Spinor normalisation & completeness

Once the Dirac equation is solved, you hold four plane-wave spinors in your hand — two describing a spin-½ particle, two describing its antiparticle — and you must answer two questions before a single cross section can be computed. First, how large is each spinor? The equation fixes their direction in the four-dimensional spinor space but leaves an overall constant $N$ free, exactly as an eigenvalue equation fixes an eigenvector only up to scale. Second, how do the four together fill that space? A matrix element is never evaluated on one spin state in isolation; real experiments average over initial spins and sum over final spins, and that sum is only tractable if the spinors close into something simple. Both questions have the same answer, and it is dictated not by taste but by relativity. The normalisation is fixed to $\bar u u = 2m$ and $u^\dagger u = 2E$ — the spin-½ echo of the scalar field's " $2E$ particles per unit volume" — so that spinor states dovetail with Lorentz-invariant phase space and the box volume cancels from every rate. The completeness relations then read $\sum_s u^{(s)}\bar u^{(s)} = \slashed p + m$ and $\sum_s v^{(s)}\bar v^{(s)} = \slashed p - m$ : the spin sum collapses to a matrix built from the momentum alone. This single fact is what turns the forbidding-looking $\overline{|\mathcal M|^2}$ into a trace of gamma matrices — Casimir's trick — and is executed inside every QED and QCD calculation ever done. This article fixes $N$ , proves the two completeness relations, builds the energy projection operators $\Lambda_\pm$ they generate, and shows why the whole machinery of spin-averaged scattering rests on them.

Key takeaways

The free-particle Dirac equation becomes algebraic, $(\slashed p - m)u = 0$ and $(\slashed p + m)v = 0$ ; it fixes each spinor's direction but leaves an overall constant $N$ free.
The relativistic convention sets $N = \sqrt{E+m}$ , giving $\bar u^{(r)} u^{(s)} = 2m\,\delta^{rs}$ (Lorentz scalar) and $u^{(r)\dagger} u^{(s)} = 2E\,\delta^{rs}$ (a density, the spin-½ version of " $2E$ per unit volume").
Particle and antiparticle spinors are orthogonal, $\bar u^{(r)} v^{(s)} = \bar v^{(r)} u^{(s)} = 0$ , and the antiparticle scalar flips sign, $\bar v^{(r)} v^{(s)} = -2m\,\delta^{rs}$ .
The completeness relations $\sum_s u^{(s)}\bar u^{(s)} = \slashed p + m$ and $\sum_s v^{(s)}\bar v^{(s)} = \slashed p - m$ close the four spinors and generate the energy projection operators $\Lambda_\pm = (\pm\slashed p + m)/2m$ .
These sums are what convert a spin-averaged $\overline{|\mathcal M|^2}$ into a trace of gamma matrices — Casimir's trick — the backbone of every unpolarised cross-section calculation.
Key idea: the Dirac equation fixes each spinor only up to a constant; setting it by $\bar u u = 2m$ (so $u^\dagger u = 2E$ ) matches relativistic phase space, and the spin sums $\sum_s u\bar u = \slashed p + m$ , $\sum_s v\bar v = \slashed p - m$ let every spin-averaged matrix element be evaluated as a gamma-matrix trace.

The core insight

The Dirac equation fixes each spinor only up to an overall constant; choosing it so that $\bar u u = 2m$ and $u^\dagger u = 2E$ matches relativistic phase space, and the completeness relations $\sum_s u^{(s)}\bar u^{(s)} = \slashed p + m$ , $\sum_s v^{(s)}\bar v^{(s)} = \slashed p - m$ turn every spin-averaged matrix element into a trace of gamma matrices.

Where this lives today

The completeness relations are not a historical curiosity — they run inside every calculation of a scattering rate. The moment an experiment at the LHC or a lepton collider does not measure the spin of its particles (which is nearly always), the theorist must average over initial spins and sum over final ones, and Casimir's trick — $\overline{|\mathcal M|^2} = \tfrac{1}{4}\sum_{\text{spins}}|\mathcal M|^2 = \tfrac14\mathrm{Tr}[\cdots]$ built from $\sum u\bar u = \slashed p + m$ — is the only practical way to do it. Every textbook QED cross section (Møller, Bhabha, Compton, pair annihilation) and every parton-level QCD rate is a gamma-matrix trace generated this way, and symbolic engines such as FORM, FeynCalc, and the automated pipelines inside MadGraph, Sherpa, and WHIZARD implement exactly these spin sums and the Clifford-algebra trace identities to churn out $|\mathcal M|^2$ for processes with dozens of external legs. When physics is spin-dependent, the same projection-operator technology carries over: polarised-beam programmes at a future ILC or FCC-ee insert the spin projector $\tfrac12(1 + \gamma^5\slashed s)$ alongside $\Lambda_\pm$ to compute helicity-dependent asymmetries, and lattice and effective-theory calculations use the massless limit $\slashed p$ and the chirality projectors $\tfrac12(1\pm\gamma^5)$ that $\Lambda_\pm$ degenerate into. The tidy fact that four normalised spinors close into $\slashed p \pm m$ is, quietly, one of the most-executed pieces of mathematics in all of particle physics.

