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One adjoint Dirac spinor contributes the displayed screening term to the four-dimensional renormalization-group beta function in the modified minimal subtraction scheme. The fermion contribution to the gluon self-energy insertion has pole $-8g^2{C}_A(k^2{\delta }_{\mu \nu }-k_{\mu }{k}_{\nu })/(3ϵ(4\pi {)}^2)$. The corresponding background gauge-field counterterm is $\delta Z_A=-8g^2{C}_A/(3ϵ(4\pi {)}^2)$; the background gauge-field renormalization identity $Z_g=Z_A^{-1/2}$ gives $g_0={\mu }^{ϵ/2}(g+4C_A{g}^3/[3ϵ(4\pi {)}^2])$. Differentiation at fixed $g_0$ yields the screening coefficient. The mass-independent subtraction includes massive fields until effective-theory threshold matching is performed.

Choose the negative-frequency Dirac spinors as the charge conjugates of the positive-frequency spinors. Transposing $\bar{u}(/p-m)=0$ and using the charge-conjugation matrix identity gives $(/p+m)C{\bar{u}}^T=0$, so the displayed choice solves the correct antiparticle equation. In the mode expansion of a Dirac field, the unitary charge-conjugation operator then exchanges particle and antiparticle annihilation operators, with conjugate phases on their creation operators. A different spinor basis can insert conventional phases or a spin-index relabeling; the displayed equality specifies a compatible choice.

For signature $+---$, use ${\gamma }^0=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ and ${\gamma }^i=\left(\begin{array}{cc}0& {\sigma }^i\\ -{\sigma }^i& 0\end{array}\right)$, where ${\sigma }^i$ are Pauli matrices. Their Clifford algebra follows directly from the Pauli anticommutator. The chirality matrix is ${\gamma }^5=\mathrm{diag}(-I_2,I_2)$, separating the two chiral components of a Dirac spinor.

For $g=\mathrm{diag}(1,-1,-1,-1)$ and the block-diagonal standard representation of the gamma matrices, write a positive-energy Dirac spinor as $(\xi ,\eta {)}^T$. The Dirac equation gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$. Choosing two orthonormal two-spinors ${\chi }_s$ yields the displayed positive-frequency solutions. Negative-frequency solutions are $v_s=\sqrt{E+m}((\mathbf{\sigma }\cdot \mathbf{p}){\eta }_s/(E+m),{\eta }_s{)}^T$ multiplying $e^{ip\cdot x}$. They obey $(/p+m)v_s=0$. Both have norm $2E$; their Dirac adjoints give ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$ and ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$. Spatial rotations act on each two-spinor with generators ${\sigma }_j/2$, explaining spin one-half.

For mode normalization $1/\sqrt{2E_{\mathbf{p}}}$, particle and antiparticle operators satisfy $\{a_{\mathbf{p}}^s,a_{\mathbf{q}}^{r†}\}=\{b_{\mathbf{p}}^s,b_{\mathbf{q}}^{r†}\}=(2\pi {)}^3{\delta }^{sr}{\delta }^3(\mathbf{p}-\mathbf{q})$. The Dirac spinor completeness relations imply ${\sum }_s[u_s(\mathbf{p})u_s^{†}(\mathbf{p})+v_s(-\mathbf{p})v_s^{†}(-\mathbf{p})]=2E_{\mathbf{p}}I$. Inserting this into the mode expansion of a Dirac field gives the displayed local anticommutator. The reversed momentum in the antiparticle completeness term is essential.

A local expression containing two fermion fields and a matrix acting on Dirac spinor indices. The choices $\Gamma ={\gamma }^{\mu }$ and $\Gamma ={\gamma }^{\mu }{\gamma }^5$ give the Dirac current and axial current.

The arrow on a charged-fermion line in a Feynman diagram tracks fermion-number flow and fixes the order of the gamma matrices between external Dirac spinors. For an external particle it agrees with the particle direction; for an external antiparticle it is opposite to the antiparticle direction. It therefore need not point along the assigned four-momentum of every external leg.

