What Happens When a Magnet Attracts Metal? (Part 2): Dirac, and the Magnetism Hidden in Spacetime

Part 2 of four. Part 1 showed that classical physics forbids magnetism (Bohr–van Leeuwen), and that the first relativistic patch (Klein–Gordon) collapses under negative energies and negative probabilities. We left Dirac holding one clue: insist on a wave equation that is first order in time yet relativistically covariant, and see what you are forced to accept.

Let me pick up exactly where we left off. We had written down Dirac’s audacious guess — a Hamiltonian that is merely linear in momentum,

\hat{H} = c(\alpha_x \hat{p}_x + \alpha_y \hat{p}_y + \alpha_z \hat{p}_z) + \beta m c^2

with four unknown coefficients $\alpha_x, \alpha_y, \alpha_z, \beta$ left dangling. The whole game now is to ask: what must these objects be for the equation to be consistent with relativity? We will not assume they are matrices. We will be forced into it.

1. Squaring the equation forces a Clifford algebra

The consistency demand is simple to state. Any solution of Dirac’s first-order equation must also solve the relativistic energy–momentum relation — that is, applying the operator twice must reproduce $E^2 = c^2\vec{p}^2 + m^2c^4$ . So let us square the Hamiltonian and insist it come out right:

(c\textstyle\sum \alpha_i p_i + \beta mc^2)^2 = c^2 \sum p_i^2 + m^2 c^4

Expanding the left side carefully — the momenta $$p_i$$ commute with one another, but we must not assume the coefficients do — gives

c^2 \sum_i \alpha_i^2 p_i^2 + c^2 \sum_{i < j} \{\alpha_i, \alpha_j\} p_i p_j + mc^3 \sum_i \{\alpha_i, \beta\} p_i + \beta^2 m^2 c^4

Now match this against the clean right-hand side, term by term. The cross terms have no business being there, so they must vanish; the square terms must have unit coefficient. That comparison hands us four algebraic commandments — the Clifford algebra:

$\alpha_i^2 = I$
$\beta^2 = I$
$\{\alpha_i, \alpha_j\} = 0 \quad (i \neq j)$
$\{\alpha_i, \beta\} = 0$

Stare at commandment 3 for a second. It says $\alpha_1 \alpha_2 = -\alpha_2 \alpha_1$ — the coefficients anticommute. No ordinary numbers do that (for numbers, $$ab = ba$$ always). So the $\alpha_i$ and $\beta$ cannot be numbers. They are forced to be matrices, and we never assumed it — relativity plus first-order-in-time compelled it. This is the moment the electron grows an internal structure.

2. Gamma matrices and the inevitability of four components

To make the Lorentz covariance manifest, it is cleaner to repackage the coefficients into the gamma matrices, $\gamma^0 = \beta$ and $\gamma^i = \beta\alpha_i$ . The four commandments above collapse into a single elegant relation,

\{\gamma^\mu, \gamma^\nu\} = 2\eta^{\mu\nu} I

where $\eta^{\mu\nu} = \text{diag}(1,-1,-1,-1)$ is the Minkowski metric, and the Dirac equation itself takes its famous compact form,

(i \gamma^\mu \partial_\mu - \tfrac{mc}{\hbar}) \psi = 0,\quad \text{where } \partial_\mu = (\tfrac{1}{c}\partial_t, \nabla)

But how big are these matrices? Here is the beautiful part: the algebra fixes the answer. We need four mutually anticommuting matrices, all with square $\pm I$ and zero trace. Run down the dimensions: 1×1 numbers can’t anticommute at all; in 2D the Pauli matrices give us only three anticommuting matrices, one short; the trace and square conditions force the dimension to be even, ruling out 3D. The first dimension that fits four is 4D. The smallest representation of the electron’s relativistic algebra is unavoidably four-dimensional.

A standard concrete choice is the Dirac representation, built from the $2\times 2$ identity and the Pauli matrices:

\begin{aligned}\gamma^0 & = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}, \quad \gamma^i = \begin{pmatrix} 0 & \sigma_i \\ -\sigma_i & 0 \end{pmatrix},\quad \\\gamma^i & = \beta \alpha_i = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix} \begin{pmatrix} 0 & \sigma_i \\ \sigma_i & 0 \end{pmatrix} = \begin{pmatrix} 0 & \sigma_i \\ -\sigma_i & 0 \end{pmatrix} \\ \sigma_x& =\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right), \sigma_y=\left(\begin{array}{cc} 0 & -i \\ +i & 0 \end{array}\right),\sigma_z=\left(\begin{array}{ll} 1 & 0 \\ 0 & -1 \end{array}\right) \end{aligned}

Let us not rush past what just happened. Demanding nothing more than a linear, relativistic wave equation has forced the wavefunction $\psi$ to carry four components — which we now read as the electron (spin up/down) and its antiparticle the positron (spin up/down). Dirac was only trying to fix the negative-probability disease; he accidentally kicked open the door to spin and antimatter. The lesson, which we will keep meeting: magnetism is not a property bolted onto matter — it is an inevitable consequence of spacetime symmetry.