The physical picture

Picture the four solutions as four arrows spanning the four-dimensional space the Dirac matrices act on. In the particle's rest frame the picture is at its simplest: a positive-energy spinor is pure "upper block", $u = \sqrt{2m}\,\binom{\chi}{0}$ , and its two-component top $\chi$ is nothing but the ordinary Pauli spin-up or spin-down state. Boost the particle and the lower block fills in, tied to the upper one by $\boldsymbol\sigma\cdot\mathbf p/(E+m)$ ; the arrow tilts into the "small components", and its length as measured by $u^\dagger u$ grows from $2m$ to $2E$ . But one measure of the arrow does not change — the Lorentz-scalar combination $\bar u u = u^\dagger\gamma^0 u$ stays pinned at $2m$ in every frame. The first figure shows this split personality: a density that grows with energy, and an invariant that does not.

Figure

A rest-frame positive-energy spinor is pure upper block, u = √(2m)(χ, 0), with u†u = 2m. Under a boost to energy E the lower block fills in through σ·p ⁄ (E+m) and the density grows to u†u = 2E, yet the Lorentz scalar ū u stays fixed at 2m in every frame — the invariant that fixes the normalisation constant N = √(E+m).

The four spinors do more than sit there — together they close. Sum the two particle spinors over spin and you do not get the identity; you get a specific matrix, $\slashed p + m$ , that projects onto the positive-energy subspace and nothing else. The antiparticle pair sums to $\slashed p - m$ . These two completeness relations are the entire practical payoff of the chapter: they let an experimentalist who never measures a single spin still compute the cross section, because the messy sum over unobserved spin states collapses into a trace. The second figure makes the normalisation quantitative for a real electron — the invariant $\bar u u = 2m$ is a flat line, while the density $u^\dagger u = 2E$ climbs without bound.

Figure

The two spinor invariants for an electron (m = 0.511 MeV) versus momentum. The Lorentz scalar ū u = 2m = 1.022 MeV is flat, independent of frame; the density u†u = 2E = 2√(p² + m²) climbs without bound. At p = 0.885 MeV the electron has E = 2m, so u†u = 2.044 MeV while ū u is still 1.022 MeV.

The core equations

Substituting a plane wave $\psi = u(p)\,e^{-ip\cdot x}$ into $(i\slashed\partial - m)\psi = 0$ sends every derivative to $-ip_\mu$ , converting the differential equation into a purely algebraic one for the constant spinor. The antiparticle wave $\psi = v(p)\,e^{+ip\cdot x}$ carries the opposite sign in the exponent, flipping the mass term.

(\slashed p - m)\,u(p) = 0,\qquad (\slashed p + m)\,v(p) = 0,\qquad \slashed p \equiv \gamma^\mu p_\mu \qquad (1)

where: $u(p)$ — particle (positive-energy) spinor; $v(p)$ — antiparticle spinor; $\slashed p$ — the momentum contracted with the gamma matrices; $p^\mu = (E,\mathbf p)$ on shell with $E = +\sqrt{\mathbf p^2 + m^2}$ . Reading: this is an eigenvalue-type equation — it fixes the direction of $u$ in spinor space but not its length, leaving the constant to be set below. Solving equation (1) in the Dirac–Pauli representation, where $\gamma^0 = \mathrm{diag}(\mathbb 1,-\mathbb 1)$ , gives the explicit two-block forms. The upper and lower two-component blocks are locked together by $\boldsymbol\sigma\cdot\mathbf p$ , and a single prefactor $N = \sqrt{E+m}$ sets the scale.

u^{(s)}(p) = \sqrt{E+m}\begin{pmatrix} \chi^{s} \\[2pt] \dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\,\chi^{s}\end{pmatrix},\qquad v^{(s)}(p) = \sqrt{E+m}\begin{pmatrix} \dfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\,\eta^{s} \\[2pt] \eta^{s}\end{pmatrix} \qquad (2)

where: $\chi^{s},\eta^{s}$ — two-component Pauli spinors ( $s = 1,2$ ), e.g. $\chi^1 = \binom{1}{0}$ , $\chi^2 = \binom{0}{1}$ ; $\boldsymbol\sigma$ — Pauli matrices; the lower block of $u$ (and upper block of $v$ ) is the "small component", suppressed by $|\mathbf p|/(E+m)$ . Reading: at rest ( $\mathbf p = 0$ ) the small components vanish and $u = \sqrt{2m}\,\binom{\chi}{0}$ is a pure Pauli spinor. The normalisation is the whole point of the article. The overall constant is fixed so that the Lorentz-scalar bilinear equals $2m$ ; the Hermitian density then automatically equals $2E$ , matching the scalar-field convention of the previous article.

\bar u^{(r)}(p)\,u^{(s)}(p) = 2m\,\delta^{rs},\qquad u^{(r)\dagger}(p)\,u^{(s)}(p) = 2E\,\delta^{rs} \qquad (3)

where: $\bar u = u^\dagger\gamma^0$ — the Dirac adjoint; $\delta^{rs}$ — Kronecker delta enforcing orthogonality of the two spin states. Reading: $\bar u u$ is a Lorentz scalar and so is frame-independent at $2m$ ; $u^\dagger u = j^0$ is a density and so scales like the energy — the two facts are the spin-½ mirror of $\rho = 2E$ for the Klein–Gordon field. The antiparticle spinors obey the same $2E$ density but a sign-flipped scalar, and particle and antiparticle spinors at the same momentum are mutually orthogonal — the four together form a complete basis of the spinor space.

