Dirac Spinors and Plane-Wave Solutions

Once the Dirac equation is in hand, the first thing any physicist does is solve it for a free particle — and the answer is where relativistic quantum mechanics stops being abstract and starts computing. A free electron of definite momentum is a plane wave, and because the Dirac equation is a four-component matrix equation, the plane wave carries a constant four-component passenger called a Dirac spinor. That spinor is not decoration: its structure is exactly what encodes spin-½, and its two energy branches are exactly what forces antimatter into the theory. The move that makes everything tractable is that a plane wave turns the differential Dirac operator into a plain number-valued matrix. Feed in $\psi = u(p)\,e^{-ip\cdot x}$ and the derivatives just pull down factors of momentum, so "being a solution" collapses to a purely algebraic condition, $(\not{p}-m)\,u=0$ . This is a matrix equation with a two-dimensional null space for each momentum — precisely the two spin states a spin-½ particle must have. The negative-energy branch, packaged into a second spinor $v(p)$ , gives two more solutions that become the positron. Master these spinors and you own the working language of every scattering calculation. The normalization $\bar{u}u=2m$ , the completeness relation $\sum_s u^s\bar{u}^s=\not{p}+m$ , and the positive density $\psi^\dagger\psi$ are not formalities: they are the exact identities inserted when you square an amplitude and sum over spins in quantum electrodynamics. We build all four spinors explicitly, verify every identity as an honest 4×4 matrix statement, and watch spin and antimatter fall out of the algebra.

Key takeaways

Free Dirac solutions factor into a phase times a constant spinor: $\psi = u^s(p)\,e^{-ip\cdot x}$ (positive energy) and $\psi = v^s(p)\,e^{+ip\cdot x}$ (negative energy), with spin label $s=1,2$ .
The plane-wave ansatz turns the differential equation into the algebraic on-shell conditions $(\not{p}-m)u=0$ and $(\not{p}+m)v=0$ , each with a two-dimensional solution space — the two spin states.
In the Dirac (standard) representation the positive-energy spinor is $u^s(p)=\bigl(\sqrt{E+m}\,\xi^s,\ \tfrac{\boldsymbol\sigma\cdot\mathbf p}{\sqrt{E+m}}\,\xi^s\bigr)^{\!\top}$ : a large upper block and a small lower block tied by $\boldsymbol\sigma\cdot\mathbf p/(E+m)$ .
Normalizations: the Lorentz scalar $\bar{u}^r u^s = 2m\,\delta^{rs}$ and the density $u^{r\dagger}u^s = 2E\,\delta^{rs}$ ; the completeness spin sums are $\sum_s u^s\bar{u}^s=\not{p}+m$ and $\sum_s v^s\bar{v}^s=\not{p}-m$ .
The upper/lower split is the large/small (nonrelativistic) decomposition; the $u$ / $v$ split is the particle/antiparticle (energy-sign) decomposition — they are NOT the same thing.
The conserved Dirac density $j^0=\psi^\dagger\psi=\sum_a|\psi_a|^2\ge 0$ is manifestly positive, curing the Klein–Gordon negative-density disease.
Key idea: a free Dirac plane wave is a phase $e^{\mp ip\cdot x}$ times a constant spinor obeying the algebraic condition $(\not{p}\mp m)w=0$ , whose two-dimensional null space per momentum encodes spin-½, with the two energy signs becoming electron and positron.

The core insight

A free Dirac plane wave factorizes into a spacetime phase $e^{\mp ip\cdot x}$ times a constant four-spinor $u(p)$ or $v(p)$ obeying the purely algebraic on-shell condition $(\not{p}\mp m)w=0$ ; its two-dimensional solution space per momentum encodes spin-½, and the positive- and negative-energy branches map to electrons and positrons.

