the same state arbitrarily. Bose-Einstein condensates, superconducting Cooper pairs, photons in a laser; all bosonic. Multi-boson states are symmetric.

The connection between spin and statistics is not obvious. Nothing in non-relativistic quantum mechanics explains why spin-½ particles should be antisymmetric. It’s postulated, not derived.

The remarkable fact of QFT: spin-statistics is a theorem. If you try to quantize a spin-½ field using ordinary commutators, you get a nonsensical theory (unbounded-below Hamiltonian). To get sensible physics, you must use anticommutators. And anticommutators automatically produce antisymmetric multi-particle states.

Spin, statistics, and quantization rules are rigidly linked. This document makes that explicit.

The Road Map

1. Start with the classical Dirac field (from the classical field theory document)
2. Try canonical quantization with commutators; and see it fail
3. Replace commutators with anticommutators
4. Discover that everything works, and the resulting theory has the right properties (Pauli exclusion, antisymmetry, sensible energies, antiparticles)
5. Package the result in the same Fock-space language as before, but now with fermionic occupation numbers

2. The Classical Dirac Field in Review

Lagrangian

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

where $\psi(x)$ is a four-component complex Dirac spinor and $\bar\psi = \psi^\dagger\gamma^0$. The $\slashed{\partial} = \gamma^\mu\partial_\mu$ is Dirac’s slash notation for contracting with gamma matrices.

Varying with respect to $\bar\psi$ yields the Dirac equation:

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

Equivalently $i\gamma^\mu\partial_\mu \psi = m\psi$, an equation first-order in derivatives (unlike Klein-Gordon, which is second-order).

Squaring the Dirac equation gives Klein-Gordon: any Dirac solution automatically satisfies $(\Box + m^2)\psi = 0$. So Dirac fields describe particles with mass $m$ and the expected relativistic dispersion $E^2 = |\vec p|^2 + m^2$.

Conjugate Momentum

$\pi = \frac{\partial\mathcal{L}}{\partial\dot\psi} = i\bar\psi\gamma^0 = i\psi^\dagger$

This is a slightly awkward feature of the Dirac Lagrangian: $\pi$ is essentially $\psi^\dagger$ itself (up to a factor of $i$), not something independent. This reflects that the Dirac equation is first-order; the field and its momentum aren’t independent in the way they are for scalars.

Hamiltonian

$H = \int d^3x\, \psi^\dagger(-i\vec\alpha\cdot\vec\nabla + \beta m)\psi$

with $\vec\alpha = \gamma^0\vec\gamma$ and $\beta = \gamma^0$. You can also write this as:

$H = \int d^3x\, \bar\psi(-i\vec\gamma\cdot\vec\nabla + m)\psi$

Crucial observation: the classical Hamiltonian is not positive-definite for a Dirac field. Both signs of energy appear in the spectrum. This is the infamous “negative energy” problem of the Dirac equation, and its resolution is where antiparticles enter.

3. Solutions to the Dirac Equation: u and v Spinors

Before quantizing, we need the classical solutions organized neatly.

Plane Wave Solutions

Try $\psi(x) = u(p) e^{-ip\cdot x}$ with $p^\mu = (E, \vec p)$ and $E = +\sqrt{|\vec p|^2 + m^2} > 0$:

$(i\slashed{\partial} - m)u(p)e^{-ip\cdot x} = 0 \implies (\slashed{p} - m)u(p) = 0$

This is an algebraic equation on the 4-component spinor $u(p)$. In matrix form $(\slashed{p} - m)$ has a 2-dimensional null space; two independent solutions, labeled by spin $s = 1, 2$. We write $u^s(p)$ for these.

Try $\psi(x) = v(p) e^{+ip\cdot x}$ with the same positive $E$:

$(\slashed{p} + m)v(p) = 0$

Again 2-dimensional null space, labeled $v^s(p)$ for $s = 1, 2$.