\bar v^{(r)} v^{(s)} = -2m\,\delta^{rs},\quad v^{(r)\dagger} v^{(s)} = 2E\,\delta^{rs},\quad \bar u^{(r)} v^{(s)} = \bar v^{(r)} u^{(s)} = 0 \qquad (4)

where: the minus sign on $\bar v v$ is not optional — it is forced by $(\slashed p + m)v = 0$ . Reading: $\bar u u > 0$ and $\bar v v < 0$ is how the scalar bilinear distinguishes matter from antimatter; the vanishing cross terms guarantee the four solutions are linearly independent. The completeness relations are the practical engine. Summing the outer product $u\bar u$ over the two spins gives not the identity but a matrix built purely from the momentum — the positive-energy content of $\slashed p + m$ .

\sum_{s=1,2} u^{(s)}(p)\,\bar u^{(s)}(p) = \slashed p + m,\qquad \sum_{s=1,2} v^{(s)}(p)\,\bar v^{(s)}(p) = \slashed p - m \qquad (5)

where: the left side is a $4\times4$ matrix (outer product $u\bar u$ ), unlike the scalar $\bar u u$ of equation (3); $m$ multiplies the $4\times4$ identity. Reading: these are the relations that make spin-averaged cross sections computable — a sum over unmeasured spins becomes a matrix you can take the trace of. Dividing the spin sums by $2m$ produces the energy projection operators, which single out the particle or antiparticle subspace and satisfy the defining properties of projectors.

\Lambda_\pm = \frac{\pm\slashed p + m}{2m},\qquad \Lambda_+ + \Lambda_- = \mathbb 1,\qquad \Lambda_\pm^2 = \Lambda_\pm,\qquad \Lambda_+\Lambda_- = 0 \qquad (6)

where: $\Lambda_+ = \tfrac{1}{2m}\sum_s u^{(s)}\bar u^{(s)}$ projects onto positive-energy (particle) states; $\Lambda_- = -\tfrac{1}{2m}\sum_s v^{(s)}\bar v^{(s)}$ projects onto antiparticle states. Reading: that $\Lambda_+ + \Lambda_- = \mathbb 1$ is the true "resolution of the identity" — the individual spin sum is only half of it.

Derivation AFixing the spinor normalisation constant

The goal is to prove that the single prefactor $N = \sqrt{E+m}$ in the explicit spinors of equation (2) is exactly what delivers both $\bar u u = 2m$ and $u^\dagger u = 2E$ , with the second following from the first with no extra freedom. We compute the two bilinears directly from the block form, using only the algebraic identity $(\boldsymbol\sigma\cdot\mathbf p)^2 = \mathbf p^2$ and the on-shell relation $\mathbf p^2 = E^2 - m^2$ , and show that the same $N$ serves the scalar and the density at once — which is why the two conventions are one convention.

Step 1. Compute the Hermitian density. With $u = \sqrt{E+m}\,\big(\chi,\ \tfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\chi\big)^{\mathsf T}$ and $\chi^\dagger\chi = 1$ , form $u^\dagger u = (E+m)\big[\chi^\dagger\chi + \chi^\dagger\tfrac{(\boldsymbol\sigma\cdot\mathbf p)^2}{(E+m)^2}\chi\big] = (E+m)\big[1 + \tfrac{\mathbf p^2}{(E+m)^2}\big]$ . Why: $u^\dagger u$ sums the squared magnitudes of all four components, and the Pauli identity $(\boldsymbol\sigma\cdot\mathbf p)^2 = \mathbf p^2\,\mathbb 1$ turns the lower block into a pure number.

Step 2. Collapse using the mass shell. Combine over a common denominator: $u^\dagger u = \tfrac{(E+m)^2 + \mathbf p^2}{E+m}$ , and substitute $\mathbf p^2 = E^2 - m^2$ so the numerator becomes $(E+m)^2 + E^2 - m^2 = 2E^2 + 2Em = 2E(E+m)$ . Hence $u^\dagger u = 2E$ . Why: the on-shell condition is what converts an expression full of $\mathbf p$ into the clean $2E$ — off shell there would be no such simplification.

Step 3. Compute the scalar. Insert $\gamma^0$ , which flips the sign of the lower block: $\bar u u = u^\dagger\gamma^0 u = (E+m)\big[1 - \tfrac{\mathbf p^2}{(E+m)^2}\big] = \tfrac{(E+m)^2 - \mathbf p^2}{E+m}$ . With $\mathbf p^2 = E^2 - m^2$ the numerator is $2Em + 2m^2 = 2m(E+m)$ , giving $\bar u u = 2m$ . Why: the single sign flip from $\gamma^0$ is the whole difference between the frame-dependent density $2E$ and the invariant scalar $2m$ — same spinor, two different bilinears.

