Local U(1) Gauge Invariance Fixes the QED Coupling

Derivation

Local U(1) Gauge Invariance Fixes the QED Coupling

D-280 Home PU-304 Threads symmetry · fields · force Depends on Dirac Equation, Antiparticles and the g=2 Prediction, Noether's Theorem for Conserved Currents

Statement

Starting from the free Dirac Lagrangian \(\mathcal{L}_0 = \bar{\psi}\left(i\gamma^\mu\partial_\mu - m\right)\psi\), we show that demanding invariance under a local phase rotation \(\psi(x)\to e^{i\alpha(x)}\psi(x)\) forces (i) the ordinary derivative to be promoted to the covariant derivative \(D_\mu = \partial_\mu + ieA_\mu\), (ii) the existence of a real vector field \(A_\mu\) transforming as \(A_\mu\to A_\mu - \frac{1}{e}\partial_\mu\alpha\), and (iii) a unique interaction term \(-e\bar{\psi}\gamma^\mu\psi\,A_\mu\) coupling the Dirac current to that field. Local U(1) symmetry therefore fixes the form of the electromagnetic coupling and the photon, up to the single number \(e\).

Why it matters

This is the prototype of the gauge principle, the organizing idea of the entire Standard Model. Rather than postulating the electromagnetic interaction, we deduce it: the requirement that a global symmetry of the free theory survive when the symmetry parameter is allowed to vary from point to point cannot be met by the free Lagrangian alone, and the minimal repair is exactly the Maxwell coupling of QED. The photon is not added by hand; it is the compensating field the symmetry demands.

The same logic, applied to the non-Abelian groups \(SU(2)\) and \(SU(3)\), generates the weak and strong interactions. Understanding the U(1) case in full detail is the entry point to Yang-Mills theory, to the concept of a connection on a fibre bundle, and to why charge is conserved and quantized within each gauge multiplet.

Assumptions

The matter field is a single Dirac spinor with a global U(1) symmetry.Without a starting global symmetry there is nothing to localize; the free Lagrangian \(\bar\psi(i\gamma^\mu\partial_\mu-m)\psi\) is invariant under \(\psi\to e^{i\alpha}\psi\) with constant \(\alpha\) precisely because \(\bar\psi\psi\) and \(\bar\psi\gamma^\mu\psi\) are phase-neutral, and this is the invariance we promote.

The gauge parameter \(\alpha(x)\) is an arbitrary smooth real function of spacetime.If \(\alpha\) were constant we would recover only the global symmetry and no new field would be needed; if it were not differentiable the transformation of the derivative term would be ill-defined and the covariant derivative could not be constructed.

The Lagrangian is built from \(\psi,\bar\psi\) and their first derivatives, is Lorentz invariant, local, and at most quadratic in derivatives.Dropping locality or the derivative-order restriction would permit non-minimal terms (e.g. a Pauli term \(\bar\psi\sigma^{\mu\nu}\psi F_{\mu\nu}\)); minimality is what makes the coupling unique, not merely possible.

The compensating field \(A_\mu\) is a genuine dynamical field with its own gauge-invariant kinetic term \(-\tfrac14 F_{\mu\nu}F^{\mu\nu}\).If we refused \(A_\mu\) dynamics it would be a mere Lagrange multiplier; requiring a gauge-invariant, Lorentz-invariant, dimension-4 kinetic term built from \(\partial_\mu A_\nu\) singles out the Maxwell term and forbids a photon mass \(m_A^2 A_\mu A^\mu\).

The representation is the fundamental charge-\(e\) representation, so the covariant derivative acts as \(D_\mu\psi=(\partial_\mu+ieA_\mu)\psi\).A field of charge \(q\) couples through \(\partial_\mu+iqA_\mu\); dropping the assumption of a definite charge leaves the coupling constant undetermined but does not change the structure of the argument.

