What Happens When a Magnet Attracts Metal? (Part 3): Why the Electron Is a Spinor

Part 3 of four. Part 2 pulled spin and magnetism out of Dirac’s equation by switching on a field and watching a new term appear. But that left us uneasy on two counts. First, spin is supposed to be intrinsic — it should be there with no field at all. Second, the dynamical argument handed us a two-component spinor, while the Dirac equation insisted on four. Where did the other two go? Answering both means dropping dynamics entirely and asking a purely kinematic question.

5. The universal covering between SU(2) and SO(3)

Here is the question that should bother us. In Part 2 we watched spin “emerge” at the level of dynamics — but only because we turned on an electromagnetic field. As an intrinsic property, spin ought to need no such crutch. At bottom its very existence must be tied to symmetry and transformation. Why must the wavefunction be acted on by \(\vec{\sigma}\) matrices at all? Why is it a two-component spinor? To answer that we have to descend to the kinematic bedrock — the group-theoretical basis of spacetime symmetry. Forget the field entirely. As a physical object, when we rotate the laboratory frame by an angle \(\theta\), the electron’s wavefunction \(\psi\) has no choice but to change. It is exactly this “transformation rule under rotation” that defines spin. Before we touch the physics, let me introduce the two most important groups in the story: SU(2) and SO(3). They are Lie groups, and they sit at the very center of modern physics.

The basic definition of SU(2) is:

If we describe it with real parameters \(x_i\in\mathbb{R}\), setting \(a=x_4+ix_3, b=x_2+ix_1\):

the constraint becomes \(x_1^2+x_2^2+x_3^2+x_4^2=1\), which tells us that SU(2), as a manifold, is \(S^3\), a 3-sphere (a hypersphere). Its \(T^2\)-fibration is described by:

You can picture this with two opposing conical surfaces for \(\chi, \varphi\) and an axis \(\theta\); at \(\theta=0\) or \(\theta=\pi/2\) one parameter degenerates. Or picture a “doughnut” (a solid torus) that scales with \(\theta\) and pinches off at the endpoints. When \(\theta=0\) the doughnut is a circle of zero width — only \(\chi\) runs along the circle, while \(\varphi\) is meaningless because the circle has no thickness. When \(\theta=\pi/2\) the doughnut swells into a sphere with no hole, so \(\chi\) drops out and only \(\varphi\) survives.

The spherical-coordinate description is:

SU(2) can also be written in terms of the Pauli matrices:

Folding in the spherical coordinates once more:

The basic definition of SO(3) is:

SO(3) is the friendlier of the two — it is the group of rotations you can see and touch. As a manifold, SO(3) is a solid ball of radius \(\pi\) swept out by the tips of \(\vec{\omega}\), with antipodal points on the surface identified. That antipodal identification is where the subtlety hides. Its origin is innocent enough — rotating \(180^\circ\) counter-clockwise about a fixed axis lands you in exactly the same place as rotating \(180^\circ\) clockwise — but the consequence is sharp: the solid ball becomes a manifold that is connected but not simply connected (not every loop in the space can be continuously shrunk to a point). This antipodally identified ball has a name: the three-dimensional real projective space, \(\mathbb{R}P^3\).

Now a word on what “representation” means, because the whole drama turns on it. Mathematically, a representation of a group \(G\) on a vector space \(V\) is a homomorphism from \(G\) into the general linear group \(GL(V)\) — the group of all invertible transformations of \(V\): \(\forall g_1, g_2 \in G, \quad D(g_1 g_2) = D(g_1)D(g_2)\). For Lie groups we also demand the map be continuous. Projective representations enter because quantum mechanics is more relaxed than that. Physical states are described by rays in Hilbert space: \(|\psi\rangle\) and \(e^{i\alpha}|\psi\rangle\) are the same physical state. So the group multiplication law only has to hold up to a phase:

where \(\omega(g_1, g_2)\) is a complex number of modulus 1, the group exponent. Bargmann’s Theorem (1954) makes this rigorous: for a Lie group \(G\) with \(H^2(\mathfrak{g}, \mathbb{R})=0\) (SO(3) and the Lorentz group both qualify), every continuous projective unitary representation can be “lifted” to an ordinary unitary representation of a central extension \(\tilde{G}\). For SO(3), where the phase ambiguity refuses to be massaged away, the cure is to find its universal covering group, SU(2), and trade SO(3)‘s projective representation for SU(2)‘s ordinary one. Hold onto that move — it is the hinge of the whole section.