Step 4. Read off the convention's logic. Both results came from one prefactor $N = \sqrt{E+m}$ ; had we instead demanded $\bar u u = 1$ we would take $N = \sqrt{(E+m)/2m}$ and be forced to reinsert factors of $2E$ by hand in every phase-space integral. Repeating Steps 1–3 for $v$ with the blocks swapped gives $v^\dagger v = 2E$ and $\bar v v = -2m$ , the sign flowing from $(\slashed p + m)v = 0$ . Why: choosing $\bar u u = 2m$ is precisely the choice that lets the spinor normalisation and the $2E$ scalar-field normalisation share one Lorentz-invariant phase space.

Go deeper · Derivation B · 1 minThe completeness relation as a sum over spins

Derivation BThe completeness relation as a sum over spins

The aim is to establish $\sum_s u^{(s)}(p)\,\bar u^{(s)}(p) = \slashed p + m$ — the outer-product spin sum, a $4\times4$ matrix, not to be confused with the scalar inner product of the previous derivation. We build the sum from the explicit block spinors, use the completeness of the Pauli two-spinors $\sum_s \chi^s\chi^{s\dagger} = \mathbb 1$ , and reassemble the four blocks into the covariant object $\slashed p + m$ . A trace check at the end confirms the normalisation is internally consistent with $\bar u u = 2m$ .

Step 1. Write the outer product block by block. Since $\bar u = u^\dagger\gamma^0 = (E+m)\big(\chi^\dagger,\ -\chi^\dagger\tfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}\big)$ , the product $u^{(s)}\bar u^{(s)}$ is a $2\times2$ array of two-component blocks, each carrying a factor $\chi^s\chi^{s\dagger}$ . Why: organising the $4\times4$ matrix into upper-left, upper-right, lower-left, lower-right blocks lets the Pauli-spinor sum act on each block independently.

Step 2. Sum over spin. Using $\sum_{s=1,2}\chi^s\chi^{s\dagger} = \mathbb 1$ , the four blocks become: upper-left $(E+m)\mathbb 1$ ; upper-right $-\boldsymbol\sigma\cdot\mathbf p$ ; lower-left $\boldsymbol\sigma\cdot\mathbf p$ ; lower-right $-\tfrac{(\boldsymbol\sigma\cdot\mathbf p)^2}{E+m} = -\tfrac{\mathbf p^2}{E+m} = -(E-m)$ . Why: the two Pauli spinors are complete in the two-dimensional spin space, so their sum erases all spin labels and leaves only momentum-dependent blocks.

Step 3. Recognise the gamma-matrix structure. Assembling the blocks, $\sum_s u\bar u = \begin{pmatrix} (E+m)\mathbb 1 & -\boldsymbol\sigma\cdot\mathbf p \\ \boldsymbol\sigma\cdot\mathbf p & -(E-m)\mathbb 1\end{pmatrix}$ . In the Dirac representation $\gamma^0 = \big(\begin{smallmatrix}\mathbb 1&0\\0&-\mathbb 1\end{smallmatrix}\big)$ and $\gamma^i = \big(\begin{smallmatrix}0&\sigma^i\\-\sigma^i&0\end{smallmatrix}\big)$ , so $E\gamma^0 - \mathbf p\cdot\boldsymbol\gamma + m\mathbb 1 = \slashed p + m$ reproduces exactly this matrix. Hence $\sum_s u^{(s)}\bar u^{(s)} = \slashed p + m$ . Why: the block matrix is not an accident — it is $\slashed p + m$ written out, so the spin sum is the covariant projector onto positive-energy states.

Step 4. Cross-check by trace and repeat for $v$ . Take the trace: $\mathrm{Tr}(\slashed p + m) = \mathrm{Tr}(\slashed p) + 4m = 0 + 4m = 4m$ , while $\mathrm{Tr}\sum_s u\bar u = \sum_s \bar u^{(s)} u^{(s)} = \sum_s 2m = 4m$ — they agree, tying completeness back to the $\bar u u = 2m$ normalisation. The identical computation with the blocks of $v$ swapped gives $\sum_s v^{(s)}\bar v^{(s)} = \slashed p - m$ . Why: the trace of an outer-product sum equals the sum of the inner products, so consistency between equations (3) and (5) is guaranteed by a single number, $4m$ .

Go deeper · Derivation C · 1 minEnergy projection operators and Casimir's trick

Derivation CEnergy projection operators and Casimir's trick

The purpose here is to show that the spin sums, divided by $2m$ , are genuine projection operators onto the particle and antiparticle subspaces, and then to demonstrate the single most important consequence — that completeness collapses a spin-averaged squared matrix element into a trace of gamma matrices, the technique known as Casimir's trick that underlies every unpolarised cross-section calculation. We verify the projector algebra directly from $\slashed p\,\slashed p = p^2 = m^2$ and then walk the sum-over-spins through a generic amplitude.

Step 1. Verify idempotency. With $\Lambda_+ = (\slashed p + m)/2m$ , compute $\Lambda_+^2 = \tfrac{(\slashed p + m)^2}{4m^2} = \tfrac{\slashed p\,\slashed p + 2m\slashed p + m^2}{4m^2}$ ; using $\slashed p\,\slashed p = p_\mu p_\nu\gamma^\mu\gamma^\nu = p^2 = m^2$ on shell, the numerator is $2m^2 + 2m\slashed p = 2m(\slashed p + m)$ , so $\Lambda_+^2 = (\slashed p + m)/2m = \Lambda_+$ . Why: $\slashed p\,\slashed p = p^2$ follows from the Clifford algebra $\{\gamma^\mu,\gamma^\nu\} = 2g^{\mu\nu}$ , and it is what makes $\Lambda_+$ idempotent — the hallmark of a projector.

