Gauge Transformations in Quantum Mechanics

A gauge transformation changes the representative wavefunction and electromagnetic potentials without changing the physical situation. For a particle of charge $q$,

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

The transformed triple $(\psi',\mathbf A',\Phi')$ is a different description of the same physics as $(\psi,\mathbf A,\Phi)$, provided observables and boundary conditions are compared correctly.

The first wave-mechanics derivation is in Gauge Transformations: First Encounter. This page is the symmetry and geometry version: gauge transformations are redundancies in description, while gauge-invariant quantities are the physical content.

Gauge Equivalence

Two descriptions are gauge equivalent when they are related by a real function $\chi(\mathbf r,t)$ as above. The physical state of a charged particle in a background electromagnetic field is not the bare wavefunction alone and not the potentials alone. It is the gauge-equivalence class of the combined description.

This is why comparing wavefunctions in different gauges can be misleading. The question is not whether

$$\psi'(\mathbf r,t) = \psi(\mathbf r,t)$$

as functions. The question is whether the full descriptions are related by the gauge rule and therefore give the same gauge-invariant predictions.

Gauge equivalence also depends on the domain and boundary conditions. On a simply connected region with ordinary single-valued gauge functions, many pure-gradient potentials are removable. On multiply connected regions, loop phases can survive as physical holonomy.

Fields Are Invariant

The electromagnetic fields are

$$\mathbf B = \nabla\times\mathbf A, \qquad \mathbf E = -\nabla\Phi-\partial_t\mathbf A.$$

Under the gauge transformation,

$$\mathbf B' = \nabla\times(\mathbf A+\nabla\chi) = \mathbf B,$$

and

$$\begin{aligned} \mathbf E' &= -\nabla(\Phi-\partial_t\chi) - \partial_t(\mathbf A+\nabla\chi) \\ &= -\nabla\Phi-\partial_t\mathbf A = \mathbf E. \end{aligned}$$

The local electric and magnetic fields therefore do not distinguish gauge-equivalent potential pairs.

Schrödinger Equation Covariance

Let

$$H[\mathbf A,\Phi] = \frac{1}{2m} \left( -i\hbar\nabla-q\mathbf A \right)^2 + q\Phi.$$

The time-dependent Schrödinger equation is

$$i\hbar\partial_t\psi = H[\mathbf A,\Phi]\psi.$$

Define the phase operator

$$U_\chi(\mathbf r,t) = e^{iq\chi(\mathbf r,t)/\hbar}.$$

Since $\psi'=U_\chi\psi$, the transformed Hamiltonian satisfies

$$H[\mathbf A',\Phi'] = U_\chi H[\mathbf A,\Phi]U_\chi^\dagger + i\hbar(\partial_tU_\chi)U_\chi^\dagger.$$

The second term is essential for time-dependent gauge transformations. It supplies the scalar-potential shift $-q\partial_t\chi$. With it included,

$$i\hbar\partial_t\psi' = H[\mathbf A',\Phi']\psi'$$

whenever the original equation holds.

This is covariance, not invariance of every written symbol. The equation keeps its form after all representatives are transformed consistently.

Invariant Quantities

The probability density is invariant:

$$\rho' = \lvert\psi'\rvert^2 = \rho.$$

The gauge-covariant probability current is invariant:

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

Expectation values of properly gauge-covariant mechanical quantities are invariant when states and operators are transformed together. For example,

$$\langle\hat{\boldsymbol\pi}\rangle = \int d^3r\, \psi^* \left( -i\hbar\nabla-q\mathbf A \right) \psi$$

agrees with the corresponding primed expression.

Closed-loop phases are also invariant:

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

Under $\mathbf A\mapsto\mathbf A+\nabla\chi$, the exponent changes by $(iq/\hbar)\oint_C\nabla\chi\cdot d\mathbf r$, which vanishes for a smooth single-valued $\chi$. More generally, the phase remains unchanged for allowed large gauge transformations because the exponential is single-valued.

Gauge Choice Versus Gauge Transformation

A gauge transformation relates equivalent descriptions. A gauge choice selects one representative for calculation.

Common gauge choices include:

• Coulomb gauge, $\nabla\cdot\mathbf A=0$;
• temporal gauge, $\Phi=0$, when compatible with the problem;
• Landau gauge for uniform magnetic fields;
• symmetric gauge for rotationally symmetric magnetic problems.

Different gauges can make different symmetries manifest. In a uniform magnetic field, Landau gauge makes one translation direction simple, while symmetric gauge makes rotations about the field axis simple. The energy spectrum is the same, but intermediate labels and wavefunction shapes differ.