That universal covering group points straight at the deep kinship between SO(3) and SU(2). In topology, the universal covering space \(\tilde{X}\) of a space \(X\) is its “upgraded version,” with two defining traits: simple connectedness (every loop in \(\tilde{X}\) shrinks to a point; no topological holes) and local isomorphism (locally \(\tilde{X}\) looks exactly like \(X\), but globally \(\tilde{X}\) is usually “larger,” covering \(X\) in an \(n:1\) fashion).

So why should SU(2) matrices generate SO(3) rotations in the first place? Here is the classic construction. Map a vector \(\mathbf{x} = (x, y, z)\) in 3D space to a second-order traceless Hermitian matrix \(X\):

(the \(\sigma_i\) being the Pauli matrices). Notice \(\det(X) = -(x^2 + y^2 + z^2) = -\|\mathbf{x}\|^2\). Now let a matrix \(U\) in SU(2) act on \(X\) by the sandwich \(X' = U X U^\dagger\). Because \(U\) is unitary with unit determinant, this transformation keeps \(X\) traceless and Hermitian, and crucially keeps the determinant fixed: \(\det(X') = \det(U X U^\dagger) = \det(X)\). That says \(\|\mathbf{x}'\|^2 = \|\mathbf{x}\|^2\) — the transformation preserves the vector’s length, so it is a 3D rotation. But now look again at \(X' = U X U^\dagger\) and replace \(U\) by \(-U\): \((-U) X (-U)^\dagger = (-1)^2 U X U^\dagger = U X U^\dagger\). We find that \(U\) and \(-U\) produce the exact same rotation. There is the algebraic root of the 2:1 cover: every rotation in SO(3) answers to two points of SU(2). This is also why rotating by \(2\pi\) does not bring you home (in SU(2) you have gone only half a lap), and why it takes \(4\pi\) to truly return (a full lap in SU(2)). Said differently: SO(3) has a “hole” (its fundamental group is \(\mathbb{Z}_2\)), while SU(2) — that is, \(S^3\) — is simply connected (fundamental group \(0\)) with no topological holes at all.

The cover’s second condition, local consistency, surfaces the other great fact about SU(2) and SO(3): near the identity they are locally isomorphic. If you only ever look at “infinitesimal” rotations — only nudge things a little — the two groups are literally indistinguishable. It is only when you rotate by a large amount (say \(2\pi\)) and probe the group’s “global picture” that you catch them differing (one comes home, the other lands on \(-I\)). Mathematically this is because they share the very same Lie algebra — an isomorphism of the tangent spaces at the identity, \(\mathfrak{su}(2) \cong \mathfrak{so}(3)\). From solving the Schrödinger equation we know SO(3) has three generators \(J_x, J_y, J_z\) (infinitesimal rotations about the \(x, y, z\) axes), obeying \([J_i, J_j] = i \epsilon_{ijk} J_k\). That single relation captures the essence of 3D rotation; the generators are simply a basis for the Lie algebra.

Now let us solve for SU(2)‘s generators and watch the isomorphism appear by hand. Take an infinitesimal transformation:

For this to live in SU(2) we need unitarity: \((I + i\epsilon S^\dagger)(I - i\epsilon S) = I \implies I - i\epsilon(S - S^\dagger) = I \implies S = S^\dagger\), so \(S\) must be Hermitian — physically observable. And there is the special-unitary constraint: using \(\det(e^A) = e^{\text{Tr}(A)}\), \(\det(U) = \det(e^{-i\epsilon S}) = e^{-i\epsilon \text{Tr}(S)} = 1 \implies \text{Tr}(S) = 0\), so \(S\) must be traceless. The matrices that fit both conditions are exactly the Pauli matrices \(\sigma_x, \sigma_y, \sigma_z\), which form a complete basis of the Lie algebra \(\mathfrak{su}(2)\). The generator \(S\) is therefore proportional to \(\sigma\). Since \([\sigma_i, \sigma_j] = 2i \epsilon_{ijk} \sigma_k\), this differs from \([L_i, L_j] = i \epsilon_{ijk} L_k\) by only a factor of \(\frac{1}{2}\) — which already proves the Lie algebras of SU(2) and SO(3) are isomorphic. Take \(S=\frac{1}{2}\sigma\) and it becomes the standard \([S_i, S_j] = i \epsilon_{ijk} S_k\), identical to \(J\)‘s relations. That hands us a self-consistent theory of angular momentum, in which the physical total angular momentum is \(L =J + S\). And here is the punchline: if spin \(S\) wants to count as “angular momentum” and add to \(J\) to form a conserved quantity, it has no choice but to obey the same algebraic rules as \(J\). Orbital angular momentum \(J = \mathbf{r} \times \mathbf{p}\) is fixed by the spatial coordinates — its commutators come straight from those of \(x\) and \(p\) and cannot be renegotiated — so we are forced into \(S=\frac{1}{2}\sigma\). Theory aside, the lab agrees: in the Stern–Gerlach experiment, measuring how electrons deflect in a field, the values that come out are \(\pm \frac{1}{2}\hbar\). That directly pins the eigenvalues of the observable \(S\) to \(\pm 1/2\), and mathematically only \(\frac{1}{2}\sigma_z\) has eigenvalues \(\pm 1/2\) (since \(\sigma_z\) has eigenvalues \(\pm 1\)).