Derivation

\[ \mathcal{L}_0 = \bar{\psi}\left(i\gamma^\mu\partial_\mu - m\right)\psi \]

Start from the free Dirac Lagrangian (prior result: Dirac equation). It is invariant under the global phase \(\psi\to e^{i\alpha}\psi\), \(\bar\psi\to e^{-i\alpha}\bar\psi\) for constant \(\alpha\). A

\[ \psi(x) \;\longrightarrow\; \psi'(x) = e^{i\alpha(x)}\psi(x), \qquad \bar\psi(x)\;\longrightarrow\;\bar\psi'(x)=e^{-i\alpha(x)}\bar\psi(x) \]

Promote the constant \(\alpha\) to a local parameter \(\alpha(x)\). This is the defining move of the gauge principle; we now test whether \(\mathcal{L}_0\) survives. A

\[ \partial_\mu\psi \;\longrightarrow\; \partial_\mu\!\left(e^{i\alpha(x)}\psi\right) = e^{i\alpha(x)}\left(\partial_\mu\psi + i\,\psi\,\partial_\mu\alpha\right) \]

Apply the product rule. The derivative does not transform covariantly: it acquires the inhomogeneous term \(i\psi\,\partial_\mu\alpha\), which vanishes only for constant \(\alpha\). A

\[ \mathcal{L}_0 \;\longrightarrow\; \mathcal{L}_0 - \bar\psi\gamma^\mu\psi\,\partial_\mu\alpha \]

Substitute Step 3 into \(\mathcal{L}_0\). The mass term \(-m\bar\psi\psi\) is invariant (phases cancel), and the \(i\gamma^\mu\partial_\mu\) term generates \(i\cdot i\,\bar\psi\gamma^\mu\psi\,\partial_\mu\alpha=-\bar\psi\gamma^\mu\psi\,\partial_\mu\alpha\). Local invariance fails by exactly this current-times-gradient term. B

\[ \partial_\mu \;\longrightarrow\; D_\mu \equiv \partial_\mu + ieA_\mu, \qquad D_\mu\psi = \left(\partial_\mu + ieA_\mu\right)\psi \]

Introduce a real vector field \(A_\mu\) and define a covariant derivative by adding a term linear in \(A_\mu\), with a coupling constant \(e\) and factor \(i\) fixed by requiring \(D_\mu\psi\) to be Hermitian-consistent and to transform like \(\psi\) itself. The ansatz is the minimal object that can cancel Step 4. B

\[ D_\mu\psi \;\longrightarrow\; \left(\partial_\mu + ieA_\mu'\right)\!\left(e^{i\alpha}\psi\right) = e^{i\alpha}\left[\partial_\mu\psi + i\psi\,\partial_\mu\alpha + ieA_\mu'\psi\right] \]

Demand \(D_\mu\psi\to e^{i\alpha}\,D_\mu\psi = e^{i\alpha}(\partial_\mu+ieA_\mu)\psi\). Expand the left side with the (as yet unknown) transformed field \(A_\mu'\), ready to match. B

\[ i\psi\,\partial_\mu\alpha + ieA_\mu'\psi = ieA_\mu\psi \;\;\Longrightarrow\;\; A_\mu' = A_\mu - \frac{1}{e}\partial_\mu\alpha \]

Match the bracket in Step 6 to \((\partial_\mu+ieA_\mu)\psi\) and cancel the common \(\partial_\mu\psi\) and factor \(\psi\). This fixes the gauge transformation of \(A_\mu\); the inhomogeneous shift \(-\tfrac1e\partial_\mu\alpha\) is exactly what is needed to absorb the offending term of Step 4. B

\[ A_\mu \to A_\mu - \tfrac{1}{e}\partial_\mu\alpha \quad\Longrightarrow\quad F_{\mu\nu}\equiv\partial_\mu A_\nu-\partial_\nu A_\mu \to F_{\mu\nu} \]

Check the field strength is gauge invariant: the shift contributes \(-\tfrac1e(\partial_\mu\partial_\nu-\partial_\nu\partial_\mu)\alpha=0\) since partial derivatives commute. Hence \(F_{\mu\nu}\) is the unique gauge-invariant object built from first derivatives of \(A_\mu\). C

\[ \mathcal{L} = \bar{\psi}\left(i\gamma^\mu D_\mu - m\right)\psi = \bar{\psi}\left(i\gamma^\mu\partial_\mu - m\right)\psi \;-\; e\,\bar{\psi}\gamma^\mu\psi\,A_\mu \]

Replace \(\partial_\mu\to D_\mu\) in \(\mathcal{L}_0\) and expand. By construction (Steps 6-7) the whole matter Lagrangian is now locally U(1) invariant. The expansion exposes the interaction \(-e\bar\psi\gamma^\mu\psi\,A_\mu = -eA_\mu J^\mu\), the current \(J^\mu=\bar\psi\gamma^\mu\psi\) being the conserved Noether current of the U(1) symmetry (prior result: Noether). B

\[ \mathcal{L}_{\text{QED}} = \bar{\psi}\left(i\gamma^\mu D_\mu - m\right)\psi \;-\; \tfrac{1}{4}F_{\mu\nu}F^{\mu\nu} \]