Where this lives today

Dirac spinors are the daily currency of high-energy physics. Every cross section computed at the LHC begins by writing external electrons, quarks, and their antiparticles as $u(p)$ and $v(p)$ spinors, then uses the spin sums $\sum u\bar u=\not p+m$ to fold the spinor algebra into gamma-matrix traces that a computer algebra system evaluates automatically. The same spinors, quantized, become the mode functions of the electron field: the $u$ -spinors ride electron annihilation operators and the $v$ -spinors ride positron creation operators. In lattice QCD, in Standard Model event generators, and in searches for physics beyond it, the four-component plane-wave spinor built here is the irreducible starting point.

The physical picture

Think of the Dirac equation as the "square root" of the Klein–Gordon equation. Taking a square root of a differential operator forces matrix-valued coefficients — the gamma matrices — and a matrix operator needs a multi-component object to act on. The four components of $\psi$ are the price of a first-order, Lorentz-covariant equation with a positive probability density. For a plane wave of definite four-momentum, all of spacetime dependence is squeezed into one universal phase $e^{\mp ip\cdot x}$ ; the only thing left to determine is a single constant column of four numbers. That constant column lives or dies by a single matrix equation. The Dirac operator, acting on a plane wave, becomes the ordinary c-number matrix $\not{p}\mp m$ , and "solving the Dirac equation" reduces to "find the null space of this 4×4 matrix." On the mass shell that null space is two-dimensional for each energy sign — exactly two spin states for a fermion, no more and no fewer. Spin is not put in by hand; it is the dimension count of a kernel.

Figure

Free Dirac plane wave: phase times spinor

A particle at rest makes the anatomy transparent. With $\mathbf p=0$ the spinor splits cleanly: the positive-energy solution $u$ lives entirely in the upper two components (two spins), the negative-energy solution $v$ lives entirely in the lower two (two spins). Boosting to a moving frame mixes the blocks, so a moving electron grows nonzero lower components of order $|\mathbf p|/(E+m)$ . That ratio is the relativistic "smallness" parameter: tiny for slow electrons, order one when the kinetic energy rivals the rest mass. The negative-energy $v$ -solutions are not a pathology to be discarded — they are mandatory for a complete basis, just as they were in the Klein–Gordon theory of "Negative-Energy Solutions and the Interpretive Crisis." Reading the extra branch $e^{+ip\cdot x}$ not as a sick electron but as an antiparticle is the conceptual hinge that turns a crisis into the prediction of the positron.

Figure

Rest frame vs boosted: where the components live

The core equations

Everything begins with the Dirac equation, the first-order relativistic wave equation whose four-component solutions we are about to construct. We work throughout in natural units $\hbar=c=1$ and the mostly-minus metric $g_{\mu\nu}=\mathrm{diag}(+,-,-,-)$ in the Dirac (standard) representation of the gamma matrices, exactly the conventions pinned in "The Klein-Gordon Equation" and used by "Dirac Gamma Matrices and Clifford Algebra."

(i\gamma^\mu\partial_\mu - m)\,\psi(x) = 0 \qquad (1)

where: $\gamma^\mu$ — the four $4\times4$ Dirac matrices satisfying $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}\mathbb{1}$ ; $\psi$ — the four-component Dirac spinor field; $m$ — the particle mass; $\partial_\mu=\partial/\partial x^\mu$ . The equation is first order in all derivatives, which is exactly why its conserved density will turn out positive. Because the equation is linear with constant coefficients, a plane-wave ansatz separates the universal phase from a constant spinor. We keep the physical energy $p^0=E_p>0$ positive in both branches and put the energy-sign information into the choice of spinor and the sign of the exponent.

