advanced33 min studyLesson 656 in Physics

Fermionic anticommutators and positive energy

Quantize the first-order Dirac field using canonical anticommutators, explain the antiparticle energy sign, and derive exclusion in a regulated mode basis.

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The question for today

What you’ll understand

State the canonical anticommutation relations with spin and momentum normalization.
Reorder the antiparticle energy term with the correct fermionic sign.
Connect repeated creation in one mode to Pauli exclusion without identifying a spinor norm with a state norm.

The Dirac equation admits negative-frequency classical solutions, but a stable quantum theory describes their coefficients with antiparticle creation operators. The first-order action and fermionic algebra together make this reinterpretation consistent with positive excitation energy and positive state norms. Choosing ordinary bosonic commutators instead fails to preserve all the desired properties. This explicit free-field construction illustrates the spin–statistics connection; it is not by itself a proof of the theorem for every relativistic interacting theory.

A first-order action has constrained canonical data

The Dirac Lagrangian is linear in the field's time derivative. Its momentum conjugate is imaginary one times the Hermitian adjoint, rather than an independent time derivative as for a real scalar. A systematic Hamiltonian treatment handles the resulting constraints with graded canonical methods. The final equal-time quantum algebra uses anticommutators between the field and its adjoint. Simply applying the scalar canonical recipe to four unconstrained commuting coordinates misses this structure.

Expand the free field in the normalized u and v modes, assigning particle annihilators to positive frequency and antiparticle creators to negative frequency. Both ladder families anticommute within and across species. The spinor completeness relations ensure that their equal-time anticommutator reconstructs the spinor identity times the spatial delta kernel. The factor inverse square root of twice energy is essential to this check, just as the Fourier measure and momentum-delta convention are.

\psi(x)=\sum_s\int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}\left[b_{\mathbf p s}u_s(p)e^{-ip\cdot x}+d_{\mathbf p s}^{\dagger}v_s(p)e^{ip\cdot x}\right],\qquad \pi_\psi=i\psi^{\dagger}

Natural units and η = diag(1,−1,−1,−1) are retained. ψ has energy dimension three-halves. u†u = v†v = 2Ep. The first-order action has constraints; πψ = iψ† is not an additional independent oscillator coordinate.↔ Scroll wide equations horizontally.

The algebra gives a positive Fock norm and exclusion

The canonical field anticommutator is understood distributionally, while each normalized discrete box mode has an anticommutator with its adjoint equal to one. A vacuum annihilated by all b and d operators then gives a unit norm for a singly created normalized mode. Applying the same creation operator twice gives zero because its self-anticommutator vanishes. Exclusion applies to a complete one-particle mode, including spin, rather than forbidding two particles with the same spatial orbital but different spin labels.

Distinct creation operators anticommute, so exchanging their order changes the sign of the state. Physical observables are even in fermionic fields and preserve fermion parity; their time evolution is still generated by an ordinary commutator with the Hamiltonian. Neither fermionic statistics nor the negative Dirac scalar bilinear makes the physical Hilbert norm negative. These are separate algebraic questions, and the positive canonical anticommutator is the relevant norm check.

\{\psi_\alpha(t,\mathbf x),\psi_\beta^{\dagger}(t,\mathbf y)\}=\delta_{\alpha\beta}\delta^3(\mathbf x-\mathbf y),\quad \{b_{\mathbf p r},b_{\mathbf q s}^{\dagger}\}=(2\pi)^3\delta_{rs}\delta^3(\mathbf p-\mathbf q)

The same ladder relation holds for d, and all other ladder anticommutators vanish. Smooth smearing and a common finite-particle domain are implicit in continuum formulas. For one normalized discrete mode, {b,b†} = 1 and (b†)² = 0.↔ Scroll wide equations horizontally.