Give \(A_\mu\) dynamics by adding the unique gauge-invariant, Lorentz-invariant, dimension-4 kinetic term \(-\tfrac14 F_{\mu\nu}F^{\mu\nu}\). A mass term \(\tfrac12 m_A^2 A_\mu A^\mu\) is forbidden because \(A_\mu A^\mu\) is not gauge invariant. The result is the full QED Lagrangian. C

Result

\[ \boxed{\;\mathcal{L}_{\text{QED}} = \bar{\psi}\left(i\gamma^\mu\partial_\mu - m\right)\psi \;-\; e\,\bar{\psi}\gamma^\mu\psi\,A_\mu \;-\; \tfrac14 F_{\mu\nu}F^{\mu\nu}\;}\qquad D_\mu=\partial_\mu+ieA_\mu \]

Reading. Demanding that the free Dirac theory be invariant under a spacetime-dependent phase forces three things at once: the derivative must be covariantized, \(\partial_\mu\to D_\mu=\partial_\mu+ieA_\mu\); a gauge field \(A_\mu\) must exist and transform as \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\); and the matter current \(J^\mu=\bar\psi\gamma^\mu\psi\) must couple to it as \(-eA_\mu J^\mu\). The photon and its coupling to charge are consequences of the symmetry, not inputs. Only the number \(e\) is left free.

Units check. In natural units (\(\hbar=c=1\), \(\mathrm{mass}=\mathrm{length}^{-1}\)) a spinor field has mass dimension \([\psi]=\tfrac32\), so \([\bar\psi\gamma^\mu\partial_\mu\psi]=4\), matching the required \([\mathcal{L}]=4\). Then \([\bar\psi\gamma^\mu\psi]=3\) and \([A_\mu]=1\), so the interaction \([\bar\psi\gamma^\mu\psi\,A_\mu]=4\) forces \(e\) to be dimensionless. Consistently \([F_{\mu\nu}]=2\Rightarrow[F_{\mu\nu}F^{\mu\nu}]=4\). The dimensionless \(e\) is the electric charge, related to the fine-structure constant by \(\alpha_{\text{em}}=e^2/4\pi\approx 1/137\).

Limiting cases

Constant \(\alpha\) (global limit): \(\partial_\mu\alpha=0\), the offending term of Step 4 vanishes, \(A_\mu\) is not required, and one recovers the free Dirac theory with its globally conserved charge \(Q=\int d^3x\,\bar\psi\gamma^0\psi\).
\(e\to 0\) (decoupling): matter and photon decouple; \(D_\mu\to\partial_\mu\) and \(\mathcal{L}\to\mathcal{L}_0-\tfrac14F^2\), i.e. free electrons and free light. Perturbation theory in \(e\) is the loop expansion of QED.
Non-relativistic, static \(A_\mu=(\phi,\mathbf{0})\): the interaction reduces to \(-e\phi\,\bar\psi\gamma^0\psi\to -e\phi\,\rho\), the Coulomb potential energy of a charge density, recovering ordinary electrostatics.
Pure gauge \(A_\mu=-\tfrac1e\partial_\mu\Lambda\): \(F_{\mu\nu}=0\), no physical field; the phase can be removed by a gauge transformation, and only the Aharonov-Bohm holonomy \(\oint A_\mu dx^\mu\) around non-contractible loops is observable.

Breaks when

A photon mass is inserted. Adding \(\tfrac12 m_A^2 A_\mu A^\mu\) breaks gauge invariance because \(A_\mu A^\mu\to A_\mu A^\mu-\tfrac2e A^\mu\partial_\mu\alpha+\dots\ne A_\mu A^\mu\). The gauge principle forbids a bare photon mass; mass can only arise via spontaneous symmetry breaking (Higgs/Stückelberg), which is a different mechanism.
The symmetry group is non-Abelian. For \(SU(N)\) the parameters \(\alpha^a\) do not commute, the field strength acquires a nonlinear term \(F^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu + g f^{abc}A^b_\mu A^c_\nu\), the gauge fields self-interact, and the simple shift transformation of Step 7 is replaced by \(A_\mu\to UA_\mu U^{-1}-\tfrac{i}{g}(\partial_\mu U)U^{-1}\). The Abelian derivation here does not capture that structure.
Chiral / anomalous couplings. If left- and right-handed components carry different charges, the classically gauge-invariant current can fail to be conserved at one loop (the Adler-Bell-Jackiw anomaly), and gauge invariance survives only if anomalies cancel between species. Pure vector QED here is anomaly-free, but the construction is not automatically consistent for arbitrary chiral charge assignments.
Gravity / curved spacetime with fermions. The plain \(\partial_\mu\) must additionally be promoted to a spin connection via the vierbein; the flat-space covariant derivative of Step 5 is then only part of the full \(D_\mu\), and global topological obstructions (nontrivial bundles) can prevent a single smooth \(A_\mu\).