\psi^{(+)}(x) = u^s(p)\,e^{-ip\cdot x}, \qquad \psi^{(-)}(x) = v^s(p)\,e^{+ip\cdot x} \qquad (2)

where: $u^s(p)$ — the positive-energy momentum-space spinor with spin label $s=1,2$ ; $v^s(p)$ — the negative-energy (antiparticle) spinor; $p\cdot x = p^0x^0-\mathbf p\cdot\mathbf x$ with $p^0=E_p=\sqrt{\mathbf p^2+m^2}$ . Writing the second branch as $e^{+ip\cdot x}$ (not $e^{-ip\cdot x}$ with $p^0<0$ ) keeps every energy positive and packages the negative-energy content into $v(p)$ — the bookkeeping that later makes the positron manifest. Substituting the ansatz collapses the differential operator into a matrix. A derivative acting on $e^{-ip\cdot x}$ pulls down $-ip_\mu$ , so $i\partial_\mu\to p_\mu$ and the whole equation becomes purely algebraic.

(\not{p}-m)\,u^s(p)=0, \qquad (\not{p}+m)\,v^s(p)=0, \qquad \not{p}\equiv\gamma^\mu p_\mu \qquad (3)

where: $\not{p}$ — the Feynman slash $\gamma^\mu p_\mu$ ; the sign flips for $v$ because $e^{+ip\cdot x}$ gives $i\partial_\mu\to -p_\mu$ . Multiplying $(\not p-m)u=0$ by $(\not p+m)$ and using $\not p\,\not p = p^2$ forces $p^2=m^2$ : the mass shell falls out automatically, reproducing the relativistic energy–momentum relation with no extra input. Now the explicit solutions. In the Dirac representation the standard-basis positive-energy spinor has a large upper block and a small lower block linked by $\boldsymbol\sigma\cdot\mathbf p/(E+m)$ .

u^s(p)=\begin{pmatrix}\sqrt{E_p+m}\,\,\xi^s\\[4pt]\dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{\sqrt{E_p+m}}\,\xi^s\end{pmatrix},\qquad v^s(p)=\begin{pmatrix}\dfrac{\boldsymbol{\sigma}\cdot\mathbf{p}}{\sqrt{E_p+m}}\,\eta^s\\[4pt]\sqrt{E_p+m}\,\,\eta^s\end{pmatrix} \qquad (4)

where: $\xi^s,\eta^s$ — two orthonormal Pauli two-spinors, $\xi^{r\dagger}\xi^s=\delta^{rs}$ ; $\boldsymbol\sigma$ — the Pauli matrices. Note the role reversal: for the antiparticle spinor $v$ the large block is the lower one — the tell that $v$ realizes the negative-energy branch, with the particle content pushed downstairs. These spinors carry two distinct normalizations, and confusing them is a classic source of factor errors. The Lorentz scalar $\bar u u$ is frame-independent; the density $u^\dagger u$ is the timelike current component and scales with energy.

\bar{u}^r(p)u^s(p)=2m\,\delta^{rs},\qquad \bar{v}^r(p)v^s(p)=-2m\,\delta^{rs},\qquad \bar u^r v^s=\bar v^r u^s=0 \qquad (5)

where: $\bar u = u^\dagger\gamma^0$ is the Dirac adjoint; the scalar $\bar u u=2m$ is Lorentz invariant, while the sign flip $\bar v v=-2m$ reflects that $v$ sits on the opposite energy branch. The particle and antiparticle spinors are orthogonal, so a general solution decomposes cleanly.

u^{r\dagger}(p)u^s(p)=2E_p\,\delta^{rs},\qquad v^{r\dagger}(p)v^s(p)=2E_p\,\delta^{rs} \qquad (6)

where: $u^\dagger u=\bar u\gamma^0 u$ is the timelike component of the conserved current — the probability density — which is positive and grows as $2E_p$ , exactly the relativistic scaling that compensates the Lorentz contraction of a normalization volume. The single most useful identities in all of relativistic scattering are the spin sums, which say that summing the spinor dyadics over spin reconstructs a projector. These are the workhorses of the "Casimir trick."

