Minimal Coupling

Minimal coupling is the rule that replaces canonical momentum by gauge-covariant kinetic momentum:

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

For a spinless nonrelativistic particle in prescribed electromagnetic potentials, the Hamiltonian is

$$H = \frac{1}{2m} \left( \hat{\mathbf p}-q\mathbf A(\hat{\mathbf r},t) \right)^2 + q\Phi(\hat{\mathbf r},t).$$

Helpful background. Minimal coupling in wave mechanics develops the differential equation. Here the covariant-momentum commutator connects that equation to magnetic-field geometry.

The Gauge-Covariant Momentum

In position representation,

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

A local phase transformation

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

does not commute cleanly with $\nabla$. The derivative produces an extra $\nabla\chi$ term. The page Local Phase Transformations derives this explicitly.

Minimal coupling introduces a vector potential so that

$$\hat{\boldsymbol\pi} = -i\hbar\nabla-q\mathbf A$$

transforms covariantly when

$$\mathbf A' = \mathbf A+\nabla\chi.$$

The key identity is

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

Thus $\hat{\boldsymbol\pi}\psi$ has the same local phase behavior as $\psi$ itself. This is the mathematical reason the kinetic momentum, not the canonical momentum alone, belongs in the Hamiltonian.

Covariant-Derivative Form

Define

$$D_i = \partial_i-\frac{iq}{\hbar}A_i, \qquad D_t = \partial_t+\frac{iq}{\hbar}\Phi.$$

Then

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

and the minimally coupled Schrödinger equation is

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

Under

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

the covariant derivatives obey

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

The equation therefore has the same form in every gauge. Gauge covariance selects the covariant derivative, but does not uniquely select this Hamiltonian: additional interactions built from $\mathbf E$, $\mathbf B$, and internal operators can also be gauge invariant. The displayed Hamiltonian is the minimal quadratic kinetic model.

Operator order matters when expanding it:

$$\boldsymbol\pi^2=\mathbf p^2 -q(\mathbf p\cdot\mathbf A+\mathbf A\cdot\mathbf p) +q^2\mathbf A^2.$$

Here $\mathbf p\cdot\mathbf A=\mathbf A\cdot\mathbf p-i\hbar\nabla\cdot\mathbf A$ as operators. Combining the two cross terms into $-2q\mathbf A\cdot\mathbf p$ therefore requires $\nabla\cdot\mathbf A=0$.

Classical Correspondence

Minimal coupling also matches the classical charged-particle Lagrangian

$$L = \frac{1}{2}m\dot{\mathbf r}^{\,2} + q\dot{\mathbf r}\cdot\mathbf A(\mathbf r,t) - q\Phi(\mathbf r,t).$$

The canonical momentum is

$$\mathbf p = \frac{\partial L}{\partial\dot{\mathbf r}} = m\dot{\mathbf r}+q\mathbf A,$$

so the mechanical, or kinetic, momentum is

$$m\dot{\mathbf r} = \mathbf p-q\mathbf A.$$

Quantization promotes this same combination to the operator $\hat{\boldsymbol\pi}$. This correspondence is useful, but the quantum reason is sharper: the combination is the one that transforms covariantly under local phase changes.

Canonical and Kinetic Momentum

Canonical momentum and kinetic momentum answer different questions.

Canonical momentum:

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

is tied to the chosen position-space phase convention and to ordinary translations. Kinetic momentum:

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

is tied to mechanical velocity:

$$\hat{\mathbf v} = \frac{\hat{\boldsymbol\pi}}{m}.$$

The commutators show the field strength:

$$[\hat r_i,\hat\pi_j] = i\hbar\delta_{ij},$$

while

$$[\hat\pi_i,\hat\pi_j] = i\hbar q \sum_k \epsilon_{ijk}B_k.$$

The noncommuting kinetic momenta produce Landau quantization. The conserved generators of magnetic translations are different combinations of position and momentum; kinetic momentum itself is not conserved in a magnetic field.

Potentials, Fields, and Curvature

The electromagnetic fields are

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

In geometric language, the potentials are connection data and the fields are curvature data. The potentials are not unique, but they organize how phases are compared at neighboring spacetime points.

This explains why the vector potential appears in the wave equation even though the Lorentz force is written in terms of $\mathbf E$ and $\mathbf B$. Locally, fields determine forces. Quantum mechanically, potentials also determine phase transport. When the accessible region has nontrivial topology, phase holonomy can remain observable even where the local magnetic field vanishes along the path.

