4.1.1 Gauge Principle#

Worked solutions for the homework problems in the 4.1.1 Gauge Principle lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Diagnosing redundancy. For each pair \((\psi, \psi')\) below, compute \(\vert\psi\vert^2\), \(\arg\psi\), and \(\langle\hat{p}_x\rangle\), and decide whether the two states represent the same physical system or different ones.

(a) \(\psi(x) = \mathrm{e}^{\mathrm{i}kx}\) and \(\psi'(x) = \mathrm{e}^{\mathrm{i}kx+\mathrm{i}\theta_0}\) (constant overall phase).

(b) \(\psi(x) = \mathrm{e}^{\mathrm{i}kx}\) and \(\psi'(x) = \mathrm{e}^{\mathrm{i}(k+k_0)x}\) (different wavevector).

(c) \(\psi(\boldsymbol{r}) = f(\boldsymbol{r})\) and \(\psi'(\boldsymbol{r}) = \mathrm{e}^{\mathrm{i}q\alpha(\boldsymbol{r})/\hbar}f(\boldsymbol{r})\), together with the simultaneous gauge transformation \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\).

State which transformation is a symmetry and which is a redundancy, and identify the operational diagnostic.

Solution.

With \(\hat{p}_x = -\mathrm{i}\hbar\,\partial_x\), a plane wave \(\mathrm{e}^{\mathrm{i}\kappa x}\) is a momentum eigenstate: \(\hat{p}_x\,\mathrm{e}^{\mathrm{i}\kappa x} = \hbar\kappa\,\mathrm{e}^{\mathrm{i}\kappa x}\), so \(\langle\hat{p}_x\rangle = \hbar\kappa\) (the constant \(\vert\psi\vert^2\) makes this the natural per-particle value).

(a) \(\vert\psi\vert^2 = \vert\psi'\vert^2 = 1\). The arguments are \(\arg\psi = kx\) and \(\arg\psi' = kx + \theta_0\), differing by the constant \(\theta_0\). For the momentum,

\[ \hat{p}_x\psi' = -\mathrm{i}\hbar(\mathrm{i}k)\mathrm{e}^{\mathrm{i}kx+\mathrm{i}\theta_0} = \hbar k\,\psi', \]

so \(\langle\hat{p}_x\rangle = \hbar k\) for both — the constant phase cancels between \(\psi'^{*}\) and \(\psi'\). Every observable agrees: the two describe the same physical system. A constant phase is the global-phase redundancy of §1.1.1.

(b) Again \(\vert\psi\vert^2 = \vert\psi'\vert^2 = 1\) — the probability densities are identical. But \(\arg\psi = kx\) and \(\arg\psi' = (k+k_0)x\) differ by the position-dependent phase \(k_0 x\). The momenta differ:

\[ \hat{p}_x\psi = \hbar k\,\psi, \qquad \hat{p}_x\psi' = \hbar(k+k_0)\,\psi', \]

so \(\langle\hat{p}_x\rangle = \hbar k\) versus \(\hbar(k+k_0)\), and the kinetic energies \(\hbar^2 k^2/2m\) versus \(\hbar^2(k+k_0)^2/2m\) differ as well. Equal \(\vert\psi\vert^2\) is not enough to identify two states: these are physically different plane-wave states (different wavevector \(k\to k+k_0\)). The map \(\psi \to \mathrm{e}^{\mathrm{i}k_0 x}\psi\) is a position-dependent phase with no compensating change of \(\boldsymbol{A}\); it relabels canonical momentum but is not a gauge redundancy of the charged-particle Hamiltonian.

(c) \(\vert\psi'\vert^2 = \vert\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\vert^2\vert f\vert^2 = \vert f\vert^2 = \vert\psi\vert^2\) — same density. The argument shifts by the position-dependent \(q\alpha(\boldsymbol{r})/\hbar\): \(\arg\psi' = \arg f + q\alpha/\hbar\). The canonical momentum does change, \(\langle\hat{p}_x\rangle \to \langle\hat{p}_x\rangle + q\langle\partial_x\alpha\rangle\), but it is accompanied by \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\), so the gauge-invariant kinetic momentum is unchanged:

\[ \langle\hat{\pi}_x'\rangle = \langle\hat{p}_x\rangle + q\langle\partial_x\alpha\rangle - q\langle A_x + \partial_x\alpha\rangle = \langle\hat{p}_x - qA_x\rangle = \langle\hat{\pi}_x\rangle. \]

Every gauge-invariant quantity — density, kinetic momentum (hence velocity), and the fields \(\boldsymbol{E},\boldsymbol{B}\) — agrees. The two describe the same physical system in two different descriptions.

Symmetry versus redundancy. Cases (a) and (c) are redundancies: (a) is the global-phase redundancy, and (c) is the local gauge redundancy in which the wavefunction and the connection transform together. Case (b) is not a redundancy — it changes a physical observable (canonical momentum and kinetic energy) without the compensating connection shift that defines a gauge transformation.