The Gordon identity rewrites a vector-current matrix element between equal-mass on-shell Dirac spinors using a momentum term and an antisymmetric spin term. In the mostly-minus convention $\{{\Gamma }^a,{\Gamma }^b\}=2{\eta }^{ab}$, $(/p-m)u=0$, and ${\Gamma }^{ab}=[{\Gamma }^a,{\Gamma }^b]/2$, it reads $2m{\bar{u}}^{\prime }{\Gamma }^{a}u={\bar{u}}^{\prime }[(p^{\prime }+p{)}^a-{\Gamma }^{ab}(p_b^{\prime }-p_b)]u$. Derive it by adding the two external Dirac equation identities. The equivalent formula in the mostly-plus Dirac convention is ${\bar{u}}^{\prime }[{\gamma }^{ab}(p_b^{\prime }-p_b)+2m{\gamma }^a+p^a+p^{\prime a}]u=0$. The identity before division by $m$ remains valid for massless equal-mass states.

For a non-Abelian simple gauge factor with $C_A>0$, pure Yang-Mills theory contributes $-11C_A{g}^3/[3(4\pi {)}^2]$. Adding $N_D$ adjoint Dirac spinors changes this to $-C_A(11-4N_D)g^3/[3(4\pi {)}^2]$. Vanishing would require $N_D=11/4$, which is not a nonnegative integer. This excludes cancellation by these fermions alone, apart from the trivial value $g=0$; other matter representations or fields change the coefficient.

Boost a massive rest Dirac spinor labelled by $\xi =(1,1{)}^T/\sqrt{2}$ along the third axis and take its massless limit. In the Weyl representation the result is the displayed equal superposition of helicities, not an eigenvector of the fixed transverse rotation generator ${\Sigma }_1/2$. For momentum $(E,0,0,E)$ the positive-energy Hamiltonian is $E{\gamma }^0{\gamma }^3$, which anticommutes with ${\Sigma }_1=\mathrm{diag}({\sigma }_1,{\sigma }_1)$. A nonzero positive-energy solution cannot also be a ${\Sigma }_1$ eigenvector. Thus a transported canonical spin label must be distinguished from literal transverse spin at zero mass; longitudinal helicity remains a good quantum number.

In the Dirac representation of the gamma matrices, the parity symmetry in quantum field theory action ${\psi }_P(x)={\gamma }^0\psi (x_P)$ and the charge conjugation action ${\psi }_C=i{\gamma }^2{\psi }^{\ast }$ obey $({\psi }_P{)}_C(x)=-{\gamma }^0{\psi }_C(x_P)$, since ${\gamma }^0$ is real and anticommutes with ${\gamma }^2$. At rest, positive-frequency Dirac spinors have nonzero upper components and negative-frequency Dirac spinors have nonzero lower components; they are respectively $+1$ and $-1$ eigenvectors of ${\gamma }^0$. With the common conventional real intrinsic-parity phase, particle and antiparticle therefore have opposite intrinsic parity. The overall sign is conventional, whereas their relative sign is fixed.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

Vary the Maxwell Lagrangian with respect to $A_{\nu }$. Since $\delta F_{\mu \nu }={\partial }_{\mu }\delta A_{\nu }-{\partial }_{\nu }\delta A_{\mu }$ and $F$ is antisymmetric,After integration by parts, the Euler-Lagrange field equation isThe gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. This gauge invariance means that potentials related in this way represent the same electromagnetic field; all four potential components are not independent physical photon degrees of freedom. In particular, the ungauged kinetic operator is singular and its temporal canonical momentum vanishes.

Choose the Lorenz gauge ${\partial }_{\mu }{A}^{\mu }=0$. The field equation then becomes $\boxed{\square A_{\mu }=0}$. Residual gauge transformations with $\square \chi =0$ preserve this condition. The correctly named Lorenz gauge condition concerns the divergence of the potential; Lorentz covariance is the separate transformation property of the equations.

For the alternative quadratic Lagrangian density, direct differentiation givesso its Euler-Lagrange field equation is also $\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction. Indeed, with $D={\partial }_{\mu }{A}^{\mu }$,The divergence identity follows by expanding ${\partial }_{\mu }{K}^{\mu }$ and commuting its second derivatives. The quadratic density alone describes four wave equations; the gauge constraint must still be incorporated in the physical states. It is not an unconstrained four-positive-polarization replacement for Maxwell equations.

Using this specified boundary-term convention, the canonical momentum conjugate to $A_{\nu }$ isThe canonical commutation relations are thereforewith field-field and momentum-momentum brackets zero. Equivalently $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The temporal component has the opposite sign to the spatial components; replacing the metric by a positive Euclidean one would destroy these brackets.