Pure Gauge and Holonomy

If

$$\mathbf A=\nabla\chi, \qquad \Phi=-\partial_t\chi,$$

then the fields vanish:

$$\mathbf E=\mathbf 0, \qquad \mathbf B=\mathbf 0.$$

On a simply connected region with a single-valued $\chi$, this potential can be removed by a gauge transformation. To remove this potential, use gauge parameter $-\chi$: $\psi'=e^{-iq\chi/\hbar}\psi$, $\mathbf A'=0$, and $\Phi'=0$. The opposite sign would generate the potential from zero rather than remove it.

The global caveat is important. In a multiply connected region, a potential can be locally pure gauge while still having nontrivial loop phase:

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

That is the geometric core of the Aharonov–Bohm Effect.

Spin and Nonminimal Terms

For a scalar nonrelativistic particle, minimal coupling gives the orbital electromagnetic coupling. A spin-$1/2$ particle also has magnetic moment coupling. In the ideal $g=2$ Pauli Hamiltonian,

$$H = \frac{\hat{\boldsymbol\pi}^{\,2}}{2m} + q\Phi - \frac{q\hbar}{2m} \boldsymbol\sigma\cdot\mathbf B.$$

The last term is not obtained by expanding the scalar Hamiltonian. It follows from the nonrelativistic limit of the minimally coupled Dirac equation, equivalently from $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2/(2m)$. Thus the ideal $g=2$ term already belongs to minimal Dirac coupling. More general particles can also have anomalous magnetic moments, electric dipole couplings, polarizability terms, and other effective interactions. Those are called nonminimal couplings because they go beyond the replacement $\hat{\mathbf p}\mapsto\hat{\mathbf p}-q\mathbf A$.

Common Mistakes

• Expanding $\left(\hat{\mathbf p}-q\mathbf A\right)^2$ while forgetting that derivatives act on $\mathbf A$ and on everything to their right.
• Treating $\hat{\mathbf p}$ as the mechanical momentum in a magnetic field.
• Changing gauge potentials without transforming the wavefunction phase.
• Forgetting the scalar potential when the gauge function depends on time.
• Assuming minimal coupling automatically includes all spin or relativistic effects.
• Calling $\mathbf A$ directly observable rather than using gauge-invariant fields, currents, and loop phases.

Exercises

1. Show that the pure-gauge potentials $\mathbf A=\nabla\chi$ and $\Phi=-\partial_t\chi$ give zero electromagnetic fields.

Solution

For the magnetic field,

$$\mathbf B = \nabla\times\mathbf A = \nabla\times\nabla\chi = \mathbf 0.$$

For the electric field,

$$\begin{aligned} \mathbf E &= -\nabla\Phi-\partial_t\mathbf A \\ &= -\nabla(-\partial_t\chi) - \partial_t(\nabla\chi) \\ &= \nabla\partial_t\chi - \partial_t\nabla\chi = \mathbf 0. \end{aligned}$$

2. Starting from $\hat\pi_i=-i\hbar\partial_i-qA_i$, compute $[\hat\pi_i,\hat\pi_j]$.

Solution

Acting on a test wavefunction and using that the $A_i$ commute as multiplication operators,

$$\begin{aligned} [\hat\pi_i,\hat\pi_j] &= [-i\hbar\partial_i,-qA_j] + [-qA_i,-i\hbar\partial_j] \\ &= i\hbar q(\partial_iA_j-\partial_jA_i). \end{aligned}$$

Since

$$\partial_iA_j-\partial_jA_i = \sum_k\epsilon_{ijk}B_k,$$

we get

$$[\hat\pi_i,\hat\pi_j] = i\hbar q \sum_k\epsilon_{ijk}B_k.$$

3. Derive the ideal Pauli magnetic term using $\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k$ and the kinetic-momentum commutator.

Solution

Only the antisymmetric part of $\pi_i\pi_j$ contributes to the epsilon contraction:

$$\begin{aligned} (\boldsymbol\sigma\cdot\boldsymbol\pi)^2 &=\boldsymbol\pi^2+ \frac{i}{2}\sum_{ijk}\epsilon_{ijk}\sigma_k[\pi_i,\pi_j]\\ &=\boldsymbol\pi^2-q\hbar\boldsymbol\sigma\cdot\mathbf B. \end{aligned}$$

Dividing by $2m$ gives the displayed Pauli kinetic and magnetic terms. The scalar minimal Hamiltonian has no Pauli matrices and cannot produce this spin term by itself. A change from the Dirac value $g=2$ requires additional physics.

References

• Y. Aharonov and D. Bohm, “Significance of electromagnetic potentials in the quantum theory,” Physical Review 115, 485–491, 1959.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• 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.