Operational diagnostic. A transformation is a redundancy if and only if every gauge-invariant observable is unchanged — in particular \(\vert\psi\vert^2\) and the kinetic momentum \(\langle\hat{\boldsymbol{p}} - q\boldsymbol{A}\rangle\) (equivalently the velocity) and the fields \(\boldsymbol{E},\boldsymbol{B}\). Equality of \(\vert\psi\vert^2\) alone is insufficient — case (b) passes that test yet is a physical change. The decisive check is the kinetic momentum: in (a) it is unchanged (no \(\boldsymbol{A}\), and \(\langle\hat{p}_x\rangle\) fixed); in (c) \(\langle\hat{p}_x\rangle\) moves but \(\langle\hat{p}_x - qA_x\rangle\) does not; in (b) \(\langle\hat{p}_x\rangle\) moves with no compensating \(\boldsymbol{A}\), so \(\langle\hat{p}_x - qA_x\rangle\) moves — the signature of a physical change.

2. Plane-wave gauge transform. Let \(\psi(\boldsymbol r,t)\) be an arbitrary state (not necessarily a plane wave). Use the single gauge function

\[ \alpha(\boldsymbol{r},t)=\boldsymbol{k}\cdot\boldsymbol{r}-\omega t, \]

with constant \(\boldsymbol{k}\) and \(\omega\), and apply

\[\begin{split} \begin{split} \psi' &= \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi,\\ \boldsymbol A' &= \boldsymbol A + \nabla\alpha,\\ \phi' &= \phi - \partial_t\alpha. \end{split} \end{split}\]

(a) Compute the shifts \(\boldsymbol{A}'-\boldsymbol{A}\) and \(\phi'-\phi\).

(b) Show directly that the fields are unchanged:

\[\begin{split} \begin{split} \boldsymbol E' &= -\nabla\phi' - \partial_t\boldsymbol A' = \boldsymbol E,\\ \boldsymbol B' &= \nabla\times\boldsymbol A' = \boldsymbol B. \end{split} \end{split}\]

(d) Give a short physical interpretation: what changes under this transformation (description/kinematic labels), and what does not (gauge-invariant fields and force law)?

Solution.

(a) The gauge function \(\alpha = \boldsymbol{k}\cdot\boldsymbol{r} - \omega t\) has

\[ \nabla\alpha = \boldsymbol{k}, \qquad \partial_t\alpha = -\omega, \]

both constant in spacetime. Hence

\[ \boldsymbol{A}' - \boldsymbol{A} = \nabla\alpha = \boldsymbol{k}, \qquad \phi' - \phi = -\partial_t\alpha = \omega. \]

The vector potential is shifted by the constant vector \(\boldsymbol{k}\) and the scalar potential by the constant \(\omega\).

(b) Using that \(\boldsymbol{k}\) and \(\omega\) are constants (\(\nabla\omega = 0\), \(\partial_t\boldsymbol{k} = 0\), \(\nabla\times\boldsymbol{k} = 0\)),

\[\begin{split} \begin{split} \boldsymbol{E}' &= -\nabla\phi' - \partial_t\boldsymbol{A}' = -\nabla(\phi+\omega) - \partial_t(\boldsymbol{A}+\boldsymbol{k}) = -\nabla\phi - \partial_t\boldsymbol{A} = \boldsymbol{E},\\ \boldsymbol{B}' &= \nabla\times\boldsymbol{A}' = \nabla\times(\boldsymbol{A}+\boldsymbol{k}) = \nabla\times\boldsymbol{A} = \boldsymbol{B}. \end{split} \end{split}\]

The result is in fact general, for any gauge function \(\alpha\): since \(\nabla\times\nabla\alpha = 0\) and mixed partials commute, \(\partial_t\nabla\alpha = \nabla\partial_t\alpha\),

\[\begin{split} \begin{split} \boldsymbol{B}' &= \nabla\times(\boldsymbol{A}+\nabla\alpha) = \boldsymbol{B} + \nabla\times\nabla\alpha = \boldsymbol{B},\\ \boldsymbol{E}' &= -\nabla(\phi-\partial_t\alpha) - \partial_t(\boldsymbol{A}+\nabla\alpha) = \boldsymbol{E} + \nabla\partial_t\alpha - \partial_t\nabla\alpha = \boldsymbol{E}. \end{split} \end{split}\]

For the present linear \(\alpha\) the shifts happen to be constants, which is the simplest possible illustration of the cancellation.

(c) The fields \(\boldsymbol{E} = -\nabla\phi-\partial_t\boldsymbol{A}\) and \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\) are functionals of the potentials \((\phi,\boldsymbol{A})\) only — the wavefunction \(\psi\) never appears in them. The potential shifts \(\nabla\alpha\) and \(-\partial_t\alpha\) are fixed by \(\alpha\) alone, and the cancellations in (b) rely solely on the identities \(\nabla\times\nabla\alpha = 0\) and \(\partial_t\nabla\alpha = \nabla\partial_t\alpha\), which are properties of \(\alpha\). The state \(\psi\) only receives the phase factor \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\); that factor does not feed back into \(\boldsymbol{E}\) or \(\boldsymbol{B}\). So the field-invariance statement holds whatever \(\psi\) is carried along.

(d) What changes is the description: the local phase convention of the wavefunction (\(\psi\to\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\)) and the kinematic labels built from the potentials — the canonical-momentum origin shifts by \(q\boldsymbol{k}\) and the zero of the electrostatic energy \(q\phi\) shifts by \(q\omega\). What does not change is anything physical: the gauge-invariant fields \(\boldsymbol{E},\boldsymbol{B}\), hence the Lorentz force \(\boldsymbol{F} = q(\boldsymbol{E}+\boldsymbol{v}\times\boldsymbol{B})\), the probability density \(\vert\psi\vert^2\), and the kinetic momentum / velocity. With \(\alpha = \boldsymbol{k}\cdot\boldsymbol{r}-\omega t\) the transformation is a pure relabeling — a uniform shift of the canonical-momentum and energy origins — with no dynamical consequence.