The wave equation gives massless frequencies ${\omega }_{\mathbf{k}}=|\mathbf{k}|$. Hermiticity pairs its positive- and negative-frequency solutions, so the Heisenberg picture field has the expansionThis is a real field expansion for each component, but its oscillator metric is indefinite. The assumed brackets, with the dagger's vector index restored, areThey verify the photon oscillator completeness and canonical brackets. Inserting ${\Pi }^{\nu }=-{\dot{A}}^{\nu }$ gives at equal timesThe other brackets cancel between the two frequency contributions. Thus both the $2\omega$ normalization and the sign $-g_{\mu \nu }$ are needed.

The Gupta-Bleuler formalism imposes the Lorenz gauge condition weakly on states:Here $(+)$ is the annihilation, positive-frequency part. Setting the full divergence to zero as an operator would conflict with the four-component canonical commutation relations. The subsidiary condition instead selects a physical pre-space inside the covariant photon Fock space, followed by a quotient of its null states.

Apply it to $|k,ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(k)|0⟩$. Commuting $k^{\mu }{a}_{\mu }$ through the creator gives a multiple of $-k_{\nu }{ϵ}^{\nu }|0⟩$. Hence the allowed photon polarization vectors obeyTake $k=(\omega ,0,0,\omega )$. The complete solution iswith complex coefficients. The last two vectors are transverse linear photon polarizations, and $(0,1,\pm i,0)/\sqrt{2}$ gives the two circular photon polarizations. The first vector is $k/\omega$, the allowed temporal-longitudinal combination. A purely spatial longitudinal vector $(0,0,0,1)$ by itself does not satisfy the subsidiary condition.

The oscillator brackets give, apart from the positive continuum normalization factor,Momentum eigenstates are distributionally normalized; a box or a wave packet makes the norm statement literal. The allowed longitudinal combination has zero norm because $k^2=0$, and it is orthogonal to every constrained state since $k\cdot ϵ=0$. It also gives a pure-gauge field wave: the field strength of $A_{\mu }\propto k_{\mu }{e}^{-ik\cdot x}$ vanishes. Thus $ϵ$ and $ϵ+ck$ represent the same physical photon polarization. The Gupta-Bleuler null-state quotient removes this null direction and leaves two positive-norm transverse photon photon polarizations. The negative-norm temporal oscillator of the unrestricted space is not an extra physical state. This is the transverse one-photon physical quotient.

Finally verify the two-photon fermion Ward identity directly for electron-positron annihilation into two photons. Define $D(r)=/r-m$ and the reduced Dirac propagatorThe equality follows from the Clifford algebra, since $(/r-m)(/r+m)=(r^2-m^2)I$. The incoming Dirac spinors obey $D(p)u(p)=0$ and $\overline{v}(q)(/q+m)=0$. On replacing one inserted photon polarization vector by $k$, the spinor factor in the two diagrams isAt the electron end, $/k=D(p)-D(p-k)$, soFor the other diagram, four-momentum conservation gives $p-k^{\prime }=k-q$. HenceThe two contributions therefore cancel exactly:Individual diagrams need not vanish; their sum implements the Ward identity. Linearity then gives invariance under $ϵ\mapsto ϵ+ck$, as required by gauge invariance. In particular, the null longitudinal photon polarization proportional to $k$ has zero physical production amplitude. Physical scattering produces the transverse photon classes, not an independent longitudinal massless photon.

For a massless supermultiplet, put the momentum along the third axis. The matrix $2\sigma \cdot p$ in the supersymmetry algebra has rank one, so only one fermionic oscillator acts nontrivially. It relates states whose helicities differ by $\frac{1}{2}$. Including the antiparticles required by CPT completion of a supermultiplet gives the following complete spectra.

A chiral multiplet contains a complex scalar $A$ and a Weyl spinor $\psi$. Its scalar particle and antiparticle have helicity $0$; its fermion particle and antiparticle have helicities $-\frac{1}{2}$ and $+\frac{1}{2}$. Therefore $n_B=2=n_F$, with maximum absolute helicity $\frac{1}{2}$.

A supersymmetric vector multiplet contains a massless gauge vector $A_{\mu }$, with helicities $+1,-1$, and a Majorana spinor gaugino $\lambda$, with helicities $+\frac{1}{2},-\frac{1}{2}$. Again $n_B=2=n_F$, now with maximum absolute helicity $1$. The chiral $F$ and vector $D$ auxiliary fields contribute no on-shell states.