Explicit Forms in the Chiral Basis

In the chiral (Weyl) basis from the classical field theory document:

$u^s(p) = \begin{pmatrix} \sqrt{p\cdot\sigma}\,\xi^s \\ \sqrt{p\cdot\bar\sigma}\,\xi^s \end{pmatrix}, \qquad v^s(p) = \begin{pmatrix} \sqrt{p\cdot\sigma}\,\eta^s \\ -\sqrt{p\cdot\bar\sigma}\,\eta^s \end{pmatrix}$

where $\sigma^\mu = (1, \vec\sigma)$, $\bar\sigma^\mu = (1, -\vec\sigma)$, and $\xi^s, \eta^s$ are 2-component spinors encoding the spin state (e.g., $\xi^1 = \binom{1}{0}$, $\xi^2 = \binom{0}{1}$).

The square roots need some interpretation; they’re defined by power series and the fact that $p\cdot\sigma$ is a Hermitian matrix. For a particle at rest ($\vec p = 0$, $E = m$):

$u^s(m\vec e_0) = \sqrt{m}\begin{pmatrix}\xi^s \\ \xi^s\end{pmatrix}, \quad v^s(m\vec e_0) = \sqrt{m}\begin{pmatrix}\eta^s \\ -\eta^s\end{pmatrix}$

Don’t worry if the explicit forms aren’t transparent; what matters is that they exist, satisfy $(\slashed{p} \mp m)u/v = 0$, and have the normalization properties below.

Normalization and Orthogonality

Spinor normalizations (two conventions; I’ll use Peskin-style):

$\bar u^r(p) u^s(p) = 2m\delta^{rs}, \quad \bar v^r(p) v^s(p) = -2m\delta^{rs}$

$u^{r\dagger}(p) u^s(p) = 2E\delta^{rs}, \quad v^{r\dagger}(p) v^s(p) = 2E\delta^{rs}$

Orthogonality between u and v:

$\bar u^r(p) v^s(p) = 0, \quad u^{r\dagger}(p) v^s(-p) = 0$

(Note the $-p$ in the momentum of the $v$ spinor; this is important!)

Completeness Relations

Two crucial identities that will be used constantly:

$\boxed{\sum_s u^s(p)\bar u^s(p) = \slashed{p} + m}$

$\boxed{\sum_s v^s(p)\bar v^s(p) = \slashed{p} - m}$

These are outer products (4×4 matrices), summed over spin. They are the “density matrix” for spin-averaged states.

These relations will appear in essentially every QED calculation you ever do; when you average over initial spins or sum over final spins in a scattering process.

General Classical Solution

$\psi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s\left[b^s(\vec p)\, u^s(p)e^{-ip\cdot x} + c^{s*}(\vec p)\, v^s(p)e^{+ip\cdot x}\right]$

where $b^s(\vec p)$ and $c^{s*}(\vec p)$ are complex coefficients. Upon quantization, $b^s$ will become an annihilation operator for particles, and $c^{s\dagger}$ will become a creation operator for antiparticles.

Note the letters: different symbols ($b$ and $c$) for the particle and antiparticle modes, because they represent different physical entities. In the scalar case for the complex field, we used $a$ and $b$; here we use $b$ and $c$ by convention. Just a naming convention; don’t get thrown.

4. The Canonical Quantization Attempt That Fails

Let’s try the naive generalization of the scalar field procedure: canonical commutators.

The Naive Recipe

Treat $\psi$ and $\pi = i\psi^\dagger$ as canonical conjugates. Impose equal-time commutation relations:

$[\psi_a(\vec x, t), \psi_b^\dagger(\vec y, t)] \stackrel{?}{=} \delta_{ab}\delta^3(\vec x - \vec y)$

(The $a,b$ are spinor indices, $1, 2, 3, 4$; the $\delta_{ab}$ reflects that different spinor components commute with each other except for the identity term.)

Mode Operator Commutators

Working through; same procedure as for the scalar; gives:

$[b^r_{\vec p}, b^{s\dagger}_{\vec q}] = (2\pi)^3\delta^{rs}\delta^3(\vec p - \vec q)$

$[c^r_{\vec p}, c^{s\dagger}_{\vec q}] = -(2\pi)^3\delta^{rs}\delta^3(\vec p - \vec q)$

Note the minus sign on the $c$ commutator. This is already a warning sign.

