Plane-wave solutions

The Dirac equation is a differential equation, but the physics of a free electron lives in a much simpler place. Once you insist that a free particle has definite energy and momentum, its whole spacetime dependence is forced into a single plane-wave exponential $e^{i(\mathbf{p}\cdot\mathbf{x}-Et)}$ , and the four-component spinor $u(E,\mathbf{p})$ riding on top of it carries everything else — spin, and the split between particle and antiparticle. Feeding that ansatz into $(i\gamma^\mu\partial_\mu-m)\psi=0$ makes every derivative pull down a factor of momentum, and the differential equation collapses to a $4\times4$ matrix equation with no calculus left: $(\gamma^\mu p_\mu-m)u=0$ . This is the whole subject of the section — a piece of linear algebra whose solution set is the raw material of every scattering amplitude computed later. The algebra pays back four independent solutions. Two have positive energy $E=+\sqrt{\mathbf{p}^2+m^2}$ and describe the electron in its two spin states; two more carry the "wrong" time dependence $e^{+iEt}$ that so alarmed Dirac in 1928, which the modern reading repackages as antiparticles of positive energy — the $v$ -spinors obeying $(\gamma^\mu p_\mu+m)v=0$ . Every solution automatically satisfies $E^2=\mathbf{p}^2+m^2$ , so the mass shell is recovered, not imposed. Each spinor has a definite internal structure — a "large" upper block and a "small" lower block coupled by $\boldsymbol{\sigma}\cdot\mathbf{p}/(E+m)$ . What follows builds those four objects explicitly, fixes their normalization, assembles their bilinears into the completeness relations that turn a spin-averaged cross section into a trace, and shows why for a general momentum the honest spin label is helicity, not $S_z$ .

Key takeaways

The plane-wave ansatz $\psi=u(E,\mathbf{p})\,e^{i(\mathbf{p}\cdot\mathbf{x}-Et)}$ removes every derivative from the Dirac equation, leaving the algebraic momentum-space equation $(\gamma^\mu p_\mu-m)u=0$ .
The spinor splits into a large upper block and a small lower block, coupled by $u_B=\dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{E+m}\,u_A$ ; resubstitution always returns the mass shell $E^2=\mathbf{p}^2+m^2$ .
There are exactly four independent solutions: two particle spinors $u_1,u_2$ with $E>0$ and two antiparticle spinors $v_1,v_2$ obeying $(\gamma^\mu p_\mu+m)v=0$ .
With normalization $N=\sqrt{E+m}$ they obey $u^\dagger u=2E$ , $\bar u u=2m$ , $\bar v v=-2m$ — the covariant conventions used throughout QED.
The spin sums $\sum_s u_s\bar u_s=\gamma^\mu p_\mu+m$ and $\sum_s v_s\bar v_s=\gamma^\mu p_\mu-m$ are the completeness relations that collapse spin-averaged $|\mathcal{M}|^2$ into Dirac traces.
Key idea: a plane-wave ansatz turns the differential Dirac equation into the algebraic $(\gamma^\mu p_\mu-m)u=0$ ; its four solutions automatically satisfy $E^2=\mathbf{p}^2+m^2$ and split into two particle and two antiparticle spinors.

The core insight

A plane-wave ansatz turns the differential Dirac equation into the algebraic $(\gamma^\mu p_\mu-m)u=0$ ; its four solutions automatically satisfy $E^2=\mathbf{p}^2+m^2$ and split into two particle and two antiparticle spinors, coupled internally by the factor $\boldsymbol{\sigma}\cdot\mathbf{p}/(E+m)$ that carries spin, boosts, and — in the massless limit — chirality.

Where this lives today

The four plane-wave spinors are typed into every automated amplitude generator at the Large Hadron Collider. When a program like MadGraph computes $e^+e^-\to\mu^+\mu^-$ or a gluon-fusion Higgs rate, it evaluates external-leg spinors $u(p),v(p)$ exactly as in equation (4), and the spin sums of equation (6) become the traces that give the unpolarized rate. The modern spinor-helicity formalism, expressing gluon amplitudes through two-component brackets $\langle ij\rangle$ , $[ij]$ , is a direct descendant of the massless limit ( $r\to1$ ) above, and it made previously hopeless multi-parton QCD amplitudes tractable. Helicity, not $S_z$ , is the label used because the electroweak interaction couples only to definite chirality, so a forward–backward asymmetry at a lepton collider is a direct read-out of the spinor structure derived here. And the antiparticle solutions have a daily clinical life: the positron of positron-emission tomography is the $v$ -spinor made real, annihilating with a tissue electron into the back-to-back 511 keV photons a PET scanner reconstructs.