Back to representations. For a Lie group, a generator’s great virtue is that any finite transformation \(D(\theta)\) can be built from the generator \(J\) by the exponential map. If \(J\) is an element of the Lie algebra \(\mathfrak{g}\), the group elements read:

where \(\mathbf{J}\) is the angular momentum operator (matrix) from before. A note on a habitual physics abuse of notation here: strictly, \(\exp\) maps abstract Lie algebra elements to abstract Lie group elements, but the physical formula \(D(\theta) = \exp(-i \theta \mathbf{n} \cdot \mathbf{J})\) is really an operation inside the representation space (the matrix space). Since \(\mathbf{J}\) here is already the matrix representation of the generator, the result \(D(\theta)\) is naturally the matrix representation of the group element.

Whether we work with SO(3) or SU(2), their Lie algebras are isomorphic — they share the one set of commutation relations \([J_i, J_j] = i \epsilon_{ijk} J_k\) (taking \(\hbar=1\)). We now want every finite-dimensional irreducible representation those rules permit. Define the ladder operators:

and introduce the \(J_z\) eigenstates \(|j, m\rangle\) with:

Compute the commutator:

which shows \(J_\pm\) act as “ladders” on the eigenvalues:

If \(m\) is an eigenvalue, so are \(m \pm 1\). But we are after finite-dimensional representations, so the spectrum must have a ceiling \(m_{max}\) and a floor \(m_{min}\):

Apply the identity \(J_- J_+ = \mathbf{J}^2 - J_z^2 - J_z\) to the highest-weight state \(|j, m_{max}\rangle\):

Labelling the top weight \(j\), i.e. \(m_{max} \equiv j\), the Casimir eigenvalue is \(j(j+1)\). Likewise, applying \(J_+ J_- = \mathbf{J}^2 - J_z^2 + J_z\) to the lowest-weight state \(|j, m_{min}\rangle\):

Solving \(m_{min}^2 - m_{min} - j(j+1) = 0\) gives two roots:

Since \(m_{min} \le m_{max} = j\), we must take \(m_{min} = -j\). Climbing from \(m_{min} = -j\) to \(m_{max} = j\) one step at a time, we have to land on top in a whole number of steps \(k\):

And there is the famous staircase: from the Lie algebra alone, the allowed values of \(j\) are \(0, 1/2, 1, 3/2, 2 \cdots\). But the Lie algebra is only a local statement. We must now test these candidates against the global group structure, and the test is single-valuedness: carry a group element around a closed loop back to the identity, and (for an ordinary representation) its representation matrix must return to the identity matrix too. SO(3) is the rotation group of 3D space, and rotating by \(2\pi\) (\(360^\circ\)) about any axis (say \(z\)) restores physical space completely: \(R_z(2\pi) = R_z(0) = \mathbf{1}\), the group identity. So an ordinary representation \(D\) of SO(3) must obey \(D(R_z(2\pi)) = D(\mathbf{1}) = I\). Plug in the Lie-algebra formula — in the \(z\)-basis \(J_z\) is diagonal with entries \(m\): \(D(2\pi) = \exp(-i 2\pi J_z) = \text{diag}(e^{-i 2\pi m}, \dots)\). For this to be the identity, every diagonal entry must be \(1\): \(e^{-i 2\pi m} = 1 \implies m \in \mathbb{Z}\). Integer \(j\) (\(0, 1, \dots\)) gives integer \(m\) and passes. But half-integer \(j\) (\(1/2, 3/2, \dots\)) gives half-integer \(m\), and \(e^{-i 2\pi m} = -1 \neq 1\). So in ordinary representations of SO(3), half-integer spins are strictly forbidden.

And now SU(2) collects what SO(3) threw out. Its geometry is different: as the universal cover of SO(3) (the 2:1 cover), the parameter \(\theta = 2\pi\) corresponds not to the identity but to \(U(2\pi)=-I\neq I\); only \(4\pi\) brings you back to the identity. So the behavior of \(D(2\pi)\) matches the behavior of the SU(2) group itself, exactly. What we have stumbled onto is a concrete instance of Bargmann’s Theorem for SO(3) and SU(2): the projective representations of the non-simply-connected SO(3) are equivalent to the ordinary representations of its universal cover SU(2). The final dictionary:

So “spin” can “emerge” from this abstract structure for one reason: quantum mechanics defines “physical state” more leniently than classical mechanics, and that leniency releases topological degrees of freedom that classical physics had masked.

