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For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

The operator $b^s(p)$ is a fermionic annihilation operator for a particle with momentum $\mathbf{p}$ and spin label $s$; $d^{s†}(p)$ is a fermionic creation operator for the corresponding antiparticle. The c-number Dirac spinors $u^s(p)$ and $v^s(p)$ solve the positive-energy particle and antiparticle equations respectively. They supply the spinor wavefunctions in the mode expansion of a Dirac field; they are not creation or annihilation operators.

Apply the specified parity transformation to the two operators in the mode expansion of a Dirac field:Relabel $k=p_P$. The momentum sum is unchanged, $p\cdot x=k\cdot x_P$, and the spinor identities give $u^s(p)={\gamma }^0{u}^s(k)$ and $v^s(p)=-{\gamma }^0{v}^s(k)$. The two minus signs in the antiparticle term cancel. ThereforeTaking the Hermitian conjugate and using $({\gamma }^0{)}^{†}={\gamma }^0$, $({\gamma }^0{)}^2=1$ in the Dirac adjoint yieldsThe unit-modulus intrinsic-parity phase cancels from all neutral fermion bilinears.

Use the printed gamma matrix conventions throughout, in particular ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. The transpose identity for ${\gamma }^0$ gives $C{\gamma }^0=-{\gamma }^{0}C$, and its Hermitian conjugate gives the corresponding identity for $C^{†}$. Thus the Dirac adjoint of the charge-conjugated field isReordering the anticommuting components of the fermion bilinear supplies a minus sign:Invariance for arbitrary fields therefore requires $C^{†}C=I$, so the charge-conjugation matrix is a unitary matrix. This is the required normalization constraint; it is not a bosonic reordering calculation.

For a spinor matrix $\Gamma$, the same calculation and the given $C^{-1}=C^T$ giveWith $V^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ and $A^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$, the supplied transpose identities implyConsequently the vector current is C-odd and the axial current is C-even: $V^{\mu }\mapsto -V^{\mu }$ and $A^{\mu }\mapsto A^{\mu }$.

For parity, put $\Lambda =\mathrm{diag}(1,-1,-1,-1)$. The gamma matrix identities ${\gamma }^0{\gamma }^{\mu }{\gamma }^0={\Lambda }^{\mu }{}_{\nu }{\gamma }^{\nu }$ and ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$ giveThus $V^0$ is P-even and $\mathbf{V}$ P-odd, whereas $A^0$ is P-odd and $\mathbf{A}$ P-even. The axial current transforms as a pseudovector.

Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Taking the Hermitian conjugation of the massive Dirac equation and using the Dirac adjoint givesTaking the matrix transpose and multiplying by the charge conjugation matrix $C$ yieldsThus the charge conjugation of a Dirac field preserves the free massive Dirac equation:The unitarity of $C$ and ${\gamma }^0{C}^{-1}{\gamma }^0=-C^{-1}{\gamma }^{0T}$ also giveConsequentlyHere $\hat{C}$ is the unitary operator acting on the relativistic quantum field, whereas $C$ acts on Dirac spinor indices.

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

Use $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and restore $\hslash$ temporarily. The free Dirac equation is $(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$, with $c=1$. Its Hermitian conjugate is $-i\hslash ({\partial }_{\mu }{\psi }^{†})({\gamma }^{\mu }{)}^{†}-m{\psi }^{†}=0$. Multiplying on the right by ${\gamma }^0$ and using $({\gamma }^{\mu }{)}^{†}{\gamma }^0={\gamma }^0{\gamma }^{\mu }$ gives the adjoint equationThe Dirac adjoint and original equations give $({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }\psi =im\bar{\psi }\psi /\hslash$ and $\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\psi =-im\bar{\psi }\psi /\hslash$. Adding gives the continuity equationThe Dirac current transforms as a four-vector; the positive density is the time component, not a Lorentz scalar. With ordinary units the number current is $c\bar{\psi }{\gamma }^{\mu }\psi$ and the electron charge current is $-e$ times it.

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.

With the gamma matrices derived in a, the momentum-space Dirac equation readsFor a positive-energy particle at rest, $E=m>0$ and $\mathbf{p}=0$, so $\xi =\eta$. The Dirac adjoint in this basis is $\bar{u}=({\eta }^{†},{\xi }^{†})$, and the specified normalization givesTherefore, for any normalized two-component spinor $\chi$,Equivalently, choosing ${\chi }_{+}=(1,0{)}^{\mathsf{T}}$ and ${\chi }_{-}=(0,1{)}^{\mathsf{T}}$ gives the two independent rest-frame spin states. A strictly massless particle has no rest frame; this step requires $m>0$.

The chiral projectors $P_{L,R}=(1\mp {\gamma }^5)/2$ obey ${\gamma }^{\mu }{P}_L=P_R{\gamma }^{\mu }$, while the Dirac adjoints are ${\bar{\psi }}_L=\bar{\psi }{P}_R$ and ${\bar{\psi }}_R=\bar{\psi }{P}_L$. Consequently the cross terms vanish and the vector current isThe vector interaction preserves fermion chirality at a vertex and couples equally to left and right components. This statement about the vertex does not assert that a fermion mass term preserves chirality. Under parity, left and right components are exchanged; their equal coefficients allow a parity-invariant contraction with a true vector field.