The physical picture

A free electron of definite energy and momentum is the most featureless thing quantum field theory allows: a wave filling all space at fixed wavelength, marching forward at a fixed rate. All the structure is bottled into the constant four-component amplitude $u$ in front of the exponential, and its allowed values are fixed by a single relativistic fact — $E^2=\mathbf{p}^2+m^2$ . The first figure shows the consequence: the energies of a free spinor lie on two disconnected hyperbolic branches, a positive-energy sheet for the particle and a mirror negative-energy sheet whose modern reading is the antiparticle.

Figure

Allowed energies of a free Dirac particle: the two branches E = ±√(p²c² + m²c⁴). The upper sheet holds the two particle spinors u₁, u₂; the lower sheet holds the two antiparticle spinors v₁, v₂. No solutions exist inside the gap of width 2mc² = 1.022 MeV for the electron, so a free electron and its positron are always separated by at least twice the rest energy

The second picture opens up the spinor itself. Write $u$ as an upper block $u_A$ and a lower block $u_B$ ; the momentum-space equation locks them together as $u_B=\boldsymbol{\sigma}\cdot\mathbf{p}\,u_A/(E+m)$ . At rest the lower block vanishes and the spinor is pure "large component" — ordinary spin. As the particle speeds up, the ratio $r=|\mathbf{p}|/(E+m)$ grows from zero toward one, mixing in the lower block, until at ultra-relativistic energies the two blocks are equal and the state is an eigenstate of chirality. That single ratio is the dial from a non-relativistic spinor to a relativistic one.

Figure

The small-component ratio r = p/(E + m) sets how much of a positive-energy spinor sits in its lower block. At rest r = 0 (pure large component, ordinary spin); at p = mc, r = 0.414 so 15 percent of the norm sits in the lower block; ultra-relativistically r → 1, the blocks equalise and helicity becomes chirality

The core equations

Start from the covariant Dirac equation and the free-particle ansatz. The whole point is that the exponential is an eigenfunction of $\partial_\mu$ , so the derivatives turn into algebra.

(i\gamma^\mu\partial_\mu-m)\psi=0,\qquad \psi(x)=u(E,\mathbf{p})\,e^{i(\mathbf{p}\cdot\mathbf{x}-Et)}=u\,e^{-ip\cdot x} \qquad (1)

where: $\gamma^\mu$ — the four Dirac matrices, $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ ; $m$ — particle mass (natural units $\hbar=c=1$ ); $p^\mu=(E,\mathbf{p})$ — the four-momentum; $u$ — a constant four-component spinor. Reading: all spacetime dependence is quarantined in the exponent, so solving the field equation means solving for the constant $u$ . Because $\partial_\mu\psi=-ip_\mu\psi$ , equation (1) becomes derivative-free. For the antiparticle ansatz $\psi\propto v\,e^{+ip\cdot x}$ the sign of the mass term flips.

(\gamma^\mu p_\mu-m)\,u=0,\qquad (\gamma^\mu p_\mu+m)\,v=0 \qquad (2)

where: $\gamma^\mu p_\mu\equiv\slashed p=\gamma^0E-\boldsymbol{\gamma}\cdot\mathbf{p}$ (Feynman slash). Reading: these are two $4\times4$ homogeneous linear systems. A non-trivial solution exists only when $\det(\slashed p\mp m)=(p^2-m^2)^2=0$ , which already forces the mass shell $E^2=\mathbf{p}^2+m^2$ . In the Dirac–Pauli representation $\gamma^0=\mathrm{diag}(I,-I)$ and $\gamma^i=\left(\begin{smallmatrix}0&\sigma^i\\-\sigma^i&0\end{smallmatrix}\right)$ , so writing $u=(u_A,u_B)^{\mathsf T}$ in two-spinor blocks makes the coupling explicit.

u_A=\frac{\boldsymbol{\sigma}\cdot\mathbf{p}}{E-m}\,u_B,\qquad u_B=\frac{\boldsymbol{\sigma}\cdot\mathbf{p}}{E+m}\,u_A \qquad (3)

where: $u_A$ — upper (large) block; $u_B$ — lower (small) block; $\boldsymbol{\sigma}$ — Pauli matrices. Reading: the two blocks are not independent — fixing one fixes the other. Substituting one into the other and using $(\boldsymbol{\sigma}\cdot\mathbf{p})^2=\mathbf{p}^2$ returns $E^2=\mathbf{p}^2+m^2$ , confirming consistency. Choosing the two simplest upper blocks $\chi_1=(1,0)^{\mathsf T}$ , $\chi_2=(0,1)^{\mathsf T}$ builds the positive-energy spinors; choosing the lower blocks builds the antiparticle spinors. Here $E=+\sqrt{\mathbf{p}^2+m^2}>0$ throughout.

