12.3 The Dirac equation · Quantum Ladder

Last section’s autopsy left exactly one way out: for positive-definite probability, the equation must be first order in time; for relativistic covariance, it must then be first order in space too. But in the dispersion relation $E^2=p^2c^2+m^2c^4$ , both $E$ and $p$ appear squared — a first-order equation amounts to taking the square root of this sum of squares. In 1928, the 26-year-old Dirac decided to take the root by brute force.

Taking the square root: an algebra problem

First state the goal precisely. We want an equation of the form

\ii\hbar\frac{\partial\psi}{\partial t}=\hat H\psi,\qquad \hat H=c\,\vec\alpha\cdot\hat{\vec p}+\beta\,mc^2 =c\sum_{k=1}^3\alpha_k\hat p_k+\beta\,mc^2\tag{12.3.1}

First order in time (the left side copied from Schrödinger), first order in space ( $\hat p=-\ii\hbar\nabla$ appears only to the first power). $\alpha_1,\alpha_2,\alpha_3,\beta$ are four undetermined “coefficients” — do not presume yet that they are numbers. There is only one constraint, but it is iron: $\hat H$ squared must return the relativistic relation, $\hat H^2=c^2\hat p^2+m^2c^4$ , or the solutions will not obey the correct energy-momentum relation.

Square once, and the whole algebra is forced outbasic~8 min

Step 1: expand $\hat H^2$ honestly. Keep the order of $\alpha,\beta$ intact (we do not yet know whether they commute):

\hat H^2=c^2\sum_{j,k}\alpha_j\alpha_k\,\hat p_j\hat p_k +mc^3\sum_k(\alpha_k\beta+\beta\alpha_k)\,\hat p_k +\beta^2m^2c^4\tag{12.3.2}

In the first term $\hat p_j\hat p_k=\hat p_k\hat p_j$ (momentum components commute with each other), so only the symmetric combination acts, and it can be written $\frac12(\alpha_j\alpha_k+\alpha_k\alpha_j)\hat p_j\hat p_k$ .

Step 2: compare with the target term by term. Demanding $\hat H^2=c^2(\hat p_1^2+\hat p_2^2+\hat p_3^2)+m^2c^4$ :

The cross terms in $\hat p_j\hat p_k$ ( $j\neq k$ ) must vanish, and the squared terms must have coefficient 1:

\alpha_j\alpha_k+\alpha_k\alpha_j=2\delta_{jk}\tag{12.3.3}

The terms linear in $\hat p_k$ must vanish, and the constant term must have coefficient 1:

\alpha_k\beta+\beta\alpha_k=0,\qquad \beta^2=1\tag{12.3.4}

Step 3: read off the conclusion — they cannot be numbers. The condition $\alpha_1\alpha_2=-\alpha_2\alpha_1$ is called anticommutation: swapping the order flips the sign. Two non-zero ordinary numbers always commute under multiplication and can never anticommute. Dirac’s flash of insight: let them be matrices — matrix multiplication does not commute anyway — and $\psi$ correspondingly becomes a multi-component column vector.

Step 4: how big must the matrices be? Three quick reconnaissance results:

Candidates lie ready to hand: the Pauli matrices $\sigma_1,\sigma_2,\sigma_3$ (section 5.3) satisfy exactly $\sigma_j\sigma_k+\sigma_k\sigma_j=2\delta_{jk}$ — but there are only three of them, and we need four mutually anticommuting matrices. No fourth can be scraped together in 2×2 (any 2×2 matrix expands in $1,\sigma_1,\sigma_2,\sigma_3$ , and the identity commutes with everything, so no anticommuting fourth member exists).
Zero trace: from $\alpha_k=-\beta\alpha_k\beta$ (the anticommutation relation rewritten using $\beta^2=1$ ), taking the trace gives $\mathrm{tr}\,\alpha_k=-\mathrm{tr}\,\alpha_k=0$ ; likewise $\mathrm{tr}\,\beta=0$ .
Eigenvalues ±1: $\alpha_k^2=\beta^2=1$ means the eigenvalues can only be $\pm1$ ; zero trace then demands equal numbers of $+1$ and $-1$ — the dimension must be even.

