The potentials are not fixed by the fields. Add the gradient of any function $\chi$ to $\mathbf{A}$ and subtract its time derivative from $\phi$, and $\mathbf{B}$ and $\mathbf{E}$ are unchanged, because a gradient has no curl. In quantum mechanics the Hamiltonian contains $\mathbf{A}$ itself, so the Schrödinger equation would change — unless the wavefunction changes too. Multiplying it by the phase $e^{iq\chi/\hbar}$, different at every point, exactly compensates. Densities, currents and energies stay the same; only the phase of the wavefunction and the canonical momentum depend on the choice. This is gauge invariance, and it can be read the other way round: demanding that a local change of phase be harmless forces the potentials into the Hamiltonian, in the combination $\hat{\mathbf{p}}-q\mathbf{A}$.

Heights on a map can be measured from sea level or from the valley floor; slopes, which are all a walker feels, do not care. Potentials are like the heights and fields like the slopes, and the wavefunction’s phase has to be re-measured along with the potentials.

The Landau gauge $\mathbf{A}=B(0,x,0)$ of a uniform field along $z$, drawn at half scale: it points along $y$, up to the right of the $y$ axis and down to its left, growing with the distance from it. It differs from the circling by the gradient of , and has the same curl.

For any smooth function $\chi(\mathbf{r},t)$, the potentials and the wavefunction may be changed together:

• $\mathbf{B}'=\mathbf{B}$ because $\nabla\times\nabla\chi=0$, and ${\mathbf{E}}^{\mathrm{\prime }}=\mathbf{E}$ because the changes of and of cancel.

The momentum acting on the phase brings down $q\nabla\chi$, which the change of $\mathbf{A}$ removes, so the kinetic momentum passes through the phase unchanged, and so does its square. The time derivative acting on the phase brings down $-q\partial_{t}\chi$, which the change of $$ removes.

The product rule with $\hat{\mathbf{p}}=-i\hbar\nabla$.

$q\nabla\chi$ cancels against $\mathbf{A}'=\mathbf{A}+\nabla\chi$.

Apply the last line twice.

$i\hbar\partial_{t}$ brings $-q\,\partial_{t}\chi$ down from the phase, which $$ cancels.

Combine the last two lines: one side vanishes exactly when the other does.

Adding $\nabla\chi$ to $\mathbf{A}$ and subtracting $\partial\chi/\partial t$ from $\phi$ changes neither $\mathbf{B}$ nor .

To keep the Schrödinger equation, the wavefunction is multiplied by a phase that varies from point to point.

Probability densities, currents and energies are the same in every gauge; only the phase of the wavefunction and the canonical momentum change.

The symmetric gauge $\tfrac12B(-y,x,0)$ and the Landau gauge $B(0,x,0)$ give the same uniform field; they differ by the gradient of .

Great work! You now know how to:

• change the potentials without changing the fields
• change the wavefunction’s phase to match
• tell what depends on the gauge from what does not