Minimal Coupling

Minimal coupling replaces the ordinary derivative by a gauge-covariant one. For a particle with signed charge $q$ in a prescribed classical four-potential,

$$A^\mu=\left(\frac{\Phi}{c},\mathbf A\right), \qquad D_\mu=\partial_\mu+\frac{iq}{\hbar}A_\mu.$$

This single substitution gives the minimally coupled Klein–Gordon and Dirac equations and reduces to the familiar kinetic momentum

$$\boldsymbol\pi=-i\hbar\nabla-q\mathbf A.$$

The electron is obtained by setting $q=-e$, with $e>0$. No sign is added by hand afterward. This page owns the covariant scalar/spinor construction and its QED boundary. The detailed nonrelativistic Schrödinger derivation remains at Minimal Coupling in Wave Mechanics.

Required background. Metric and Units, Four-Vectors, the Klein–Gordon Equation, and the Covariant Dirac Equation supply the convention ledger and free operators to be coupled. Classical electromagnetic potentials and fields are assumed.

One gauge package, fixed completely

The electromagnetic fields are

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

With the mostly-minus metric,

$$A_\mu=\left(\frac{\Phi}{c},-\mathbf A\right).$$

Define the field-strength tensor by

$$F_{\mu\nu} = \partial_\mu A_\nu-\partial_\nu A_\mu.$$

Its components include

$$F_{0i}=\frac{E_i}{c}, \qquad F_{ij}=-\epsilon_{ijk}B_k.$$

For a real scalar function $\chi(x)$, the complete convention package is

$$A_\mu'=A_\mu-\partial_\mu\chi, \qquad \psi'=e^{iq\chi/\hbar}\psi,$$

or, equivalently,

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

The signs of $D_\mu$, the potential transformation, and the matter phase are one linked choice. Changing only one of them destroys gauge covariance.

Directly,

$$\begin{aligned} D_\mu'\psi' &= \left( \partial_\mu+\frac{iq}{\hbar}A_\mu' \right)e^{iq\chi/\hbar}\psi \\ &= e^{iq\chi/\hbar} \left( \partial_\mu+\frac{iq}{\hbar}A_\mu \right)\psi \\ &= e^{iq\chi/\hbar}D_\mu\psi. \end{aligned}$$

Thus $D_\mu\psi$ transforms with the same local phase as $\psi$.

Curvature of the covariant derivative

Ordinary partial derivatives commute on the flat coordinate patch, but covariant derivatives do not:

$$[D_\mu,D_\nu]\psi = \frac{iq}{\hbar}F_{\mu\nu}\psi.$$

The electromagnetic field strength is therefore the local obstruction to commuting two gauge-covariant derivatives. In spatial language, with $\pi_i=-i\hbar\partial_i-qA_i$,

$$\begin{aligned} [\pi_i,\pi_j] &= iq\hbar (\partial_iA_j-\partial_jA_i) \\ &= iq\hbar\epsilon_{ijk}B_k. \end{aligned}$$

Here $A_i$ in the three-vector expression is the Euclidean component of $\mathbf A$, not the lowered spatial component of the four-potential. The kinetic momenta commute only where the relevant magnetic field vanishes.

Three minimally coupled wave equations

The nonrelativistic Schrödinger Hamiltonian is

$$H_S = \frac{\boldsymbol\pi^2}{2m}+q\Phi.$$

Its full probability-current and operator-ordering analysis belongs to the released wave-mechanics owner linked above.

For a charged complex scalar amplitude, the minimally coupled equation is

$$\left( D_\mu D^\mu +\frac{m^2c^2}{\hbar^2} \right)\phi=0.$$

The order matters when the potential varies: $D_\mu D^\mu$ is an operator composition, not an algebraic square of commuting symbols. A real scalar cannot carry this nontrivial continuous $U(1)$ charge representation by itself; a charged scalar field is complex.

For a spinor,

$$\left( i\hbar c\,\gamma^\mu D_\mu-mc^2 \right)\psi=0.$$

All three formulas use the same signed $q$, potentials, and gauge phase.

Dirac Hamiltonian in an electromagnetic background

Use

$$D_0 = \frac{1}{c}\partial_t +\frac{iq}{\hbar c}\Phi,$$

and the lowered spatial potential $A_i^{(4)}=-A_i^{(3)}$. Separating the time derivative in the covariant Dirac equation gives

$$i\hbar\partial_t\psi = H_D\psi,$$

where

$$\boxed{ H_D = c\boldsymbol\alpha\cdot\boldsymbol\pi +\beta mc^2 +q\Phi }.$$

This formula makes the canonical/kinetic distinction explicit:

$$\mathbf p=-i\hbar\nabla, \qquad \boldsymbol\pi=\mathbf p-q\mathbf A.$$

The canonical momentum depends on the representation and gauge choice; the kinetic momentum is gauge covariant and is tied to mechanical velocity. When $\mathbf A(\mathbf x,t)$ varies in space, derivatives act on the potential as well as on the spinor, so operator order cannot be inferred from classical commuting symbols.

For an electron, $q=-e$ gives

$$\boldsymbol\pi=\mathbf p+e\mathbf A, \qquad q\Phi=-e\Phi.$$

Both signs follow from the same substitution.

