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In quantum electrodynamics, gauge invariance connects local phase freedom, electric-charge conservation and the restrictions on photon interactions. For definiteness takeThe gauge transformationhas $D_{\mu }\psi \mapsto e^{-ie\alpha }{D}_{\mu }\psi$: the derivative of the local phase cancels the added gauge potential. Also $F_{\mu \nu }$ is unchanged because mixed partial derivatives commute. Consequently every term in the Lagrangian density is invariant. This derives the minimal electromagnetic coupling of a Dirac field, including the interaction $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$. A fermion mass is compatible with the symmetry, whereas a local Proca photon mass $m_{\gamma }^2{A}_{\mu }{A}^{\mu }/2$ is not.

The classical Maxwell equations are ${\partial }_{\mu }{F}^{\mu \nu }=j^{\nu }$, with $j^{\nu }=e\bar{\psi }{\gamma }^{\nu }\psi$. Taking a divergence gives ${\partial }_{\nu }{j}^{\nu }=0$ because $F^{\mu \nu }$ is antisymmetric. The same conservation law follows directly from the charged Dirac equation and its adjoint. For vanishing flux at spatial infinity, $Q=\int d^{3}x{j}^0$ is conserved. Gauss law identifies this charge with the electric flux through a surrounding surface. Particle-antiparticle creation is allowed, but its net electric charge is zero.

Local gauge invariance is also a redundancy in the potentials, rather than an extra propagating degree of freedom. The canonical momentum of $A_0$ vanishes, and its field equation enforces the Gauss law constraint in gauge theory, $\nabla \cdot \mathbf{E}=j^0$. After imposing the constraint and quotienting gauge freedom, a massless photon has two transverse physical polarizations. Gauge transformations vanishing at the boundary identify equivalent descriptions; constant phase transformations at infinity can act on charged states and generate their conserved charge. This distinction explains why gauge redundancy and electric charge are compatible.

In perturbative QED, the quadratic photon operator cannot be inverted before gauge fixing. Adding $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$ gives the covariant gauge propagatorThis propagator contains longitudinal and timelike components, but these are not external physical photon states. A Gupta-Bleuler null-state quotient, or the equivalent BRST construction, removes them. In linear Lorenz gauge, $\delta ({\partial }_{\mu }{A}^{\mu })=\square \alpha$, so the Faddeev-Popov determinant is independent of $A_{\mu }$ and the Faddeev-Popov ghost fields decouple. This last statement concerns the linear gauge; Abelian gauge groups alone do not guarantee decoupling in a nonlinear gauge condition.

At the quantum level, the vector symmetry gives a Ward identity. One way to obtain it is to change variables by a localized phase in a time-ordered correlation function containing $\psi (y)$ and $\bar{\psi }(z)$. The action variation becomes a current divergence; the two field insertions contribute opposite contact terms, since their charges are opposite. Fourier transformation turns the divergence into the photon momentum, and amputating the external fermion propagators turns the two contact terms into the difference of their inverse propagators. With the charge removed from the proper vertex and the overall propagator factor $i$ removed from its inverse, the result is the proper-vertex Ward-Takahashi identity in QEDAt tree level this is simply $k_{\mu }{\gamma }^{\mu }=(/p+/k-m)-(/p-m)$. Between on-shell external spinors, the inverse propagators vanish, so the full physical scattering amplitude satisfies $k_{\mu }{\mathcal{M}}^{\mu }=0$. Therefore replacing a photon polarization vector by ${\epsilon }_{\mu }+c{k}_{\mu }$ leaves the amplitude unchanged. The identity applies to the complete required sum of diagrams, not necessarily each diagram separately. It also ensures cancellation of the gauge parameter from physical observables when all terms at the chosen perturbative order are included.

For the photon two-point function, the Ward identity implies transverse photon vacuum polarization:Thus ultraviolet subtraction renormalizes the gauge-invariant kinetic term, rather than requiring a local photon-mass counterterm. Perturbative unbroken QED retains a massless photon. Transversality by itself would not exclude a nonlocal massless pole in a different theory; the assertion here includes the perturbative QED setting.