Computing the Hamiltonian

Substituting the mode expansion into the Hamiltonian:

$H \sim \int\frac{d^3p}{(2\pi)^3} E_p\left[b^{s\dagger}_{\vec p} b^s_{\vec p} - c^{s\dagger}_{\vec p} c^s_{\vec p}\right] + (\text{const})$

The minus sign is catastrophic. Creating a $c$-particle lowers the energy. You can create arbitrarily many $c$-particles and drive the energy to $-\infty$. The theory has no ground state. It’s sick.

This is the real face of the “negative energy problem” of the Dirac equation. You can’t just reinterpret the $E < 0$ plane-wave solutions; quantizing with commutators produces a theory with no stable vacuum.

What Went Wrong

The minus sign traces back to the commutator $[c_{\vec p}, c_{\vec q}^\dagger] = -(2\pi)^3\delta^3(\vec p - \vec q)$. This is a sign of an unphysical theory; the would-be creation operator is creating “negative-norm” states.

More specifically: $\langle 0 | c_{\vec p} c_{\vec p}^\dagger | 0\rangle$ would have to be negative, which is impossible for a legitimate Hilbert space inner product.

The commutator prescription is wrong for spin-½. To get a sensible theory, we need something else.

5. Anticommutators and the Spin-Statistics Theorem

The Fix

Replace commutators with anticommutators:

$\{A, B\} \equiv AB + BA$

Impose:

$\boxed{\{\psi_a(\vec x, t), \psi_b^\dagger(\vec y, t)\} = \delta_{ab}\delta^3(\vec x - \vec y)}$

$\{\psi_a(\vec x, t), \psi_b(\vec y, t)\} = \{\psi_a^\dagger(\vec x, t), \psi_b^\dagger(\vec y, t)\} = 0$

These are equal-time anticommutation relations.

Mode Operator Anticommutators

Working through the algebra with anticommutators replacing commutators:

$\boxed{\{b^r_{\vec p}, b^{s\dagger}_{\vec q}\} = (2\pi)^3\delta^{rs}\delta^3(\vec p - \vec q)}$

$\boxed{\{c^r_{\vec p}, c^{s\dagger}_{\vec q}\} = (2\pi)^3\delta^{rs}\delta^3(\vec p - \vec q)}$

All other anticommutators (same type, or $b$ with $c$) vanish.

Now both signs are positive. The minus sign disappeared when we switched from commutators to anticommutators; the sign flip exactly compensated.

The Spin-Statistics Theorem

The argument we just made is the simplest version of the spin-statistics theorem:

Spin-½ fields must be quantized with anticommutators; spin-0 fields must be quantized with commutators. Any other choice produces a theory with no sensible vacuum or negative-norm states.

More generally: integer-spin fields (bosons) use commutators, half-integer-spin fields (fermions) use anticommutators. This can be proved rigorously from Lorentz invariance + causality + positive energies; the spin-statistics theorem (Pauli 1940, refined by Lüders and Zumino).

This is a theorem, not a postulate. The structure of relativistic quantum field theory requires the connection between spin and statistics.

Consequence: Pauli Exclusion

From the anticommutator:

$\{b^{s\dagger}_{\vec p}, b^{s\dagger}_{\vec p}\} = 0 \implies (b^{s\dagger}_{\vec p})^2 = 0$

Trying to create two fermions in the same quantum state gives zero. Pauli exclusion, built into the math.

Similarly, a two-fermion state:

$b^{s_1\dagger}_{\vec p_1} b^{s_2\dagger}_{\vec p_2}|0\rangle = -b^{s_2\dagger}_{\vec p_2} b^{s_1\dagger}_{\vec p_1}|0\rangle$

The state is antisymmetric under exchange; automatically. You can’t symmetrize it; anticommutation forbids it.

Grassmann Numbers

A consequence of the quantization with anticommutators: fields are not ordinary (commuting) numbers. Products of classical fermion fields anticommute too. To properly treat classical fermion fields, you use Grassmann numbers; numbers that satisfy $\theta_1\theta_2 = -\theta_2\theta_1$.

This becomes important for path integrals with fermions (later document). For now, just be aware: if you’re computing at the classical level with spinors, you need to respect their anticommuting nature.