u_s=\sqrt{E+m}\begin{pmatrix}\chi_s\\[2pt]\dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{E+m}\,\chi_s\end{pmatrix},\qquad v_s=\sqrt{E+m}\begin{pmatrix}\dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{E+m}\,\chi_s\\[2pt]\chi_s\end{pmatrix} \qquad (4)

where: $s=1,2$ labels the two spin states; $\sqrt{E+m}$ — the covariant normalization factor. Reading: for the particle the large block is on top; for the antiparticle it is on the bottom. The coupling factor $\boldsymbol{\sigma}\cdot\mathbf{p}/(E+m)$ is the small-component ratio that vanishes at rest and saturates at one for $E\gg m$ . The normalization $\sqrt{E+m}$ is chosen so the bilinears are clean Lorentz-covariant quantities. These orthogonality relations are the ones every amplitude calculation uses.

u_r^\dagger u_s=2E\,\delta_{rs},\quad \bar u_r u_s=2m\,\delta_{rs},\quad \bar v_r v_s=-2m\,\delta_{rs},\quad \bar u_r v_s=\bar v_r u_s=0 \qquad (5)

where: $\bar u=u^\dagger\gamma^0$ — the Dirac adjoint. Reading: $u^\dagger u=2E$ is a density (it transforms as the time component of the current $j^0$ ), while $\bar u u=2m$ is a Lorentz scalar. Particle and antiparticle spinors of the same momentum are orthogonal under $\bar u v$ . Finally, summing the outer products $u\bar u$ over spin gives the completeness relations — projectors onto the positive- and negative-energy subspaces of the mass shell.

\sum_{s=1,2}u_s(p)\,\bar u_s(p)=\gamma^\mu p_\mu+m,\qquad \sum_{s=1,2}v_s(p)\,\bar v_s(p)=\gamma^\mu p_\mu-m \qquad (6)

where: the sum runs over the two spin states at fixed $p$ . Reading: these are the workhorses of QED. A spin-averaged $\overline{|\mathcal{M}|^2}$ contains factors $\sum u\bar u$ , each replaced by $\slashed p+m$ , turning the spin sum into a single Dirac trace — no explicit spinor components needed.

Derivation AFrom the differential equation to the algebraic mass shell

The goal is to show, without hand-waving, that the free-particle plane-wave ansatz turns the differential Dirac equation into a purely algebraic $4\times4$ matrix equation, and that this equation contains the relativistic energy–momentum relation as a built-in consistency condition rather than an extra assumption. We push the ansatz through the derivatives, reduce the equation to two-spinor blocks, and close the loop by resubstituting one block into the other so the mass shell falls out.

Step 1. Insert $\psi=u\,e^{-ip\cdot x}$ with $p\cdot x=Et-\mathbf{p}\cdot\mathbf{x}$ into $(i\gamma^\mu\partial_\mu-m)\psi=0$ . Each derivative acts only on the exponential: $\partial_\mu(e^{-ip\cdot x})=-ip_\mu e^{-ip\cdot x}$ . So $i\gamma^\mu\partial_\mu\psi=\gamma^\mu p_\mu\,\psi$ , and cancelling the common exponential leaves $(\gamma^\mu p_\mu-m)u=0$ . Why: the ansatz is an eigenfunction of every $\partial_\mu$ , so calculus is replaced by multiplication by momentum — the defining move that makes the problem finite-dimensional.

Step 2. Evaluate $\slashed p=\gamma^\mu p_\mu=\gamma^0E-\boldsymbol{\gamma}\cdot\mathbf{p}$ in the Dirac–Pauli basis. With $\gamma^0=\mathrm{diag}(I,-I)$ and $\gamma^i=\left(\begin{smallmatrix}0&\sigma^i\\-\sigma^i&0\end{smallmatrix}\right)$ , the matrix $\slashed p-m$ has blocks $\left(\begin{smallmatrix}(E-m)I&-\boldsymbol{\sigma}\cdot\mathbf{p}\\ \boldsymbol{\sigma}\cdot\mathbf{p}&-(E+m)I\end{smallmatrix}\right)$ . Splitting $u=(u_A,u_B)^{\mathsf T}$ gives the coupled pair $(E-m)u_A=(\boldsymbol{\sigma}\cdot\mathbf{p})u_B$ and $(\boldsymbol{\sigma}\cdot\mathbf{p})u_A=(E+m)u_B$ . Why: block form exposes that the four equations are really two two-component equations, and it is the off-diagonal $\boldsymbol{\sigma}\cdot\mathbf{p}$ that ties spin to momentum.

