The Dirac equation, too, becomes the equation of a field. Expand $\hat{\psi}(\mathbf{r},t)$ in its plane-wave spinors: the positive-energy solutions come with operators $\hat{b}$ that remove electrons, the negative-energy ones with operators $\hat{d}^{\dagger}$ that make positrons. Put this into the energy and it comes out as $\sum E_{p}(\hat{b}^{\dagger}\hat{b}-\hat{d}\hat{d}^{\dagger})$. The order of $\hat{d}\hat{d}^{\dagger}$ now decides everything. If $\hat{d}$ and $\hat{d}^{\dagger}$ obeyed a commutator, $-\hat{d}\hat{d}^{\dagger}$ would be $-\hat{d}^{\dagger}\hat{d}-1$ and each positron would lower the energy by $E_{p}$: states of ever lower energy, and no ground state. With an anticommutator, $-\hat{d}\hat{d}^{\dagger}=\hat{d}^{\dagger}\hat{d}-1$ and every quantum, electron or positron, adds $E_{p}$. So the Dirac field must be quantised with anticommutators: its quanta are fermions, and the exclusion principle comes with them. That is the connection between spin and statistics for spin one-half; the scalar field goes the other way, since its energy counts quanta only with commutators. The Dirac sea is no longer needed: the vacuum has no electrons and no positrons, and a positron is a quantum of the same field as the electron, with the opposite charge.

A ledger kept with the wrong sign for one column shows a firm growing richer with every debt. The Dirac field kept with commutators is that ledger; anticommutators put the sign right.

The energy above the vacuum, in units of $E_{p}$, of a state with $n$ positrons of energy $E_{p}$, across. With anticommutators each adds , the positrons in different modes. With commutators, dashed, each would take away, and of them in one mode would lie below the vacuum for every .

The Dirac field is expanded in the normalised plane-wave spinors $u_{\mathbf{p}s}$ of positive energy and $v_{\mathbf{p}s}$ of negative energy, with $p\cdot x=E$, and with operators for electrons and for positrons:

• Before its operators are put in order, the energy is $\hat{H}=\sum_{\mathbf{p},s}E_{p}\bigl(\hat{b}^{\dagger}_{\mathbf{p}s}\hat{b}_{\mathbf{p}s}-\hat{d}_{\mathbf{p}s}\hat{d}^{\dagger}_{\mathbf{p}s}\bigr)$ (stated).

The negative-energy solutions enter the energy as $-E_{p}\hat{d}\hat{d}^{\dagger}$. Reorder it with each rule. With the anticommutator each positron adds , and the vacuum is the state of lowest energy. With the commutator each positron would take away, and a state of positrons in one mode would lie below the vacuum for every : there would be no ground state.

The Dirac Hamiltonian between the field operators, with the spinors orthonormal (stated).

Each positron adds $E_{p}$; the constant is absorbed into the zero of energy.

Each positron would take $E_{p}$ away.

With commutators, $n$ positrons in one mode, the vacuum’s energy taken as zero: $-nE_{p}$ for every $n$, with no lower bound.

Only the anticommutator gives a ground state. The scalar field, with energy $\sum\hbar\omega(\hat{a}^{\dagger}\hat{a}+\frac12)$, needs commutators: bosons.

In the quantised Dirac field the positive-energy spinors come with operators that remove electrons, and the negative-energy spinors with operators that make positrons.

The positron term of the energy is $-E_{p}\hat{d}\hat{d}^{\dagger}$. With an anticommutator it becomes ; with a commutator it would be .

Quantised with commutators, the Dirac field would have states of ever lower energy; with anticommutators its energy has a floor. So spin-$\tfrac12$ quanta are fermions.

With commutators, $n$ positrons of energy $E_{p}$ in one mode would lie $nE_{p}$ below the vacuum, for every .

The vacuum of the Dirac field holds no electrons and no positrons; the negative-energy states of the one-particle theory no longer need filling.

Great work! You now know how to:

• quantise the Dirac field with electrons and positrons
• show that it needs anticommutators, and so fermions
• explain why no Dirac sea is needed