• Symmetry (root): the universe has rotational symmetry, so the Lie algebra \(\mathfrak{so}(3) \cong \mathfrak{su}(2)\) exists.
• Quantization (opportunity): the nature of probability waves permits “projective representations,” letting the “half-integer parts” the Lie algebra allowed — but classical physics forbade — survive.
• Intrinsic nature (formation): these surviving half-integer representations cannot correspond to any motion through space, so they can only be read as the particle’s innate intrinsic angular momentum.

That third bullet is the deepest one: the half-integer representations have nowhere to live in ordinary space, so the electron has to carry them internally. That is spin, with no field anywhere in sight — exactly the intrinsic origin Part 2 left us wanting.

Let us now actually compute the representations for different \(j\). Use the exponential map:

where \(\hat{n}\) is the rotation axis (unit vector) and \(\theta\) the rotation angle.

The simplest case, \(j=0\), the scalar representation: dimension \(d=2(0)+1=1\). The basis is the single state \(|0,0\rangle\). Since \(m\) can only be \(0\), the generator is \(J_z = [0]\); the ladder operators annihilate the highest/lowest weights, so \(J_+ = [0], J_- = [0]\), and thus \(J_x = 0, J_y = 0, J_z = 0\). The exponential map gives \(D(\theta) = e^{-i\theta \mathbf{n} \cdot \mathbf{0}} = 1\), the trivial representation. This is a scalar: rotate however you like, the value is multiplied by \(1\) and never changes.

For \(j=\frac{1}{2}\), the dimension is \(2j+1=2\). We already know the generator is \(\mathbf{J}=\frac{1}{2} \sigma \Longrightarrow \hat{n} \cdot \mathbf{J}=\frac{1}{2}(\hat{n} \cdot \sigma)\). To get higher powers of \((\hat{n} \cdot \mathbf{J})\), recall the Pauli property \((\hat{n} \cdot \vec{\sigma})^2 = I\). So the power law for the generators is:

with general term:

Summing the series by splitting the exponential into even and odd parts:

The even coefficient is \(\sum \frac{(-1)^m}{(2m)!} (\frac{\theta}{2})^{2m} = \cos(\frac{\theta}{2})\) and the odd one is \(-i \sum \frac{(-1)^m}{(2m+1)!} (\frac{\theta}{2})^{2m+1} = -i \sin(\frac{\theta}{2})\). Finally:

This is the \(j=1/2\) representation, mapping rotations to \(2 \times 2\) complex matrices. Checking \(2\pi\): set \(\theta=2\pi\), so \(\cos(\pi)=-1, \sin(\pi)=0\), and the result is \(-I\) — the half-lap signature we predicted.

For \(j=1\), the dimension is \(2j+1=3\), so we need \(3 \times 3\) matrices. In the angular-momentum basis physicists favor (the Cartesian basis), the generators satisfy \((J_k)_{ab} = -i \epsilon_{kab}\). For instance the \(z\)-rotation generator \(J_z\):

For an arbitrary axis \(\hat{n}\), let \(K = \hat{n} \cdot \mathbf{J}\). Computing powers of \(J_z\) directly (other directions go the same way):

The pattern: for \(j=1\) generators the characteristic equation is \((\hat{n} \cdot \mathbf{J})^3 = (\hat{n} \cdot \mathbf{J})\). That means: odd terms (\(k=1, 3, 5 \dots\)) give \((\hat{n} \cdot \mathbf{J})^k = (\hat{n} \cdot \mathbf{J})\); even terms (\(k=2, 4, 6 \dots\)) give \((\hat{n} \cdot \mathbf{J})^k = (\hat{n} \cdot \mathbf{J})^2\); and the \(k=0\) term is \(I\). Expand the Taylor series again, this time pulling \(I\) out separately because \(J^2 \neq I\):

The odd coefficient is \(-i(\theta - \frac{\theta^3}{3!} + \dots) = -i \sin\theta\) and the even one is \((\frac{-\theta^2}{2!} + \frac{\theta^4}{4!} - \dots) = \cos\theta - 1\). The result:

This is the \(j=1\) representation (Rodrigues’ rotation formula in physics dress), mapping rotations to \(3 \times 3\) real matrices (the \(J\) carry an \(i\), but \(i \cdot J\) is real). Checking \(2\pi\): set \(\theta=2\pi\), so \(\sin(2\pi)=0, \cos(2\pi)=1\), and \(D^{(1)} = I - 0 + (1-1)(\dots) = I\). Integer spin comes home in one lap, exactly as the single-valuedness test demanded.