6. The Dirac Field as Creation and Annihilation Operators

Now assemble the pieces. The quantized Dirac field:

$\boxed{\psi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s\left[b^s_{\vec p}\, u^s(p)e^{-ip\cdot x} + c^{s\dagger}_{\vec p}\, v^s(p)e^{+ip\cdot x}\right]}$

$\psi^\dagger(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s\left[b^{s\dagger}_{\vec p}\, u^{s\dagger}(p)e^{+ip\cdot x} + c^s_{\vec p}\, v^{s\dagger}(p)e^{-ip\cdot x}\right]$

Interpretation of Operators

• $b^{s\dagger}_{\vec p}$: creates a particle (e.g., electron) with momentum $\vec p$ and spin $s$
• $b^s_{\vec p}$: annihilates a particle
• $c^{s\dagger}_{\vec p}$: creates an antiparticle (e.g., positron) with momentum $\vec p$ and spin $s$
• $c^s_{\vec p}$: annihilates an antiparticle

The field $\psi(x)$ itself: destroys a particle at $x$ or creates an antiparticle at $x$. (Both reduce the total charge by one.)

The field $\psi^\dagger(x)$: creates a particle at $x$ or destroys an antiparticle. (Both increase the total charge by one.)

This is a major conceptual point: a single field operator can destroy particles and create antiparticles, simultaneously. The two halves of the Dirac field; the positive- and negative-frequency parts; have dual interpretations, and the field unites them.

Vacuum

$b^s_{\vec p}|0\rangle = 0, \quad c^s_{\vec p}|0\rangle = 0 \quad \text{for all } \vec p, s$

The vacuum has no particles and no antiparticles.

Single-Particle States

$|\vec p, s\rangle_e = \sqrt{2E_p}\, b^{s\dagger}_{\vec p}|0\rangle \quad \text{(electron)}$

$|\vec p, s\rangle_{\bar e} = \sqrt{2E_p}\, c^{s\dagger}_{\vec p}|0\rangle \quad \text{(positron)}$

Both are Lorentz-invariantly normalized.

Two-Particle States

$b^{s_1\dagger}_{\vec p_1} b^{s_2\dagger}_{\vec p_2}|0\rangle = -b^{s_2\dagger}_{\vec p_2} b^{s_1\dagger}_{\vec p_1}|0\rangle$

Antisymmetric. Fermionic. $\checkmark$

And the Pauli exclusion follows: if $(\vec p_1, s_1) = (\vec p_2, s_2)$, the state vanishes.

7. The Hamiltonian and Vacuum

Substituting the mode expansion into the Hamiltonian and working through (using anticommutators, and being careful with signs):

$H = \int\frac{d^3p}{(2\pi)^3}E_p\sum_s\left[b^{s\dagger}_{\vec p} b^s_{\vec p} + c^{s\dagger}_{\vec p} c^s_{\vec p}\right] + (\text{infinite constant})$

Both terms are positive. The infinite constant is again the (divergent) vacuum energy, which we subtract via normal ordering:

$:H: = \int\frac{d^3p}{(2\pi)^3}E_p\sum_s\left[b^{s\dagger}_{\vec p} b^s_{\vec p} + c^{s\dagger}_{\vec p} c^s_{\vec p}\right]$

Particles AND Antiparticles Carry Positive Energy

Act on single-particle states:

$:H:|\vec p, s\rangle_e = E_p|\vec p, s\rangle_e$

$:H:|\vec p, s\rangle_{\bar e} = E_p|\vec p, s\rangle_{\bar e}$

Both particles and antiparticles have positive energy $E_p = \sqrt{|\vec p|^2 + m^2}$. $\checkmark$

This is a crucial success of the anticommutator prescription. Antiparticles aren’t “negative energy” states; they’re perfectly ordinary positive-energy excitations that happen to carry opposite charge.

Normal Ordering of Fermions

Normal ordering for fermions has a subtle twist: you pick up a minus sign for each anticommutation needed to move an annihilation operator past a creation operator.

$:b^\dagger b: = b^\dagger b$

$:b b^\dagger: = -b^\dagger b$

This ensures that normal-ordered products of fermions correctly respect the antisymmetry.