Squaring exposes spin–field coupling

Minimal coupling makes the Dirac operator sensitive to the noncommutativity of $D_\mu$. With $\sigma^{\mu\nu}=i[\gamma^\mu,\gamma^\nu]/2$,

$$(\gamma^\mu D_\mu)^2 = D_\mu D^\mu +\frac{q}{2\hbar}\sigma^{\mu\nu}F_{\mu\nu}.$$

Therefore every minimally coupled Dirac solution obeys the second-order equation

$$\left( D_\mu D^\mu +\frac{m^2c^2}{\hbar^2} +\frac{q}{2\hbar}\sigma^{\mu\nu}F_{\mu\nu} \right)\psi=0.$$

The extra field-strength term is absent for a minimally coupled scalar. Its magnetic part is the algebraic seed of the Pauli spin coupling and the tree-level Dirac value $g=2$.

SI and Gaussian-unit translations

This page uses SI-normalized potentials, for which

$$i\hbar D_\mu=i\hbar\partial_\mu-qA_\mu.$$

Some Gaussian-unit texts write a substitution of the form

$$p_\mu\longrightarrow p_\mu-\frac{q}{c}A_\mu^{\mathrm G}.$$

The displayed $1/c$ is tied to how the Gaussian four-potential and charge are defined. It must not be inserted into this page’s SI formula while retaining the SI $A^\mu=(\Phi/c,\mathbf A)$. Translate the entire electromagnetic unit system, including field definitions and charge units, rather than moving a single factor of $c$.

Minimal does not mean most general

Minimal coupling is the interaction generated by replacing ordinary derivatives with $D_\mu$. It fixes the tree-level magnetic moment of an elementary Dirac particle to $g=2$. Symmetry also permits nonminimal effective operators. The leading spinor example has the tensor structure

$$\bar\psi\sigma^{\mu\nu}F_{\mu\nu}\psi,$$

with a coefficient determined by microscopic physics or experiment. In QED, radiative corrections generate an anomalous magnetic moment. Writing that term is not a failure of gauge covariance; it is a statement that the low- energy theory contains more than the minimal operator.

The coefficient and its SI signs are matched explicitly on Magnetic Moment.

This page treats $A_\mu$ as a prescribed classical background. When the electromagnetic field itself fluctuates quantum mechanically, photons carry energy and momentum and radiative corrections require QED. Pair creation already becomes possible when the matter field is quantized in a prescribed classical electric background; quantized photons are not a prerequisite. The first-quantized equation used here does not itself provide the vacuum and particle-number observables needed to calculate that process.

Common pitfalls

Replacing $q$ by positive $e$ everywhere. The electron has $q=-e$. Substitute once into the full Hamiltonian so scalar and vector terms retain consistent signs.

Mixing SI and Gaussian formulas. A lone $1/c$ cannot be translated in isolation. Fix the four-potential and charge normalization first.

Expanding $\boldsymbol\pi^2$ as commuting algebra. Momentum derivatives act on $\mathbf A$, and different kinetic-momentum components fail to commute in a magnetic field.

Changing the gauge phase but not $D_\mu$. Gauge covariance is a package of three correlated signs: the potential shift, matter phase, and covariant derivative.

Exercises

1. Gauge covariance

Verify the transformation of $D_\mu\psi$ and explain why the ordinary derivative does not transform in the same way.

Solution

Differentiating the local phase produces an extra term $iq(\partial_\mu\chi)\psi/\hbar$. The potential shift in $D_\mu'$ cancels it, leaving

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

The ordinary derivative contains no compensating potential term, so $\partial_\mu\psi$ acquires the extra derivative of the phase.

2. Kinetic-momentum commutator

Derive $[\pi_i,\pi_j]$ in position representation.

Solution

Using $[p_i,f]=-i\hbar\partial_i f$,

$$\begin{aligned} [\pi_i,\pi_j] &= -q[p_i,A_j]-q[A_i,p_j] \\ &= iq\hbar(\partial_iA_j-\partial_jA_i) \\ &= iq\hbar\epsilon_{ijk}B_k. \end{aligned}$$

3. Aharonov–Bohm phase

For a closed path $C$ in a region where $\mathbf B=0$ along the path, show that the gauge-invariant phase difference is determined by the enclosed magnetic flux.

Solution

The vector-potential contribution to the phase is

$$\Delta\theta = \frac{q}{\hbar}\oint_C\mathbf A\cdot d\boldsymbol\ell.$$

Stokes’ theorem gives

$$\Delta\theta = \frac{q}{\hbar}\int_S\mathbf B\cdot d\mathbf S = \frac{q\Phi_B}{\hbar}.$$

A single path phase is gauge dependent, while the closed-loop difference is unchanged by $\mathbf A\mapsto\mathbf A+\nabla\chi$ for single-valued $\chi$.

4. Squared Dirac operator

Derive the coefficient of $\sigma^{\mu\nu}F_{\mu\nu}$ in $(\gamma^\mu D_\mu)^2$.

Solution

Separate symmetric and antisymmetric products:

$$\gamma^\mu\gamma^\nu D_\mu D_\nu = D_\mu D^\mu +\frac14[\gamma^\mu,\gamma^\nu][D_\mu,D_\nu].$$

Using $[\gamma^\mu,\gamma^\nu]=-2i\sigma^{\mu\nu}$ and $[D_\mu,D_\nu]=iqF_{\mu\nu}/\hbar$ gives

$$(\gamma^\mu D_\mu)^2 = D_\mu D^\mu +\frac{q}{2\hbar}\sigma^{\mu\nu}F_{\mu\nu}.$$

References

• J. D. Jackson and L. B. Okun, “Historical Roots of Gauge Invariance,” Reviews of Modern Physics 73, 663–680, 2001, doi:10.1103/RevModPhys.73.663.
• J. J. Sakurai, Advanced Quantum Mechanics, Addison–Wesley, 1967.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, doi:10.1017/9781139540940.
• F. Schwabl, Advanced Quantum Mechanics, 3rd ed., Springer, 2005, doi:10.1007/3-540-28528-8.