3. Switching on the electromagnetic field

Beautiful as that is, it is still a free electron. To see magnetism we have to let the electron feel a field. The recipe is the minimal coupling substitution: replace momentum by the kinetic momentum $\vec{p} \to \vec{p} - q\vec{A}$ , and the energy operator by $i\hbar\partial_t \to i\hbar\partial_t - q\Phi$ , where $\vec{A}$ and $\Phi$ are the vector and scalar potentials. The Dirac equation becomes

i\hbar \frac{\partial \psi}{\partial t} = [c \vec{\alpha} \cdot (\vec{p} - q\vec{A}) + \beta m c^2 + q\Phi] \psi

or, in fully covariant dress with the covariant derivative $D_\mu = \partial_\mu + \frac{iq}{\hbar}A_\mu$ ,

(i\hbar \gamma^\mu D_\mu - mc) \psi = 0\quad \text{or}\quad \gamma^\mu (p_\mu - qA_\mu) \psi = mc \psi

We do not need the relativistic regime here — a fridge magnet is hardly relativistic. We need the non-relativistic limit, where the rest energy $$mc^2$$ dwarfs everything else. Split the four-spinor into a “large component” $\phi$ and a “small component” $\chi$ (each a 2-spinor), and peel off the fast rest-energy phase:

\psi(\vec{r}, t) = \begin{pmatrix} \phi(\vec{r}, t) \\ \chi(\vec{r}, t) \end{pmatrix} e^{-imc^2t/\hbar}

Differentiating in time by the product rule,

i\hbar \frac{\partial \psi}{\partial t} = \left( i\hbar \frac{\partial}{\partial t} \begin{pmatrix} \phi \\ \chi \end{pmatrix} + mc^2 \begin{pmatrix} \phi \\ \chi \end{pmatrix} \right) e^{-imc^2t/\hbar}

and substituting into the Dirac equation (in the standard representation), the exponential cancels on both sides and we are left with

i\hbar \frac{\partial}{\partial t} \begin{pmatrix} \phi \\ \chi \end{pmatrix} + mc^2 \begin{pmatrix} \phi \\ \chi \end{pmatrix} = c \begin{pmatrix} 0 & \vec{\sigma} \\ \vec{\sigma} & 0 \end{pmatrix} \cdot \vec{\pi} \begin{pmatrix} \phi \\ \chi \end{pmatrix} + \begin{pmatrix} mc^2 & 0 \\ 0 & -mc^2 \end{pmatrix} \begin{pmatrix} \phi \\ \chi \end{pmatrix} + q\Phi \begin{pmatrix} \phi \\ \chi \end{pmatrix}

where $\vec{\pi} = \vec{p} - q\vec{A}$ is the kinetic momentum. This is really two coupled equations:

\begin{aligned} & i \hbar \frac{\partial \phi}{\partial t}+m c^2 \phi=c(\vec{\sigma} \cdot \vec{\pi}) \chi+m c^2 \phi+q \Phi \phi \\ & i \hbar \frac{\partial \chi}{\partial t}+m c^2 \chi=c(\vec{\sigma} \cdot \vec{\pi}) \phi-m c^2 \chi+q \Phi \chi \end{aligned}

2mc^2 \chi \approx c(\vec{\sigma} \cdot \vec{\pi})\phi \implies \chi \approx \frac{\vec{\sigma} \cdot \vec{\pi}}{2mc} \phi

which incidentally explains the name: $\chi$ is smaller than $\phi$ by a factor of order $$v/c$$ .