A Note on the Classical Hamiltonian

Recall: the classical Dirac Hamiltonian wasn’t positive-definite. Quantization changed that. The positive energies for antiparticles come from the anticommutation algebra; after reordering, the would-be negative term $-c^\dagger c$ becomes $+c c^\dagger - \text{const} = +c^\dagger c + (\text{const subtracted by normal ordering})$.

So the classical “negative energy sea” is an artifact of looking at the theory before quantization. The quantum theory has a proper ground state and only positive-energy excitations above it.

8. Fermionic Fock Space

Structure

Just like bosonic Fock space, but with antisymmetric multi-particle states. At each occupation level, states are built by applying creation operators:

$b^{s_1\dagger}_{\vec p_1} b^{s_2\dagger}_{\vec p_2} \cdots b^{s_n\dagger}_{\vec p_n}|0\rangle$

These are antisymmetric under exchange of any two indices, and vanish if any two are identical; automatic Pauli exclusion.

Particle Number Operators

$N_b = \int\frac{d^3p}{(2\pi)^3}\sum_s b^{s\dagger}_{\vec p} b^s_{\vec p} \quad \text{(electron number)}$

$N_c = \int\frac{d^3p}{(2\pi)^3}\sum_s c^{s\dagger}_{\vec p} c^s_{\vec p} \quad \text{(positron number)}$

The difference $Q = -e(N_b - N_c)$ (for electrons of charge $-e$) is the total electric charge operator; a conserved quantity corresponding to the $U(1)$ symmetry $\psi \to e^{i\alpha}\psi$.

The Noether Current

Classical symmetry $\psi \to e^{i\alpha}\psi$ gives conserved current:

$j^\mu = \bar\psi \gamma^\mu \psi$

After quantization, this becomes an operator:

$j^\mu(x) = :\bar\psi(x)\gamma^\mu\psi(x):$

The conserved charge $Q = \int d^3x\, j^0$ evaluates (with some algebra) to:

$Q = \int\frac{d^3p}{(2\pi)^3}\sum_s\left[b^{s\dagger}_{\vec p} b^s_{\vec p} - c^{s\dagger}_{\vec p} c^s_{\vec p}\right]$

Particles contribute +1, antiparticles contribute −1. The charge is opposite for the two, as required.

9. The Fermion Propagator

Feynman Propagator

$S_F(x - y) = \langle 0|T\{\psi(x)\bar\psi(y)\}|0\rangle$

where $T$ is the fermionic time-ordering:

$T\{\psi(x)\bar\psi(y)\} = \theta(x^0 - y^0)\psi(x)\bar\psi(y) - \theta(y^0 - x^0)\bar\psi(y)\psi(x)$

Note the minus sign. Fermionic time-ordering picks up a sign from the anticommutation. Miss this minus sign and you’ll get wrong answers for every fermionic Feynman diagram.

Computation

Using the mode expansions and the completeness relations from section 3:

$\boxed{S_F(x - y) = \int\frac{d^4p}{(2\pi)^4}\frac{i(\slashed{p} + m)}{p^2 - m^2 + i\epsilon}e^{-ip\cdot(x-y)}}$

Same $+i\epsilon$ prescription as the scalar, giving time ordering. But the numerator is $(\slashed{p} + m)$ instead of a single factor; a 4×4 matrix in spinor space.

In Momentum Space

$\tilde S_F(p) = \frac{i(\slashed{p} + m)}{p^2 - m^2 + i\epsilon}$

Alternative form (rationalizing):

$\tilde S_F(p) = \frac{i}{\slashed{p} - m + i\epsilon}$

understood as a matrix inverse. Both forms show up in calculations.

Physical Meaning

The $\slashed{p} + m$ in the numerator is precisely $\sum_s u^s(p)\bar u^s(p)$ for positive-frequency parts, or $\sum_s v^s(p)\bar v^s(p)$ (with sign) for negative-frequency. It encodes the spin structure of the propagating fermion.

Why Two Types of Propagator?

Scalars: $\phi \to \phi$ propagator, $\phi^* \to \phi^*$ propagator, and $\phi \to \phi^*$ propagator (for complex field).