In general we can build generators and representations for every \(j\). The construction rests on three matrix-element formulas for the angular momentum operators of quantum mechanics; with these three in hand we can write down the matrices for \(j=0, 3/2, 2\) or even \(j=100\). First, \(J_z\) is diagonal:

Then \(J_+\) (the raising operator) is superdiagonal:

\(J_-\) (the lowering operator) is subdiagonal — the transpose of \(J_+\) in the real case. And \(J_x, J_y\) are assembled from \(J_\pm\):

Take \(j=3/2\) as an example: the spin-3/2 representation. This is a fermion, electron-like, but with 4 components — it shows up in \(\Delta\) baryons or in the gravitino of supergravity. Dimension: \(d = 2(3/2) + 1 = 4\). Build \(J_z\) (diagonal):

The \(J_+\) coefficients require \(\sqrt{j(j+1) - m(m+1)}\) with \(j=3/2\), i.e. \(\sqrt{3.75 - m(m+1)}\). \(m=1/2 \to 3/2\): \(\sqrt{3.75 - 0.75} = \sqrt{3}\); \(m=-1/2 \to 1/2\): \(\sqrt{3.75 - (-0.25)} = \sqrt{4} = 2\); \(m=-3/2 \to -1/2\): \(\sqrt{3.75 - 0.75} = \sqrt{3}\). So:

And with \(J_x = \frac{1}{2}(J_+ + J_+^\dagger)\):

This is a 4×4 unitary matrix, and under a \(2\pi\) rotation, because the diagonal entries are half-integers, it becomes \(-I_{4\times 4}\). So \(j=3/2\) is, again, a faithful representation of SU(2) and a projective representation of SO(3) — same half-integer signature, four components this time.

So we have told the whole story of spin from symmetry and group theory, and notice what we did not need: relativistic corrections, the engine of Part 2. But notice, too, what is now missing in the other direction. Part 2 started from the Dirac equation and landed directly on a 4-component wavefunction; here we started from spatial rotation (SU(2)/SO(3)) and recovered only the 2-component spinor (the Pauli spinor) at \(j=1/2\). Where did the other two components go? The answer, once again, is relativity. Our symmetry analysis so far has only allowed for spatial rotations — never Lorentz boosts. Only when we promote the symmetry group to the full Lorentz group can we explain why the electron must be the direct sum of a “left-handed” and a “right-handed” SU(2) representation (\(2+2=4\)), closing the loop perfectly back to the structure of the Dirac equation.

6. The Lorentz group

To get spin in full, we have to upgrade to the real symmetry group — the Lorentz group \(SO(1,3)\), which contains both rotations and boosts. And when we go hunting for its “fundamental representation,” something genuinely strange happens: the algebraic structure splits down the middle.

The Lorentz group is defined by:

At root it is the group of linear transformations that preserve the metric of Minkowski spacetime. From the metric-preserving condition \(\Lambda^{\mathrm{T}} g \Lambda=g \rightarrow g_{\mu \nu} \Lambda^\mu{ }_\rho \Lambda^\nu{ }_\sigma=g_{\rho \sigma}\) we can squeeze out a constraint on the component \(\Lambda^0_0\):

So a Lorentz transformation must have either \(\Lambda^0_0 \geq 1\) or \(\Lambda^0_0 \leq -1\) — already the group is disconnected. That lets us cut it into two manifolds, \(O^+(1,3)\) and \(O^-(1,3)\). The former is the orthochronous Lorentz group; the latter contains no identity element, so it is not a group at all, only the antichronous branch.

The metric condition also fixes the determinant:

Transformations with \(|\Lambda|=1\) are written \(SO(1,3)\), the proper Lorentz group; those with \(|\Lambda|=-1\) form the improper branch. Putting the two cuts together carves \(\mathrm{O}(1,3)\) into four connected manifolds — but in practice we only ever need the proper orthochronous branch \(SO^+(1,3)\). That is because the other three are reachable from \(SO^+(1,3)\) by acting with two specific Lorentz transformations: time reversal \(\mathcal{T}=\mathcal{T}^{-1}=\operatorname{diag}(-1,1,1,1)\) and parity \(\mathcal{P}=\mathcal{P}^{-1}=\operatorname{diag}(1,-1,-1,-1)\). And the frame transformations of the real world are, strictly, orthochronous and proper.