4. The payoff: magnetism falls out

Now substitute $\chi$ back into the large-component equation. Everything reduces to a single equation for $\phi$ :

i\hbar \frac{\partial \phi}{\partial t} = \left[ \frac{(\vec{\sigma} \cdot \vec{\pi})^2}{2m} + q\Phi \right] \phi

And here, in that innocent-looking $(\vec{\sigma}\cdot\vec{\pi})^2$ , is where the magnet has been hiding all along. Use the Pauli identity $(\vec{\sigma}\cdot\vec{A})(\vec{\sigma}\cdot\vec{B}) = \vec{A}\cdot\vec{B} + i\vec{\sigma}\cdot(\vec{A}\times\vec{B})$ :

(\vec{\sigma} \cdot \vec{\pi})^2 = \vec{\pi} \cdot \vec{\pi} + i\vec{\sigma} \cdot (\vec{\pi} \times \vec{\pi})

If $\vec{\pi}$ were an ordinary vector this second term would be zero — any vector crossed with itself vanishes. But $\vec{\pi} = \vec{p} - q\vec{A}$ is an operator, and $\vec{p}$ does not commute with $\vec{A}$ . Acting on a test function $$f$$ ,

(\vec{\pi} \times \vec{\pi})f = (\vec{p} - q\vec{A}) \times (\vec{p} - q\vec{A})f= (\vec{p} \times \vec{p} - q\vec{p} \times \vec{A} - q\vec{A} \times \vec{p} + q^2\vec{A} \times \vec{A})f

With $\vec{p}\times\vec{p}=0$ and $\vec{A}\times\vec{A}=0$ , the survivors are

\begin{aligned}-q(\vec{p} \times \vec{A} & + \vec{A} \times \vec{p})f = -q \left[ (-i\hbar \nabla) \times (\vec{A}f) + \vec{A} \times (-i\hbar \nabla f) \right]\\&= i\hbar q \left[ (\nabla \times \vec{A})f - \vec{A} \times (\nabla f) + \vec{A} \times (\nabla f) \right]= i\hbar q (\nabla \times \vec{A})f = i\hbar q \vec{B} f\end{aligned}

because $\nabla\times\vec{A} = \vec{B}$ . So the self-cross-product, far from vanishing, is the magnetic field:

(\vec{\sigma} \cdot \vec{\pi})^2 = \vec{\pi}^2 + i\vec{\sigma} \cdot (i\hbar q \vec{B}) = \vec{\pi}^2 - \hbar q \vec{\sigma} \cdot \vec{B}

Put it back, and we have derived the Pauli equation, with a brand-new term sitting in the Hamiltonian:

i\hbar \frac{\partial \phi}{\partial t} = \left[ \frac{(\vec{p}-q\vec{A})^2}{2m} + q\Phi - \frac{q\hbar}{2m} \vec{\sigma} \cdot \vec{B} \right] \phi

H = \frac{(\vec{p}-q\vec{A})^2}{2m} + q\Phi - \frac{q\hbar}{2m} \vec{\sigma} \cdot \vec{B}

That last term, $-\frac{q\hbar}{2m}\vec{\sigma}\cdot\vec{B}$ , is a Zeeman energy $U = -\vec{\mu}\cdot\vec{B}$ — an intrinsic magnetic moment coupling directly to the field. Compare its size to the magnetic moment from orbital motion and you find the moment is twice as large as a naive classical estimate: the famous gyromagnetic ratio $$g = 2$$ , predicted with no extra input.

It is worth dwelling on why it appeared. Had we never introduced the $\vec{\sigma}$ matrices — that is, had we stayed with the spinless Klein–Gordon world — the kinetic term would have been a plain $\vec{\pi}^2$ , which produces no linear coupling to $\vec{B}$ , and the Hamiltonian would have degenerated to the spinless $H_{\text{Spin-0}} = \frac{(\vec{p}-q\vec{A})^2}{2m} + q\Phi$ . In other words: the $\vec{\sigma}$ matrix is what gives the particle an internal axis. Without it the electron is an isotropic point that a field can only push around from the outside. With it, the non-commutativity of $\vec{\pi}$ — which is just $\vec{B}$ in disguise — gets piped directly into the electron’s internal dimensions and becomes a magnetic energy. The matrix structure relativity forced upon us in Section 1 is, quite literally, the source of the magnet.

Where this leaves us

So we have our culprit. The magnet a child plays with is powered by an intrinsic magnetic moment that has no classical cause whatsoever; it is the low-speed residue of the electron’s relativistic, four-component nature. Bohr and van Leeuwen were right that classical physics forbids magnetism — they were simply working in a world that was missing two of its dimensions.

But notice that we obtained spin almost by a side door, by switching on a field and watching a term appear. That should leave us slightly uneasy. Spin is supposed to be intrinsic — it ought to be there with no field at all, written into the very way the electron responds when we rotate our laboratory. And there is a loose thread we tied off and ignored: the dynamical argument gave us a two-component spinor, yet the Dirac equation insisted on four. Where did the other two go?

Answering that means leaving dynamics behind and asking a purely kinematic question: how must a wavefunction transform when we rotate space — and rotate spacetime? That is the subject of Part 3, where SU(2), SO(3), and the Lorentz group will explain why the electron is a spinor in the first place. To be continued.