Failure modes

Sign/charge-convention slip. Writing \(D_\mu=\partial_\mu-ieA_\mu\) then also \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\); the two conventions must be paired consistently or the inhomogeneous terms fail to cancel. The pairing is \(D_\mu=\partial_\mu+ieA_\mu\) with \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\) (or both signs flipped together).
Forgetting the product-rule term. Treating \(\partial_\mu(e^{i\alpha}\psi)=e^{i\alpha}\partial_\mu\psi\) drops the \(i\psi\,\partial_\mu\alpha\) piece and makes the free theory look locally invariant, hiding the entire reason the gauge field is needed.
Claiming the mass term breaks the symmetry. \(-m\bar\psi\psi\) is fully U(1) invariant (the phases of \(\psi\) and \(\bar\psi\) cancel); it is the derivative (kinetic) term that fails. Confusing which term breaks leads to the false conclusion that gauge invariance requires massless fermions.
Thinking \(A_\mu\) itself is observable. Only \(F_{\mu\nu}\) (and holonomies) are gauge invariant; quoting a numerical value of \(A_\mu\) as physical ignores the freedom \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\).
Assuming \(e\) is predicted. The gauge principle fixes the form of the coupling and that \(e\) is a single dimensionless constant, but not its value; \(e\) (equivalently \(\alpha_{\text{em}}\)) is measured, and it runs with energy scale.

Discussion

The deep content of this derivation is that a redundancy of description forces a physical interaction. The local phase \(\alpha(x)\) is unobservable at every point independently, yet insisting that physics not depend on this arbitrary choice means the phases at neighbouring points must be comparable, and comparison requires a connection. \(A_\mu\) is precisely that connection: it tells us how to parallel-transport the internal phase from \(x\) to \(x+dx\). The covariant derivative \(D_\mu\psi\) is the difference between \(\psi(x+dx)\) and the parallel-transported \(\psi(x)\), which is why it, and not \(\partial_\mu\psi\), transforms homogeneously.

Geometrically the electron field is a section of a complex line bundle over spacetime, \(A_\mu\) is the bundle connection, and \(F_{\mu\nu}\) is its curvature. The gauge transformation \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\) is a change of local frame in the fibre; the invariance of \(F_{\mu\nu}\) is the statement that curvature is frame-independent. This picture is what generalizes verbatim to Yang-Mills theory once the fibre carries a non-Abelian group.

The coupling that emerges, \(-eA_\mu J^\mu\), ties the interaction directly to the Noether current \(J^\mu=\bar\psi\gamma^\mu\psi\) of the very symmetry being gauged. Global U(1) invariance guarantees \(\partial_\mu J^\mu=0\) (charge conservation); localizing that symmetry makes \(J^\mu\) the source of the field, \(\partial_\mu F^{\mu\nu}=eJ^\nu\), which is just Maxwell's equations with the Dirac current as source. Conservation of charge and the coupling of charge to the photon are thus two faces of the same symmetry.

At the quantum level the redundancy must be handled carefully: naive path integration over all \(A_\mu\) overcounts gauge-equivalent configurations, requiring gauge fixing and Faddeev-Popov ghosts, while the residual global symmetry is encoded in the Ward-Takahashi identities that relate the electron self-energy to the vertex and guarantee, for instance, that charge renormalization is universal (\(Z_1=Z_2\)) so that the photon stays massless to all orders. The masslessness we imposed classically is thereby protected quantum-mechanically by the same U(1) that generated the coupling.

Common misconceptions. (1) "Gauge symmetry is a physical symmetry relating distinct states" — it is a redundancy; gauge-related configurations are the same physical state. (2) "The photon exists because of a conservation law" — the causal arrow runs the other way here: localizing the symmetry requires the field, and the associated global symmetry implies the conservation law. (3) "Local invariance uniquely predicts \(e\)" — it constrains the structure, not the magnitude.

Worked examples

Verify explicitly that \(\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi\) is locally U(1) invariant.