Step 3. Close the loop. From the second equation $u_B=\boldsymbol{\sigma}\cdot\mathbf{p}\,u_A/(E+m)$ ; substitute into the first: $(E-m)u_A=(\boldsymbol{\sigma}\cdot\mathbf{p})\dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{E+m}u_A$ . Use the identity $(\boldsymbol{\sigma}\cdot\mathbf{a})(\boldsymbol{\sigma}\cdot\mathbf{b})=\mathbf{a}\cdot\mathbf{b}+i\boldsymbol{\sigma}\cdot(\mathbf{a}\times\mathbf{b})$ with $\mathbf{a}=\mathbf{b}=\mathbf{p}$ to get $(\boldsymbol{\sigma}\cdot\mathbf{p})^2=\mathbf{p}^2$ . Then $(E-m)(E+m)u_A=\mathbf{p}^2u_A$ , i.e. $E^2-m^2=\mathbf{p}^2$ . Why: the requirement that the two blocks be mutually consistent is exactly the relativistic dispersion relation — the mass shell is not imposed, it is the condition for a non-trivial spinor to exist.

Step 4. Read off the spectrum. $E^2=\mathbf{p}^2+m^2$ has two roots $E=\pm\sqrt{\mathbf{p}^2+m^2}$ . The positive root with $u_A=\chi_s$ gives the particle spinors $u_{1,2}$ ; the $(\slashed p+m)v=0$ system with $E>0$ gives the antiparticle spinors $v_{1,2}$ — four independent solutions, as a fourth-order characteristic polynomial demands. Why: the doubling of the spinor from two spin states to four is the fingerprint of antimatter — the price of a first-order relativistic equation.

Go deeper · Derivation B · 1 minFixing the normalization and the covariant bilinears

Derivation BFixing the normalization and the covariant bilinears

The aim is to pin the spinors' overall constant to $N=\sqrt{E+m}$ , and to prove that this single choice makes $u^\dagger u=2E$ (a density) and $\bar u u=2m$ (a scalar) simultaneously — the standard QED conventions, whose correct factors make the flux and phase-space factors of a cross section come out right. We compute both bilinears from the block form and use the mass shell to simplify.

Step 1. Write the positive-energy spinor as $u_s=N(\chi_s,\ r\,\hat{n}\!\cdot\!\boldsymbol{\sigma}\,\chi_s)^{\mathsf T}$ where $r\,\hat n\cdot\boldsymbol\sigma=\boldsymbol{\sigma}\cdot\mathbf{p}/(E+m)$ , and form $u_s^\dagger u_s=|N|^2[\chi^\dagger\chi+\chi^\dagger(\boldsymbol{\sigma}\cdot\mathbf{p})^2\chi/(E+m)^2]$ . With $(\boldsymbol{\sigma}\cdot\mathbf{p})^2=\mathbf{p}^2$ and $\chi^\dagger\chi=1$ , this is $|N|^2[1+\mathbf{p}^2/(E+m)^2]$ . Why: $u^\dagger u$ is the probability density $j^0$ of the plane wave, so its value sets the number of particles per unit volume the normalization describes.

Step 2. Simplify using $\mathbf{p}^2=E^2-m^2=(E-m)(E+m)$ . The bracket becomes $\dfrac{(E+m)^2+(E-m)(E+m)}{(E+m)^2}=\dfrac{(E+m)\,[\,(E+m)+(E-m)\,]}{(E+m)^2}=\dfrac{2E}{E+m}$ . Hence $u^\dagger u=|N|^2\,\dfrac{2E}{E+m}$ . Choosing $N=\sqrt{E+m}$ gives $u^\dagger u=2E$ . Why: the factor $\sqrt{E+m}$ is exactly what cancels the $E+m$ in the denominator, leaving the clean covariant density $2E$ that pairs with the $1/2E$ in the invariant phase-space measure.

Step 3. Now the scalar. $\bar u u=u^\dagger\gamma^0u=|N|^2[\chi^\dagger\chi-\chi^\dagger(\boldsymbol{\sigma}\cdot\mathbf{p})^2\chi/(E+m)^2]=|N|^2[1-\mathbf{p}^2/(E+m)^2]$ . The same algebra gives $1-\mathbf{p}^2/(E+m)^2=2m/(E+m)$ , so $\bar u u=|N|^2\,2m/(E+m)=2m$ . For the antiparticle spinor the identical computation gives $\bar v v=-2m$ , the extra sign coming from $\gamma^0$ acting on a lower-block-dominant spinor. Why: $\bar u u$ is a Lorentz scalar, so it must be frame-independent — and indeed $2m$ carries no momentum, unlike the density $2E$ . The relative minus sign in $\bar v v$ is the algebraic seed of the antiparticle's opposite charge.