3. Covariance of derivatives. Verify by direct computation that under \(\psi\to\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi,\;\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha,\;\phi\to\phi-\partial_t\alpha\),

\[\begin{split} \begin{split} D'_{i}\psi' &= \mathrm{e}^{\mathrm{i}q\alpha/\hbar}D_{i}\psi,\\ D'_{t}\psi' &= \mathrm{e}^{\mathrm{i}q\alpha/\hbar}D_{t}\psi. \end{split} \end{split}\]

(a) Carry out the spatial calculation explicitly in 1D, showing which \(\partial_{x}\alpha\) terms cancel.

(b) Repeat for the time derivative, showing how the shift \(\phi\to\phi-\partial_{t}\alpha\) cancels the \(\partial_{t}\alpha\) term from differentiating \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\).

(c) Note that the spatial gauge shift uses \(+\nabla\alpha\) while the temporal one uses \(-\partial_{t}\alpha\). Explain in one sentence why these signs are opposite — what does that have to do with the opposite signs in \(D_{i}\) vs \(D_{t}\)?

Solution.

(a) In 1D the spatial covariant derivative is \(D_x = \partial_x - \mathrm{i}(q/\hbar)A_x\). Under the gauge transformation \(A_x \to A_x + \partial_x\alpha\) and \(\psi \to \psi' = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\),

\[ D'_x\psi' = \left(\partial_x - \mathrm{i}\frac{q}{\hbar}(A_x+\partial_x\alpha)\right)\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi. \]

The product rule gives \(\partial_x\bigl(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\bigr) = \mathrm{i}(q/\hbar)(\partial_x\alpha)\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi + \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\partial_x\psi\), so

\[\begin{split} \begin{split} D'_x\psi' &= \mathrm{i}\frac{q}{\hbar}(\partial_x\alpha)\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi + \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\partial_x\psi - \mathrm{i}\frac{q}{\hbar}A_x\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi - \mathrm{i}\frac{q}{\hbar}(\partial_x\alpha)\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\\ &= \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\left(\partial_x - \mathrm{i}\frac{q}{\hbar}A_x\right)\psi = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}D_x\psi. \end{split} \end{split}\]

The first term (\(+\mathrm{i}(q/\hbar)\partial_x\alpha\), from \(\partial_x\) acting on the phase factor) cancels the last term (\(-\mathrm{i}(q/\hbar)\partial_x\alpha\), from the shift of \(A_x\)). The two \(\partial_x\alpha\) pieces are equal and opposite — that is the whole point of the construction.

(b) The temporal covariant derivative is \(D_t = \partial_t + \mathrm{i}(q/\hbar)\phi\). Under \(\phi \to \phi - \partial_t\alpha\),

\[ D'_t\psi' = \left(\partial_t + \mathrm{i}\frac{q}{\hbar}(\phi-\partial_t\alpha)\right)\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi. \]

Using \(\partial_t\bigl(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\bigr) = \mathrm{i}(q/\hbar)(\partial_t\alpha)\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi + \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\partial_t\psi\),

\[\begin{split} \begin{split} D'_t\psi' &= \mathrm{i}\frac{q}{\hbar}(\partial_t\alpha)\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi + \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\partial_t\psi + \mathrm{i}\frac{q}{\hbar}\phi\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi - \mathrm{i}\frac{q}{\hbar}(\partial_t\alpha)\,\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\psi\\ &= \mathrm{e}^{\mathrm{i}q\alpha/\hbar}\left(\partial_t + \mathrm{i}\frac{q}{\hbar}\phi\right)\psi = \mathrm{e}^{\mathrm{i}q\alpha/\hbar}D_t\psi. \end{split} \end{split}\]

The \(+\mathrm{i}(q/\hbar)\partial_t\alpha\) generated by differentiating \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\) is cancelled by the \(-\mathrm{i}(q/\hbar)\partial_t\alpha\) carried in by the shift \(\phi \to \phi - \partial_t\alpha\) inside the term \(+\mathrm{i}(q/\hbar)\phi\).

(c) In each covariant derivative the connection term must generate a piece \(-\mathrm{i}(q/\hbar)(\partial\alpha)\) that cancels the \(+\mathrm{i}(q/\hbar)(\partial\alpha)\) produced when the derivative hits the phase factor; since the spatial connection sits in \(D_i = \partial_i - \mathrm{i}(q/\hbar)A_i\) with a minus sign while the temporal connection sits in \(D_t = \partial_t + \mathrm{i}(q/\hbar)\phi\) with a plus sign (opposite signs mirroring the Lorentzian signature of spacetime), the potential shifts must carry opposite signs too — \(+\nabla\alpha\) for \(\boldsymbol{A}\) and \(-\partial_t\alpha\) for \(\phi\) — so that \((-)\times(+)\) and \((+)\times(-)\) both deliver the same cancelling term.