\sum_{s=1,2}u^s(p)\,\bar{u}^s(p)=\not{p}+m,\qquad \sum_{s=1,2}v^s(p)\,\bar{v}^s(p)=\not{p}-m \qquad (7)

where: $\not p\pm m = 2m\,\Lambda_\pm$ are twice the positive/negative energy projectors $\Lambda_\pm=(\pm\not p+m)/2m$ . When you square a QED amplitude and average over initial spins and sum over final ones, these collapse the spinor algebra into traces of gamma matrices — spinor bookkeeping becomes trace technology. Finally, the payoff that motivated the whole four-component apparatus: the Dirac density is manifestly non-negative, the exact defect the Klein–Gordon current suffered from.

j^\mu=\bar\psi\gamma^\mu\psi,\qquad \partial_\mu j^\mu=0,\qquad j^0=\psi^\dagger\psi=\sum_{a=1}^{4}|\psi_a|^2\ge 0 \qquad (8)

where: $j^\mu$ — the conserved Dirac four-current; $j^0=\psi^\dagger\psi$ is a sum of squared moduli, so it can never be negative — the first-order structure of the Dirac equation buys a positive probability density where "Klein-Gordon Probability Current and Negative Densities" could not.

Derivation ABuilding the spinors at rest and boosting to reveal the small components

We construct the positive-energy spinor from scratch: solve the momentum-space equation $(\not p-m)u=0$ in the rest frame where it is trivial, then read off the general-momentum solution by keeping the full $\not p$ and solving block by block. The strategy is to split $u=(\phi,\chi)^\top$ into upper ( $\phi$ ) and lower ( $\chi$ ) Pauli two-spinors and let the lower block of the matrix equation express $\chi$ in terms of $\phi$ ; the watch-out is that the normalization must be fixed at the end so that $u^\dagger u=2E$ and $\bar u u=2m$ both come out right, and one must not confuse this large/small split with the particle/antiparticle split.

Step 1. In the rest frame $p^\mu=(m,\mathbf 0)$ , so $\not p=\gamma^0 m$ and the equation reads $(\gamma^0-1)m\,u=0$ . In the Dirac representation $\gamma^0=\mathrm{diag}(\mathbb 1,-\mathbb 1)$ , so this projects onto the $+1$ eigenspace: the upper two components. The rest-frame solution is $u=\sqrt{2m}\,(\xi,0)^\top$ with $\xi$ any Pauli two-spinor. Why the upper block: $\gamma^0$ has eigenvalue $+1$ on the top two slots and $-1$ on the bottom two, so demanding $\gamma^0 u=u$ kills the lower block entirely at rest.

Step 2. For general momentum write $\not p = \gamma^0 E_p - \boldsymbol\gamma\cdot\mathbf p$ . With $u=(\phi,\chi)^\top$ and $\gamma^i=\begin{pmatrix}0&\sigma^i\\-\sigma^i&0\end{pmatrix}$ , the lower-block row of $(\not p - m)u=0$ becomes $-\boldsymbol\sigma\cdot\mathbf p\,\phi-(E_p+m)\chi=0$ .

Step 3. Solve for the small component: $\chi=\dfrac{\boldsymbol\sigma\cdot\mathbf p}{E_p+m}\,\phi$ . The lower components are locked to the upper ones; they vanish at rest and grow with momentum, with magnitude ratio $|\mathbf p|/(E_p+m)$ . Why "small": for $|\mathbf p|\ll m$ we have $E_p+m\approx 2m$ , so $\chi\approx (\boldsymbol\sigma\cdot\mathbf p/2m)\phi$ is suppressed by $|\mathbf p|/2m\ll 1$ — the nonrelativistic hierarchy.

Step 4. Fix the normalization. With $\phi=\sqrt{E_p+m}\,\xi$ one finds $u^\dagger u=(E_p+m)\bigl[1+\mathbf p^2/(E_p+m)^2\bigr]\xi^\dagger\xi$ . Using $E_p^2-m^2=\mathbf p^2$ this simplifies to $2E_p$ , and the scalar to $\bar u u=(E_p+m)-\mathbf p^2/(E_p+m)=2m$ , giving the boxed spinor of equation (4). Verified: at $\mathbf p\to 0$ the spinor returns to $\sqrt{2m}(\xi,0)^\top$ ; a symbolic check for $p^\mu=(\sqrt{13},0,0,3),\,m=2$ gives $u^\dagger u=2\sqrt{13}=7.211$ and $\bar u u=4=2m$ exactly (Bjorken–Drell §3.2; Peskin–Schroeder §3.3).

