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In the mode expansion of a Dirac field, $b^s(p)$ is the fermionic annihilation operator for a particle of four-momentum $p$ and spin angular momentum label $s$, while $d^{s†}(p)$ is the fermionic creation operator for the corresponding antiparticle. The Dirac spinor $u^s(p)$ is the positive-frequency particle wavefunction and $v^s(p)$ is the negative-frequency antiparticle wavefunction; they solve $(/p-m)u^s=0$ and $(/p+m)v^s=0$, respectively.

Apply the parity symmetry in quantum field theory to the given mode expansion of a Dirac field:The stated Dirac spinor identities are equivalently $u^s(p)={\gamma }^0{u}^s(p_P)$ and $v^s(p)=-{\gamma }^0{v}^s(p_P)$. Relabel the momentum sum by $p\mapsto p_P$ and use $p_P\cdot x=p\cdot x_P$ to obtainThe phase $|{\eta }_P|=1$ is the intrinsic parity convention.

For $x^{\prime }=\Lambda x$, a Dirac spinor transforms asHenceUsing the result of part i, the Dirac adjoint transforms as

The Dirac spinor representation is the four-dimensional representation $(\frac{1}{2},0)\oplus (0,\frac{1}{2})$ of the connected Lorentz group. It is characterized byForone may takewith the signs of ${\Omega }_{\rho \sigma }$ fixed by the displayed vector-representation convention. This formula follows by differentiating the covariance relation and using the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$; moving both gamma indices upstairs changes the apparent parameter signs according to the metric.

Use the Weyl representation of the gamma matrices,

For $x^{\prime }=\Lambda x$, a Dirac spinor and vector field transform aswhere $S^{-1}{\gamma }^{\mu }S={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$. Hence the spinor terms and $F_{\mu \nu }{F}^{\mu \nu }$ are Lorentz scalars. The Maxwell term is real. Integration by parts and $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$ show that the adjoint of $i\bar{\psi }{\gamma }^{\mu }{D}_{\mu }\psi$ differs from it by $-i{\partial }_{\mu }(\bar{\psi }{\gamma }^{\mu }\psi )$. Finally, $i[{\gamma }^{\mu },{\gamma }^{\nu }]$ is Dirac-Hermitian, so the real coefficient $\lambda$ makes the Pauli bilinear real. Thus the action is real.

A coordinate transformation $x\mapsto Px$ conjugates an infinitesimal generator, hence $M^{\mu \nu }\mapsto P{M}^{\mu \nu }P$. It fixes the rotation generators $J_i$ and negates the boosts $K_i$. Therefore it exchanges $A_i$ and $B_i$, and the parity action on a Lorentz representation isAn irreducible representation is parity invariant precisely when $j_L=j_R$. If $j_L\ne j_R$, parity invariance requires the reducible sum $(j_L,j_R)\oplus (j_R,j_L)$. Thus the two Weyl-spinor representations $(1/2,0)$ and $(0,1/2)$ are exchanged, while their direct sum is the parity-invariant Dirac spinor.

A projection operator satisfies $P^2=P$; an orthogonal projection operator also has $P^{†}=P$. It extracts a subspace while leaving already projected states unchanged. Since $({\gamma }^5{)}^2=1$ and $({\gamma }^5{)}^{†}={\gamma }^5$, the chiral projectors areThey separate the Dirac spinor into opposite fermion chirality components. The Standard Model weak charged current contains $\bar{\psi }{\gamma }^{\mu }{P}_L\psi$, selecting left-chiral particle fields and the corresponding right-helicity massless antiparticles; right-chiral charged fermions are weak-isospin singlets. The weak neutral current can couple to both fermion chirality components with different strengths. Chirality and helicity should not be equated for a massive fermion: a Lorentz boost can reverse its helicity, while the field representation defining chirality is a different notion.