Go deeper · Derivation C · 1 minThe spin sums as mass-shell projectors

Derivation CThe spin sums as mass-shell projectors

The goal is to establish the completeness relations $\sum_s u_s\bar u_s=\slashed p+m$ and $\sum_s v_s\bar v_s=\slashed p-m$ , and to show they are projection operators onto the positive- and negative-energy solutions of the mass shell. This is the single most-used identity in perturbative QED: it lets a spin-averaged cross section be computed as a trace of gamma matrices, never touching an explicit spinor. We build the sum, verify it annihilates the correct equation, and normalize by taking a trace.

Step 1. Define $P_+\equiv\sum_{s=1,2}u_s(p)\bar u_s(p)$ . Because each $u_s$ satisfies $(\slashed p-m)u_s=0$ , it follows that $(\slashed p-m)P_+=0$ : the operator $P_+$ maps everything into the space of positive-energy solutions and is killed by $\slashed p-m$ on the left. The only $4\times4$ matrix (built from $p$ and $m$ ) with this property, up to scale, is $\slashed p+m$ , since $(\slashed p-m)(\slashed p+m)=\slashed p\,\slashed p-m^2=p^2-m^2=0$ on shell. Why: $\slashed p\,\slashed p=p_\mu p_\nu\gamma^\mu\gamma^\nu=p^2$ collapses because the symmetric part of $\gamma^\mu\gamma^\nu$ is $g^{\mu\nu}$ — so $\slashed p+m$ is automatically annihilated by $\slashed p-m$ , exactly like $P_+$ .

Step 2. Fix the normalization with a trace. $\mathrm{Tr}(\slashed p+m)=p_\mu\mathrm{Tr}\,\gamma^\mu+4m=4m$ , since $\mathrm{Tr}\,\gamma^\mu=0$ . Independently, $\mathrm{Tr}\,P_+=\sum_s\mathrm{Tr}(u_s\bar u_s)=\sum_s\bar u_su_s=\sum_s 2m=4m$ , using the cyclic property of the trace and $\bar u_su_s=2m$ from the previous derivation. The two traces agree, fixing the constant to unity: $P_+=\slashed p+m$ . Why: matching one number (the trace) is enough because both matrices already share the same annihilator, so they can differ only by an overall scalar — and $4m=4m$ sets that scalar to one.

Step 3. Repeat for the antiparticles. The $v_s$ obey $(\slashed p+m)v_s=0$ , so $P_-\equiv\sum_s v_s\bar v_s$ is annihilated by $\slashed p+m$ and must be proportional to $\slashed p-m$ . Its trace is $\sum_s\bar v_sv_s=\sum_s(-2m)=-4m$ , matching $\mathrm{Tr}(\slashed p-m)=-4m$ , so $P_-=\slashed p-m$ . Why: the sign flip relative to the particle case is precisely the $\bar v v=-2m$ normalization, and it is what makes antiparticle loops and crossed diagrams carry the correct relative sign in QED amplitudes.

Limiting cases

Case	Result	Meaning
Rest frame $\mathbf{p}=0$	$u_s=\sqrt{2m}\,(\chi_s,0)^{\mathsf T}$ , $v_s=\sqrt{2m}\,(0,\chi_s)^{\mathsf T}$	lower block vanishes; the four spinors are the standard basis, pure spin up/down for particle and antiparticle
Non-relativistic $\|\mathbf{p}\|\ll m$	small-component ratio $r\approx\|\mathbf{p}\|/2m=v/2c$	the lower block is the "small component" of order $v/c$ ; recovers Pauli two-spinor QM plus the spin–orbit correction
Ultra-relativistic $E\gg m$	$r\to\|\mathbf{p}\|/E\to1$ ; upper and lower blocks equal	helicity eigenstates become chirality eigenstates; wrong-helicity amplitudes suppressed by $m/E$
Massless $m=0$	$\slashed p\,u=0$ , blocks decouple into left/right Weyl spinors	chirality is exactly conserved; the two-component Weyl equation governs each handedness independently
Momentum along quantisation axis $\mathbf{p}\parallel\hat z$	$u_1,u_2$ are eigenstates of $S_z$ and of helicity	the only case where the spin label and the helicity label coincide

Common trapThe four plane-wave spinors are eigenstates of Sz for any momentum

They are not. Building the spinors from $\chi_1=(1,0)^{\mathsf T}$ and $\chi_2=(0,1)^{\mathsf T}$ makes them eigenstates of $S_z$ only when $\mathbf{p}$ points along the $z$ axis. For a general momentum, the spin operator $\mathbf{S}=\tfrac12\boldsymbol{\Sigma}$ does not commute with the free Dirac Hamiltonian — only the total angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ is conserved — so a state of definite $S_z$ is not a state of definite energy for a moving particle. The label that is good is helicity, $h=\boldsymbol{\Sigma}\cdot\hat{\mathbf{p}}$ , the projection of spin along the direction of motion, because $\boldsymbol{\Sigma}\cdot\mathbf{p}$ does commute with $\slashed p$ . This is why every serious calculation of a polarized cross section is done in the helicity basis, not the $S_z$ basis: helicity is Lorentz-covariant along the momentum, $S_z$ is not.

