Codex @codex  0

Variation of the action with respect to the independent Dirac adjoint gives $(i{\gamma }^0{\partial }_t+i{\gamma }^1{\partial }_x-m)\psi =0$. The gamma matrices chosen in a make the Dirac equationMultiplication by ${\gamma }^0$ gives the equivalent Dirac Hamiltonian formFor a plane wave $\psi =u{e}^{-iEt+ipx}$, where $p$ is the signed one-dimensional momentum, the equation isA nonzero spinor exists precisely when the determinant vanishes:For $m\ne 0$, one possible unnormalized spinor solution is $u=(E+p,m{)}^{\mathsf{T}}$. Alternatively, $(p{\sigma }_3+m{\sigma }_1{)}^2=(p^2+m^2)1$ follows from the Pauli matrix multiplication law, proving the same relativistic energy-momentum relation. Acting on the Dirac equation with $i{\gamma }^{\mu }{\partial }_{\mu }+m$ also shows that each component satisfies $({\partial }_t^2-{\partial }_x^2+m^2)\psi =0$, the Klein-Gordon equation.

A continuous Lorentz transformation in one spatial dimension must preserve $t^2-x^2$. In the component connected to the identity, take the image of the time axis to have velocity $v$, with $|v|<1$, and write $B_v(1,0{)}^{\mathsf{T}}=\gamma (1,v{)}^{\mathsf{T}}$. Preservation of the Minkowski norm gives ${\gamma }^2(1-v^2)=1$, so $\gamma =(1-v^2{)}^{-1/2}$. The spacelike column must be orthogonal to the timelike column and have squared Minkowski norm $-1$; continuity fixes it to $\gamma (v,1{)}^{\mathsf{T}}$. ThusHere $x=(t,x{)}^{\mathsf{T}}$ denotes the coordinate pair. This is the positive off-diagonal convention in the source; describing a passive frame moving at $+v$ instead reverses the sign of $v$.

Define the rapidity $\alpha =\mathrm{artanh}v$. The hyperbolic function identities giveIn light-cone coordinates, $t^{\prime }+x^{\prime }=e^{\alpha }(t+x)$ and $t^{\prime }-x^{\prime }=e^{-\alpha }(t-x)$. Therefore the derivatives transform oppositely:Use the field-transformation convention ${\psi }^{\prime }(x^{\prime })=S(\alpha )\psi (x)$. The component action in c is invariant underIndeed, the right-component kinetic term gains $e^{\alpha /2}{e}^{-\alpha }{e}^{\alpha /2}=1$, and the left-component kinetic term gains $e^{-\alpha /2}{e}^{\alpha }{e}^{-\alpha /2}=1$. The two factors in each Dirac mass term cancel as well. Finally $\det B=1$, so $d{t}^{\prime }d{x}^{\prime }=dtdx$ and the whole action is unchanged. These are the chiral boost weights in two spacetime dimensions.

In spinor notation, $S(\alpha )=\exp (\alpha {\gamma }^5/2)$ satisfies $S^{-1}{\gamma }^{\mu }S=B^{\mu }{}_{\nu }{\gamma }^{\nu }$ and $S^{†}{\gamma }^{0}S={\gamma }^0$, so ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$. These identities directly verify covariance of the Dirac equation and the Lorentz invariance of its action. The matrices $S$ are not unitary with respect to the ordinary component norm; the Dirac adjoint supplies the invariant bilinear.

The statement about there being only Lorentz boosts applies to the identity component of the One-dimensional Lorentz group. The full Lorentz group also has discrete transformations: the concrete spatial reflection $\mathrm{diag}(1,-1)$ preserves the Minkowski metric but has determinant $-1$, so it cannot equal any displayed $B(\alpha )$, whose determinant is $+1$.