Set up: transform \(\psi\to e^{i\alpha}\psi\), \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\), and show \(\mathcal{L}\to\mathcal{L}\). B

\[ D_\mu\psi \to \left(\partial_\mu+ie\big(A_\mu-\tfrac1e\partial_\mu\alpha\big)\right)e^{i\alpha}\psi \]

Substitute both transformations into \(D_\mu\psi\). B

\[ = e^{i\alpha}\left[\partial_\mu\psi + i(\partial_\mu\alpha)\psi + ieA_\mu\psi - i(\partial_\mu\alpha)\psi\right] = e^{i\alpha}\left(\partial_\mu+ieA_\mu\right)\psi = e^{i\alpha}D_\mu\psi \]

Expand with the product rule; the \(+i(\partial_\mu\alpha)\psi\) from differentiating \(e^{i\alpha}\) cancels the \(-i(\partial_\mu\alpha)\psi\) from the shift of \(A_\mu\). B

\[ \bar\psi\gamma^\mu D_\mu\psi \to \bar\psi e^{-i\alpha}\gamma^\mu e^{i\alpha}D_\mu\psi = \bar\psi\gamma^\mu D_\mu\psi \]

The conjugate carries \(e^{-i\alpha}\); phases cancel, and the mass term likewise. Every term is invariant. B

\[ \mathcal{L}\;\longrightarrow\;\mathcal{L}\qquad\text{(locally U(1) invariant)} \]

Reading. The covariant derivative transforms exactly like the field it acts on, which is the whole point of introducing \(A_\mu\); local invariance is then automatic term by term.

Units check. Purely a symmetry statement; every term retains mass dimension 4 throughout, unchanged by the transformation.

Compute the fine-structure constant and the Coulomb energy scale from the dimensionless coupling.

Set up: use \(\alpha_{\text{em}}=e^2/4\pi\varepsilon_0\hbar c\) (SI) with measured \(e=1.602\times10^{-19}\,\mathrm{C}\), and find the potential energy of two electrons at \(r=1\,\mathrm{nm}\). A

\[ \alpha_{\text{em}} = \frac{e^2}{4\pi\varepsilon_0\hbar c} = \frac{(1.602\times10^{-19})^2}{4\pi(8.854\times10^{-12})(1.055\times10^{-34})(2.998\times10^{8})} \]

Insert numbers with units \(\mathrm{C},\,\mathrm{F\,m^{-1}},\,\mathrm{J\,s},\,\mathrm{m\,s^{-1}}\); the combination is dimensionless. A

\[ \alpha_{\text{em}} \approx 7.297\times10^{-3} = \frac{1}{137.0} \]

Evaluate. This is the value of the coupling \(e\) that the gauge principle left undetermined; it is measured, not derived. A

\[ U(r) = \frac{e^2}{4\pi\varepsilon_0 r} = \alpha_{\text{em}}\,\frac{\hbar c}{r} = 7.297\times10^{-3}\times\frac{197.3\,\mathrm{eV\,nm}}{1\,\mathrm{nm}} \]

Use the convenient constant \(\hbar c=197.3\,\mathrm{eV\,nm}\) to turn the coupling into an energy at separation \(r=1\,\mathrm{nm}\). A

\[ \alpha_{\text{em}}\approx\frac{1}{137},\qquad U(1\,\mathrm{nm})\approx 1.44\,\mathrm{eV} \]

Reading. The single free number of the gauge construction sets the strength of all electromagnetic phenomena; two electrons a nanometre apart repel with about \(1.4\,\mathrm{eV}\) of potential energy, comparable to a chemical bond scale.

Units check. \(\hbar c\) has units \(\mathrm{eV\,nm}\); dividing by \(r\) in \(\mathrm{nm}\) gives \(\mathrm{eV}\), and \(\alpha_{\text{em}}\) is dimensionless, so \(U\) is in \(\mathrm{eV}\). Correct.

