Gauge Covariance

Gauge covariance means that potentials, amplitudes, and differential operators change together while physical predictions agree. A time-dependent gauge transformation changes the Hamiltonian by more than unitary conjugation, so its instantaneous energy labels need not be unchanged. This page works through that distinction and its consequences for evolution kernels and global phases. The covariant-derivative proof belongs to Minimal Coupling.

Required background. Minimal Coupling fixes the gauge phase and signed charge. Helpful background. The Dirac Hamiltonian as an Operator explains operator domains and unitary evolution.

A time-dependent phase changes the Hamiltonian

Retain $\hbar,c$ and the SI potential convention. For a real, sufficiently regular function $\chi(\mathbf x,t)$, use

$$U_\chi=e^{iq\chi/\hbar},\qquad \psi'=U_\chi\psi,\qquad \mathbf A'=\mathbf A+\nabla\chi,\qquad \Phi'=\Phi-\partial_t\chi.$$

If $i\hbar\partial_t\psi=H\psi$, differentiation of $\psi'=U_\chi\psi$ gives

$$H'=U_\chi H U_\chi^\dagger +i\hbar(\partial_tU_\chi)U_\chi^\dagger.$$

The last term is $-q\partial_t\chi$. For the Dirac Hamiltonian, conjugation replaces $\mathbf p-q\mathbf A$ by $\mathbf p-q(\mathbf A+\nabla\chi)$, and the time term changes $q\Phi$ to $q\Phi'$. This verifies the entire time-evolution equation in the new gauge.

For a time-independent $U_\chi$, the Hamiltonians are unitarily equivalent, including their domains $D(H')=U_\chi D(H)$. For a time-dependent transformation, omitting the derivative term is incorrect. Boundary conditions must transform too; a phase that changes a periodicity condition cannot silently be treated as a unitary map on the original periodic domain.

One uniform electric field in two gauges

Let $\mathbf E=E_0\hat{\mathbf z}$ and $\mathbf B=0$. Two potential choices are

$$\begin{array}{c|cc} &\Phi&\mathbf A\\ \hline \text{static gauge}&-E_0z&0\\ \text{temporal gauge}&0&-E_0t\,\hat{\mathbf z} \end{array}$$

They are related by $\chi=-E_0tz$. Hence $U_\chi=e^{-iqE_0tz/\hbar}$ and

$$\begin{aligned} H_{\rm s} &=c\boldsymbol\alpha\cdot\mathbf p+\beta mc^2-qE_0z,\\ H_{\rm t} &=c\boldsymbol\alpha\cdot (\mathbf p+qE_0t\,\hat{\mathbf z})+\beta mc^2. \end{aligned}$$

In the temporal gauge the canonical $p_z$ commutes with the Hamiltonian, but the kinetic momentum is $\pi_z=p_z+qE_0t$. Thus $d\langle\pi_z\rangle/dt=qE_0$. In the static gauge $\pi_z=p_z$ and the commutator with $-qE_0z$ produces the same force.

The field has not stopped accelerating the charge merely because a conserved canonical momentum exists in one gauge. A stationary energy eigenfunction in the static gauge likewise need not remain a single-frequency function after multiplication by the space- and time-dependent phase.

These are relations between external-field wave equations. A uniform electric field over all space has additional physical idealizations; interpreting vacuum production requires a quantum matter field and a choice of state.

Gauge shifts of energy labels

Take $\chi=-Ct$ for a constant potential offset $C$. Then $\Phi'=\Phi+C$ and $H'=H+qC$. A stationary solution $e^{-iEt/\hbar}\psi_E(\mathbf x)$ becomes

$$\psi'=e^{-i(E+qC)t/\hbar}\psi_E(\mathbf x).$$

The new energy label is $E'=E+qC$, while $E'-q\Phi'=E-q\Phi$. Transition energies between states of the same charge and the mechanical dynamics are unchanged. An absolute scalar-potential offset is therefore not a gauge-independent test of whether a nonrelativistic expansion is valid. Relevant potential differences, field strengths, and gauge-covariant residual energies must be compared with the rest-energy gap.