Taking $k\to 0$ in the proper-vertex identity gives ${\Gamma }^{\mu }(p,p)=\partial S^{-1}(p)/\partial p_{\mu }$. Matching the divergent coefficients of the derivative and vertex terms yields the Ward identity for QED renormalization constants, $Z_1=Z_2$, in a gauge-preserving prescription. If ${\psi }_0=\sqrt{Z_2}\psi$ and $A_0=\sqrt{Z_3}A$, comparison of the interaction coefficients givesHence charge renormalization is fixed by photon wave-function renormalization. Gauge invariance constrains counterterms and relates renormalizations; it does not make the electric charge independent of scale.

The Soft photon theorem gives another physical manifestation. Attaching a photon of small momentum $k$ to the external charged legs gives the leading factorwhere $Q_i$ is charge in units of $e$ and ${\eta }_i=+1$ for outgoing and $-1$ for incoming legs. Replacing $\epsilon$ by $k$ makes this factor proportional to ${\sum }_i{\eta }_i{Q}_i$, which vanishes by electric charge conservation. Massless photons also cause infrared divergences; physical inclusive measurements combine unresolved real emission with virtual corrections, rather than treating either contribution alone as an observable.

Finally, the vector gauge symmetry of a Dirac fermion in QED is anomaly-free, so it can be maintained by regularization and renormalization. This must be distinguished from the axial current: its divergence has explicit mass terms, as in Question 3, and can additionally have a quantum chiral anomaly. Conservation of the gauge current does not imply axial-current conservation. The resulting picture is that gauge invariance simultaneously organizes the physical photon states, enforces electric-charge conservation and ties together amplitudes and ultraviolet subtractions.

Use gamma matrices satisfying the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with ${\gamma }^{0†}={\gamma }^0$ and ${\gamma }^{i†}=-{\gamma }^i$. The Dirac adjoint is $\bar{\Psi }={\Psi }^{†}{\gamma }^0$. Take the Dirac actionIt differs from the manifestly Hermitian form with $(i/2)\bar{\Psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\Psi$ only by a boundary term. Treat $\Psi$ and $\bar{\Psi }$ as independent variables when applying the principle of stationary action. Varying $\bar{\Psi }$ givesVarying $\Psi$ and integrating by parts gives the adjoint Dirac equation, $i({\partial }_{\mu }\bar{\Psi }){\gamma }^{\mu }+m\bar{\Psi }=0$. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ also gives $(\square +m^2)\Psi =0$, so its dispersion relation is Lorentz invariant.

For a Lorentz transformation $x^{\prime }=\Lambda x$, the field transforms in the Spinor representation of the Lorentz group:Since ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$, this identity gives the Lorentz covariance of the Dirac operator:Thus every solution is carried to another solution. The field is a spinor rather than a four-vector; the transformation of the gamma matrices supplies the necessary covariance. The Dirac action is Lorentz invariant because its integrand is a scalar and $d^{4}x$ is invariant.

For electric charge $q$, use the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$. The minimal electromagnetic coupling of a Dirac field isThe local gauge transformations in this convention areDirect substitution gives $D_{\mu }^{\prime }{\Psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\Psi$, which proves gauge covariance of the charged Dirac equation. The mass and kinetic terms are invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$, so the full action has gauge invariance. The conserved current is $j^{\mu }=q\bar{\Psi }{\gamma }^{\mu }\Psi$. It transforms as a four-vector under Lorentz transformations, and varying $A_{\mu }$ gives ${\partial }_{\nu }{F}^{\nu \mu }=j^{\mu }$.

To make the infinitesimal spinor transformation explicit, write ${\Lambda }^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+{\omega }^{\mu }{}_{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. DefineThe Lorentz generators from gamma-matrix commutators have precisely the needed algebra: $[{\sigma }^{\mu \nu },{\gamma }^{\rho }]=2i({\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu })$, which verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }$ to first order. The infinitesimal transformation of a Dirac field is thereforeAt the same coordinate argument, the orbital change must also be included:Both formulas describe the same transformation; their arguments differ.

Finally, the gamma adjoint identities imply ${\sigma }^{\mu \nu †}={\gamma }^0{\sigma }^{\mu \nu }{\gamma }^0$ and hence the pseudo-unitarity of the spinor Lorentz representation, $S^{†}{\gamma }^{0}S={\gamma }^0$. It follows thatThus the Dirac scalar bilinear $\bar{\Psi }\Psi$ is a Lorentz scalar. Infinitesimally the spin terms in $\delta (\bar{\Psi }\Psi )$ cancel; at fixed coordinates only the ordinary scalar orbital transformation remains.