Local Phase Transformations

A local phase transformation changes the phase convention of a wavefunction by an amount that depends on position and possibly time:

$$\psi(\mathbf r,t) \mapsto \psi'(\mathbf r,t) = e^{iq\chi(\mathbf r,t)/\hbar} \psi(\mathbf r,t).$$

If $\chi$ is constant, this is just a global phase. If $\chi$ varies in space or time, ordinary derivatives detect the change. The basic lesson is that local phase freedom requires a compensating connection. In electromagnetic wave mechanics that connection is the scalar and vector potential.

The practical gauge-transformation rules for charged-particle Schrödinger equations are introduced in Gauge Transformations: First Encounter. This page focuses on the structural reason those rules are necessary.

From Global to Local Phase

For one isolated normalized state, multiplying by $e^{i\alpha}$ changes only the representative of the same ray. The page Global Phase and Physical States explains that ray viewpoint.

A local phase transformation applies a different phase at each spacetime point. It does not merely multiply the whole wavefunction by one number. In one spatial dimension, writing $\theta(x)=q\chi(x)/\hbar$,

$$\psi'(x) = e^{i\theta(x)}\psi(x).$$

The probability density is unchanged:

$$\lvert\psi'(x)\rvert^2 = \lvert\psi(x)\rvert^2.$$

But the derivative is not simply phase rotated:

$$\frac{d\psi'}{dx} = e^{i\theta(x)} \left( \frac{d\psi}{dx} + i\frac{d\theta}{dx}\psi \right).$$

The extra term is the whole story. A derivative compares wavefunction values at neighboring points. If the phase convention can change from point to point, comparison requires a rule for how to align those phases.

Why Ordinary Momentum Is Not Covariant

The canonical momentum operator in position representation is

$$\hat{\mathbf p} = -i\hbar\nabla.$$

Under the local phase transformation

$$\psi' = e^{iq\chi/\hbar}\psi,$$

ordinary momentum gives

$$\hat{\mathbf p}\psi' = e^{iq\chi/\hbar} \left( \hat{\mathbf p}\psi + q(\nabla\chi)\psi \right).$$

The extra $q\nabla\chi$ term means $\hat{\mathbf p}\psi$ does not transform in the same simple way as $\psi$. This is not a problem for global phases, because $\nabla\chi=0$ when $\chi$ is constant. It becomes unavoidable for local phases.

The canonical momentum is therefore not the right gauge-covariant object for a charged particle in electromagnetic potentials. The covariant, mechanical momentum is

$$\hat{\boldsymbol\pi} = \hat{\mathbf p}-q\mathbf A.$$

Covariant Spatial Derivative

Define the spatial covariant derivative by

$$D_i = \partial_i - \frac{iq}{\hbar}A_i.$$

Then

$$\hat\pi_i = -i\hbar D_i.$$

The electromagnetic gauge transformation is

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

With these two transformations together,

$$D_i'\psi' = e^{iq\chi/\hbar}D_i\psi.$$

Equivalently,

$$\hat\pi_i'\psi' = e^{iq\chi/\hbar} \hat\pi_i\psi.$$

This is gauge covariance. The object may change representative, but it changes by the same local phase as the wavefunction itself. That is exactly what is needed for probabilities, currents, and equations of motion to be independent of phase convention.

Time-Dependent Phase

If the phase convention also depends on time, the scalar potential participates. The standard transformation is

$$\Phi' = \Phi-\frac{\partial\chi}{\partial t}.$$

A useful time covariant derivative is

$$D_t = \frac{\partial}{\partial t} + \frac{iq}{\hbar}\Phi.$$

It obeys

$$D_t'\psi' = e^{iq\chi/\hbar}D_t\psi.$$

With these definitions, the minimally coupled Schrödinger equation can be written compactly as

$$i\hbar D_t\psi = -\frac{\hbar^2}{2m} \sum_i D_iD_i\psi.$$

This equation has the same form in every gauge. The wavefunction, scalar potential, and vector potential have changed representative; the physics has not.

Gauge Redundancy Versus Physical Symmetry

A local phase transformation in this setting is a change of description, not a new physical state. The transformed pair

$$(\psi,\mathbf A,\Phi) \quad\longmapsto\quad (\psi',\mathbf A',\Phi')$$

represents the same charged-particle physics when all three objects are transformed consistently.