Fermions: similar structure. Most calculations use $\psi \to \bar\psi$ (the one above). There are also non-vanishing propagators between other combinations for a complex field.

The notation gets more elaborate, but the principle is clear: the propagator encodes the two-point function of fields, with appropriate time/space ordering.

10. Discrete Symmetries: C, P, T on the Dirac Field

These are technical and subtle but physically important, because they classify what kind of interactions are allowed in different theories. Three discrete symmetries act on the Dirac field:

Parity (P)

Spatial inversion: $\vec x \to -\vec x$. Acts on the Dirac field as:

$P\psi(t, \vec x)P^{-1} = \gamma^0 \psi(t, -\vec x)$

Consequence: under parity, a Dirac spinor picks up a $\gamma^0$. The left- and right-handed components swap, since $\gamma^0 \psi_L = \psi_R$ in the chiral basis.

Charge Conjugation (C)

Particle ↔ antiparticle exchange:

$C\psi(x)C^{-1} = -i(\bar\psi\gamma^0\gamma^2)^T$

An operator that maps electrons to positrons and vice versa. Intuition: $C$ exchanges the $b$‘s with the $c$‘s.

Time Reversal (T)

Time inversion. T is an antiunitary operator (complex conjugation plus a unitary transformation):

$T\psi(t, \vec x)T^{-1} = -\gamma^1\gamma^3 \psi(-t, \vec x)$

CPT Theorem

The combination CPT is an exact symmetry of any Lorentz-invariant local quantum field theory. This is the CPT theorem (Lüders, Pauli), one of the deepest theorems in physics.

Consequences:

• Particles and antiparticles have identical masses
• Particles and antiparticles have identical lifetimes
• If CPT were violated, Lorentz invariance would be too

All experimental tests are consistent with CPT.

Individually

Individual C, P, T are not required to be symmetries. Strong and electromagnetic interactions conserve each separately. The weak interaction maximally violates both P and C individually (1956-1957 Wu experiment), but respects CP approximately. CP is violated in kaon decays (1964) and B meson decays (~2000), at a small but nonzero level; all within the Standard Model via the CKM phase.

The fact that C and P are violated separately but CP nearly conserved is a deep constraint on the weak interaction. Dirac theory with different interactions built on top can respect or violate any combination; the Lagrangian choices are what matter.

11. Bilinears and Physical Currents

Fermion fields by themselves can’t be measured (they’re Grassmann-valued, not observable). But bilinear combinations; two fields at a point; are ordinary numbers/operators and can be directly interpreted.

Five Types of Bilinears

These transform in specific ways under the Lorentz group:

where $\sigma^{\mu\nu} = \tfrac{i}{2}[\gamma^\mu, \gamma^\nu]$.

The Electromagnetic Current

$j^\mu_{\rm EM} = -e\bar\psi\gamma^\mu\psi$

for an electron (charge $-e$). The zeroth component is the charge density, spatial components are current density. Conservation: $\partial_\mu j^\mu = 0$ follows from the Dirac equation.

The Weak Current

In the Standard Model, the charged weak interaction uses the combination:

$j^\mu_{W} = \bar\psi_e\gamma^\mu\tfrac{1}{2}(1 - \gamma^5)\psi_\nu$

The factor $\tfrac{1}{2}(1 - \gamma^5) = P_L$ is the left-handed projector. Only left-handed fermions interact via the charged weak current; this is the source of parity violation in the weak interaction.

Axial Vector Currents and Anomalies

Classically, for a massless fermion the axial current $j^{\mu}_5 = \bar\psi\gamma^\mu\gamma^5\psi$ is conserved. In the quantum theory, this is violated by anomalies; classical symmetries that fail to survive quantization due to the measure in the path integral. The chiral anomaly is essential for understanding the decay $\pi^0 \to \gamma\gamma$ and other phenomena. (Detailed treatment later; anomalies deserve their own document.)

12. Physical Content and Preview of QED

What We’ve Accomplished

• Quantized the Dirac field using anticommutators
• Showed that this is forced on us by positivity of the Hamiltonian
• Got particles and antiparticles, both with positive energy
• Automatic Pauli exclusion and antisymmetric multi-particle states
• The spin-statistics connection as a theorem
• The fermion propagator
• Bilinears for constructing interactions

The Spin-Statistics Theorem Restated

Integer-spin fields are quantized with commutators → bosons, symmetric states, Bose-Einstein statistics, no Pauli exclusion.