Go deeper · Derivation B · 1 minThe spin-sum completeness relation as a genuine matrix identity

Derivation BThe spin-sum completeness relation as a genuine matrix identity

We prove $\sum_s u^s\bar u^s=\not p+m$ not by manipulating symbols but by building the explicit 4×4 dyadic and simplifying it with the Pauli-spinor completeness $\sum_s\xi^s\xi^{s\dagger}=\mathbb 1$ . The route is to form $\sum_s u^s u^{s\dagger}$ block by block, collapse the spin sum in each block, and then right-multiply by $\gamma^0$ to convert $u^{s\dagger}$ into the Dirac adjoint $\bar u^s$ ; the payoff is the projector $\not p+m$ , and the trace of both sides gives an independent numerical cross-check.

Step 1. Form the outer-product sum. With $u^s=(\sqrt{E+m}\,\xi^s,\,\tfrac{\boldsymbol\sigma\cdot\mathbf p}{\sqrt{E+m}}\xi^s)^\top$ and $\sum_s\xi^s\xi^{s\dagger}=\mathbb 1$ , the upper-left block of $\sum_s u^s u^{s\dagger}$ is $(E+m)\mathbb 1$ , the lower-right is $\tfrac{(\boldsymbol\sigma\cdot\mathbf p)^2}{E+m}=\tfrac{\mathbf p^2}{E+m}\mathbb 1=(E-m)\mathbb 1$ , and the off-diagonal blocks are $\boldsymbol\sigma\cdot\mathbf p$ .

Step 2. Right-multiply by $\gamma^0=\mathrm{diag}(\mathbb 1,-\mathbb 1)$ to build $\sum_s u^s\bar u^s=(\sum_s u^s u^{s\dagger})\gamma^0$ . This flips the sign of the two right-hand blocks, turning the lower-right into $-(E-m)\mathbb 1=(m-E)\mathbb 1$ and the lower-left into $-\boldsymbol\sigma\cdot\mathbf p$ .

Step 3. Assemble. The result is $\begin{pmatrix}(E+m)\mathbb 1 & -\boldsymbol\sigma\cdot\mathbf p\\ \boldsymbol\sigma\cdot\mathbf p & (m-E)\mathbb 1\end{pmatrix}$ , which is exactly $\not p+m$ in the Dirac representation, where $\not p=\gamma^0 E-\boldsymbol\gamma\cdot\mathbf p=\begin{pmatrix}E\,\mathbb 1 & -\boldsymbol\sigma\cdot\mathbf p\\ \boldsymbol\sigma\cdot\mathbf p & -E\,\mathbb 1\end{pmatrix}$ and $m=m\,\mathbb 1_4$ . Why $\gamma^0$ enters: the completeness relation is a statement about $u\bar u$ , and $\bar u=u^\dagger\gamma^0$ , so the $\gamma^0$ is not optional — dropping it gives the (non-covariant) $\sum u u^\dagger$ instead.

Step 4. Cross-check by trace: $\mathrm{Tr}(\not p+m)=\mathrm{Tr}(\not p)+4m=0+4m=4m$ , and independently $\sum_s\bar u^s u^s=\sum_s 2m=4m$ . The two agree. Verified: for $p^\mu=(\sqrt{13},0,0,3),\,m=2$ , direct 4×4 construction gives $\sum_s u^s\bar u^s-(\not p+m)=0$ to machine precision, and $\sum_s v^s\bar v^s=\not p-m$ likewise (Peskin–Schroeder Eq. 3.66–3.67).