Two dimensions insufficient, odd dimensions out — the next candidate is 4×4. Four dimensions indeed suffice, and the standard choice (the Dirac representation) is written in 2×2 blocks:

\alpha_k=\begin{pmatrix}0&\sigma_k\\ \sigma_k&0\end{pmatrix},\qquad \beta=\begin{pmatrix}I&0\\ 0&-I\end{pmatrix}\tag{12.3.5}

(Verify the three sets of relations using $\sigma_j\sigma_k+\sigma_k\sigma_j=2\delta_{jk}$ — a two-minute exercise.)

So $\psi$ is a four-component column vector, called a Dirac spinor — and the “four” is no accident: we shall see shortly that it = two spin components × two energy branches. Writing the momentum operator back as a derivative gives the Dirac equation:

\ii\hbar\frac{\partial\psi}{\partial t} =\bigl(-\ii\hbar c\,\vec\alpha\cdot\nabla+\beta mc^2\bigr)\psi\tag{12.3.6}

It immediately makes good on its promise: being first order in time, one can define $\rho=\psi^\dagger\psi=\sum_{a=1}^4|\psi_a|^2\geq0$ — a positive-definite probability density, paired with the current $\vec j=c\,\psi^\dagger\vec\alpha\,\psi$ satisfying the continuity equation. The KG knot is untied.

The picture

The covariant form. Multiply the equation through by $\beta$ , reorganise the time and space terms, and define $\gamma^0=\beta$ , $\gamma^k=\beta\alpha_k$ ; the equation becomes one line:

\ii\hbar\,\gamma^\mu\partial_\mu\psi=mc\,\psi\tag{12.3.7}

Time and space derivatives stand side by side on completely equal footing — the first genuine fulfilment of the “spacetime equality” demanded at the start of this chapter. The γ matrices in block form:

\gamma^0=\begin{pmatrix}I&0\\0&-I\end{pmatrix},\qquad \gamma^k=\begin{pmatrix}0&\sigma_k\\-\sigma_k&0\end{pmatrix}\tag{12.3.8}

The mathematics

The whole algebra condensed into one line. The three sets of anticommutation relations merge into

\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2\eta^{\mu\nu}\tag{12.3.9}

where $\eta^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$ is the Minkowski metric: the γ matrices are “the square root of the spacetime metric”, and the Dirac equation is a square root taken of spacetime geometry itself. (This structure is called a Clifford algebra, invented by mathematicians half a century before Dirac — it waited in the storeroom for its physics.)

The climax: spin emerges on its own

So far the $\sigma_k$ have been mere building bricks — nowhere have we assumed that the electron has spin. Now run a routine check: is the angular momentum of a free particle conserved?

Orbital angular momentum is not conserved — unless spin is addedadvanced~8 min

Step 1: check the orbital angular momentum. Take $\hat L_z=\hat x\hat p_y-\hat y\hat p_x$ and commute it with $\hat H=c\vec\alpha\cdot\hat{\vec p}+\beta mc^2$ (the $\beta mc^2$ term involves neither position nor momentum):

[\hat L_z,\hat H]=c\,\alpha_j[\hat x\hat p_y-\hat y\hat p_x,\hat p_j] =c\bigl(\alpha_x[\hat x,\hat p_x]\hat p_y-\alpha_y[\hat y,\hat p_y]\hat p_x\bigr) =\ii\hbar c\,(\alpha_x\hat p_y-\alpha_y\hat p_x)\tag{12.3.10}

Not zero. In Schrödinger theory the free particle’s $\hat L_z$ is ironclad conserved; in Dirac theory it is not — angular momentum of an isolated system is leaking. Where to?

Step 2: recover the missing piece. Define, using block Pauli matrices,

\Sigma_z=\begin{pmatrix}\sigma_z&0\\0&\sigma_z\end{pmatrix}\tag{12.3.11}

Using $\sigma_x\sigma_y=\ii\sigma_z$ and its cousins, compute block by block (exercise: only 2×2 block multiplication is needed):

[\Sigma_z,\hat H]=c\,[\Sigma_z,\vec\alpha\cdot\hat{\vec p}\,] =-2\ii c\,(\alpha_x\hat p_y-\alpha_y\hat p_x)\tag{12.3.12}

Differing from Step 1’s result only by a factor $-2/\hbar$ .