The Higgs mechanism is compatible with unbroken supersymmetry. A supersymmetric vacuum may have nonzero scalar vacuum expectation values that break the internal gauge symmetry while obeying both F-flatness and D-flatness. For example, oppositely charged chiral superfields with $W=0$, no Fayet–Iliopoulos term and equal nonzero scalar magnitudes have $F=D=0$ but break the Abelian gauge group. In the supersymmetric Higgs mechanism, a massless vector multiplet combines with the chiral multiplet along the broken gauge direction. One real scalar becomes the longitudinal vector polarization; the other remains a physical real scalar, and the gaugino combines with the Higgsino into a Dirac spinor. The resulting massive vector multiplet has $3+1=4$ bosonic and $4$ fermionic states with a common mass, as in part (i). Any other uneaten chiral multiplets remain separately in the spectrum.

The irreducible massive N=1 vector multiplet consists of one massive vector, one real scalar, and two massive spin-$\frac{1}{2}$ fermions. Thus its on-shell boson and fermion counts areThe two fermions can be written as two Majorana spinors or one Dirac spinor. All these states have the same mass when supersymmetry is unbroken; auxiliary fields are not extra propagating states.

To obtain rather than assume this spectrum, use the rest-frame supersymmetry algebra $\{Q_{\alpha },Q_{\beta }^{†}\}=2M{\delta }_{\alpha \beta }$. The two normalized fermionic creation operators form a spin-$\frac{1}{2}$ doublet. Start with a spin-$\frac{1}{2}$ fermionic Clifford vacuum annihilated by the two lowering operators. The zero- and two-creation-operator levels each have spin $\frac{1}{2}$, whereas the one-creation-operator level hasIt is bosonic, giving exactly the scalar and vector above. A higher-spin Clifford vacuum would introduce spin above $1$, while a spin-zero Clifford vacuum gives maximum spin only $\frac{1}{2}$.

Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and Feynman slash notation $/p={\gamma }^{\mu }{p}_{\mu }$. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$. In the Dirac representation of the gamma matrices, direct block multiplication givesandThus the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, proving the Clifford relationsSince partial derivatives commute, their symmetric product selects the symmetric part of the gamma matrices. Multiplying the Dirac equation by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesHence every spinor component satisfies the Klein-Gordon equation with the same mass.

For parity, let $x_P=(t,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_j\psi (x_P)=-({\partial }_j\psi )(x_P)$. Combining these signs with ${\gamma }^j{\gamma }^0=-{\gamma }^0{\gamma }^j$ provesThis establishes parity symmetry in quantum field theory for the free Dirac equation, rather than merely asserting its covariance.

A positive-energy plane wave $\psi =u(p)e^{-ip\cdot x}$ solves $(/p-m)u=0$. Write $u=(\chi ,\eta {)}^T$. The two block equations areThe second gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\chi /(E_p+m)$. In the first, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$ and $E_p^2-{\mathbf{p}}^2=m^2$ make the remaining coefficient zero. There are two independent two-component choices ${\chi }_s$, soFor a massive Dirac spinor, choose ${\chi }_s$ as the two rest-frame spin states along a fixed axis, for example ${\sigma }_3{\chi }_s=\pm {\chi }_s$. The label $s$ distinguishes these canonical spin states after boosting; it is not automatically a helicity label.

Now $p\cdot x_P=p_P\cdot x$ and ${\gamma }^0{u}_s(p)=u_s(p_P)$ for the same fixed ${\chi }_s$. ThereforeThe parity action on canonical massive spin states leaves the canonical spin label unchanged while reversing the momentum. Spin is an axial vector under parity; its orientation is unchanged. Consequently helicity, the projection of spin along momentum, reverses. If $s$ had instead been chosen to denote helicity, the transformed label would have the opposite sign.

The Dirac adjoint is $\bar{\psi }={\psi }^{†}{\gamma }^0$, with ${\gamma }^{0†}={\gamma }^0$. ThusThe chirality matrix has ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ in this representation, so $({\gamma }^5{)}^2=I_4$ and it anticommutes with every ${\gamma }^{\mu }$. In particular ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. The parity transformation of a Dirac bilinear therefore yieldsThe first bilinear is a scalar under parity, whereas the second is a pseudoscalar.