4. Gauge on punctured plane. Consider a particle restricted to the punctured plane \(\mathbb{R}^{2}\setminus\{0\}\) (the origin removed). Let the gauge function be

\[ \alpha(\rho,\varphi)=\kappa\,\varphi, \]

where \(\varphi\) is the azimuthal angle and \(\kappa\) is a constant.

(a) Compute \(\nabla\alpha\) in polar coordinates and write the transformed vector-potential shift \(\Delta\boldsymbol{A}=\nabla\alpha\).

(b) Show that \(\nabla\times\nabla\alpha=0\) only away from the origin, so \(\boldsymbol{B}\) is unchanged on the punctured plane. Then evaluate \(\oint \nabla\alpha\cdot \mathrm{d}\boldsymbol{\ell}\) around a loop enclosing the hole and use Stokes/distribution language to explain why a singular magnetic flux \(\Phi_{\mathrm{hole}}\) at the hole is required.

(c) Explain why this gauge function is not globally single-valued when \(\varphi\to\varphi+2\pi\). What condition on \(\kappa\) (in terms of \(q\) and \(\hbar\)) makes the wavefunction phase factor \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar}\) single-valued after one full loop?

(d) Summarize the physical picture: outside the hole, local fields are unchanged; at the hole, a singular flux \(\Phi_{\mathrm{hole}}\) carries the nontrivial holonomy. Explain why both statements are needed for a consistent global description.

Solution.

(a) In polar coordinates \(\nabla = \boldsymbol{e}_\rho\,\partial_\rho + \boldsymbol{e}_\varphi\,\rho^{-1}\partial_\varphi\) (with \(\boldsymbol{e}_\rho, \boldsymbol{e}_\varphi\) the polar basis unit vectors). Since \(\alpha = \kappa\varphi\) depends only on \(\varphi\), \(\partial_\rho\alpha = 0\) and \(\partial_\varphi\alpha = \kappa\), so

\[ \nabla\alpha = \frac{\kappa}{\rho}\,\boldsymbol{e}_\varphi, \qquad \Delta\boldsymbol{A} = \nabla\alpha = \frac{\kappa}{\rho}\,\boldsymbol{e}_\varphi. \]

The shift is purely azimuthal and falls off as \(1/\rho\).

(b) The \(z\)-component of the curl of a vector field \(\boldsymbol{V} = V_\rho\boldsymbol{e}_\rho + V_\varphi\boldsymbol{e}_\varphi\) is \((\nabla\times\boldsymbol{V})_z = \rho^{-1}\bigl[\partial_\rho(\rho V_\varphi) - \partial_\varphi V_\rho\bigr]\). Here \(V_\varphi = \kappa/\rho\) and \(V_\rho = 0\), so \(\rho V_\varphi = \kappa\) is constant and

\[ (\nabla\times\nabla\alpha)_z = \frac{1}{\rho}\,\partial_\rho(\kappa) = 0 \qquad (\rho > 0). \]

Thus \(\nabla\times\nabla\alpha = 0\) everywhere on the punctured plane, and \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\) is unchanged there. Yet the circulation around a loop enclosing the hole does not vanish. On a circle of radius \(\rho\), \(\mathrm{d}\boldsymbol{\ell} = \boldsymbol{e}_\varphi\,\rho\,\mathrm{d}\varphi\), so

\[ \oint \nabla\alpha\cdot\mathrm{d}\boldsymbol{\ell} = \int_0^{2\pi}\frac{\kappa}{\rho}\,(\rho\,\mathrm{d}\varphi) = 2\pi\kappa \neq 0. \]

By Stokes’ theorem this circulation equals \(\int(\nabla\times\nabla\alpha)\cdot\mathrm{d}\boldsymbol{S}\) over any surface spanning the loop. The integrand vanishes everywhere the surface stays on the punctured plane, yet the integral is \(2\pi\kappa\) — a contradiction that is resolved only if \(\nabla\times\nabla\alpha\) carries a singular contribution at the excluded origin,

\[ \nabla\times\nabla\alpha = 2\pi\kappa\,\delta^2(\boldsymbol{r})\,\boldsymbol{e}_z. \]

The curl of the gradient is not globally zero — it is a point source. In field language, \(\Delta\boldsymbol{A} = \nabla\alpha\) is the vector potential of a singular magnetic flux line \(\Phi_{\mathrm{hole}} = \oint\Delta\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{\ell} = 2\pi\kappa\) threading the hole. A would-be “pure gauge” \(\nabla\alpha\) built from a multivalued \(\alpha\) is really a flux-carrying configuration: a consistent global description must place that flux somewhere, and the only location available is the excluded origin.

(c) The function \(\alpha = \kappa\varphi\) jumps by \(2\pi\kappa\) under \(\varphi\to\varphi+2\pi\), so it is multivalued for any \(\kappa\neq0\). The object that must be well defined is the wavefunction phase factor \(\mathrm{e}^{\mathrm{i}q\alpha/\hbar} = \mathrm{e}^{\mathrm{i}q\kappa\varphi/\hbar}\); after one full loop it is multiplied by \(\mathrm{e}^{2\pi\mathrm{i}q\kappa/\hbar}\). Single-valuedness of the wavefunction requires

\[ \mathrm{e}^{2\pi\mathrm{i}q\kappa/\hbar} = 1, \]

equivalent to \(q\kappa/\hbar = n\), i.e. \(\kappa = n\hbar/q\) for \(n\in\mathbb{Z}\).