So we focus on the proper orthochronous Lorentz group \(SO^+(1,3)\). In this connected component any transformation can be written as an exponential map from the identity. Just as \(SO(3)\) has 3 rotation generators, \(SO^+(1,3)\) has 6 degrees of freedom (3 rotations + 3 boosts) and so 6 generators. Take an infinitesimal \(\Lambda \approx I - i\epsilon X\) and, just as before, two sets of generators fall out: the rotation generators \(\vec{J} = (J_1, J_2, J_3)\) for spatial rotations (our familiar angular momentum operators), and the boost generators \(\vec{K} = (K_1, K_2, K_3)\) for velocity transformations along the \(x, y, z\) axes. These 6 satisfy the Lorentz Lie algebra \(\mathfrak{so}(1,3)\):

Pure rotations are closed (the \(SO(3)\) subalgebra):

Rotations and boosts (the boosts rotate among themselves like a vector):

Boosts are not closed among themselves (composing two boosts in different directions yields not just a boost but also a rotation — Thomas precession; the minus sign is a fingerprint of the spacetime metric \(g=\text{diag}(1,-1,-1,-1)\), and is what separates this from the algebra of \(SO(4)\)):

At this point \(J\) and \(K\) are still tangled together. To find irreducible representations we make a non-unitary change of basis (complexification), defining two new sets of operators \(\vec{N}^+\) and \(\vec{N}^-\):

Let us work out their commutators. Inside \(\vec{N}^+\) first:

Likewise one checks \([N_i^-, N_j^-] = i\epsilon_{ijk} N_k^-\). And the truly startling result is the cross-commutator between \(\vec{N}^+\) and \(\vec{N}^-\):

They commute. After complexification, the Lie algebra of the Lorentz group splits into the direct sum of two completely independent \(\mathfrak{su}(2)\) algebras:

This is an enormous simplification. Since we now know the representations of \(\mathfrak{su}(2)\) inside and out (labelled by a spin \(j\)), the irreducible representations of the Lorentz group can be labelled uniquely by a pair of half-integers or integers \((j_L, j_R)\). By this decomposition the most fundamental spinor representation is no longer unique: there are two most basic choices (the fundamental representations), got by taking \(j=1/2\) in one \(\mathfrak{su}(2)\) and \(j=0\) in the other. This is exactly where chirality is born.

The left-handed Weyl spinor carries the label \((1/2, 0)\): spin \(1/2\) under \(\vec{N}^-\), a scalar under \(\vec{N}^+\). It is a 2-component complex vector, written \(\psi_L\). Here \(\vec{N}^- = \frac{1}{2}\vec{\sigma}\) and \(\vec{N}^+ = 0\), and solving for the physical generators gives:

The right-handed Weyl spinor carries \((0, 1/2)\): a scalar under \(\vec{N}^-\), spin \(1/2\) under \(\vec{N}^+\). Also a 2-component complex vector, written \(\psi_R\). Here \(\vec{N}^- = 0\) and \(\vec{N}^+ = \frac{1}{2}\vec{\sigma}\), with physical generators:

Look hard at the sign of \(\vec{K}\). It tells us that although \(\psi_L\) and \(\psi_R\) behave identically under spatial rotation (\(\vec{J}\)) — both are spin \(1/2\) — their behavior under a Lorentz boost (\(\vec{K}\)) is exactly opposite. That sign is the entire physical content of “handedness.”

And now we can finally answer the question Part 2 left hanging. If \(\psi_L\) and \(\psi_R\) are each 2-component objects, why does the electron need 4 components? The answer is parity (\(\mathcal{P}\)). Parity flips the spatial coordinates, \(\vec{x} \to -\vec{x}\). Now \(\vec{J}\) is an axial vector (\(\vec{r} \times \vec{p}\)), so it is unchanged: \(\vec{J} \to \vec{J}\); while \(\vec{K}\) is a polar vector (\(\sim \vec{v}\)), so it flips sign: \(\vec{K} \to -\vec{K}\). Feed that into the definitions of \(\vec{N}^\pm\) and you find parity swaps the two algebras:

So parity turns the left-handed representation \((1/2, 0)\) into the right-handed one \((0, 1/2)\). If we want to describe a particle like the electron — one with both spin and mass, obeying parity conservation under electromagnetism — we cannot pick just one. We are obliged to take their “direct sum.” The Dirac spinor \(\Psi\), as a representation of \(SO^+(1,3)\) extended by parity, is precisely that direct sum of the two fundamental representations:

And there is the missing two components, recovered at last: two come from the left-handed sector, two from the right-handed sector, bound tightly together by the mass term and parity. The Pauli spinor we met earlier with \(SU(2)\) was only a silhouette of this relativistic object — its shadow in the rest frame (the non-relativistic limit). The thread Part 2 tied off and ignored is now fully knotted.