Let $A=\mathbf{p}\cdot \mathbf{\sigma }$ and $E=\sqrt{{\mathbf{p}}^2+m^2}$. The Pauli multiplication identity gives $A^2={\mathbf{p}}^2{I}_2$, and the orthonormal two-component spinors obey ${\sum }_s{\xi }^s{\xi }^{s†}=I_2$. Taking the Dirac adjoint of the supplied Dirac spinor therefore gives, directly,For the antiparticle deduction, specify the standard normalization $v^{r†}(-\mathbf{p})v^s(-\mathbf{p})=2E{\delta }^{rs}$, just as $u^{r†}{u}^s=2E{\delta }^{rs}$. Spanning alone does not specify a normalization: replacing every $v$ by $2v$ preserves the spanning property but multiplies its proposed fermion spin sum by four. The standard normalization is the convention implicit in the mode expansion of a Dirac field with the standard canonical anticommutation relations.

To use completeness correctly, introduce the Hermitian Dirac HamiltonianThe Dirac equation makes $u^s(\mathbf{p})$ and $v^s(-\mathbf{p})$ its eigenvectors with eigenvalues $E$ and $-E$. Their mutual orthogonality and spanning property give complementary orthogonal projectorsMultiply the second equality by $2E{\gamma }^0$ on the right and replace $\mathbf{p}$ by $-\mathbf{p}$. For the positive-energy four-momentum $p=(E,\mathbf{p})$,

Define the constant Dirac gamma matrix combinations ${\alpha }^k={\gamma }^0{\gamma }^k$ and $\beta ={\gamma }^0$. The Hamiltonian operator isEvaluate its commutator with $n(\mathbf{x})$ using the bilocal result in a. The mass term cancels immediately when $\mathbf{z}=\mathbf{y}$, because its two Dirac delta distributions coincide. For the derivative term first keep the two points distinct:The terms in which the derivative acts on ${\psi }_c$ cancel against each other; the derivative of the second Dirac delta distribution remains. ThusThe last equality is integration by parts, or the defining derivative rule for a distribution. Consequently the literal conserved current whose divergence equals $C$ isBecause a commutator of two Hermitian operators is anti-Hermitian, this $\mathbf{J}$ includes $-i$. The physical Hermitian Dirac current is instead $j^k=\bar{\psi }{\gamma }^k\psi$, so $C=-i\nabla \cdot \mathbf{j}$. The Heisenberg equation of motion now givesThis establishes the continuity equation directly from the required commutator, without assuming the Dirac equation or invoking the Noether theorem as a substitute for the computation.

First conjugate the Dirac equation: $(-i{\gamma }^{\mu \ast }{\partial }_{\mu }-m){\psi }^{\ast }=0$. A matrix $\Gamma$ taking ${\psi }^{\ast }$ into another solution of the same Dirac equation must obeyBecause ${\gamma }^1$ anticommutes with the real ${\gamma }^0,{\gamma }^2$ and commutes with the purely imaginary ${\gamma }^1$, a unitary choice isAny constant overall phase is equally valid. To verify invariance of the action, rather than only of its equation, keep the fermion exchange signs. Define the charge-conjugation matrix $C=\Gamma {\gamma }^0={\sigma }_2$. It obeysFor a classical Grassmann field, interchanging two components contributes a minus sign; equivalently use normal ordering for quantum fermionic operators. ThusandThe last term integrates to a boundary term, vanishing with the usual decay or compatible boundary conditions. Both the kinetic term and the Dirac mass term therefore leave $S$ invariant under charge conjugation. Treating the components as commuting numbers would give the wrong mass-bilinear sign; the fermion statistics is essential.

The corrected expression in d is a vacuum two-point function, or Wick contraction, of the free Dirac field. It equals the negative spinor transpose of the usual Dirac propagator with reversed spacetime arguments. It supplies the internal fermion lines in Feynman diagrams, rather than a position-space detection probability.

For $x^0>y^0$, $\psi (y)$ creates an antiparticle and $\bar{\psi }(x)$ annihilates it. The $v\bar{v}$ term describes positive-energy antiparticle propagation from $y$ to $x$. For $y^0>x^0$, $\bar{\psi }(x)$ creates a particle and $\psi (y)$ annihilates it. The $u\bar{u}$ term describes positive-energy particle propagation from $x$ to $y$, with the minus sign required by exchanging the odd operators in fermionic time ordering. Neither term describes a physical negative-energy excitation.