Equivalently the inserted flux \(\Phi_{\mathrm{hole}} = 2\pi\kappa = 2\pi n\hbar/q = n\,\Phi_0\) is an integer multiple of the flux quantum for charge \(q\),

\[ \Phi_0 = \frac{2\pi\hbar}{q} = \frac{h}{q} \]

(for an electron with \(q=-e\) this is \(h/e\); in a superconductor the Cooper-pair charge \(2e\) gives the familiar \(h/2e\)). Only for \(\kappa = n\hbar/q\) is the multivalued gauge function an admissible (genuinely redundant) transformation; for other \(\kappa\) it shifts the physical Aharonov–Bohm phase and is not a redundancy.

(d) Outside the hole the transformation is locally indistinguishable from a trivial gauge change: \(\nabla\times\nabla\alpha = 0\), so \(\boldsymbol{B}\) and every local field measurement are untouched. At the hole, the multivaluedness is concentrated as a singular flux \(\Phi_{\mathrm{hole}} = 2\pi\kappa\) carrying the nontrivial holonomy \(\oint\boldsymbol{A}\cdot\mathrm{d}\boldsymbol{\ell}\). Both statements are needed. The local statement says that no local probe can detect the change. The global statement says that a non-local probe — an interference experiment on a loop encircling the hole — does register it, through the Aharonov-Bohm phase \(q\Phi_{\mathrm{hole}}/\hbar = 2\pi q\kappa/\hbar\). Reporting only the local fact would falsely conclude “nothing happened”; reporting only the flux would miss that the particle’s accessible region is field-free. A topologically nontrivial domain — the puncture — is precisely where “gauge-equivalent” stops meaning “physically identical,” unless the flux is quantized to \(n\Phi_0\), in which case even the holonomy is trivial and the transformation is a true redundancy.

5. What gauge transformations cannot do. A particle moves in a static gravitational potential \(V(\boldsymbol{r}) = mgz\), where \(\boldsymbol{r}=(x,y,z)\) and \(z\) is the vertical (height) coordinate. One might claim: “A spatial gauge transformation \(\psi \to \mathrm{e}^{\mathrm{i}q\alpha(\boldsymbol{r})/\hbar}\psi\) should remove \(V\) from the Schrödinger equation, just as it removes a static \(\boldsymbol{A}\).”

(a) Try to find \(\alpha(\boldsymbol{r})\) that eliminates \(V\). Where does the attempt fail?

(b) Allow \(\alpha\) to depend on \(t\). The only way to cancel \(V\) in the \(q\phi\) slot of the Hamiltonian is \(-\partial_{t}\alpha = -V/q\), which forces \(\alpha(\boldsymbol{r}, t) = (V(\boldsymbol{r})/q)\,t + g(\boldsymbol{r})\) for some \(g\). Compute the resulting \(\boldsymbol{A}' = \boldsymbol{A} + \nabla\alpha\) and show it is now time-dependent. Conclude that \(V\) has not been removed — it has been shuffled into a time-dependent vector potential.

(c) Conclude: gauge transformations can shuffle physical content between \(\phi\) and \(\boldsymbol{A}\) but cannot erase it. The energy landscape of an external \(V(\boldsymbol{r})\) is gauge-invariant — gravity does not arise from a \(U(1)\) gauge redundancy.

Solution.

(a) A static gauge function \(\alpha(\boldsymbol{r})\) has \(\partial_t\alpha = 0\), so it shifts only the vector potential, \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\), and leaves the scalar potential entirely untouched, \(\phi\to\phi-\partial_t\alpha = \phi\). In the Hamiltonian

\[ \hat{H} = \frac{(\hat{\boldsymbol{p}}-q\boldsymbol{A})^2}{2m} + q\phi, \]

the gravitational potential sits in the \(q\phi\) slot, \(\phi = V/q = mgz/q\) — and a static \(\alpha\) cannot reach that slot. One might instead hope to bury \(V\) inside the kinetic term by a clever \(\boldsymbol{A}\): but \((\hat{\boldsymbol{p}}-q\boldsymbol{A})^2/2m\) is quadratic in momentum, and no choice of \(\boldsymbol{A}(\boldsymbol{r})\) produces a momentum-independent term \(-V(\boldsymbol{r})\) — completing the square always leaves a residual term linear in \(\hat{\boldsymbol{p}}\). The attempt fails because the redundancy generated by a static \(\alpha\) is confined to \(\boldsymbol{A}\), whereas \(V\) lives in \(\phi\).

(b) Allowing \(\alpha(\boldsymbol{r},t)\), cancelling \(V\) in the \(q\phi\) slot requires \(\phi' = \phi - \partial_t\alpha = 0\) with \(q\phi = V\), i.e. \(\partial_t\alpha = V/q\). Integrating,

\[ \alpha(\boldsymbol{r},t) = \frac{V(\boldsymbol{r})}{q}\,t + g(\boldsymbol{r}), \]

for an arbitrary static \(g(\boldsymbol{r})\). Starting from \(\boldsymbol{A} = 0\), the resulting vector potential is

\[ \boldsymbol{A}' = \boldsymbol{A} + \nabla\alpha = \frac{t}{q}\,\nabla V + \nabla g = \frac{mg\,t}{q}\,\boldsymbol{e}_z + \nabla g, \]

using \(\nabla V = \nabla(mgz) = mg\,\boldsymbol{e}_z\). The \(\boldsymbol{e}_z\)-term (with \(\boldsymbol{e}_z\) the unit vector along the \(z\)-axis) grows linearly in \(t\): \(\boldsymbol{A}'\) is explicitly time-dependent. The potential energy has not been removed — it has reappeared as a time-dependent vector potential. Consistently, the gauge-invariant field is unchanged: before, \(\boldsymbol{E} = -\nabla\phi = -(mg/q)\boldsymbol{e}_z\); after,

\[ \boldsymbol{E}' = -\nabla\phi' - \partial_t\boldsymbol{A}' = 0 - \frac{mg}{q}\,\boldsymbol{e}_z = \boldsymbol{E}. \]