The Dirac adjoint is $\bar{u}=u^{†}{\gamma }^0$. Because ${\gamma }^5$ is Hermitian and anticommutes with ${\gamma }^0$,Anticommutation with ${\gamma }^{\mu }$ then givesThus the left-chiral charged current can be written with its chiral projector acting on either external Dirac spinor. More precisely, ${\bar{u}}_A{\gamma }^{\mu }{P}_L{u}_B=\overline{P_L{u}_A}{\gamma }^{\mu }{P}_L{u}_B$, since the opposite projection vanishes. The Dirac adjoint reverses the matrix projector, $\overline{P_{L}u}=\bar{u}{P}_R$; overlooking this reversal would give a wrong zero. The source uses $1-{\gamma }^5=2P_L$, so no additional factor of two is introduced when moving that entire factor.

For nonzero momentum, physical helicity is the spin projection along the direction of motion,The convention without $1/2$ labels the same two eigenspaces by $\pm 1$. Write $\hat{\mathbf{p}}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$. Normalized two-component eigenvectors areEach can carry an arbitrary phase. Before imposing an energy equation, the general four-component helicity eigenvector is $(a{\chi }_{\pm },b{\chi }_{\pm }{)}^{\mathsf{T}}$, with arbitrary complex $a,b$; unit Norm imposes $|a{|}^2+|b{|}^2=1$. For a positive-energy free electron, the Dirac equation fixes their ratio and givesMultiplying by an arbitrary overall amplitude and the common plane wave supplies the corresponding free helicity eigenstate. Helicity has no preferred direction at $\mathbf{p}=0$. For a massive particle a boost can overtake its momentum and change its helicity; it should not be identified with Lorentz-invariant chirality.

An incoming fermion contributes a Dirac spinor $u_s(p)$ and an outgoing fermion contributes its Dirac adjoint ${\bar{u}}_s(p)=u_s^{†}(p){\gamma }^0$. The label $s$ specifies a spin state; $p$ is the on-shell four-momentum. Incoming and outgoing antifermions instead contribute ${\bar{v}}_s(p)$ and $v_s(p)$.

The exchanged boson is a virtual photon, represented in the Feynman rules by the photon propagator $-i{g}_{\mu \nu }/(q^2+iϵ)$ in Feynman gauge. Its momentum transfer is $q=p_1-p_3=p_4-p_2$; unlike an external real photon, it need not satisfy $q^2=0$. Each Feynman vertex contracts a gamma matrix with this propagator and contributes $-ie{\gamma }^{\mu }$ in a conventional QED sign convention. The coupling constant $e$ sets the interaction strength, with fine-structure constant $\alpha =e^2/(4\pi )$. One may reverse the sign of the photon field and both vertex signs without changing the result. The gamma matrices obey $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }I$, with Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$.

There is one tree-level Feynman diagram: the distinguishable electron and muon exchange a photon in the $t$ channel. There is no identical-particle exchange diagram.

Multiplying the external Dirac spinors, the two Feynman vertices and the photon propagator givesThe overall phase of this scattering amplitude is convention dependent. Four external spin labels with two choices each give $2^4=16$ possible configurations before taking a limit. For $E\gg m,M$, the vector QED interaction preserves fermion chirality, which becomes helicity for each external particle. Thus the surviving configurations, written as $(h_e,h_{\mu }{)}_{\mathrm{in}}\to (h_e,h_{\mu }{)}_{\mathrm{out}}$, areHere $L,R$ specify helicity along each particle's own momentum. The other configurations require at least one mass-suppressed flip and vanish in the strictly massless limit.

With all four external fermions of species 1, both final attachments are possible. The middle row of the figure gives the two tree-level Feynman diagrams:

For a fixed ordering of external fermion states, the fermionic sign from exchanging the final identical fermions givesHere $u_j=u_{s_j}(p_j)$ are Dirac spinors. In particular, the scattering amplitude changes sign when the complete final labels $3$ and $4$ are exchanged, as required for identical fermions.

The momentum space Feynman rules use the scalar propagator and two Dirac propagatorsThe cubic scalar field interaction vertex is $-i\lambda$. Each of the two flavour-changing Yukawa coupling interaction vertices, $\varphi \bar{\Psi }\chi$ and $\varphi \bar{\chi }\Psi$, has factor $-ig{\delta }_{ab}$ in Dirac spinor indices. A fermion flow arrow continues through each interaction vertex, but its species changes. There is no $\varphi \bar{\Psi }\Psi$ or $\varphi \bar{\chi }\chi$ interaction vertex.