Worked examples

1 · Building an explicit electron spinor (fully worked)

Construct the spin-up positive-energy spinor $u_1$ for an electron ( $m=0.511\ \mathrm{MeV}$ ) moving along $+\hat z$ with momentum $|\mathbf{p}|=1.00\ \mathrm{MeV}$ . Verify its normalization and identify its helicity.

Step 1. Energy on the mass shell: $E=\sqrt{\mathbf{p}^2+m^2}=\sqrt{(1.00)^2+(0.511)^2}=\sqrt{1.2611}=\mathbf{1.1230\ MeV}$ . Then $E+m=1.6340\ \mathrm{MeV}$ .

Step 2. Small-component ratio: $r=|\mathbf{p}|/(E+m)=1.00/1.6340=\mathbf{0.6120}$ . Because $\mathbf{p}=|\mathbf{p}|\hat z$ , we have $\boldsymbol{\sigma}\cdot\mathbf{p}=|\mathbf{p}|\sigma_z$ , and acting on $\chi_1=(1,0)^{\mathsf T}$ gives $\sigma_z\chi_1=+\chi_1$ . So the lower block is $r\chi_1$ .

Step 3. Assemble with $N=\sqrt{E+m}=\sqrt{1.6340}=1.2783\ \mathrm{MeV}^{1/2}$ : $u_1=N\,(1,\,0,\,r,\,0)^{\mathsf T}=(\mathbf{1.278},\ 0,\ \mathbf{0.782},\ 0)^{\mathsf T}\ \mathrm{MeV}^{1/2}$ . The lower ("small") component is 61 percent of the upper — not small here, because $|\mathbf{p}|\approx2m$ puts the electron squarely in the relativistic regime.

Step 4. Check the norm: $u_1^\dagger u_1=N^2(1+r^2)=1.6340\times(1+0.3745)=1.6340\times1.3745=\mathbf{2.246}$ , and $2E=2\times1.1230=2.246$ . The normalization $u^\dagger u=2E$ holds exactly, and the fraction of the norm in the lower block is $r^2/(1+r^2)=0.2725$ — 27 percent of the probability lives in the lower two components.

Step 5. Helicity: spin is along $+\hat z$ and momentum is along $+\hat z$ , so $h=\boldsymbol{\Sigma}\cdot\hat{\mathbf{p}}=+1$ — this is a positive-helicity (right-handed) electron. Had we chosen $\chi_2$ , we would have built the negative-helicity partner.

2 · The small component in the two limits

How large is the lower block for (a) a 1 eV conduction electron and (b) a 45.6 GeV LEP-beam electron?

Step 1. Non-relativistic (1 eV). With $T=1\ \mathrm{eV}\ll m=0.511\ \mathrm{MeV}$ , $|\mathbf{p}|\approx\sqrt{2mT}$ and $E+m\approx2m$ , so $r\approx|\mathbf{p}|/2m=\sqrt{T/2m}=\sqrt{1/(2\times5.11\times10^5)}\approx\mathbf{9.9\times10^{-4}}$ . The general scaling is $r\approx v/2c$ — the small block is one part in a thousand here, which is why ordinary chemistry runs on the two-component Pauli spinor and the Dirac lower block appears only as the tiny spin–orbit correction.

Step 2. Ultra-relativistic (45.6 GeV). Now $E\gg m$ and $r\to|\mathbf{p}|/E\to1$ : upper and lower blocks are essentially equal, and the spinor is a chirality eigenstate to precision $m/E$ . Numerically the wrong-chirality contamination is $m/E=0.511\ \mathrm{MeV}/45.6\ \mathrm{GeV}=\mathbf{1.1\times10^{-5}}$ . This is why the electron behaves as a nearly massless, definite-chirality particle at collider energies — the same $m/E$ helicity-flip suppression that governs, for example, the rate of $\pi^-\to e^-\bar\nu_e$ relative to the muon channel.

3 · Counting states from the spin sum

Show that exactly two particle states exist at each momentum, directly from the completeness relation.

Step 1. The positive-energy projector is $\Lambda_+=(\slashed p+m)/2m$ , so $\sum_s u_s\bar u_s=2m\,\Lambda_+$ ; a projector's rank equals its trace.