The transformation moved the physics out of \(\phi\) and into \(\boldsymbol{A}\) without erasing it.

(c) A gauge transformation can shuffle physical content between \(\phi\) and \(\boldsymbol{A}\) but cannot erase it: the invariant combination \(\boldsymbol{E} = -\nabla\phi - \partial_t\boldsymbol{A}\) — here the uniform \(-(mg/q)\boldsymbol{e}_z\) — is fixed under every gauge choice. The energy landscape of an external \(V(\boldsymbol{r})\) is gauge-invariant physical input. Unlike the phase of \(\psi\), which is local internal data and therefore a redundancy, the gravitational potential is not a redundancy of the \(U(1)\) description and cannot be gauged away. Gravity does not arise from a \(U(1)\) phase redundancy. (Gravity is governed by a different local invariance — general covariance / diffeomorphisms — which lies outside the \(U(1)\) gauge principle of this section.)

★ 6. Second-order bilinear covariance. Consider the same local phase redundancy

\[ \psi' = \mathrm{e}^{\mathrm{i}q\alpha(\boldsymbol r)/\hbar}\psi. \]

For second spatial derivatives, study the bilinear combination

\[ \partial_{ij}[\psi]\equiv \psi\,\partial_i\partial_j\psi-(\partial_i\psi)(\partial_j\psi). \]

(a) Expand \(\partial_{ij}[\psi']\) explicitly and show that the non-covariant remainder equals \((\mathrm{i}q/\hbar)(\partial_i\partial_j\alpha)\psi^2 \cdot \mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\).

(b) Introduce a symmetric rank-2 gauge field \(A_{ij}=A_{ji}\) and define

\[ \mathcal{D}_{ij}[\psi]\equiv \psi\,\partial_i\partial_j\psi-(\partial_i\psi)(\partial_j\psi)-\mathrm{i}\frac{q}{\hbar}A_{ij}\psi^2. \]

Find the transformation law for \(A_{ij}\) such that

\[ \mathcal{D}'_{ij}[\psi'] = \mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\,\mathcal{D}_{ij}[\psi]. \]

(d) Build one gauge-invariant scalar density from \(\mathcal{D}_{ij}[\psi]\) and \(\psi\) (for example using index contraction and complex conjugation), and state in one sentence what physical type of constrained motion this kind of second-derivative gauge structure is designed to capture.

Solution.

Write \(\Lambda \equiv q\alpha/\hbar\), so \(\psi' = \mathrm{e}^{\mathrm{i}\Lambda}\psi\) and \(\partial_i\psi' = \mathrm{e}^{\mathrm{i}\Lambda}\bigl(\partial_i\psi + \mathrm{i}(\partial_i\Lambda)\psi\bigr)\).

(a) Differentiating once more,

\[ \partial_i\partial_j\psi' = \mathrm{e}^{\mathrm{i}\Lambda}\Bigl[\partial_i\partial_j\psi + \mathrm{i}(\partial_i\Lambda)\partial_j\psi + \mathrm{i}(\partial_j\Lambda)\partial_i\psi + \mathrm{i}(\partial_i\partial_j\Lambda)\psi - (\partial_i\Lambda)(\partial_j\Lambda)\psi\Bigr]. \]

The two terms of \(\partial_{ij}[\psi']\) are therefore

\[ \psi'\,\partial_i\partial_j\psi' = \mathrm{e}^{2\mathrm{i}\Lambda}\,\psi\Bigl[\partial_i\partial_j\psi + \mathrm{i}(\partial_i\Lambda)\partial_j\psi + \mathrm{i}(\partial_j\Lambda)\partial_i\psi + \mathrm{i}(\partial_i\partial_j\Lambda)\psi - (\partial_i\Lambda)(\partial_j\Lambda)\psi\Bigr], \]

\[ (\partial_i\psi')(\partial_j\psi') = \mathrm{e}^{2\mathrm{i}\Lambda}\Bigl[(\partial_i\psi)(\partial_j\psi) + \mathrm{i}(\partial_j\Lambda)\psi\,\partial_i\psi + \mathrm{i}(\partial_i\Lambda)\psi\,\partial_j\psi - (\partial_i\Lambda)(\partial_j\Lambda)\psi^2\Bigr]. \]