Step 2. Confirm the pair resolves the identity. Add the two: $\Lambda_+ + \Lambda_- = \tfrac{(\slashed p + m) + (-\slashed p + m)}{2m} = \tfrac{2m}{2m} = \mathbb 1$ , and multiply: $\Lambda_+\Lambda_- = \tfrac{(\slashed p+m)(-\slashed p+m)}{4m^2} = \tfrac{m^2 - \slashed p\,\slashed p}{4m^2} = 0$ . Why: complementary, orthogonal, idempotent — the three properties say the four spinors split the space cleanly into a particle half and an antiparticle half, which is the precise meaning of "completeness".

Step 3. Set up a matrix element. A typical amplitude is $\mathcal M = \bar u(p')\,\Gamma\,u(p)$ for some string of gammas $\Gamma$ ; its modulus squared is $|\mathcal M|^2 = [\bar u(p')\Gamma u(p)]\,[\bar u(p')\Gamma u(p)]^{*} = \bar u(p')\Gamma u(p)\,\bar u(p)\bar\Gamma u(p')$ , where $\bar\Gamma = \gamma^0\Gamma^\dagger\gamma^0$ . Why: writing the complex conjugate as a second Dirac sandwich is what lets the two $u(p)$ factors sit next to each other, ready to be summed.

Step 4. Average and sum over spins. Summing over the final spin and averaging over the two initial spins, $\overline{|\mathcal M|^2} = \tfrac12\sum_{s,s'}\bar u^{s'}(p')\Gamma u^{s}(p)\,\bar u^{s}(p)\bar\Gamma u^{s'}(p')$ ; replacing $\sum_s u^s\bar u^s = \slashed p + m$ and $\sum_{s'} u^{s'}\bar u^{s'} = \slashed p' + m$ turns the spinor sandwiches into $\overline{|\mathcal M|^2} = \tfrac12\,\mathrm{Tr}\big[(\slashed p' + m)\,\Gamma\,(\slashed p + m)\,\bar\Gamma\big]$ . Why: this is Casimir's trick — the completeness relations remove every explicit spinor, leaving a trace of gamma matrices that gamma-algebra identities (or a computer) evaluate mechanically. Every unpolarised QED and QCD rate is computed this way.

Limiting cases

Case	Result	Meaning
Rest frame ( $\mathbf p = 0$ )	$u = \sqrt{2m}\,\binom{\chi}{0}$ , small components vanish	spinor is a pure Pauli two-spinor; $\bar u u = u^\dagger u = 2m$ coincide
Non-relativistic ( $\lvert\mathbf p\rvert \ll m$ )	lower block $\approx \tfrac{\boldsymbol\sigma\cdot\mathbf p}{2m}\,\chi$ , order $v/c$	"large" and "small" components; the Pauli equation and its $g = 2$ moment emerge
Ultra-relativistic ( $E \gg m$ )	upper and lower blocks equal in magnitude	helicity $\to$ chirality; $\bar u u = 2m$ negligible beside $u^\dagger u = 2E$
Massless ( $m \to 0$ )	$\sum_s u^{(s)}\bar u^{(s)} \to \slashed p$ ; $\bar u u \to 0$	Weyl spinors; $\Lambda_\pm$ degenerate and chirality projectors $\tfrac12(1\pm\gamma^5)$ take over
Spin sum vs identity	$\sum_s u\bar u = \slashed p + m \ne \mathbb 1$	the sum is $2m\Lambda_+$ , a projector onto particles, not the full completeness $\Lambda_+ + \Lambda_-$

Common trapThe completeness relation is a resolution of the identity, so the spin sum equals 1

It does not. The relation $\sum_{s} u^{(s)}(p)\,\bar u^{(s)}(p) = \slashed p + m$ is a sum of outer products — a $4\times4$ matrix — and it equals $2m\,\Lambda_+$ , the projector onto the positive-energy subspace, emphatically not the identity $\mathbb 1$ . Two distinct things get conflated here. First, do not confuse the outer product $u\bar u$ (a matrix) with the inner product $\bar u u = 2m$ (a scalar); they are different objects with different ranks, and only the latter is a normalisation. Second, the true resolution of the identity needs both particle and antiparticle sums: $\Lambda_+ + \Lambda_- = \tfrac{1}{2m}\big(\sum_s u\bar u - \sum_s v\bar v\big) = \mathbb 1$ . Drop the $v$ half and you have projected away the antiparticle subspace, which is exactly why the spin sum is a projector, not unity. (A related trap: the value $2m$ is a convention — some texts, notably older editions of Griffiths, normalise to $\bar u u = 1$ — but whichever you pick, the accompanying phase-space and flux factors must use the matching convention, or stray factors of $2E$ will corrupt the cross section.)

