advanced29 min studyLesson 48 in Physics

Dirac fields and fermionic excitations

Relate spinor equations to relativistic dispersion, antiparticles, and the anticommutation algebra behind Pauli exclusion.

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What you’ll understand

  • Use the Clifford algebra to check the relativistic dispersion of the Dirac equation.
  • Identify what fermionic anticommutation implies for a single mode’s occupation.
  • Distinguish spinor components, physical spin states, and particle/antiparticle excitations.

A scalar field cannot describe an electron’s spin. The Dirac field adds a spinor transformation law and a first-order relativistic equation. Quantizing it also changes the algebra of excitations: creation operators anticommute. These changes fit together under the assumptions of local relativistic quantum theory. They do not mean that fermions simply repel, nor that a four-component spinor represents four independently observable spin directions.

Linear in derivatives, relativistic in content

Use natural units and metric η=(+−−−)\eta=(+---). Gamma matrices obey the Clifford algebra, while ψˉ=ψ†γ0\bar\psi=\psi^{\dagger}\gamma^0 is the Dirac adjoint. The free Lagrangian is ψˉ(iγμ∂μ−m)ψ\bar\psi(i\gamma^\mu\partial_\mu-m)\psi. Its equation is first order in time and space. The matrices are not arbitrary decoration: their anticommutation is exactly what makes the squared differential operator reproduce the scalar relativistic mass-shell condition.

Multiply the Dirac equation by the conjugate differential operator. Since ordinary partial derivatives commute, the antisymmetric part of the gamma-matrix product drops out and every spinor component satisfies the Klein–Gordon equation. This demonstrates the same energy-momentum dispersion, not an equivalence of spin-zero and spin-one-half theories. The allowed spinor solutions still satisfy the additional first-order constraints and transform differently under rotations and boosts.

{γμ,γν}=2ημν1,(iγμ∂μ−m)ψ=0 ⟹ (□+m2)ψ=0\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf1,\qquad (i\gamma^\mu\partial_\mu-m)\psi=0\ \Longrightarrow\ (\Box+m^2)\psi=0
The implication follows by left-multiplying by iγ·∂ + m and then removing an overall minus sign. The gamma matrices are representation dependent; the Clifford relation is not.↔ Scroll wide equations horizontally.

Anticommutation changes the state space

The Dirac field is expanded in positive-frequency spinors multiplying particle annihilation operators and negative-frequency spinors multiplying antiparticle creation operators. Each massive on-shell particle has two spin states, with a corresponding antiparticle sector. The four components of the field organize this relativistically; they are not four simultaneous outcomes of measuring one spin component. Antiparticles appear as positive-energy excitations with opposite conserved charge, not as accessible negative-energy states.

For a discrete normalized fermionic mode, {b,b†}=1\{b,b^{\dagger}\}=1 and {b†,b†}=0\{b^{\dagger},b^{\dagger}\}=0 imply (b†)2=0(b^{\dagger})^2=0. One mode can be empty or occupied once. Two fermions may still share a spatial region when their full one-particle modes differ, for example through spin. Pauli exclusion is a constraint on states, not a classical force that forbids any spatial overlap.

{ψα(t,x),ψβ†(t,y)}=δαβδ(3)(x−y),{b,b†}=1,(b†)2=0\{\psi_\alpha(t,\mathbf x),\psi_\beta^{\dagger}(t,\mathbf y)\}=\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y),\qquad \{b,b^{\dagger}\}=1,\qquad (b^{\dagger})^2=0
The field relation uses ψ dagger, not the Dirac adjoint ψ bar. The single-mode relation assumes discrete normalization, such as a finite box.↔ Scroll wide equations horizontally.

Signs carry physical information

Exchanging two fermionic creation operators reverses the sign of the state. That minus sign generates antisymmetric multiparticle amplitudes and exchange interference. In time-ordered fermion correlators, swapping two fermion operators also inserts a minus sign. The momentum-space free propagator is i(γμpμ+m)/(p2−m2+iϵ)i(\gamma^\mu p_\mu+m)/(p^2-m^2+i\epsilon), a matrix in spinor indices. Replacing it by the scalar propagator discards the numerator that carries spin information.

The spin-statistics connection is a structural theorem with assumptions including relativistic locality and a physically suitable Hilbert space and spectrum. It is not derived here merely by naming a field “spinor.” A useful practical check is that the correctly quantized free theory has positive particle and antiparticle excitation energies. The detailed proof and extensions in lower dimensions demand more care than an oscillator analogy supplies.

The occupation operator of one fermionic mode

Let N=b†bN=b^{\dagger}b, with b∣0⟩=0b|0\rangle=0, {b,b†}=1\{b,b^{\dagger}\}=1, and b2=(b†)2=0b^2=(b^{\dagger})^2=0. Show that its only eigenvalues are zero and one.

  1. Compute N2=b†bb†bN^2=b^{\dagger}bb^{\dagger}b. Replace bb†bb^{\dagger} by 1−b†b1-b^{\dagger}b, obtaining N2=N−b†b†bb=NN^2=N-b^{\dagger}b^{\dagger}bb=N.

  2. If N∣n⟩=n∣n⟩N|n\rangle=n|n\rangle, then N2=NN^2=N implies n2=nn^2=n. The possible eigenvalues are therefore 00 and 11.

  3. The state ∣1⟩=b†∣0⟩|1\rangle=b^{\dagger}|0\rangle is normalized because ⟨0∣bb†∣0⟩=1\langle0|bb^{\dagger}|0\rangle=1. Trying to create another identical excitation gives b†∣1⟩=0b^{\dagger}|1\rangle=0, rather than a normalized two-particle state in that mode.

The result

Exclusion follows from the operator algebra. A second particle can occupy another orthogonal mode, but not the same complete one-particle state.

Explore the model

Interactive lab

Interactive lab · quantum

A two-level spin analogy, with its boundary stated

Vary the mixing angle and relative phase of a two-level pure state. Compare unchanged basis probabilities under a phase change with the moving Bloch coordinates, as a reminder that spin amplitudes contain more information than a pair of populations.

A two-level spin analogy, with its boundary statedA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis is a two-level quantum-state model. It does not represent the full Dirac spinor, antiparticle creation, fermionic Fock-space anticommutation, or relativistic spinor transformations.

Check your understanding

What does Pauli exclusion forbid in this free-field construction?

Try it yourself

For two different orthogonal modes, compare b1†b2†∣0⟩b_1^{\dagger}b_2^{\dagger}|0\rangle and b2†b1†∣0⟩b_2^{\dagger}b_1^{\dagger}|0\rangle. Does their sign difference make one forbidden?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the anticommutator of different creation operators and distinguish an overall phase from a relative phase.

Compare with the solution

They differ by a minus sign because {b1†,b2†}=0\{b_1^{\dagger},b_2^{\dagger}\}=0. Both expressions describe the same occupied pair up to an overall phase, so neither is forbidden. The sign matters when alternatives are combined coherently in a larger amplitude. Exclusion appears when the mode labels coincide, making the repeated creation operator zero.

Take it with you

  • The Clifford algebra enforces relativistic dispersion while retaining a spinor transformation law.
  • Anticommuting mode operators produce exclusion and exchange signs.
  • A two-level spin illustration does not replace the full particle–antiparticle Dirac field.

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