Reordering reverses the apparent energy problem

Before normal ordering, one particle mode and its antiparticle partner contribute a positive particle number term minus an antiparticle annihilator followed by its creator. The fermionic relation turns that latter product into one minus antiparticle number. Both occupation coefficients are therefore positive, with a negative vacuum constant. Removing the constant gives the expected sum of positive particle and antiparticle energies. This is a sign determined by algebra, not an arbitrary manual change of a negative-energy spectrum.

If a commuting Dirac field is attempted, the negative-frequency sector forces a conflict between the canonical field algebra, positive norms, and a Hamiltonian bounded below. Different incorrect conventions expose the conflict as negative norms or unbounded energy. A finite-mode demonstration makes the issue visible, but locality and Lorentz covariance are also essential to the full spin–statistics theorem. Interactions further require regulated composite Hamiltonians and renormalization; the free anticommutation relations do not construct that continuum dynamics automatically.

H_{\mathrm{pair}}=E(b^{\dagger}b-dd^{\dagger})=E(N_b+N_d-1),\qquad :H_{\mathrm{pair}}:=E(N_b+N_d)

This is one retained particle mode and one retained antiparticle mode, both of energy E > 0, with unit discrete anticommutators. Nᵦ,Nd take values zero or one. The continuum Hamiltonian sums this structure over momentum and spin, with a regulated vacuum constant.↔ Scroll wide equations horizontally.

Work it through

Check three finite-mode energies

A retained particle–antiparticle mode pair has E = 3. Compute unsubtracted and normal-ordered energies for its vacuum, its state with only the antiparticle occupied, and its state with both modes occupied.

For the vacuum, both numbers vanish, so $H_{\mathrm{pair}}=3(0+0-1)=-3$ , while the normal-ordered energy is zero.
With only the antiparticle occupied, the unsubtracted energy is $3(0+1-1)=0$ and the subtracted energy is 3. Its excitation gap above vacuum is positive.
With both occupied, the unsubtracted energy is $3(1+1-1)=3$ and the subtracted energy is 6. Adding either allowed quantum costs 3.

The result

The unsubtracted energies are −3, 0, and 3; the normal-ordered energies are 0, 3, and 6. All excitation gaps are positive.

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Interactive lab

Interactive lab · probability

Independent occupied modes have Bernoulli counts

Reset and set BOTH Success probability p to 0.5 and Trials n to 1; predict mean 0.5 and variance 0.25. Then Reset and set BOTH p to 0.5 and n to 2; predict mean 1 and variance 0.5. This binomial law can describe independent fermion modes each occupied with probability p in a diagonal product mixture. It does not implement anticommuting operators, exchange amplitudes, a pure Fock state, or correlated occupations.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forA classical distribution of occupation counts can agree with one mixed preparation without proving its quantum operator algebra.

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Read the checkpoint explanation

The canonical relation {b†,b†} = 0 gives 2(b†)² = 0. This excludes duplicate occupation of the same complete mode while allowing distinct modes to be occupied.

Your turn

Try it yourself

Two orthogonal normalized fermion modes have energies 2 and 5. Define |12⟩ = b₁†b₂†|0⟩. Find its norm, normal-ordered energy, and the sign of b₂†b₁†|0⟩ relative to it. What is b₁†b₁†|0⟩?

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 unit anticommutators for each mode and zero cross anticommutators.

Compare with the solution

The norm is one: moving annihilators through distinct creators and using each unit anticommutator leaves ⟨0|0⟩ = 1. The energy is 2 + 5 = 7. Exchanging the creators gives $b_2^{\dagger}b_1^{\dagger}|0\rangle=-|12\rangle$ . Repeating one creator gives zero. The exchange sign changes amplitudes between histories, while a lone global minus sign does not change a state’s probabilities.

Keep the idea

Take it with you

The first-order Dirac action leads to constrained fermionic canonical data.
Canonical anticommutators give positive normalized states and exclude duplicate mode occupation.
Fermionic reordering makes both particle and antiparticle excitation energies positive.

Read further

Original sources and deeper explanations to follow at your own pace.

David Tong, Quantum Field Theory, 5.1–5.2.1: failed commuting quantization, canonical anticommutators and positive energy

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