Limiting cases

The spinor structure simplifies sharply in the natural kinematic regimes, and each limit isolates one physical idea.

Case	Result	Meaning
Rest frame, $\mathbf p=0$	$u^s=\sqrt{2m}\,(\xi^s,0)^\top$ , $v^s=\sqrt{2m}\,(0,\eta^s)^\top$	Upper and lower blocks decouple; $\xi^s,\eta^s$ are ordinary Pauli two-spinors. Particle sits "up," antiparticle "down."
Nonrelativistic, $\|\mathbf p\|\ll m$	small block $\approx \tfrac{\boldsymbol\sigma\cdot\mathbf p}{2m}\,$ (large block)	Lower components suppressed by $\|\mathbf p\|/2m$ ; the large components obey the Pauli equation — nonrelativistic spin physics recovered.
Ultrarelativistic / massless, $E\gg m$	$\bar u u\to 0$ ; helicity $=\pm$ chirality, $P_{L,R}=\tfrac12(1\mp\gamma^5)$	The scalar normalization collapses; helicity becomes a good quantum number and left/right Weyl components decouple.

Worked examples

Example 1 — Construct an explicit spinor and check its density. Take a particle of mass $m=2$ (natural units) moving along $z$ with $p_z=3$ ; build $u^1(p)$ and verify $u^\dagger u = 2E$ .

Step 1. On-shell energy: $E=\sqrt{p_z^2+m^2}=\sqrt{9+4}=\sqrt{13}=3.6056$ . The normalization factor is $\sqrt{E+m}=\sqrt{\sqrt{13}+2}=\sqrt{5.6056}=2.3676$ .

Step 2. Large (upper) block for spin-up $\xi^1=(1,0)^\top$ : $\sqrt{E+m}\,\xi^1=(2.3676,\,0)^\top$ .

Step 3. Small (lower) block: $\dfrac{\boldsymbol\sigma\cdot\mathbf p}{\sqrt{E+m}}\xi^1$ . Here $\boldsymbol\sigma\cdot\mathbf p=p_z\sigma_z=3\,\sigma_z$ , and $\sigma_z\xi^1=\xi^1$ , so the block is $\dfrac{3}{2.3676}(1,0)^\top=(1.2671,\,0)^\top$ .

Step 4. Assemble: $u^1(p)=(2.3676,\,0,\,1.2671,\,0)^\top$ . Its density is $u^\dagger u = 2.3676^2 + 1.2671^2 = 5.6056 + 1.6055 = 7.2111$ . The answer $u^{1\dagger}u^1 = 7.2111 = 2\sqrt{13} = 2E_p$ , confirming the density normalization $u^\dagger u=2E$ exactly for a concrete spinor. The small-to-large ratio is $1.2671/2.3676 = 0.5352 = |\mathbf p|/(E+m)$ , an order-one number — this particle is genuinely relativistic.

Example 2 — Verify the on-shell condition kills the spinor. For the same $u^1(p)$ , check $(\not p - m)u^1 = 0$ .

Step 1. Build $\not p = \gamma^0 E - \gamma^3 p_z$ . In block form $\not p = \begin{pmatrix} E\,\mathbb 1 & -p_z\sigma_z\\ p_z\sigma_z & -E\,\mathbb 1\end{pmatrix}$ with $E=\sqrt{13},\,p_z=3$ .

Step 2. Act on the large block. The upper two rows of $(\not p-m)u^1$ give $(E-m)\cdot 2.3676 + (-p_z\sigma_z)\cdot(1.2671,0)^\top$ . Numerically $(E-m)=1.6056$ , so the first term is $1.6056\times 2.3676=3.8017$ ; the second is $-3\times 1.2671=-3.8013$ (up to rounding). They cancel: $3.8017-3.8013\approx 0$ .

Step 3. Act on the small block. The lower two rows give $p_z\sigma_z\cdot(2.3676,0)^\top + (-E-m)\cdot(1.2671,0)^\top = 3\times 2.3676 - 5.6056\times 1.2671 = 7.1028 - 7.1028 = 0$ .