For more general time-dependent gauges, the Hamiltonian need not even have a common stationary eigenbasis with its old representative. Gauge invariance of predictions is a statement about transformed states and observables, not equality of two bare matrices at a chosen instant.

Evolution kernels acquire endpoint phases

Let $\mathcal U(t_f,t_i)$ be the evolution operator. Its gauge transformation is

$$\mathcal U'(t_f,t_i) =U_\chi(t_f)\mathcal U(t_f,t_i)U_\chi(t_i)^\dagger.$$

It follows either from the transformed Hamiltonian or by applying both sides to the transformed initial state. In position representation the kernel obeys

$$K'(x_f,t_f;x_i,t_i) =e^{iq\chi_f/\hbar} K(x_f,t_f;x_i,t_i)e^{-iq\chi_i/\hbar}.$$

The endpoint phases cancel against transformed input and output states in a physical transition probability. A kernel at distinct endpoints is gauge covariant; it need not be a gauge-invariant complex number.

The same endpoint rule appears in the prescribed-path electromagnetic phase

$$\exp\left[ \frac{iq}{\hbar}\int (\mathbf A\cdot d\mathbf x-\Phi\,dt) \right].$$

The integrand changes by $d\chi$. This is a useful phase-transport object, not by itself a complete quantum propagator for arbitrary dynamics.

Zero local curvature does not remove global holonomy

On a simply connected regular patch, vanishing field strength permits a locally pure-gauge potential to be removed. On a region with an excluded flux tube, the accessible magnetic field can vanish while

$$\exp\left(\frac{iq}{\hbar}\oint \mathbf A\cdot d\boldsymbol\ell\right)$$

remains nontrivial. A single-valued gauge phase cannot change this holonomy. The flux-phase derivation is on Minimal Coupling; the lesson here is the distinction between a local differential statement and a globally allowed change of amplitude.

On an angular coordinate $\varphi\sim\varphi+2\pi$, a candidate $\chi=a\varphi$ produces $U_\chi(\varphi+2\pi)/U_\chi(\varphi) =e^{2\pi iqa/\hbar}$. Preserving the same periodic wavefunction domain requires $qa/\hbar\in\mathbb Z$. A formal gradient can otherwise transfer the information into a twisted boundary condition rather than remove it.

Gauge covariance in all these examples acts on prescribed classical potentials. It neither quantizes those potentials nor introduces photon states.

Exercises

1. Derive the Hamiltonian transformation from $\psi'=U_\chi\psi$ and explain which term would be missed by treating $U_\chi$ as time independent.

Solution

$i\hbar\partial_t\psi' =i\hbar\dot U_\chi\psi+U_\chi H\psi$. Replace $\psi$ by $U_\chi^\dagger\psi'$ to obtain the displayed formula. The missed term is $i\hbar\dot U_\chi U_\chi^\dagger =-q\partial_t\chi$, precisely the scalar-potential shift.

2. In the uniform-field example, compute $U_\chi p_z U_\chi^\dagger$ and show that the extra time term cancels the old scalar potential.

Solution

$U_\chi p_zU_\chi^\dagger =p_z-q\partial_z\chi=p_z+qE_0t$. Also $i\hbar\dot U_\chi U_\chi^\dagger=qE_0z$, which cancels the $-qE_0z$ in $H_{\rm s}$.

3. Does multiplication of a kernel by endpoint phases change a transition probability between consistently transformed states?

Solution

No. In $\langle\psi_f'|\mathcal U'|\psi_i'\rangle$, the endpoint unitaries cancel pairwise, leaving $\langle\psi_f|\mathcal U|\psi_i\rangle$. Changing the kernel without changing the endpoint states would compare inconsistent descriptions.

References

• J. D. Jackson and L. B. Okun, “Historical Roots of Gauge Invariance,” Reviews of Modern Physics 73, 663–680, 2001, doi:10.1103/RevModPhys.73.663 — gauge transformations and their quantum interpretation.
• J. J. Sakurai, Advanced Quantum Mechanics, Addison–Wesley, 1967 — electromagnetic gauge covariance.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940 — local gauge transformations and field-theory context.