This differs from an ordinary symmetry operation that maps one physical state to another distinct physical state. For example, a spatial translation can move a localized wavepacket to a different position. A gauge transformation instead changes the phase convention and potentials used to describe the same situation.

The distinction is subtle because constant phase rotations also appear in Noether’s theorem and charge conservation in field theory. The bridge page From Phase Symmetry to Gauge Theory separates those roles. In this chapter, the main point is operational: local phase convention is redundant, while gauge-invariant phase holonomy can be observable.

Probability Current

The gauge-covariant probability current for a charged particle is

$$\mathbf j = \frac{1}{m} \operatorname{Re} \left[ \psi^* \left( -i\hbar\nabla-q\mathbf A \right) \psi \right].$$

Because the covariant momentum transforms by the same phase as $\psi$, the current is invariant:

$$\mathbf j' = \mathbf j.$$

In polar form,

$$\psi = R e^{iS/\hbar},$$

the local phase transformation changes

$$S \mapsto S+q\chi.$$

The gauge-invariant velocity field depends on

$$\nabla S-q\mathbf A,$$

not on $\nabla S$ or $\mathbf A$ separately.

Closed Loops

For an open path, a phase integral can depend on the chosen gauge and endpoints. For a closed loop, the electromagnetic phase

$$\exp \left( \frac{iq}{\hbar} \oint_C \mathbf A\cdot d\mathbf r \right)$$

is invariant under smooth single-valued gauge transformations:

$$\oint_C \nabla\chi\cdot d\mathbf r = 0.$$

This is why local phase transformations naturally lead to holonomy. The Aharonov–Bohm Effect is the prototype: the phase around a loop can be measurable even when $\mathbf B=\nabla\times\mathbf A$ vanishes along the particle’s path.

Common Mistakes

• Changing the vector potential without changing the wavefunction phase.
• Treating a local phase transformation as if it were only a harmless global phase.
• Calling $\mathbf A$ observable by itself rather than using gauge-invariant combinations.
• Forgetting the scalar potential when the gauge function depends on time.
• Comparing canonical momentum labels from different gauges as if they were gauge-invariant observables.

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• Y. Aharonov and D. Bohm, “Significance of electromagnetic potentials in the quantum theory,” Physical Review 115, 485-491, 1959.

Exercises

1. Show explicitly that $D_i'\psi'=e^{iq\chi/\hbar}D_i\psi$ for $D_i=\partial_i-iqA_i/\hbar$.

Solution

Use

$$A_i' = A_i+\partial_i\chi, \qquad \psi' = e^{iq\chi/\hbar}\psi.$$

Then

$$\begin{aligned} D_i'\psi' &= \left( \partial_i-\frac{iq}{\hbar}A_i' \right) \left( e^{iq\chi/\hbar}\psi \right) \\ &= e^{iq\chi/\hbar} \left[ \partial_i\psi + \frac{iq}{\hbar}(\partial_i\chi)\psi - \frac{iq}{\hbar} \left( A_i+\partial_i\chi \right)\psi \right] \\ &= e^{iq\chi/\hbar} \left( \partial_i-\frac{iq}{\hbar}A_i \right)\psi \\ &= e^{iq\chi/\hbar}D_i\psi. \end{aligned}$$

2. Verify the time covariant derivative transformation for $D_t=\partial_t+iq\Phi/\hbar$.

Solution

Use

$$\Phi' = \Phi-\partial_t\chi, \qquad \psi' = e^{iq\chi/\hbar}\psi.$$

Then

$$\begin{aligned} D_t'\psi' &= \left( \partial_t+\frac{iq}{\hbar}\Phi' \right) \left( e^{iq\chi/\hbar}\psi \right) \\ &= e^{iq\chi/\hbar} \left[ \partial_t\psi + \frac{iq}{\hbar}(\partial_t\chi)\psi + \frac{iq}{\hbar} \left( \Phi-\partial_t\chi \right)\psi \right] \\ &= e^{iq\chi/\hbar}D_t\psi. \end{aligned}$$

3. Let $\psi=R e^{iS/\hbar}$. Show that $\nabla S-q\mathbf A$ is gauge invariant.

Solution

The local phase transformation gives

$$S' = S+q\chi, \qquad \mathbf A' = \mathbf A+\nabla\chi.$$

Therefore

$$\nabla S'-q\mathbf A' = \nabla S+q\nabla\chi - q\mathbf A - q\nabla\chi = \nabla S-q\mathbf A.$$