For the requested gamma matrix trace identities, use the cyclic property of the trace. Conjugation of a product of $n$ gamma matrices by ${\gamma }^5$ multiplies it by $(-1{)}^n$, but leaves its trace unchanged. All odd products therefore have zero trace. Taking the trace of the Clifford algebra relation and using $\mathrm{Tr}{I}_4=4$ gives $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. For four factors, move the first gamma matrix successively past the next three using the anticommutators. If $A=\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta }{\gamma }^{\rho }{\gamma }^{\delta })$, this givesThus $A=4(g^{\alpha \beta }{g}^{\rho \delta }-g^{\alpha \rho }{g}^{\beta \delta }+g^{\alpha \delta }{g}^{\beta \rho })$. Contracting with the relevant four-vectors proves all four answers:

Use the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplying the block gamma matrices and using the Pauli matrix multiplication law givesThus these matrices satisfy the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$. Since partial derivatives commute, the antisymmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ drops out of its contraction with ${\partial }_{\mu }{\partial }_{\nu }$. Multiplying the Dirac equation on the left by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesTherefore every component of the Dirac spinor obeys the Klein-Gordon equation with mass $m$.

For Lorentz covariance, take $x^{\prime }=Lx$ and ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$, with constant $L$ and $S$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. The given intertwining relation then impliesThis proves invariance of the Dirac equation under the joint coordinate and spinor transformation.

The infinitesimal identity follows directly from the spinor Lorentz generator commutator. To derive it, first move ${\gamma }^{\rho }$ through a two-matrix product using the Clifford algebra:Subtract the expression with $\mu$ and $\nu$ interchanged and multiply by $i/2$. ThusWriting $S=1-\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }$ gives $S^{-1}=1+\frac{i}{4}{\sigma }^{\mu \nu }{\omega }_{\mu \nu }+O({\omega }^2)$, and henceThe last step uses ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, including the metric signs in raising the first index.

For parity symmetry in quantum field theory, set $y=x_P=(t,-\mathbf{x})$. The time derivative of $\psi (y)$ is unchanged and each spatial derivative changes sign. Also ${\gamma }^0$ commutes with itself and anticommutes with every ${\gamma }^j$. ConsequentlyThis verifies the specified parity symmetry in quantum field theory action explicitly.

For charge conjugation, put $B=i{\gamma }^2$, so ${\psi }_C=B{\psi }^{\ast }$. In the Dirac representation of the gamma matrices, ${\gamma }^0,{\gamma }^1,{\gamma }^3$ are real and ${\gamma }^2$ is imaginary. The Clifford algebra therefore gives $B{\gamma }^{\mu \ast }{B}^{-1}=-{\gamma }^{\mu }$ for all four indices. Complex conjugation of the Dirac equation, followed by multiplication by $B$, yieldsThis is the antilinear charge-conjugation intertwiner property. No electromagnetic background is being held fixed in this free-field calculation.

Apply the specified parity symmetry in quantum field theory action before charge conjugation. Since ${\gamma }^0$ is real and $B{\gamma }^0=-{\gamma }^{0}B$,Thus the induced parity action on the charge-conjugate spinor has the opposite sign. In particular, if a spinor has intrinsic parity $\eta =\pm 1$, its charge conjugate has intrinsic parity $-\eta$. Equivalently, at rest the positive-frequency Dirac spinors occupy the upper two components and have ${\gamma }^0$ eigenvalue $+1$, whereas the negative-frequency Dirac spinors occupy the lower two components and have eigenvalue $-1$. The corresponding particle and antiparticle creation operators inherit these opposite signs under the specified parity convention. Hence electrons and positrons have opposite intrinsic parities; conventionally one assigns $+1$ to the electron and $-1$ to the positron. The common real overall parity sign may be reversed without changing this relative sign, as in opposite intrinsic parities of a Dirac particle and antiparticle.

Use $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Vary the Maxwell action with respect to the electromagnetic four-potential. Antisymmetry of the electromagnetic field tensor gives $\delta \mathcal{L}=-F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. After integration by parts, the Euler-Lagrange field equation isA gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. It therefore leaves the Maxwell action and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for canonical quantization: before gauge fixing, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.

Choose the Lorenz gauge $D={\partial }_{\mu }{A}^{\mu }=0$, attainable by solving $\square \chi =-D$ with suitable boundary data. The Euler-Lagrange field equation then reduces toResidual Lorenz gauge symmetry consists of transformations with $\square \chi =0$. The gauge condition and this residual redundancy ultimately leave two transverse polarizations of a free photon.

For the specified kinetic density ${\mathcal{L}}_C=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$, one has $\partial {\mathcal{L}}_C/\partial ({\partial }_{\mu }{A}_{\nu })=-{\partial }^{\mu }{A}^{\nu }$. Its Euler-Lagrange field equation is therefore $-\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction, sinceThe divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a Feynman gauge action; the additional physical-state condition will implement the Lorenz gauge. One must not impose $D=0$ inside the original action before varying it.