Subtracting, the three pieces \(\mathrm{i}(\partial_i\Lambda)\psi\,\partial_j\psi\), \(\mathrm{i}(\partial_j\Lambda)\psi\,\partial_i\psi\), and \(-(\partial_i\Lambda)(\partial_j\Lambda)\psi^2\) are common to both lines and cancel, leaving

\[ \partial_{ij}[\psi'] = \mathrm{e}^{2\mathrm{i}\Lambda}\Bigl[\psi\,\partial_i\partial_j\psi - (\partial_i\psi)(\partial_j\psi) + \mathrm{i}(\partial_i\partial_j\Lambda)\psi^2\Bigr] = \mathrm{e}^{2\mathrm{i}\Lambda}\Bigl[\partial_{ij}[\psi] + \mathrm{i}(\partial_i\partial_j\Lambda)\psi^2\Bigr]. \]

With \(\Lambda = q\alpha/\hbar\), so \(\partial_i\partial_j\Lambda = (q/\hbar)\partial_i\partial_j\alpha\),

\[ \partial_{ij}[\psi'] = \mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\,\partial_{ij}[\psi] + \frac{\mathrm{i}q}{\hbar}(\partial_i\partial_j\alpha)\,\psi^2\,\mathrm{e}^{2\mathrm{i}q\alpha/\hbar}. \]

Every term containing a first derivative of \(\alpha\) has cancelled; the non-covariant remainder is exactly \((\mathrm{i}q/\hbar)(\partial_i\partial_j\alpha)\psi^2\cdot\mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\), as claimed.

(b) Using \(\psi'^2 = \mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\psi^2\) together with the result of (a),

\[ \mathcal{D}'_{ij}[\psi'] = \partial_{ij}[\psi'] - \frac{\mathrm{i}q}{\hbar}A'_{ij}\psi'^2 = \mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\Bigl[\partial_{ij}[\psi] + \frac{\mathrm{i}q}{\hbar}(\partial_i\partial_j\alpha)\psi^2 - \frac{\mathrm{i}q}{\hbar}A'_{ij}\psi^2\Bigr]. \]

This equals \(\mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\,\mathcal{D}_{ij}[\psi] = \mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\bigl[\partial_{ij}[\psi] - \mathrm{i}(q/\hbar)A_{ij}\psi^2\bigr]\) if and only if \((\partial_i\partial_j\alpha) - A'_{ij} = -A_{ij}\), i.e.

\[ A_{ij} \to A'_{ij} = A_{ij} + \partial_i\partial_j\alpha. \]

The inhomogeneous shift \(\partial_i\partial_j\alpha\) is automatically symmetric in \(i,j\), consistent with \(A_{ij} = A_{ji}\); it is the rank-2 analogue of \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\), carrying one extra derivative.

(c) Neither term transforms cleanly on its own. From the expansion in (a), \(\psi\,\partial_i\partial_j\psi\) alone picks up \(\mathrm{i}(\partial_i\Lambda)\psi\,\partial_j\psi + \mathrm{i}(\partial_j\Lambda)\psi\,\partial_i\psi - (\partial_i\Lambda)(\partial_j\Lambda)\psi^2 + \mathrm{i}(\partial_i\partial_j\Lambda)\psi^2\), and \((\partial_i\psi)(\partial_j\psi)\) alone picks up the same first three pieces. Each separately carries artifacts proportional to \(\partial\Lambda\) and \((\partial\Lambda)^2\) that no single linear shift of one field could absorb. Only their difference cancels all of those, leaving the lone inhomogeneous piece \(\mathrm{i}(\partial_i\partial_j\Lambda)\psi^2\) — and an inhomogeneous shift linear in \(\partial_i\partial_j\alpha\) is precisely what a connection field \(A_{ij}\) can soak up. Pairing the second-derivative term with the product-of-first-derivatives term is the rank-2 echo of why \(D_i = \partial_i - \mathrm{i}(q/\hbar)A_i\) pairs a derivative with a connection: the combination is engineered so that the redundancy artifact collapses to a pure connection shift.

(d) The object \(\mathcal{D}_{ij}[\psi]\) carries phase weight \(\mathrm{e}^{2\mathrm{i}q\alpha/\hbar}\) (the same weight as \(\psi^2\)), and its complex conjugate carries \(\mathrm{e}^{-2\mathrm{i}q\alpha/\hbar}\). Contracting the symmetric index pair with \(\delta^{ij}\) to form a scalar and taking the modulus squared,

\[ \mathcal{S}[\psi] \equiv \sum_{i,j}\bigl\vert\mathcal{D}_{ij}[\psi]\bigr\vert^2 = \sum_{i,j}\overline{\mathcal{D}_{ij}[\psi]}\;\mathcal{D}_{ij}[\psi], \]

transforms by \(\mathrm{e}^{-2\mathrm{i}q\alpha/\hbar}\,\mathrm{e}^{+2\mathrm{i}q\alpha/\hbar} = 1\) — a gauge-invariant scalar density. (Equivalently, \(\delta^{ij}\,\bar{\psi}^2\,\mathcal{D}_{ij}[\psi]\) is invariant, the factor \(\bar{\psi}^2\) supplying the cancelling \(\mathrm{e}^{-2\mathrm{i}q\alpha/\hbar}\).) This rank-2 connection pattern is the same algebraic idea used in more advanced models with restricted-mobility matter, where second-derivative couplings enforce restricted mobility — matter that carries charge and dipole quantum numbers and cannot move as freely as in ordinary electromagnetism. That application lies beyond the minimal \(U(1)\) gauge principle of §4.1.1, but the bilinear pairing in (c) is exactly what makes the construction work here.