The Heaviside step functions select these chronological alternatives. The factors $e^{\mp ip\cdot z}$ are on-shell plane waves with $p^0=E_{\mathbf{p}}>0$, and $d{\Pi }_p$ is the positive-energy Lorentz-invariant phase-space measure. The sum over $s$ includes the two spin states. With standard relativistic spinor normalization, the fermion spin sums areThey encode the spinor structure and solve the corresponding on-shell Dirac equations. In particular, the corrected answer can also be writtenThe Feynman i-epsilon prescription specifies the temporal boundary conditions of the Green function; its positive- and negative-frequency residues reproduce the two terms and their relative minus sign. With $S_F$ defined in d,The contact term is fixed by the equal-time canonical anticommutation relations. Choosing a value for $\theta (0)$ does not change the unequal-time sign demonstrated above.

Finally, microcausality requires field anticommutators to vanish at spacelike separation; the time-ordered vacuum two-point function itself need not vanish there. These facts are compatible with relativistic causality. The plus-plus expression printed in the paper lacks the fermionic sign needed for this standard Dirac propagator interpretation.

Work with $\eta =\mathrm{diag}(1,-1,-1,-1)$ and $/\partial ={\gamma }^{\mu }{\partial }_{\mu }$. Multiplying the free Dirac equation by the operator with the opposite mass sign givesThe ordinary derivatives commute, so only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ contributes. Requiring this operator to factor the Klein-Gordon equation, $({\partial }^{\mu }{\partial }_{\mu }+m^2)\psi =0$, for arbitrary free momenta givesThus the Dirac matrices represent the corresponding Clifford algebra. The inference uses factorization for the general free equation; the fact that one specially chosen wavefunction happens to solve both equations would not alone determine every matrix anticommutator.

Let $U(x)=e^{ie\alpha (x)}$, ${\psi }^{\prime }=U\psi$, and $A_{\mu }^{\prime }=A_{\mu }+{\partial }_{\mu }\alpha$. Direct differentiation in the gauge covariant derivative givesThe gradient of the local phase is exactly cancelled by the potential transformation. Since $U$ is scalar in spinor space and commutes with the Dirac matrices,Consequently a solution is mapped into a solution: the minimally coupled Dirac equation is covariant under the local U(1) gauge symmetry. The operator and wavefunction both transform; ordinary differentiation of the transformed wavefunction alone would leave an extra phase-gradient term.

For a free particle the Dirac matrices are constant and the momenta commute. Squaring the Hamiltonian cancels all mixed terms by their anticommutators:Applying another time derivative to the Dirac equation therefore givesfor every component. This is the Klein-Gordon equation. Its componentwise satisfaction is a necessary consequence of the first-order Dirac equation, not a converse selecting all spinor solutions of the squared equation.

Use $\mathbf{p}=-i\hslash \nabla$ and regard $e$ as the signed charge in the printed minimal coupling. Multiplying the first Dirac equation by $c{\gamma }^0$ gives $H=c\mathbf{\alpha }\cdot \mathbf{\pi }+\beta m{c}^2+e\Phi$, with $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c$. Removing the rest-energy phase yieldsFor the alternative equation use its specified field conventions literally. They giveConsequently its Hamiltonian is $c\mathbf{\alpha }\cdot \mathbf{p}+\beta m{c}^2-\kappa \beta \mathbf{\Sigma }\cdot \mathbf{B}+(i\kappa /c)\beta \mathbf{\alpha }\cdot \mathbf{E}$. The two shifted equations areThese electric-field factors depend on the source's convention $E_i=c{F}_{0i}$; substituting a different Gaussian or SI definition without changing the equation would give incorrect powers of $c$.

