Why does an ordinary derivative miss the same physical state?

A derivative produces an extra term

Let $\psi(x)$ be a complex scalar field of charge $q$ in continuous spacetime. Use natural units $\hbar=c=1$ . For a real position-dependent function $\alpha(x)$ , multiply the field by $e^{iq\alpha(x)}$ . Differentiation acts on that exponential as well as on the field.

The extra term comes entirely from the new phase scale. Using the ordinary derivative in an energy would then count a change of description as a change of state. A comparison rule must change with the local scale.

\partial_\mu(e^{iq\alpha}\psi)=e^{iq\alpha}\bigl(\partial_\mu\psi+iq(\partial_\mu\alpha)\psi\bigr)

The index

\mu

selects time or a spatial direction. The extra term measures the change of scale in that direction.

The potential supplies the correction

Introduce a potential $A_\mu$ and define $D_\mu=\partial_\mu-iqA_\mu$ . Adding $\partial_\mu\alpha$ to the potential cancels the unwanted term. Substitution shows that the derivative transforms by the same phase as the field.

This is covariance: both objects change together. Consequently $|D_\mu\psi|^2$ is gauge invariant and can enter the matter kinetic term. Comparing two directions in opposite orders reveals the field strength.

A_\mu^\prime=A_\mu+\partial_\mu\alpha,\qquad D_\mu^\prime\psi^\prime=e^{iq\alpha}D_\mu\psi

Defining the charge and field convention first fixes the sign of the potential transformation.

[D_\mu,D_\nu]\psi=-iqF_{\mu\nu}\psi,\qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu

The bracket subtracts the two orders of operation. Mixed second derivatives of the gauge function cancel, leaving the field strength unchanged.

U_{ij}=\exp\!\left[-iq\int_i^j A_\mu\,dx^\mu\right]

This link transports from j to i; the integral runs from i to j. The discrete phase angle equals charge times the continuous gauge function, reproducing chapter 1’s endpoint transformation.

Source: David Tong

Choosing a convenient representative

Many potentials describe the same field strength. Gauge fixing selects a useful representative; the Lorenz condition is $\partial_\mu A^\mu=0$ . It cannot remove a physical field strength. Residual transformations satisfying $\Box\alpha=0$ preserve this condition.

For a uniform magnetic field perpendicular to a plane, take $A_x=0$ and $A_y=Bx$ . Choosing $\alpha=-Bxy/2$ gives the symmetric potential below. The vector pictures differ, yet their curls agree.

(A_x,A_y)=(0,Bx)\longrightarrow\left(-\frac{By}{2},\frac{Bx}{2}\right),\qquad \partial_xA_y-\partial_yA_x=B

Both potentials describe the same uniform magnetic field and the same oriented flux through every closed path.

Analytic reference at one point. Units follow the chosen magnetic-field and length units.
Position	First potential	Second potential	Magnetic field
$(x,y)=(2,1)$	$(0,2B)$	$(-B/2,B)$	$B$

Local transformations imply conservation relations

The coupling of current $j^\mu$ to the potential contains $\int d^4x\,A_\mu j^\mu$ . Vary the potential and integrate the extra term by parts. When the boundary term vanishes, the variation is proportional to the divergence of the current. Local gauge invariance therefore requires current conservation.

Quantum correlation functions satisfy related Ward–Takahashi identities. A useful consequence concerns an external photon: adding a component parallel to its four-momentum to its polarization leaves the scattering amplitude unchanged.

Write the photon amplitude as $\mathcal A=\epsilon_\mu\mathcal M^\mu$ with momentum $k^\mu$ . For an on-shell photon and physical on-shell remaining external states, a complete calculation with a nonanomalous conserved current obeys $k_\mu\mathcal M^\mu=0$ , so $\epsilon_\mu\to\epsilon_\mu+\beta k_\mu$ changes nothing. This checks whether diagrams are missing. All contributions at the stated order are needed for the cancellation.

\delta S_{\mathrm{int}}=\int d^4x\,(\partial_\mu\alpha)j^\mu=-\int d^4x\,\alpha\,\partial_\mu j^\mu

This is a classical argument. Its quantum counterpart also requires the regulator and integration measure to preserve the relevant symmetry, a condition revisited in the anomaly chapter.

Try the reasoning in an experiment

Bring the continuous argument back to links

Predict how a slowly varying local phase scale changes adjacent links.
Change local phases and record link angles and the loop. Can every local value change while the loop stays fixed?
Use the product rule to explain why changing an amplitude without changing its links introduces an extra physical phase difference.

Exercises and explanations

Does a constant $\alpha(x)$ produce the extra derivative term?

Read the explanation

No. Its derivative vanishes. The problem arises for position-dependent phase scales, which explains why local transformations require additional structure.

Which potential in the table is closer to the “real potential”?

Read the explanation

Either can be used. Their field strengths and loop phases agree. Boundary conditions and convenience guide the choice; allowed gauge transformations must respect those boundaries.

Does finding $k_\mu\mathcal M^\mu\ne0$ establish a quantum anomaly?

Read the explanation

No. Check omitted diagrams, signs, external-state conditions and regularization first. An anomaly argument must identify the preserved current and calculate the remaining quantum violation with a specified regulator.

Sources and reproduced scope

Derivations, tables and the plot are authored for this course. The reading locations below separate source results from our teaching extensions. Numerical examples are model references or labeled synthetic observations. No experimental dataset or source figure is copied.

David Tong, Gauge Theory (2018 lecture notes)

Sections 2.1–2.1.3 (connections and Wilson loops), 2.7.3 (Higgs phase), 3.1.1–3.2.2 (two-dimensional anomaly and Ward identities), 4.2.1 (link endpoints) and 4.3.1–4.3.3 (lattice fermions). Printed pages 26–37, 105–108, 122–139, 207–210, 220–228.

Aneesh V. Manohar, Introduction to Effective Field Theories, arXiv:1804.05863v1 (2018)

Sections 3.5 and 4 (integrating out and power counting), 5.5–5.7 (matching), 5.10.1, PDF equations (5.51)–(5.60) (QCD running and operator mixing), and 6.2–6.3 (field redefinitions). The scalar-channel examples are derived for this course. The known-log boundary expansion checks truncation of the published running factors; it does not calculate finite hard matching.