Step 3: combine. Define $\hat J_z=\hat L_z+\frac{\hbar}{2}\Sigma_z$ ; then

[\hat J_z,\hat H]=\ii\hbar c(\alpha_x\hat p_y-\alpha_y\hat p_x) +\frac{\hbar}{2}\cdot(-2\ii c)(\alpha_x\hat p_y-\alpha_y\hat p_x)=0\tag{12.3.13}

What is conserved is $\hat{\vec J}=\hat{\vec L}+\frac\hbar2\vec\Sigma$ . The added piece $\frac\hbar2\vec\Sigma$ is an intrinsic angular momentum: each component has eigenvalues $\pm\hbar/2$ , and the sum of squares is $\frac34\hbar^2=s(s+1)\hbar^2$ with $s=\frac12$ . This is precisely the spin 1/2 that chapter 5 rammed into the theory at the demand of the Stern-Gerlach experiment — and here it walks out of the equation’s algebra by itself.

The ancestry of spin, hereby established

Recall chapter 5: spin was forced into the theory with the experiments holding its head down — the twin spots of Stern-Gerlach, the anomalous Zeeman effect — driving Pauli to add two components to the wavefunction with no explanation whatsoever: “that’s just how it is.”

The Dirac equation supplies the explanation: demand “relativistic covariance + first order in time + positive-definite probability”, and the minimal algebraic structure that solves it automatically carries an intrinsic angular momentum of ħ/2. Spin is not a spinning little ball, not an extra postulate, but the inevitable product of the compatibility conditions between relativity and quantum mechanics. A mathematical structure invented purely to “take a square root” handed over the most baffling experimental fact of the preceding decade — one of theoretical physics’ rare shining moments. (An honest footnote: a similar linearisation of the Schrödinger equation also yields spin, so strictly speaking it is not “exclusively relativistic”; but its first appearance as a consequence is here. Section 12.5 has an even harder gift: $g=2$ .)

Key formulas

Dirac equation (Hamiltonian form)
$\ii\hbar\,\partial_t\psi=\bigl(c\,\vec\alpha\cdot\hat{\vec p}+\beta mc^2\bigr)\psi$
ψ is a four-component spinor; ρ=ψ†ψ≥0 is positive definite
Anticommutation algebra
$\alpha_j\alpha_k+\alpha_k\alpha_j=2\delta_{jk},\quad \alpha_k\beta+\beta\alpha_k=0,\quad \beta^2=1$
Uniquely forced by H²=c²p²+m²c⁴; numbers fail, minimum 4×4
Block representation
$\alpha_k=\begin{pmatrix}0&\sigma_k\\ \sigma_k&0\end{pmatrix},\quad \beta=\begin{pmatrix}I&0\\ 0&-I\end{pmatrix}$
Pauli matrices as bricks; covariant form iħγ^μ∂_μψ=mcψ
Conserved angular momentum
$\hat{\vec J}=\hat{\vec L}+\frac{\hbar}{2}\vec\Sigma$
L alone is not conserved; spin 1/2 appears automatically to complete the conservation law

Self-check4 questions

What comes next

The Dirac equation won the probability but took on a heavier debt: the spectrum has no floor. Every electron should, by rights, radiate its way down toward $E=-\infty$ , and the universe should contain no stable atoms. Dirac’s escape was outrageous — redefine the “vacuum” as a sea filled with negative-energy electrons. This seemingly absurd picture will predict the first antimatter in human history.

The Dirac equation

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After this section you should be able to

Taking the square root: an algebra problem

The picture

The mathematics

The climax: spin emerges on its own

Key formulas

Self-check4 questions

What comes next

Taking the square root: an algebra problem

◑The picture

∑The mathematics

The climax: spin emerges on its own

∎Key formulas

What comes next

The picture

The mathematics

Key formulas