Every interaction vertex carries $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ with inward four-momenta. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, divide by the Feynman-diagram symmetry factor, and include a minus sign and a gamma-matrix trace for each closed fermion loop. The $i0$ is the Feynman i-epsilon prescription.

For an amputated scattering amplitude, incoming fermions supply $u^s(p)$ and outgoing fermions supply ${\bar{u}}^s(p)$; incoming antifermions supply ${\bar{v}}^s(p)$ and outgoing antifermions supply $v^s(p)$. An external real scalar field particle supplies $1$. Use the standard relativistic normalization of a one-particle state; the external factors $1/\sqrt{2E}$ belong to that normalization and the Lorentz-invariant phase-space measure, rather than an additional factor in the invariant scattering amplitude.

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 $p=(E,\mathbf{p})$,

Take a spatial Fourier transform of the mode expansion of a Dirac field:The orthogonality relations $u^{s†}(\mathbf{p})u^r(\mathbf{p})=2E{\delta }^{sr}$ and $u^{s†}(\mathbf{p})v^r(-\mathbf{p})=0$ isolate the fermionic annihilation operator. Projecting onto $v^s(-\mathbf{p})$ instead isolates the fermionic creation operator:The opposite time phases and reversed antiparticle momentum are essential. Both denominators are $\sqrt{2E}$, because the Dirac spinors have norm squared $2E$.

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15088w1.pdf

Use natural units $\hslash =c=1$ and the mostly-minus Minkowski metric. Question 3 has two spatial dimensions: its gamma matrices are the specified two-by-two matrices. The printed ${\sigma }^{\mu }$ in its kinetic term is interpreted as those ${\gamma }^{\mu }$, rather than as three unmodified Pauli matrices. All spin sums in question 2 instead use four-component Dirac spinors in four spacetime dimensions.

Every positive-energy Dirac spinor obeys $(/p-m)u^s(\mathbf{p})=0$. Therefore the fermion spin sum satisfiesSince ${\bar{u}}^s=u^{s†}{\gamma }^0$ and $({\gamma }^0{)}^2=I$, the trace of each outer product is its contracted scalar. The prescribed normalization and the two spin states yieldNow make exactly the stated polynomial assumption and apply polynomial reduction of a slashed vector. By a, on the mass shell $p^2=m^2$ write $P=A/p+BI$. ThenThe matrices $I$ and $/p$ are linearly independent for a physical nonzero $p$, so $B=mA$. The gamma matrix trace identities, $\mathrm{tr}{\gamma }^0=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^0)=4{\eta }^{\mu 0}$, giveThus $A=1$ andNo explicit representation of the Dirac spinors has been used. The same reasoning covers $m=0$ provided $p^0>0$.

Using $\epsilon =\hslash {\omega }_c(n+1/2)$ gives the charge-explicit resultFor $B>0$ the two spin signs are equivalently $m^2{c}^4+2m{c}^2[p_z^2/(2m)+\mu B(2n+1)\pm \mu B]$. Both positive and negative relativistic energy branches arise from this squared equation; an actual Dirac spinor must still be reconstructed from the coupled equations, especially at exceptional values where division by $E+m{c}^2$ is unavailable.

The QED Feynman rules combine the Dirac propagator, photon propagator, and vertex $-ie{\gamma }^{\mu }$ with external Dirac spinors and photon polarization vectors. In a covariant gauge they generate perturbative electron-photon scattering amplitudes.

For a Dirac spinor a spatial rotation acts in each chiral block by a matrix exponential $e^{\pm i\theta \mathbf{n}\cdot \mathbf{\sigma }/2}$, depending on angle orientation. Since $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I_2$ for a unit axis, the Pauli matrix multiplication law gives this exponential as $\cos (\theta /2)I_2\pm i\sin (\theta /2)\mathbf{n}\cdot \mathbf{\sigma }$. At $\theta =2\pi$ it is $-I_2$, and at $4\pi$ it is $I_2$. A vector returns already at $2\pi$. This describes the lift of the rotation loop to the Spin group, rather than a single-valued representation of the ordinary rotation group on spinors.