7. Gauge connection on links. Consider a one-dimensional tight-binding model with hopping between neighboring sites \(n\) and \(n+1\):

\[ \hat{H}_{\text{hop}}=-t\sum_n \left(U_{n,n+1}\,\hat{c}_{n+1}^{\dagger}\hat{c}_n+\text{h.c.}\right), \]

where \(U_{n,n+1}=\mathrm{e}^{\mathrm{i}q a A_n/\hbar}\) is a link variable and \(a\) is the lattice spacing. Under a local phase redefinition,

\[ \hat{c}_n\to \mathrm{e}^{\mathrm{i}q\alpha_n/\hbar}\hat{c}_n. \]

(a) Find how \(U_{n,n+1}\) must transform so that each hopping term remains invariant.

(b) Translate your answer into a transformation law for \(A_n\). Show that, in the continuum limit, it becomes \(A\to A+\partial_x\alpha\).

(c) Explain why the link variable is the lattice version of a connection: what does it compare between neighboring sites, and why is it forced by local phase redundancy?

Solution.

(a) Under \(\hat{c}_n \to \mathrm{e}^{\mathrm{i}q\alpha_n/\hbar}\hat{c}_n\) the creation operator transforms as \(\hat{c}_n^\dagger \to \mathrm{e}^{-\mathrm{i}q\alpha_n/\hbar}\hat{c}_n^\dagger\), so the bare neighbour bilinear becomes

\[ \hat{c}_{n+1}^\dagger\hat{c}_n \to \mathrm{e}^{-\mathrm{i}q\alpha_{n+1}/\hbar}\,\mathrm{e}^{\mathrm{i}q\alpha_n/\hbar}\,\hat{c}_{n+1}^\dagger\hat{c}_n = \mathrm{e}^{-\mathrm{i}q(\alpha_{n+1}-\alpha_n)/\hbar}\,\hat{c}_{n+1}^\dagger\hat{c}_n. \]

For the hopping term \(U_{n,n+1}\,\hat{c}_{n+1}^\dagger\hat{c}_n\) to stay invariant, the link variable must absorb the inverse factor:

\[ U_{n,n+1} \to U'_{n,n+1} = \mathrm{e}^{\mathrm{i}q\alpha_{n+1}/\hbar}\,U_{n,n+1}\,\mathrm{e}^{-\mathrm{i}q\alpha_n/\hbar}. \]

Check: \(U'_{n,n+1}\,\hat{c}_{n+1}^{\prime\dagger}\hat{c}'_n = \mathrm{e}^{\mathrm{i}q\alpha_{n+1}/\hbar}U_{n,n+1}\mathrm{e}^{-\mathrm{i}q\alpha_n/\hbar}\cdot\mathrm{e}^{-\mathrm{i}q\alpha_{n+1}/\hbar}\mathrm{e}^{\mathrm{i}q\alpha_n/\hbar}\,\hat{c}_{n+1}^\dagger\hat{c}_n = U_{n,n+1}\,\hat{c}_{n+1}^\dagger\hat{c}_n\), and the Hermitian conjugate term is then invariant automatically. The link variable picks up the phase of the site it points to and the inverse phase of the site it points from — a parallel-transport law.

(b) Writing \(U_{n,n+1} = \mathrm{e}^{\mathrm{i}qaA_n/\hbar}\) and \(U'_{n,n+1} = \mathrm{e}^{\mathrm{i}qaA'_n/\hbar}\), all factors are c-numbers, so exponents add:

\[ \frac{qa}{\hbar}A'_n = \frac{qa}{\hbar}A_n + \frac{q}{\hbar}(\alpha_{n+1}-\alpha_n), \]

so \(A'_n = A_n + (\alpha_{n+1}-\alpha_n)/a\).

In the continuum limit \(a\to0\), label sites by position \(x\) with \(\alpha_n = \alpha(x)\) and \(\alpha_{n+1} = \alpha(x+a)\):

\[ \frac{\alpha_{n+1}-\alpha_n}{a} = \frac{\alpha(x+a)-\alpha(x)}{a} \xrightarrow[a\to0]{} \partial_x\alpha, \]

which is the definition of the derivative. Hence \(A \to A + \partial_x\alpha\) — exactly the continuum gauge transformation \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\) restricted to one dimension.

(c) The link variable \(U_{n,n+1}\) lives on the bond between sites \(n\) and \(n+1\), not on a site. It is a parallel transporter: it specifies how to carry the phase convention of site \(n\) over to site \(n+1\) so that the two can be compared. The phase of \(\hat{c}_n\) at each site is local internal data with an independently choosable convention, so the bare neighbour comparison \(\hat{c}_{n+1}^\dagger\hat{c}_n\) has no convention-independent meaning — under a local rephasing it changes by \(\mathrm{e}^{-\mathrm{i}q(\alpha_{n+1}-\alpha_n)/\hbar}\). The link variable is forced to exist precisely because only a quantity transforming by exactly the endpoint phases can soak up that relative phase and render the hopping term gauge-invariant. That is the defining property of a connection: it compares internal data at neighbouring points. Thus \(U_{n,n+1}\) is the lattice connection, \(A_n\) is its potential, the Peierls phase \(\mathrm{e}^{\mathrm{i}qaA_n/\hbar}\) is the lattice realization of minimal coupling, and \(\partial_x\alpha\) emerges as the continuum gauge-transformation law.