As an aside, these Weyl spinors are deeply tied to the \(\gamma^5\) matrix from the gamma-matrix discussion. The Hermitian operator \(\gamma^5 \equiv i \gamma^0 \gamma^1 \gamma^2 \gamma^3\) is exactly the operator that “identifies” and “defines” the Weyl spinors. Without \(\gamma^5\) there would be no mathematical way to tell “left-handed” from “right-handed.” In Dirac theory \(\gamma^5\) is the chirality operator, with one crucial algebraic property: it anticommutes with every \(\gamma^\mu\) (\(\{\gamma^5, \gamma^\mu\} = 0\)), yet it commutes with the Lorentz generators \(S^{\mu\nu} = \frac{i}{4}[\gamma^\mu, \gamma^\nu]\) (\([\gamma^5, S^{\mu\nu}] = 0\)). So \(\gamma^5\) is a conserved quantity of the Lorentz representation (for massless particles), and we can classify spinors by its eigenvalues. A right-handed Weyl spinor has \(\gamma^5\) eigenvalue +1 (\(\gamma^5 \psi_R = +\psi_R\)), a left-handed one has eigenvalue −1 (\(\gamma^5 \psi_L = -\psi_L\)). So what physics calls “left-handedness” and “right-handedness” is mathematically just whether the \(\gamma^5\) eigenvalue is −1 or +1. And since the Dirac spinor \(\Psi\) is a mixture (direct sum) of left and right, how do we sift one handedness out of a mixed \(\Psi\)? With the projection operators built from \(\gamma^5\):

These have the standard projector properties (\(P^2=P, P_L P_R = 0, P_L+P_R=1\)), and their job is to kill one handedness and keep the other:

In particle physics — weak interactions especially — you will keep meeting the combination \(\frac{1-\gamma^5}{2}\), and it is telling you: “this interaction plays only with left-handed spinors; right-handed ones, step aside.” (That is the mathematical face of parity violation.) To make the structure leap off the page, one can pick a special set of gamma matrices called the Weyl (or chiral) representation. Unlike the Dirac representation from earlier, here the gamma matrices are block-diagonal, so the Dirac spinor splits explicitly into its upper and lower Weyl pieces \(\Psi = (\psi_L, \psi_R)^T\) — the perspective of choice in high-energy physics. In low-energy condensed matter, by contrast, we usually keep the Dirac (standard) representation, where \(\psi_L\) and \(\psi_R\) are deeply mixed and the “large component / small component” non-relativistic split is the more natural reading.

7. The magnetic gradient force

We have come a long way through the mathematics. From deriving the Dirac equation to representing the Lorentz group, we have established that the electron must carry spin and is a 4-component relativistic object. Now let us walk back to the paradox we started this whole series with: if the Lorentz force does no work, then who is doing the work when a magnet lifts a nail? To answer, we need to connect microscopic spin to a macroscopic force.

Taking the non-relativistic limit of the Dirac equation handed us, naturally, the Pauli equation. Let us revisit the “Zeeman term” that seemed to appear from nowhere:

In classical physics we relate the magnetic moment \(\vec{\mu}\) to angular momentum \(\vec{L}\) through the gyromagnetic ratio. For orbital angular momentum:

If we try the same logic for the “spin magnetic moment,” we have to tie the spin operator \(\vec{S}\) to a moment. Recall the definition of the spin operator from Section 5:

Substitute \(\vec{\sigma}\) back into the Zeeman term:

Writing magnetic potential energy in its general form \(U = -\vec{\mu}_S \cdot \vec{B}\) and comparing, we read off the electron’s spin magnetic moment \(\vec{\mu}_S\):

And in the general Landé g-factor form \(\vec{\mu} = g \frac{e}{2m} \vec{S}\), the conclusion is immediate:

This \(g=2\) is no fudge factor tuned to fit experiment; it is a direct mathematical consequence of the spacetime symmetry of the Dirac equation. It says: electron spin produces a magnetic moment twice as efficiently as classical orbital motion. The same factor we got in Part 2 by switching on a field, we have now earned from pure symmetry.