Step 2. $\mathrm{Tr}\,\Lambda_+=\dfrac{1}{2m}\mathrm{Tr}(\slashed p+m)=\dfrac{1}{2m}(0+4m)=\mathbf{2}$ , so exactly two independent particle spinors exist at each momentum — the two spin states — with no reference to explicit components. The same trace machinery that counts states evaluates $\overline{|\mathcal{M}|^2}$ , which is why cross sections are computed with traces, not columns of numbers.

Physics of the portrait

The plane-wave solutions handed physics a gift it did not want: two of the four states carried negative energy. Dirac's first patch was the "sea" of filled negative-energy states; the cleaner idea — that these solutions are ordinary positive-energy antiparticles running backward in time — was worked out by a Swiss physicist a decade before Feynman made it famous.

Origins · 1 minErnst Stückelberg and how this was discovered

Ernst Stückelberg

1905–1984 · relativistic quantum theory and the backward-in-time reading of antiparticles

Ernst Carl Gerlach Stückelberg was one of the most original and least celebrated physicists of the twentieth century — a Basel physicist who solved problems years ahead of his contemporaries and then buried the results in dense, idiosyncratic papers few read in time. Confronting exactly the negative-energy plane-wave solutions of this article, he proposed in 1941 that they need not describe negative-energy particles at all: a negative-energy solution running one way in time is mathematically identical to a positive-energy antiparticle running the other way. The "wrong" time dependence $e^{+iEt}$ of the $v$ -spinors is not a disease but a relabelling — the same insight Feynman would popularise in 1949 as the Feynman–Stückelberg interpretation, the foundation of how antiparticles appear in every Feynman diagram. Stückelberg went further, formulating a covariant perturbation theory and, with his student Petermann, discovering the renormalization group in 1953. Slowed by illness and a working style that hid his depth, he watched others win Nobel Prizes for ideas he had reached first — Feynman himself said Stückelberg "did the work and walks alone toward the twilight, and here I am covered in all the glory, which rightfully should be his."

History

1928

— Dirac's relativistic equation yields four plane-wave solutions, two with negative energy

1930

— Dirac's hole theory recasts the negative-energy solutions as an unseen filled sea

1931

— Dirac predicts the anti-electron, a positive particle of the electron's mass

1932

— Anderson finds the positron in cosmic-ray cloud-chamber tracks

1941

— Stückelberg reinterprets negative-energy waves as antiparticles moving backward in time

1949

— Feynman builds the same interpretation into the diagrammatic rules of QED

Going further

Texts. Thomson, Modern Particle Physics, §4.6–4.7 (plane-wave spinors, normalization, helicity) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.2 (solutions of the Dirac equation) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.3 (free-particle solutions and spin sums) · Halzen & Martin, Quarks and Leptons, §5.3–5.4 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 11 (spinor solutions and helicity).

Provenance. The differential equation and its Clifford algebra are owned by aq-pp-dirac-equation and aq-pp-covariant-dirac; the negative-energy story and the mass shell $E^2=\mathbf{p}^2+m^2$ trace back to aq-pp-klein-gordon-equation and are recalled here by name, not re-derived. A verifier should confirm: (1) the ansatz reduces $(i\gamma^\mu\partial_\mu-m)\psi=0$ to $(\slashed p-m)u=0$ ; (2) the block coupling $u_B=\boldsymbol{\sigma}\cdot\mathbf{p}\,u_A/(E+m)$ returns $E^2=\mathbf{p}^2+m^2$ ; (3) $N=\sqrt{E+m}$ gives $u^\dagger u=2E$ and $\bar u u=2m$ , $\bar v v=-2m$ ; (4) $\sum_s u_s\bar u_s=\slashed p+m$ with $\mathrm{Tr}=4m$ ; (5) the worked electron ( $m=0.511$ MeV, $|\mathbf{p}|=1.00$ MeV) gives $E=1.1230$ MeV, $r=0.612$ , $u^\dagger u=2.246=2E$ .

Prerequisites. The Dirac equation · Klein–Gordon equation · Covariance of the Dirac equation. Leads to. Antiparticles · Spin and helicity states · Dirac probability current.

Ask the tutor: Ask me to build any spinor explicitly for a momentum you choose, or to walk the spin sums through a full $e^+e^-\to\mu^+\mu^-$ trace.

Why does the differential equation collapse to pure algebra so cleanly?