Step 4. Both blocks vanish, so $(\not p - m)u^1 = 0$ . The spinor sits exactly in the null space of $\not p - m$ — it is a genuine positive-energy solution. (The tiny residuals are decimal rounding; the exact symbolic check gives identically zero, verified in the accompanying script.)

The story behind the physics

The four-component spinor is the object Dirac was forced into when, in 1928, he demanded a wave equation first order in time whose square reproduced the relativistic energy relation. The two extra "negative-energy" solutions embarrassed him until, in 1930, he reinterpreted them as an infinite filled sea whose holes are antiparticles — a picture soon replaced by Feynman and Stückelberg's cleaner reading but vindicated in spirit when Anderson found the positron in a cloud chamber in 1932. The two-component Pauli spinors $\xi$ that sit inside the large block are the very objects Pauli had introduced in 1927 to describe nonrelativistic spin; Dirac's construction shows spin was relativity all along.

Origins · 1 mindirac spinor and how this was discovered

dirac spinor

the four-component object of relativistic QM · 1928– · Cambridge · Zürich · Pasadena

The Dirac spinor is the four-component wavefunction on which the gamma matrices act. Its upper and lower blocks are not four particles but a spin-½ particle and its antiparticle partner; its structure encodes spin, the electron's magnetic moment, and — through the negative-energy branch reinterpreted via charge conjugation — the positron. It emerged from Paul Dirac's 1928 demand for a first-order relativistic equation, absorbed Pauli's 1927 two-component spin object into its large block, and was vindicated when Anderson's 1932 positron realized its negative-energy solutions.

History

1927

— Pauli introduces two-component spinors for nonrelativistic electron spin

1928

— Dirac writes the first-order relativistic equation, forcing four-component spinors

1930

— Dirac proposes the hole theory reinterpreting the negative-energy v-spinors

1932

— Anderson observes the positron in a cloud chamber, realizing the v-solutions

1948

— Feynman–Stückelberg reinterpretation recasts v-spinors as forward-in-time antiparticles

Going further

References. Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 7 · Sakurai & Napolitano, Modern Quantum Mechanics, Ch. 8 (Relativistic QM) · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 20 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.3 · Bjorken & Drell, Relativistic Quantum Mechanics, §3.2.

Provenance. All equations are standard relativistic QM. Independently verified in the accompanying script: (1) $(\not p\mp m)w=0$ follows from the plane-wave ansatz and $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ ; (2) the Dirac-representation spinors give $\bar u u=2m$ , $u^\dagger u=2E_p$ for $p^\mu=(\sqrt{13},0,0,3),\,m=2$ ; (3) the spin sums $\sum u\bar u=\not p+m$ , $\sum v\bar v=\not p-m$ hold as exact 4×4 identities with $\mathrm{Tr}(\not p+m)=4m$ ; (4) the projectors $\Lambda_\pm=(\pm\not p+m)/2m$ are idempotent, complete, and orthogonal. Base repair: the source article stated the explicit spinors in the chiral (Weyl) basis while framing the derivation in the Dirac (standard) representation and quoting $\bar v v=-2m$ — an internal basis inconsistency. This article pins everything to the Dirac representation of "The Klein-Gordon Equation," so every explicit component, block structure, and sign is self-consistent and machine-checked in one basis.

Builds on: The Dirac Equation · Dirac Gamma Matrices and Clifford Algebra · The Klein-Gordon Equation · Negative-Energy Solutions and the Interpretive Crisis. Leads to: Spin from the Dirac Equation · The Dirac Sea and Antiparticles · Charge Conjugation and the Positron · Bilinear Covariants and Dirac Adjoint · Nonrelativistic Reduction and the Pauli Equation.

Why does the plane-wave ansatz turn a differential equation into an algebraic one?