Problems

Show that the mass term \(-m\bar\psi\psi\) is invariant under the local transformation \(\psi\to e^{i\alpha(x)}\psi\), and explain in one sentence why this does not save the full Lagrangian.
Solution
Under the transformation \(\bar\psi\psi\to\bar\psi e^{-i\alpha}e^{i\alpha}\psi=\bar\psi\psi\), so \(-m\bar\psi\psi\) is invariant for any \(\alpha(x)\) because the field and its conjugate carry opposite phases that cancel with no derivative acting. It does not save the theory because the kinetic term \(i\bar\psi\gamma^\mu\partial_\mu\psi\) contains a derivative that acts on \(e^{i\alpha(x)}\), producing \(-\bar\psi\gamma^\mu\psi\,\partial_\mu\alpha\); it is this term, not the mass term, that spoils local invariance.
Starting from \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\), prove that \(F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\) is gauge invariant, and state what physical quantity this guarantees is observable.
Solution
\(F_{\mu\nu}\to\partial_\mu\!\big(A_\nu-\tfrac1e\partial_\nu\alpha\big)-\partial_\nu\!\big(A_\mu-\tfrac1e\partial_\mu\alpha\big)=F_{\mu\nu}-\tfrac1e(\partial_\mu\partial_\nu-\partial_\nu\partial_\mu)\alpha\). Since mixed partials commute for smooth \(\alpha\), the last bracket is zero, so \(F_{\mu\nu}\to F_{\mu\nu}\). This guarantees that the electric and magnetic fields, \(E^i=F^{i0}\) and \(B^i=-\tfrac12\epsilon^{ijk}F_{jk}\), are observable while \(A_\mu\) itself is not.
A field of charge \(q\) (rather than \(e\)) is to be coupled to the same photon \(A_\mu\). Write its covariant derivative and its required gauge transformation, and verify the interaction term.
Solution
For charge \(q\) the transformation is \(\psi\to e^{iq\alpha/e}\psi\) if we keep \(A_\mu\to A_\mu-\tfrac1e\partial_\mu\alpha\); more cleanly, use the field's own phase \(\psi\to e^{iq\Lambda}\psi\) with \(A_\mu\to A_\mu-\partial_\mu\Lambda\), giving \(D_\mu=\partial_\mu+iqA_\mu\). Then \(D_\mu\psi\to(\partial_\mu+iq(A_\mu-\partial_\mu\Lambda))e^{iq\Lambda}\psi=e^{iq\Lambda}(\partial_\mu+iqA_\mu)\psi\), so it transforms covariantly. Expanding \(\bar\psi(i\gamma^\mu D_\mu-m)\psi\) gives interaction \(-q\bar\psi\gamma^\mu\psi\,A_\mu\): the coupling strength is the charge \(q\), demonstrating that charge is the U(1) representation label.
Show that the Euler-Lagrange equation for \(A_\nu\) from \(\mathcal{L}_{\text{QED}}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-\tfrac14F_{\mu\nu}F^{\mu\nu}\) yields \(\partial_\mu F^{\mu\nu}=eJ^\nu\) with \(J^\nu=\bar\psi\gamma^\nu\psi\).
Solution
Write the \(A\)-dependent part: \(\mathcal{L}\supset -e\bar\psi\gamma^\nu\psi A_\nu-\tfrac14F_{\alpha\beta}F^{\alpha\beta}\). The Euler-Lagrange equation is \(\partial_\mu\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}-\frac{\partial\mathcal{L}}{\partial A_\nu}=0\). Using \(\frac{\partial}{\partial(\partial_\mu A_\nu)}\big(-\tfrac14F_{\alpha\beta}F^{\alpha\beta}\big)=-F^{\mu\nu}\) and \(\frac{\partial\mathcal{L}}{\partial A_\nu}=-e\bar\psi\gamma^\nu\psi\), we get \(-\partial_\mu F^{\mu\nu}+e\bar\psi\gamma^\nu\psi=0\), i.e. \(\partial_\mu F^{\mu\nu}=eJ^\nu\). These are the inhomogeneous Maxwell equations with the Dirac current as source; antisymmetry of \(F^{\mu\nu}\) then forces \(\partial_\nu J^\nu=0\), recovering charge conservation.
Estimate the QED interaction energy scale for the coupling \(e\) at separation equal to the electron Compton wavelength \(\lambda_C=\hbar/m_ec=386\,\mathrm{fm}\), and compare it to the electron rest energy \(m_ec^2=0.511\,\mathrm{MeV}\).
Solution
\(U(\lambda_C)=\alpha_{\text{em}}\frac{\hbar c}{\lambda_C}=\alpha_{\text{em}}\frac{\hbar c}{\hbar/m_ec}=\alpha_{\text{em}}\,m_ec^2\). Numerically \(U=7.30\times10^{-3}\times0.511\,\mathrm{MeV}=3.73\times10^{-3}\,\mathrm{MeV}=3.73\,\mathrm{keV}\). The ratio \(U/(m_ec^2)=\alpha_{\text{em}}\approx1/137\), showing that electromagnetic binding is weak compared to the rest mass: this smallness of \(\alpha_{\text{em}}\) is precisely why QED perturbation theory converges well, and it is a measured input the gauge principle does not predict.