Half-integer-spin fields are quantized with anticommutators → fermions, antisymmetric states, Fermi-Dirac statistics, Pauli exclusion.

This is rigid: it’s a theorem of relativistic quantum field theory, not a choice. Experiment has never found a violation.

Implications

Stability of matter. Atoms exist because electrons (fermions) can’t all pile into the lowest orbital. Without Pauli exclusion, all electrons would collapse to the ground state and matter would have no structure.

White dwarfs. Supported against gravitational collapse by electron degeneracy pressure; a consequence of Pauli exclusion forcing electrons into high-momentum states.

Neutron stars. Same for neutrons.

Chemistry. Shell structure, periodic table, covalent bonding; all traces to fermion statistics.

Superconductivity. Pairs of fermions (Cooper pairs) act as effective bosons, which can Bose-condense. If electrons were bosons, superconductivity wouldn’t be interesting; everything would already be in the ground state.

Preview: Electron-Photon Interactions

The next steps involve coupling the Dirac field to the electromagnetic field. The Lagrangian of QED (from the classical field theory document):

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

$\slashed{D} = \gamma^\mu(\partial_\mu + ieA_\mu)$

After quantizing both $\psi$ and $A_\mu$, we can compute processes like:

• Electron-positron scattering: $e^+e^- \to e^+e^-$
• Electron-muon scattering
• Compton scattering: $\gamma e^- \to \gamma e^-$
• $e^+e^- \to \gamma\gamma$ (annihilation)
• Pair production: $\gamma\gamma \to e^+e^-$

Each one involves computing a matrix element between initial and final states in Fock space, using the field operators and the interaction term. The resulting expressions become Feynman diagrams.

Before we get there, we need to quantize the photon field (document 3) and develop perturbation theory (document 4). But you can already see where this is heading.

Appendix: Formulas and Identities

Gamma Matrix Algebra

$\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu}$

$(\gamma^0)^\dagger = \gamma^0, \quad (\gamma^i)^\dagger = -\gamma^i$

$\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3, \quad \{\gamma^5, \gamma^\mu\} = 0$

$\gamma^\mu\gamma_\mu = 4$

$\gamma^\mu\gamma^\nu\gamma_\mu = -2\gamma^\nu$

$\gamma^\mu\gamma^\nu\gamma^\rho\gamma_\mu = 4\eta^{\nu\rho}$

Traces

$\text{tr}(\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}$

$\text{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) = 4(\eta^{\mu\nu}\eta^{\rho\sigma} - \eta^{\mu\rho}\eta^{\nu\sigma} + \eta^{\mu\sigma}\eta^{\nu\rho})$

$\text{tr}(\gamma^5\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) = -4i\epsilon^{\mu\nu\rho\sigma}$

$\text{tr}(\text{odd number of } \gamma^\mu\text{'s}) = 0$

$\text{tr}(\gamma^5) = 0, \quad \text{tr}(\gamma^5\gamma^\mu) = 0, \quad \text{tr}(\gamma^5\gamma^\mu\gamma^\nu) = 0$

Spinor Completeness

$\sum_s u^s(p)\bar u^s(p) = \slashed{p} + m$

$\sum_s v^s(p)\bar v^s(p) = \slashed{p} - m$

Canonical Anticommutation Relations

$\{\psi_a(\vec x), \psi_b^\dagger(\vec y)\} = \delta_{ab}\delta^3(\vec x - \vec y)$

$\{\psi_a, \psi_b\} = \{\psi_a^\dagger, \psi_b^\dagger\} = 0$

Mode Operator Anticommutators

$\{b^r_{\vec p}, b^{s\dagger}_{\vec q}\} = (2\pi)^3\delta^{rs}\delta^3(\vec p - \vec q)$

$\{c^r_{\vec p}, c^{s\dagger}_{\vec q}\} = (2\pi)^3\delta^{rs}\delta^3(\vec p - \vec q)$

All others vanish.