Worked examples

1 · An explicit electron spinor and its two norms (fully worked)

Build the spin-up positive-energy spinor for an electron ( $m = 0.511$ MeV) moving along $\hat z$ with energy $E = 2m$ , and verify $u^\dagger u = 2E$ and $\bar u u = 2m$ numerically.

Step 1. Fix the kinematics. With $E = 2m = 1.022$ MeV, the momentum is $|\mathbf p| = \sqrt{E^2 - m^2} = \sqrt{(2m)^2 - m^2} = \sqrt3\,m = \mathbf{0.885}$ MeV, so $\gamma = E/m = 2$ and $\beta = |\mathbf p|/E = \sqrt3/2 = 0.866$ . The prefactor is $N = \sqrt{E+m} = \sqrt{3m} = \sqrt{1.533} = \mathbf{1.238}\ \mathrm{MeV}^{1/2}$ , and the small-component ratio is $\tfrac{|\mathbf p|}{E+m} = \tfrac{\sqrt3 m}{3m} = \tfrac{1}{\sqrt3} = \mathbf{0.577}$ .

Step 2. Write the spinor. For spin up, $\chi = \binom{1}{0}$ , and $\boldsymbol\sigma\cdot\mathbf p = |\mathbf p|\,\sigma_z$ acts on $\chi$ to give $|\mathbf p|\binom{1}{0}$ . Thus

u^{(1)} = \sqrt{E+m}\begin{pmatrix} 1 \\ 0 \\ \tfrac{|\mathbf p|}{E+m} \\ 0\end{pmatrix} = 1.238\begin{pmatrix} 1 \\ 0 \\ 0.577 \\ 0\end{pmatrix} = \begin{pmatrix} 1.238 \\ 0 \\ 0.715 \\ 0\end{pmatrix}\ \mathrm{MeV}^{1/2}.

The third entry $0.715 = 1.238\times0.577$ is the "small component" — here not so small, because the electron is already relativistic.

Step 3. Verify both norms. The density is $u^\dagger u = 1.238^2\,(1 + 0.577^2) = 1.533\times(1 + 0.333) = 1.533\times1.333 = \mathbf{2.044}$ MeV, which equals $2E = 2\times1.022 = 2.044$ MeV. The scalar inserts $\gamma^0$ , flipping the sign of the lower block: $\bar u u = 1.533\times(1 - 0.333) = 1.533\times0.667 = \mathbf{1.022}$ MeV, which equals $2m = 2\times0.511 = 1.022$ MeV. Both check exactly: $u^\dagger u = 2E = 2.044$ MeV (frame-dependent) and $\bar u u = 2m = 1.022$ MeV (invariant), the two points read off the punchline figure at $p = 0.885$ MeV.

2 · Completeness two ways: the spin-summed $\lvert\bar u u\rvert^2$

Evaluate $\displaystyle S = \sum_{r,s=1,2}\big|\bar u^{(r)}(p)\,u^{(s)}(p)\big|^2$ first from orthonormality and then from the completeness relation, and confirm the two agree.

Step 1. Direct route via orthonormality. Equation (3) gives $\bar u^{(r)} u^{(s)} = 2m\,\delta^{rs}$ , so only the diagonal terms $r = s$ survive: $S = \sum_{r=s}(2m)^2 = 2\times(2m)^2 = \mathbf{8m^2}$ . The factor of two counts the two spin states. This uses nothing but the normalisation.

Step 2. Trace route via completeness. Write $|\bar u^{r} u^{s}|^2 = \bar u^{r} u^{s}\,\bar u^{s} u^{r}$ and sum: $S = \sum_{r,s}\bar u^{r} u^{s}\bar u^{s} u^{r} = \mathrm{Tr}\big[\big(\sum_s u^s\bar u^s\big)\big(\sum_r u^r\bar u^r\big)\big] = \mathrm{Tr}\big[(\slashed p + m)^2\big]$ . Expand: $(\slashed p + m)^2 = \slashed p\,\slashed p + 2m\slashed p + m^2 = m^2 + 2m\slashed p + m^2 = 2m^2 + 2m\slashed p$ . Take the trace, using $\mathrm{Tr}\,\mathbb 1 = 4$ and $\mathrm{Tr}\,\slashed p = 0$ : $S = \mathrm{Tr}(2m^2\mathbb 1) + 2m\,\mathrm{Tr}(\slashed p) = 8m^2 + 0 = \mathbf{8m^2}$ .

Step 3. Compare and read the lesson. Both routes give $S = 8m^2$ ; for the electron, $8m^2 = 8\times0.511^2 = \mathbf{2.09\ MeV^2}$ . The agreement is the whole point of Casimir's trick in miniature: the completeness relation reproduces, as a mechanical gamma-matrix trace, an answer you could also get by hand-summing orthonormal spinors — and for any realistic operator sandwiched between the spinors, only the trace route stays tractable.

Physics of the portrait

The spinor, its four solutions, and the negative-energy pair that became antimatter are one person's legacy. The completeness relation $\sum_s v\bar v = \slashed p - m$ is the modern, tidy descendant of an idea that once seemed like a disaster: an infinite sea of filled negative-energy states.