The page Landau Gauge and Symmetric Gauge owns that worked comparison.

Gauge Redundancy Is Not an Ordinary Symmetry

An ordinary physical symmetry maps a state to another physically possible state, often with different labels. A spatial translation can move a localized packet. A spin rotation can rotate a spin polarization.

A gauge transformation, in the present electromagnetic sense, changes description. It should not create a new physical state. This is why gauge-dependent quantities are not observables by themselves.

The distinction becomes richer in field theory, where global phase symmetry is tied to charge conservation and local gauge redundancy is built into the field variables. The nonrelativistic lesson remains simple: predictions must be expressed in gauge-invariant or gauge-covariant form.

Small and Large Transformations

On simple domains, $\chi$ can often be chosen as an ordinary single-valued smooth function. Then

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

for every closed loop $C$.

On spaces with holes or nontrivial boundary conditions, allowed gauge transformations can have winding. The phase factor $e^{iq\chi/\hbar}$ must be single-valued, but $\chi$ itself may change by

$$\Delta\chi = \frac{2\pi\hbar}{q}n, \qquad n\in\mathbb Z,$$

around a closed cycle. Such large gauge transformations can shift line integrals by flux quanta while leaving the exponential phase unchanged.

This is the topology behind flux periodicity and the Aharonov–Bohm effect. The invariant object is not the raw line integral alone in all conventions, but the phase modulo $2\pi$.

Practical Checklist

When checking a calculation:

• transform $\psi$, $\mathbf A$, and $\Phi$ together;
• use kinetic momentum $\hat{\boldsymbol\pi}=\hat{\mathbf p}-q\mathbf A$ for mechanical velocity;
• compare probability densities, currents, spectra, transition probabilities, or loop phases;
• treat gauge-dependent labels as bookkeeping unless tied to an invariant statement;
• keep boundary conditions and single-valuedness conditions explicit.

Common Mistakes

• Comparing $\psi$ and $\psi'$ directly without transforming the potentials.
• Calling two gauge choices physically different because the wavefunctions look different.
• Forgetting the extra Hamiltonian term for time-dependent $U_\chi$.
• Treating canonical momentum eigenvalues in one gauge as universal observables.
• Saying potentials are “unphysical” in a way that erases gauge-invariant holonomy.
• Ignoring large gauge transformations and flux periodicity on multiply connected spaces.

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. Derive the Hamiltonian transformation law with a time-dependent gauge function.

Solution

Let $\psi'=U_\chi\psi$ and suppose

$$i\hbar\partial_t\psi = H\psi.$$

Then

$$\begin{aligned} i\hbar\partial_t\psi' &= i\hbar(\partial_tU_\chi)\psi + U_\chi i\hbar\partial_t\psi \\ &= \left[ i\hbar(\partial_tU_\chi)U_\chi^\dagger + U_\chi H U_\chi^\dagger \right]\psi'. \end{aligned}$$

Therefore

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

For $U_\chi=e^{iq\chi/\hbar}$, the second term is $-q\partial_t\chi$, matching $\Phi'=\Phi-\partial_t\chi$.

2. Show that the loop phase is invariant under a smooth single-valued gauge transformation.

Solution

Under $\mathbf A'=\mathbf A+\nabla\chi$,

$$\oint_C\mathbf A'\cdot d\mathbf r = \oint_C\mathbf A\cdot d\mathbf r + \oint_C\nabla\chi\cdot d\mathbf r.$$

For smooth single-valued $\chi$,

$$\oint_C\nabla\chi\cdot d\mathbf r = \chi(\text{final})-\chi(\text{initial}) = 0.$$

Thus the exponential phase is unchanged.

3. In polar form $\psi=Re^{iS/\hbar}$, verify that the current depends on $\nabla S-q\mathbf A$.

Solution

Acting on $\psi=Re^{iS/\hbar}$,

$$\left( -i\hbar\nabla-q\mathbf A \right)\psi = e^{iS/\hbar} \left[ -i\hbar\nabla R + R(\nabla S-q\mathbf A) \right].$$

Multiplying by $\psi^*=Re^{-iS/\hbar}$ and taking the real part gives

$$\mathbf j = \frac{R^2}{m} \left( \nabla S-q\mathbf A \right).$$

The imaginary term involving $\nabla R$ drops out. Since $S\mapsto S+q\chi$ and $\mathbf A\mapsto\mathbf A+\nabla\chi$, the combination is gauge invariant.