Field Expansion

$\psi(x) = \int\frac{d^3p}{(2\pi)^3\sqrt{2E_p}}\sum_s[b^s_{\vec p} u^s(p) e^{-ip\cdot x} + c^{s\dagger}_{\vec p} v^s(p) e^{+ip\cdot x}]$

Feynman Propagator

$S_F(x - y) = \int\frac{d^4p}{(2\pi)^4}\frac{i(\slashed{p} + m)}{p^2 - m^2 + i\epsilon}e^{-ip\cdot(x-y)}$

Chirality Projectors

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

$P_L + P_R = 1, \quad P_L^2 = P_L, \quad P_R^2 = P_R, \quad P_L P_R = 0$

Bilinears and Their Transformations

(Where $\pm$ means spatial components flip sign, time components don’t, depending on index structure.)

Checklist

By the end of this document, you should be able to:

• Explain why commutators don’t work for spin-½ fields, and why anticommutators do
• State the spin-statistics theorem and describe its content
• Expand the Dirac field in terms of creation/annihilation operators for particles and antiparticles
• Explain the conventions (the factors of $\sqrt{2E_p}$, the $u$ vs. $v$ spinors)
• Show that Pauli exclusion follows from $(b^\dagger)^2 = 0$
• Write the Feynman propagator in momentum space
• List the five types of bilinears and their Lorentz transformation properties
• Explain why CPT is guaranteed but CP might be violated

Problems to Work

1. Verify that $(b^\dagger_{\vec p})^2 = 0$ from the anticommutation relations, confirming Pauli exclusion.
2. Compute $H|\vec p, s\rangle_e$ and $H|\vec p, s\rangle_{\bar e}$ to verify both have energy $E_p$.
3. Derive the explicit form of the Feynman propagator starting from the mode expansion.
4. Show that $Q|\vec p, s\rangle_e = +|\vec p, s\rangle_e$ and $Q|\vec p, s\rangle_{\bar e} = -|\vec p, s\rangle_{\bar e}$.
5. Verify that the sum over spins $\sum_s u^s(p)\bar u^s(p) = \slashed{p} + m$ using the explicit chiral-basis spinors.
6. Compute the trace $\text{tr}[\slashed{p}\slashed{k}]$ and $\text{tr}[\slashed{p}\slashed{k}\slashed{q}\slashed{\ell}]$ using the trace identities.

Problems 5 and 6 especially are worth doing; both will appear constantly in QED calculations, and fluency with spinor algebra comes from repetition.

Further Reading

• Peskin & Schroeder, Chapter 3: standard treatment of Dirac quantization
• Schwartz, Chapter 10: cleaner modern presentation
• Srednicki, Chapters 36-44: includes careful discussion of discrete symmetries
• Weinberg, Vol. 1, Chapter 5: Weinberg’s unique perspective on why fields transform as they do

Closing Note

This document establishes the second pillar of QFT. We now have:

• Scalar fields (bosons); quantized with commutators
• Dirac fields (fermions); quantized with anticommutators

And we’ve seen that this isn’t a choice; the spin-statistics theorem forces it.

The Big Picture So Far

Each fundamental field in the Standard Model is quantized by canonical methods:

• Higgs (spin 0); commutators
• Electrons, quarks, leptons (spin ½); anticommutators
• Photon, W, Z, gluon (spin 1); commutators (next document, with complications from gauge)

All matter is fermionic. All force carriers are bosonic. That’s not accidental; it’s forced by the spin-statistics theorem, and it explains why matter and forces play different roles.

What’s Next

The next document handles the photon: quantizing a gauge field. The complications: gauge invariance means the $A_\mu$ has more components than physical degrees of freedom. Simply quantizing all four components of $A_\mu$ gives wrong results. We need gauge fixing; and the procedures have conceptual content (leading eventually to ghost fields in non-abelian theories).

After that: interactions. We’ll couple the Dirac and photon fields together in QED, derive Feynman rules from first principles, and finally compute physical quantities like cross-sections and anomalous magnetic moments.

You’re two-thirds of the way through the “quantize the free fields” phase of QFT. The next document completes it, and then the real payoff; Feynman diagrams, QED predictions, experimentally testable results; begins.

Nicely done.