Origins · 1 minPaul Dirac and how this was discovered

Paul Dirac

1902–1984 · the relativistic electron and the birth of antimatter

Paul Adrien Maurice Dirac was an English theoretical physicist of famously few words and unmatched formal instinct, who in 1928 set out to find a relativistic wave equation that was first order in time — repairing the negative-probability defect of the Klein–Gordon equation. Demanding that his operator square to $E^2 = \mathbf p^2 + m^2$ forced the coefficients to be anticommuting matrices, the gamma matrices, and forced the wavefunction to have four components: the Dirac spinor. Two of its four plane-wave solutions had positive energy and carried spin up and down — the electron, with the correct magnetic moment $g = 2$ falling out for free, a triumph nobody had engineered. But two solutions stubbornly had negative energy. Rather than discard them, Dirac proposed in 1930 that the vacuum is a filled sea of negative-energy states, and that a hole in that sea would appear as a positive-energy particle of opposite charge. In 1931 he predicted this "anti-electron", and in 1932 Carl Anderson found the positron in cosmic-ray cloud-chamber tracks — one of the great predictive triumphs of theoretical physics, and the reason the $v$ -spinors of this article carry the label "antiparticle". The Dirac-sea picture is now superseded by quantum field theory, where the negative-energy solutions are simply antiparticle creation operators and the completeness relation $\sum_s v^{(s)}\bar v^{(s)} = \slashed p - m$ encodes their spin content cleanly, but the physics Dirac extracted from an apparent embarrassment — that every charged fermion has an antiparticle — remains exact. He shared the 1933 Nobel Prize with Schrödinger, and held Newton's Lucasian chair at Cambridge; his credo, that a physical law must possess mathematical beauty, is nowhere better vindicated than in the equation that bears his name.

History

1928

— Dirac writes his first-order relativistic wave equation and its four-component spinor solutions

1930

— Dirac proposes the filled negative-energy sea and the hole interpretation

1931

— Dirac predicts an anti-electron of equal mass and opposite charge

1932

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

1949

— Feynman and Dyson's covariant perturbation theory makes the spin sums u ū = p-slash + m and Casimir's trace trick routine

Going further

Texts. Thomson, Modern Particle Physics, §4.6–4.7 (solutions of the Dirac equation, spin sums and completeness) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.3 (Dirac-field spinors, $\bar u u = 2m$ , $\sum u\bar u = \slashed p + m$ ) · Halzen & Martin, Quarks and Leptons, §5.3–5.4 (normalisation and projection operators) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.3 and Casimir's trick in §7.7 · Schwartz, Quantum Field Theory and the Standard Model, §11.1 · Bjorken & Drell, Relativistic Quantum Mechanics, §3.2 (energy projection operators). The key results to keep in mind: $N = \sqrt{E+m}$ gives $u^\dagger u = 2E$ and $\bar u u = 2m$ , with the electron numbers $E = 2m = 1.022$ MeV, $|\mathbf p| = 0.885$ MeV, small component $0.577$ ; the spin sum $\sum_s u\bar u = \slashed p + m$ holds block by block with trace $4m$ ; the projectors satisfy $\Lambda_\pm^2 = \Lambda_\pm$ , $\Lambda_+ + \Lambda_- = \mathbb 1$ , and $\Lambda_+\Lambda_- = 0$ ; and the spin-summed check gives $\sum_{r,s}|\bar u^r u^s|^2 = 8m^2 = 2.09\ \mathrm{MeV}^2$ , recoverable both directly from orthonormality and by the gamma-matrix trace route.

Prerequisites. Dirac equation · Gamma matrices and the Clifford algebra · Dirac solutions · Spin from the Dirac equation. Leads to. Spin and helicity states · Antiparticles · Matrix elements and spin averaging · Dirac adjoint and bilinear covariants.

Ask the tutor: Ask me to run Casimir's trick end to end on a real process — say electron–muon scattering — and show exactly where $\sum_s u\bar u = \slashed p + m$ turns the spin sum into the trace that gives the cross section.

Why normalise to ū u = 2m rather than the tidier ū u = 1?

Because the spinor normalisation cannot be chosen in isolation — it has to mesh with the way the whole scattering machinery counts particles. The relativistic convention sets " $2E$ particles per unit volume" for scalar fields, which is what makes Lorentz-invariant phase space and the flux factor come out clean, and $u^\dagger u = 2E$ is the spin-½ version of exactly that density. Its Lorentz-scalar partner is then forced to be $\bar u u = 2m$ . If you instead normalise to $\bar u u = 1$ (as some older texts do), nothing is wrong, but you must divide each external leg's phase-space factor by $2E$ by hand to compensate — you have only moved the $2E$ elsewhere. Keeping $\bar u u = 2m$ lets one invariant phase space serve particles of every spin, which is why Peskin, Thomson, Halzen–Martin, and Schwartz all adopt it.

What is the difference between ū u and u† u?

They are different bilinears of the same spinor and they transform differently. The combination $u^\dagger u$ is the time component of the probability current $j^\mu = \bar u\gamma^\mu u$ , i.e. a density; it is not Lorentz invariant, and it equals $2E$ , growing with the particle's energy exactly as a density must under length contraction. The combination $\bar u u = u^\dagger\gamma^0 u$ inserts a $\gamma^0$ and is a genuine Lorentz scalar; it equals $2m$ in every frame. The single factor of $\gamma^0$ — which flips the sign of the lower two components — is the whole difference between the frame-dependent $2E$ and the invariant $2m$ . In figures and formulas, whenever a quantity must be the same to all observers (like a mass term in a Lagrangian) it is built from $\bar u u$ ; whenever it is a counting density it is built from $u^\dagger u$ .