An axial vector must couple to the axial current to obtain the analogous parity-invariant interaction. Since ${\gamma }^5{\psi }_L=-{\psi }_L$ and ${\gamma }^5{\psi }_R={\psi }_R$,The relative sign changes; the overall sign is conventional. A pure axial interaction can itself conserve parity when $A_{\mu }$ transforms as an axial vector. Simply contracting the original vector current with an axial-vector field would instead produce a parity-odd interaction.

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 $g=\mathrm{diag}(1,-1,-1,-1)$.

With $\eta =\mathrm{diag}(1,-1)$, the Clifford algebra gives $({\gamma }^0{)}^2=1$, $({\gamma }^1{)}^2=-1$, and ${\gamma }^0{\gamma }^1=-{\gamma }^1{\gamma }^0$. Henceso $\{{\gamma }^5,{\gamma }^0\}=\{{\gamma }^5,{\gamma }^1\}=0$. Its square isChoose the following gamma matrices, expressed using the Pauli matrices:These satisfy $({\gamma }^0{)}^{†}={\gamma }^0$, $({\gamma }^1{)}^{†}=-{\gamma }^1$ and $({\gamma }^5{)}^{†}={\gamma }^5$. The chirality eigenvalues are therefore $\boxed{+1,-1}$, and we order the spinor as $\psi =({\psi }_R,{\psi }_L{)}^{\mathsf{T}}$ throughout. Its Dirac adjoint is $\bar{\psi }=({\psi }_L^{†},{\psi }_R^{†})$. No additional factor of $i$ multiplies ${\gamma }^0{\gamma }^1$ in this spacetime dimension. This realizes a Dirac field in two spacetime dimensions.

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 $+1$.

In the convention of iii, the invariant decay amplitude is ${\mathcal{M}}_{s_1{s}_2}=-g{\bar{u}}_{\chi }^{s_2}(p_2)u_{\Psi }^{s_1}(p_1)$. Including the common overall $i$ in the definition changes no squared decay amplitude. An unpolarized ensemble is an incoherent equal mixture of the two initial spin states; hence take a spin average over $s_1$ and a spin sum over the unobserved final $s_2$. Different final spin states are orthogonal outcomes, so their probabilities, rather than their decay amplitudes, add.

Using the Dirac adjoint and the fermion spin sum givesThe gamma-matrix trace identities used here are $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$. The mass-shell conditions $p_1^2=M^2$ and $p_2^2=p_3^2=m^2$, together with four-momentum conservation, implyThereforeIn the parent's rest frame this also says $E_2=E_3=M/2$.

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

Put $x_P=(t,-x,y)$ and $U_P={\gamma }^1$. Since $U_P^{†}=-{\gamma }^1$ and ${\gamma }^0{\gamma }^1{\gamma }^0=-{\gamma }^1$, the Dirac adjoint transforms asThe chain rule gives ${\partial }_{\mu }\psi (x_P)=r_{\mu }({\partial }_{\mu }\psi )(x_P)$, with $(r_0,r_1,r_2)=(1,-1,1)$ and no sum in this formula. The Clifford algebra givesMultiplication by $r_{\mu }$ therefore restores each kinetic term:However, the Dirac scalar bilinear changes sign because $({\gamma }^1{)}^2=-I_2$:Changing variables in the action has unit absolute Jacobian determinant. HenceA nonzero Dirac mass term for this single two-component field breaks parity symmetry in quantum field theory with the specified transformation.

The specified time reversal includes complex conjugating $i{\gamma }^{\mu }{\partial }_{\mu }$; omitting that operation changes the answer. Put $x_T=(-t,x,y)$, $U_T={\gamma }^2$, and $(r_0,r_1,r_2)=(-1,1,1)$. As in c, the Dirac adjoint in the source's transformed bilinear is ${\bar{\psi }}^{\prime }(x)=\bar{\psi }(x_T){\gamma }^2$. The conjugated Dirac operator has coefficient matricesCombining these coefficients with the chain rule signs and the Clifford algebra gives, without summing over $\mu$,For example the time coefficient is ${\gamma }^2(-i{\gamma }^0){\gamma }^2(-1)=i{\gamma }^0$, and the second spatial coefficient is ${\gamma }^2(-i{\gamma }^2){\gamma }^2=i{\gamma }^2$. Thus the kinetic term is invariant. The real Dirac mass term again reverses sign since ${\bar{\psi }}^{\prime }{\psi }^{\prime }=\bar{\psi }({\gamma }^2{)}^2\psi =-\bar{\psi }\psi$. The spacetime integration has unit absolute Jacobian determinant, soEach operation reverses the sign of the mass, while their composition reverses it twice and preserves the kinetic term. In an operator formulation the complex conjugation of coefficients expresses the antiunitary operator character of time reversal; the transformation must not be treated as an ordinary unitary substitution with all $i$'s held fixed. These statements concern the stated free action.