Because $\partial_\mu$ acting on $e^{-ip\cdot x}$ just multiplies by $-ip_\mu$ . So $i\gamma^\mu\partial_\mu$ becomes $\gamma^\mu p_\mu=\not p$ , a constant matrix. The Dirac equation $(i\gamma^\mu\partial_\mu-m)\psi=0$ then reduces to $(\not p-m)u=0$ — no derivatives left, just a 4×4 matrix acting on a constant column. This is the universal trick for constant-coefficient linear equations: plane waves diagonalize translations, converting calculus into linear algebra.

What fixes the on-shell condition p² = m²?

Multiply $(\not p-m)u=0$ on the left by $(\not p+m)$ : you get $(\not p^2-m^2)u=0$ . Since $\not p^2=\tfrac12\{\gamma^\mu,\gamma^\nu\}p_\mu p_\nu=g^{\mu\nu}p_\mu p_\nu=p^2$ , this reads $(p^2-m^2)u=0$ . For a nonzero spinor you need $p^2=m^2$ , i.e. $E=\pm\sqrt{\mathbf p^2+m^2}$ . The mass shell is not an extra assumption; it is forced by the gamma algebra, and it is exactly why relativistic energy–momentum emerges automatically.

Why two spin states rather than one or four?

Once you are on shell, $(\not p-m)$ is a rank-2 matrix: its null space is two-dimensional. Equivalently, $\Lambda_+=(\not p+m)/2m$ is a projector with trace 2, so it projects onto a two-dimensional subspace. Those two dimensions are the two spin states. The other two dimensions of the four-component space are the null space of $(\not p+m)$ — the two $v$ -spinors. So $4 = 2\,(\text{spins}) \times 2\,(\text{energy signs})$ .

Is the "small component" the same as the antiparticle part?

No — this is the central pitfall. The small (lower) component is present in every moving electron and is suppressed by $|\mathbf p|/(E+m)$ ; it is a nonrelativistic-limit statement. The antiparticle is the entirely separate $v$ -spinor with the opposite energy sign. A pure electron is a $u$ -spinor with both large and small components; it contains zero antiparticle content. Watch the two decompositions carefully: upper/lower is large/small, while $u/v$ is particle/antiparticle.

Why does the density carry a factor 2E while the scalar carries 2m?

$\bar u u=u^\dagger\gamma^0 u$ is a Lorentz scalar, so it must be the same number in every frame — and at rest that number is $2m$ . The density $u^\dagger u$ is the time component of the four-vector current $j^\mu=\bar u\gamma^\mu u$ , so it transforms and grows with energy as $2E$ . Physically, a relativistic normalization volume Lorentz-contracts by $\gamma=E/m$ , so the density must scale up by $E/m$ to keep the total probability (density × volume) invariant. Hence $2m\to 2E$ .

What happens to these spinors when m → 0?

The scalar normalization $\bar u u=2m\to 0$ , so it stops being useful; you switch to $u^\dagger u=2E$ . More importantly, chirality (eigenvalue of $\gamma^5$ ) becomes conserved and coincides with helicity, so the left- and right-handed pieces $P_{L,R}u=\tfrac12(1\mp\gamma^5)u$ decouple and each satisfies a two-component Weyl equation. This is why massless (or ultrarelativistic) fermions are naturally described by helicity/chirality, and why the weak interaction — coupling only to left-handed fields — produces strongly polarized particles at high energy.

How do these plane waves become the quantized Dirac field?

You promote the expansion coefficients to operators. The free field is $\psi(x)=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\sum_s(a^s_{\mathbf p}u^s(p)e^{-ip\cdot x}+b^{s\dagger}_{\mathbf p}v^s(p)e^{+ip\cdot x})$ . The $u$ -spinors multiply electron annihilation operators $a$ ; the $v$ -spinors multiply positron creation operators $b^\dagger$ . The plane-wave spinors built here are precisely the mode functions of second quantization, and the anticommutators of $a,b$ enforce Fermi statistics.