Because a plane wave is an eigenfunction of every derivative operator. When $\psi=u\,e^{-ip\cdot x}$ , applying $\partial_\mu$ just multiplies the wave by $-ip_\mu$ — the shape of $\psi$ never changes, only its amplitude is scaled. So the operator $i\gamma^\mu\partial_\mu$ turns into the matrix $\gamma^\mu p_\mu$ , and the field equation, a coupled system of partial differential equations, becomes a single $4\times4$ matrix equation for the constant column $u$ . It is the same trick that turns $\hat{\mathbf{p}}\to\hbar\mathbf{k}$ for a free Schrödinger particle, merely carrying four components instead of one. The catch: it works only for a free particle of definite momentum — switch on an interaction and the momentum is no longer sharp, so you must return to the differential equation.

What exactly is the difference between the u and v spinors?

The $u$ -spinors solve $(\slashed p-m)u=0$ and describe particles; the $v$ -spinors solve $(\slashed p+m)v=0$ and describe antiparticles. In the plane wave, a $u$ rides on $e^{-ip\cdot x}$ (positive-frequency, $e^{-iEt}$ ) while a $v$ rides on $e^{+ip\cdot x}$ (negative-frequency, $e^{+iEt}$ ). Structurally, the $u$ -spinor has its large block on top and its small block on the bottom; the $v$ -spinor is flipped, large block on the bottom. Both are built with the same positive energy $E=+\sqrt{\mathbf{p}^2+m^2}$ and coupling factor $\boldsymbol{\sigma}\cdot\mathbf{p}/(E+m)$ — the antiparticle is not a negative-energy particle in the modern reading but an ordinary positive-energy object of opposite charge, which is why $\bar v v=-2m$ carries the tell-tale minus sign.

Why is the lower block called the "small component"?

Because for a slow particle it is small. The lower block is $\boldsymbol{\sigma}\cdot\mathbf{p}\,u_A/(E+m)$ , and non-relativistically $E\approx m$ so the prefactor is $\approx|\mathbf{p}|/2m=v/2c$ . A slow electron therefore has a lower block suppressed by its speed in units of $c$ , which is why the two-component Pauli spinor is an excellent approximation and the Dirac lower block shows up only as the fine-structure and spin–orbit corrections. The name is a misnomer at high energy: when $|\mathbf{p}|\gtrsim m$ the ratio approaches one and the "small" component is just as large as the "large" one — in the worked example at $|\mathbf{p}|\approx2m$ it already holds 27 percent of the norm.

How do the spin sums save so much work in a cross-section calculation?

An unpolarized cross section averages $|\mathcal{M}|^2$ over initial spins and sums over final spins. Written out, $|\mathcal{M}|^2$ contains products like $[\bar u(p')\Gamma u(p)][\bar u(p)\bar\Gamma u(p')]$ , and the spin sum turns $\sum_s u_s(p)\bar u_s(p)$ into the matrix $\slashed p+m$ . Once every external spinor pair is replaced this way, the expression becomes a trace of gamma matrices and slashed momenta, evaluated by a short set of trace theorems ( $\mathrm{Tr}$ of an odd number of gammas is zero, $\mathrm{Tr}(\gamma^\mu\gamma^\nu)=4g^{\mu\nu}$ , and so on). You never write down a single explicit spinor component — the plane-wave solutions enter tree-level QED only through their bilinear sums.

Are the negative-energy solutions real, or just a mathematical artefact?

They are as real as the positive-energy ones — you cannot throw them away. The four plane-wave spinors form a complete basis; keeping only the two positive-energy ones would make it impossible to expand a general solution or build a properly (anti)commuting quantum field. What changed over history is their interpretation. Dirac first imagined a filled "sea" whose holes are antiparticles, which works for fermions but fails for bosons. The Stückelberg–Feynman reading, standard today, keeps the mathematics but relabels a negative-energy solution running forward in time as a positive-energy antiparticle running backward — the same object, cleaner bookkeeping. In quantum field theory the issue dissolves: the $v$ -spinor multiplies a creation operator for an antiparticle, and every physical energy is positive.

Why must I use helicity instead of S_z for a moving particle?

Because $S_z$ is not conserved for a moving free particle, while helicity is. The spin operator $\mathbf{S}=\tfrac12\boldsymbol{\Sigma}$ does not commute with the Dirac Hamiltonian $H=\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m$ unless $\mathbf{p}$ lies along the quantisation axis; only the total $\mathbf{J}=\mathbf{L}+\mathbf{S}$ commutes. Helicity, $h=\boldsymbol{\Sigma}\cdot\hat{\mathbf{p}}$ , projects spin onto the direction of motion, and $\boldsymbol{\Sigma}\cdot\mathbf{p}$ does commute with $\slashed p$ , so a plane wave can be a simultaneous eigenstate of energy, momentum, and helicity. That makes it the natural Lorentz-friendly spin label — and since the weak interaction couples only to left-handed chirality (coinciding with helicity in the massless limit), it is also the physically measured quantity in electroweak processes.