Why does the spin sum give a matrix, but the normalisation gives a number?

Because they multiply the spinors in opposite orders. The normalisation $\bar u u$ is an inner product: a row spinor $\bar u$ times a column spinor $u$ contracts all four components into a single number, $2m$ . The completeness sum $\sum_s u^{(s)}\bar u^{(s)}$ is an outer product: a column times a row produces a full $4\times4$ matrix, and summing the two spin states gives $\slashed p + m$ . Both use the same spinors; one contracts them, the other spreads them into a matrix. The inner product answers "how big is this spinor?", the outer product answers "what subspace do these spinors span?" — which is why the first is a normalisation and the second is completeness.

What do the u and v spinors mean physically?

The $u$ spinors are the two positive-energy solutions and describe a spin-½ particle (say, an electron) with spin up or down; the $v$ spinors are the two solutions attached to the $e^{+ip\cdot x}$ plane wave and describe its antiparticle (a positron). Historically the $v$ spinors are Dirac's negative-energy solutions reinterpreted: rather than a filled sea with a hole, modern quantum field theory treats them as antiparticle states outright. The tell-tale sign is the scalar bilinear — $\bar u u = +2m$ for particles, $\bar v v = -2m$ for antiparticles — so the sign of $\bar\psi\psi$ literally distinguishes matter from antimatter. Both carry the same positive density $u^\dagger u = v^\dagger v = 2E$ , because both describe real, positive-energy quanta once the theory is quantised.

How exactly does completeness make a cross section computable?

Through Casimir's trick. A matrix element is a Dirac sandwich $\bar u(p')\Gamma u(p)$ ; its square, summed over the spins nobody measured, contains $\sum_s u^{(s)}(p)\bar u^{(s)}(p)$ and $\sum_{s'} u^{(s')}(p')\bar u^{(s')}(p')$ . Replacing each sum by $\slashed p + m$ (or $\slashed p' + m$ ) eliminates every explicit spinor and leaves $\overline{|\mathcal M|^2} = \tfrac14\mathrm{Tr}[(\slashed p' + m)\Gamma(\slashed p + m)\bar\Gamma]$ — a trace of gamma matrices. Traces of gamma-matrix strings are evaluated by a short list of Clifford-algebra identities (trace of an odd number vanishes; trace of two is $4g^{\mu\nu}$ ; and so on), so what looked like an intractable sum over four or eight spin combinations becomes a mechanical algebraic exercise. Without completeness you would have to build every spinor explicitly and sum by brute force; with it, the spins are gone before you start.

Is the choice of the Pauli spinors χ¹, χ² physical?

No — it is a choice of basis, exactly like choosing the $z$ -axis to label ordinary spin. The Dirac equation determines a two-dimensional space of positive-energy solutions at each momentum, and any orthonormal pair of two-component spinors $\chi^1,\chi^2$ picks a basis for it. The conventional choice $\chi^1 = \binom{1}{0}$ , $\chi^2 = \binom{0}{1}$ labels spin along $z$ , but a helicity basis (spin along $\mathbf p$ ) or any rotated pair is equally valid; a general physical state is a linear combination. What is not a choice is the completeness relation: $\sum_s u^{(s)}\bar u^{(s)} = \slashed p + m$ holds for any orthonormal basis, because the sum over a complete basis is basis-independent. So the individual spinors are conventional, but everything physical — cross sections, the projector $\slashed p + m$ — is not.

Why does the antiparticle scalar ū u come out negative, and is that a problem?

It is not a problem; it is a feature that labels antimatter. The sign is forced by the defining equation: $u$ satisfies $(\slashed p - m)u = 0$ while $v$ satisfies $(\slashed p + m)v = 0$ , and running the normalisation computation with that opposite-sign mass term flips the scalar to $\bar v v = -2m$ . Nothing negative happens to a probability — the density $v^\dagger v = 2E$ is perfectly positive, so there is always a positive number of antiparticles per unit volume. Only the Lorentz-scalar bilinear changes sign, and that sign is precisely what makes $\bar\psi\psi$ a useful order parameter: a mass term $m\bar\psi\psi$ in the Lagrangian treats particles and antiparticles symmetrically because the sign flip is compensated by the way $v$ enters the field expansion. The negative $\bar v v$ is the harmless bookkeeping trace of Dirac's negative-energy solutions, now understood as antiparticles.

Spinor normalisation & completeness

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

1 · An explicit electron spinor and its two norms (fully worked)

2 · Completeness two ways: the spin-summed $\lvert\bar u u\rvert^2$

Physics of the portrait

Going further

Before this

Where this leads

Spinor normalisation & completeness

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

1 · An explicit electron spinor and its two norms (fully worked)

2 · Completeness two ways: the spin-summed ∣uˉu∣2\lvert\bar u u\rvert^2∣uˉu∣2

Physics of the portrait

Going further

Before this

Where this leads

2 · Completeness two ways: the spin-summed $\lvert\bar u u\rvert^2$