With the moment \(\vec{\mu}\) in hand, we can finally name who does the work. The classical Lorentz force \(\vec{F}_{Lorentz} = q(\vec{v} \times \vec{B})\) does indeed do no work. But an object with an intrinsic magnetic moment is governed by its potential energy \(U\), and in Hamiltonian mechanics force is the negative gradient of potential energy:

Since spin \(\vec{\mu}\) is an intrinsic property, constant under spatial differentiation, we get:

There is the force that does the work — the gradient force. In a uniform field (\(\nabla \vec{B} = 0\)) it vanishes and only a torque remains. But the field of a real magnet is non-uniform (the field lines fan out), so \(\nabla \vec{B} \neq 0\). It is a conservative force, converting the potential energy of the field–moment coupling into the object’s kinetic energy. So a magnet attracting iron is, at bottom, a quantized spin magnetic moment being pulled by the gradient force in a non-uniform field. The Lorentz force handles the bending; the gradient force does the work. The energy bill from Part 1 is finally paid — and not by the Lorentz force at all.

So there really are two kinds of magnetic force, and from the standpoint of symmetry they answer to two completely different mechanisms — which is where we brush up against the edge of quantum field theory, where interactions are dictated by symmetry:

• The Lorentz force springs from local gauge symmetry. The electromagnetic interaction exists to preserve the \(U(1)\) local phase invariance of the wavefunction, \(\psi \to e^{i\alpha(x)}\psi\). To compensate for the phase varying with position, \(\partial_\mu \alpha(x)\), we must introduce the gauge field \(A_\mu\) and the covariant derivative \(D_\mu = \partial_\mu - ieA_\mu\). The resulting equation of motion (the Lorentz force) is essentially a statement of how the gauge field couples to the current. This geometric constraint forces the force to be perpendicular to the four-velocity; projected into 3D space, that is \(\vec{v} \times \vec{B}\). Its “no work” property is a direct expression of the geometry of gauge symmetry.

• The gradient force springs from broken spatial translational symmetry. By Noether’s theorem, force and work are bound up with spacetime symmetry: momentum conservation \(\longleftrightarrow\) spatial translational symmetry. In a uniform field the Hamiltonian \(H = -\vec{\mu} \cdot \vec{B}\) does not contain position \(\vec{r}\); the system has translational symmetry, momentum is conserved (\(\dot{\vec{p}} = -\partial H / \partial \vec{r} = 0\)), and there is no net force. But in a non-uniform field (near a magnet) \(\vec{B}(\vec{r})\) depends on position: \(\frac{\partial H}{\partial \vec{r}} = -\vec{\mu} \cdot \frac{\partial \vec{B}}{\partial \vec{r}} \neq 0\). The magnet’s mere presence breaks the homogeneity of space, and it is precisely this breaking of the spacetime background symmetry that forces the electron to change its momentum in response to the energy gradient. So the Lorentz force is the price of keeping an internal gauge symmetry, while the work-doing gradient force is the product of an external spacetime symmetry being broken.

Now that we stand on quantum field theory’s doorstep, honesty compels one last remark: Dirac’s \(g=2\), glorious as it is, is not the final word. In the Dirac equation we treat the electron as a classical field coupled to the electromagnetic field. But in full quantum electrodynamics (QED), the vacuum is not empty. As an electron propagates it ceaselessly emits and absorbs virtual photons, and even spawns electron–positron pairs. So the interaction vertex between electron and field is no longer a clean point (tree level) but carries infinitely many loop corrections. Julian Schwinger computed the first-order (one-loop) correction in 1948, giving the famous formula:

where \(\alpha \approx 1/137\) is the fine-structure constant. This nudges the theoretical value to \(g \approx 2.002319...\), in astonishing agreement with experiment (to twelve decimal places). That tiny deviation — the anomalous magnetic moment — not only confirms the relativistic origin of \(g=2\) but reveals the deeper physics underneath magnetism: when we feel a magnet pull, we are not only witnessing the geometric attributes of spacetime (spin), we are reaching into the seething ocean of virtual particles that fills the vacuum.

Where this leaves us

So the loop is closed. The electron is a spinor not because a field made it one, but because rotating space — and boosting spacetime — leaves it no other option. Half-integer representations have nowhere to live except inside the particle; parity then forces left and right together; and the four components the Dirac equation demanded turn out to be \((1/2,0)\oplus(0,1/2)\) all along. The fridge magnet’s pull is the gradient force acting on the quantized moment that all of this entails, with \(g=2\) written into the geometry of spacetime itself.

But every magnet you have ever held contains something like \(10^{23}\) of these little spins, and so far we have studied exactly one. Why do they line up? A single electron’s moment is feeble; a refrigerator magnet is not. The leap from one spin to a cooperating multitude is a story about many bodies — the Heisenberg model, the Ising model, and the spontaneous symmetry breaking that, astonishingly, ties a block of iron on your fridge to the Higgs mechanism that gives particles their mass. That is Part 4. To be continued.