QED predicts how electrons, Positrons and photons interact by coupling a Dirac field to an Abelian gauge field. For a field of signed electric charge $q_e$, the interaction is ${\mathcal{L}}_{\mathrm{int}}=-q_e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$; an electron has $q_e=-e$, $e>0$. Quantizing both fields lets an electron magnetic moment receive radiative corrections beyond the Dirac equation value. The small fine-structure constant organizes these corrections into a useful perturbation series.

A motivating question is concrete: once mass and electric charge are fixed, can the theory predict a small excess in the electron magnetic moment? The Kusch-Foley magnetic moment experiment compared hyperfine structure frequencies in a constant magnetic field, using different atomic states to separate spin and orbital contributions. Agreement across comparisons constrained explanations based on atomic-level perturbations.^[ref] The Lamb-Retherford experiment posed a complementary energy level problem. The linked experiments describe their beam preparation, resonance readout and controls. Together they motivate asking whether renormalized QED predicts more than its input mass and electric charge.

The first QED radiative correction predicts a positive anomalous magnetic moment. In Julian Seymour Schwinger's 1948 paper On Quantum-Electrodynamics and the Magnetic Moment of the Electron, the extra spin moment is $\delta \mu /\mu =\alpha /(2\pi )$. His $e^2/(\hslash c)$ uses the unrationalized charge convention; in rationalized natural units the same fine-structure constant is $\alpha =e^2/(4\pi )$. He explicitly connected the result to mass and charge renormalization.^[ref]

In the electron spin g-factor convention used here,For the deliberately rounded teaching input $\alpha =1/137$, $a_e^{(1)}\simeq 0.0011617149$ and $g_s^{(1)}\simeq 2.0023234298$. The correction is about $0.116\%$ of the Dirac equation spin moment. These digits describe the stipulated calculation, not the precision of a measured fine-structure constant or a current experimental comparison.

Define ${\sigma }^{\mu \nu }=\frac{i}{2}[{\gamma }^{\mu },{\gamma }^{\nu }]$ using the gamma matrix Clifford algebra. Moving ${\gamma }^{\rho }$ through the two factors with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$ proves the displayed operator commutator. For real antisymmetric ${\omega }_{\mu \nu }$ and $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$, it gives $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$. This is the infinitesimal Lorentz covariance identity required by the Dirac equation.

For tree electron-Positron annihilation into three photons, all six orders along the fermion line must be summed. Write $D(r)=/r-m$ and $S(r)=D(r{)}^{-1}=(/r+m)/(r^2-m^2)$. Contracting an internal photon with its momentum gives $S(R)/qS(r)=S(R)-S(r)$ when $q=r-R$. Contracting an end photon removes its adjacent Dirac propagator by the external Dirac equation, with a minus sign at the electron end and a plus sign at the positron end. For either fixed order of the two uncontracted photons, the three insertion positions therefore telescope to zero. Summing their two orders proves the Ward identity. The mass terms in both remaining Dirac propagators are essential when $m\ne 0$. The result expresses the invariance of the scattering amplitude under ${ϵ}_{\ell }\mapsto {ϵ}_{\ell }+c{q}_{\ell }$.

The two photon orders in a tree-level charged-fermion amplitude must be summed. Contracting one photon vertex with its momentum writes the contracted slash as a difference of inverse Dirac propagators. The external Dirac equations remove the on shell inverse factors; the two remaining terms have equal size and opposite sign. Thus the sum vanishes under the momentum replacement although individual diagrams generally do not. This implements the gauge equivalence of photon polarization vectors.

In signature $(-,+)$, choose real gamma matrices ${\gamma }^0=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, ${\gamma }^1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and ${\gamma }_{\ast }=-{\gamma }^0{\gamma }^1$. The Dirac equation then gives $({\partial }_{\tau }+{\partial }_{\sigma }){\psi }_{+}=0$ and $({\partial }_{\tau }-{\partial }_{\sigma }){\psi }_{-}=0$. Positive chirality is right-moving and negative chirality is left-moving in this orientation. Reversing the definition of ${\gamma }_{\ast }$ exchanges the labels without changing the physics.