Quantum Field Theory Course

Total: 440+ pages covering classical field theory through the Standard Model

Part I, Chapter 1 | Page 7 of 8

The Dirac Field

Lagrangian for spin-1/2 fermions

1.24 The Dirac Equation

To describe spin-1/2 particles (electrons, quarks, neutrinos), we need a spinor field ψ(x). The Dirac field is a 4-component complex object:

$$\psi(x) = \begin{pmatrix} \psi_1(x) \\ \psi_2(x) \\ \psi_3(x) \\ \psi_4(x) \end{pmatrix}$$

The Dirac Lagrangian is:

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

where:

  • $\bar{\psi} = \psi^\dagger \gamma^0$ is the Dirac adjoint
  • γμ are the Dirac gamma matrices (4×4 matrices)
  • m is the fermion mass
  • We use natural units with ℏ = c = 1

Gamma Matrices

The gamma matrices satisfy the Clifford algebra:

$$\boxed{\{\gamma^\mu, \gamma^\nu\} = 2g^{\mu\nu}\mathbb{I}}$$

where { , } denotes the anticommutator. In the Dirac (standard) representation:

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

where σi are the Pauli matrices.

1.25 Deriving the Dirac Equation

Treating ψ and $\bar{\psi}$ as independent fields, the Euler-Lagrange equation for $\bar{\psi}$ gives:

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

This is the Dirac equation. Multiplying by (iγν∂ν + m):

$$(i\gamma^\nu \partial_\nu + m)(i\gamma^\mu \partial_\mu - m)\psi = 0$$

Using the Clifford algebra:

$$-\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu - m^2 = -\frac{1}{2}\{\gamma^\nu, \gamma^\mu\}\partial_\nu \partial_\mu - m^2 = -\partial_\mu \partial^\mu - m^2$$

Therefore each component satisfies the Klein-Gordon equation:

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

But the Dirac equation is first-order in time, unlike Klein-Gordon. This allows for a positive-definite probability density.

1.26 Dirac Adjoint and Conserved Current

Varying with respect to ψ gives the adjoint Dirac equation:

$$\bar{\psi}(i\overleftarrow{\partial}_\mu \gamma^\mu + m) = 0$$

or equivalently:

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

Probability Current

Multiplying the Dirac equation by $\bar{\psi}$ from the left and the adjoint equation by ψ from the right, then adding:

$$\partial_\mu(\bar{\psi}\gamma^\mu \psi) = 0$$

Therefore the Dirac current:

$$\boxed{j^\mu = \bar{\psi}\gamma^\mu \psi}$$

is conserved: ∂μjμ = 0. The time component:

$$j^0 = \psi^\dagger \psi \geq 0$$

is positive definite, solving the negative probability problem of the Klein-Gordon equation!

1.27 Global U(1) Symmetry

The Dirac Lagrangian is invariant under the global phase transformation:

$$\psi \to e^{i\alpha}\psi, \quad \bar{\psi} \to e^{-i\alpha}\bar{\psi}$$

where α is a constant. This U(1) symmetry leads to the conserved current jμ = $\bar{\psi}\gamma^\mu \psi$ by Noether's theorem.

The conserved charge:

$$Q = \int d^3x \, j^0 = \int d^3x \, \psi^\dagger \psi$$

is the total fermion number (electric charge for electrons).

Lorentz Invariance

Under a Lorentz transformation Λ, the Dirac field transforms as:

$$\psi(x) \to S(\Lambda)\psi(\Lambda^{-1}x)$$

where S(Λ) is a 4×4 matrix representation of the Lorentz group satisfying:

$$S(\Lambda)^{-1}\gamma^\mu S(\Lambda) = \Lambda^\mu_{\phantom{\mu}\nu}\gamma^\nu$$

This ensures the Dirac Lagrangian is Lorentz invariant. Spinors transform under the (1/2, 0) ⊕ (0, 1/2) representation of the Lorentz group.

1.28 Energy-Momentum Tensor for Dirac Field

The canonical energy-momentum tensor is:

$$T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)}\partial^\nu \psi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \bar{\psi})}\partial^\nu \bar{\psi} - g^{\mu\nu}\mathcal{L}$$

For the Dirac Lagrangian:

$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu \psi)} = i\bar{\psi}\gamma^\mu$$

Therefore:

$$\boxed{T^{\mu\nu} = i\bar{\psi}\gamma^\mu \partial^\nu \psi}$$

This is not symmetric. The symmetric (Belinfante) tensor is:

$$\Theta^{\mu\nu} = \frac{i}{2}\bar{\psi}\gamma^\mu \overleftrightarrow{\partial}^\nu \psi$$

where $\overleftrightarrow{\partial}^\nu = \overrightarrow{\partial}^\nu - \overleftarrow{\partial}^\nu$.

Hamiltonian

The Hamiltonian density is:

$$\mathcal{H} = T^{00} = i\bar{\psi}\gamma^0 \partial_0 \psi = i\psi^\dagger \dot{\psi}$$

Using the Dirac equation, this can be rewritten as:

$$\mathcal{H} = \bar{\psi}(-i\gamma^i \partial_i + m)\psi$$

Key Concepts (Page 7)

  • • Dirac Lagrangian: $\mathcal{L} = \bar{\psi}(i\gamma^\mu \partial_\mu - m)\psi$
  • • Dirac equation: (iγμ∂μ - m)ψ = 0
  • • Clifford algebra: {γμ, γν} = 2gμν
  • • Conserved current: jμ = $\bar{\psi}\gamma^\mu \psi$
  • • Positive probability density: ρ = ψ†ψ ≥ 0
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