Charged Particle in an Electromagnetic Field: Minimal Coupling

How does a quantum particle feel a magnetic field? The magnetic force $q\mathbf{v}\times\mathbf{B}$ does no work and has no potential energy, so it cannot simply be added to the Hamiltonian as a term $V(\mathbf{r})$ . The resolution — inherited wholesale from classical Hamiltonian mechanics — is not to change the potential but to change the momentum. Wherever the canonical momentum $\mathbf{p}$ appears in the free Hamiltonian, replace it by the kinetic momentum $\mathbf{p}-\tfrac{q}{c}\mathbf{A}$ , where $\mathbf{A}$ is the magnetic vector potential. This single substitution — minimal coupling — is how electromagnetism enters the Schrödinger equation, and it is dictated, not invented. The move looks almost too cheap, and its first consequence seems alarming: the potentials $\mathbf{A}$ and $\phi$ are not unique. A gauge transformation $\mathbf{A}\to\mathbf{A}+\nabla\chi$ , $\phi\to\phi-\tfrac1c\partial_t\chi$ leaves the physical fields $\mathbf{E}$ and $\mathbf{B}$ untouched but changes $\mathbf{A}$ everywhere. If the Hamiltonian contains $\mathbf{A}$ , does the physics depend on an arbitrary choice? The answer is the deepest structural fact in this article: the wavefunction absorbs the change as a local phase $\psi\to e^{iq\chi/\hbar c}\psi$ , and every observable — probability density, current, energy — comes out identical. Gauge freedom in the potentials is exactly matched by phase freedom in the wavefunction. This article establishes the coupled equation, proves its gauge invariance, works out the modified probability current, and unpacks the kinetic term into its paramagnetic and diamagnetic pieces, recovering the Zeeman coupling $-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ in the symmetric gauge. It gives the cyclotron frequency and the Landau-level ladder $E_n=\hbar\omega_c(n+\tfrac12)$ — the entry point to the quantum Hall effect and the Aharonov–Bohm phase, each developed in its own home. What is not here is spin: minimal coupling of the scalar Schrödinger equation captures orbital magnetism only; the intrinsic magnetic moment and $g=2$ arrive with the Pauli and Dirac equations, signposted by name.

Key takeaways

Minimal coupling: replace $\mathbf{p}\to\mathbf{p}-\tfrac{q}{c}\mathbf{A}$ and add $q\phi$ , giving $\hat H=\tfrac{1}{2m}\big(\mathbf{p}-\tfrac{q}{c}\mathbf{A}\big)^2+q\phi$ .
Under a gauge transformation $\mathbf{A}\to\mathbf{A}+\nabla\chi,\ \phi\to\phi-\tfrac1c\partial_t\chi$ , the solution picks up a local phase $\psi\to e^{iq\chi/\hbar c}\psi$ ; all observables are unchanged.
The combination $\mathbf{D}=-i\hbar\nabla-\tfrac{q}{c}\mathbf{A}$ is a covariant derivative: it transforms homogeneously, $\mathbf{D}'\psi'=e^{iq\chi/\hbar c}\mathbf{D}\psi$ , which is why the equation is form-invariant.
The probability current gains a gauge term, $\mathbf{j}=\tfrac{\hbar}{m}\mathrm{Im}(\psi^*\nabla\psi)-\tfrac{q}{mc}\mathbf{A}|\psi|^2$ , and is itself gauge-invariant.
Expanding the kinetic term gives a paramagnetic piece linear in $\mathbf{A}$ (the $-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ Zeeman coupling in the symmetric gauge) and a diamagnetic piece $\propto A^2$ ; a uniform field quantizes motion into Landau levels $E_n=\hbar\omega_c(n+\tfrac12)$ with $\omega_c=qB/mc$ .
Key idea: electromagnetism enters quantum mechanics by promoting the ordinary derivative to a gauge-covariant one; the arbitrariness of the potentials is exactly compensated by the local phase freedom of the wavefunction, so no observable ever depends on the gauge.

The core insight

Minimal coupling ( $\mathbf{p}\to\mathbf{p}-\tfrac{q}{c}\mathbf{A}$ ) turns the ordinary gradient into the covariant derivative $-i\hbar\nabla-\tfrac{q}{c}\mathbf{A}$ , and the gauge freedom of $(\mathbf{A},\phi)$ is precisely cancelled by a local phase rotation $e^{iq\chi/\hbar c}$ of the wavefunction — making the coupled Schrödinger equation gauge-invariant by construction.

Where this lives today

Minimal coupling is the template for how all matter couples to all gauge fields. The rule "replace the ordinary derivative by the gauge-covariant derivative" reappears verbatim in the Dirac equation (where it produces the $g=2$ magnetic moment), in quantum electrodynamics (where the covariant derivative defines the photon–electron vertex), and — with matrix-valued potentials — in the electroweak and strong interactions. Experimentally, its consequences are everywhere: cyclotron resonance measures effective masses in semiconductors; the integer and fractional quantum Hall effects, built on Landau quantization, provide the resistance standard and won two Nobel Prizes; nuclear magnetic resonance and MRI rest on the orbital and spin couplings this Hamiltonian organizes; and the Aharonov–Bohm effect, now routinely observed in mesoscopic rings, is the cleanest demonstration that the gauge potential is physically real in quantum mechanics. Superconductivity's Meissner effect and flux quantization are the diamagnetic term taken to its collective extreme. The single substitution $\mathbf{p}\to\mathbf{p}-\tfrac{q}{c}\mathbf{A}$ is thus not a technical convenience but the seed of gauge theory itself.

The physical picture

A charged particle in an electromagnetic field is governed classically by the Hamiltonian $H=\tfrac{1}{2m}(\mathbf{p}-\tfrac{q}{c}\mathbf{A})^2+q\phi$ , where $\mathbf{p}$ is the canonical momentum conjugate to position — not the kinetic momentum $m\mathbf{v}$ . The two differ precisely by the vector potential: $m\mathbf{v}=\mathbf{p}-\tfrac{q}{c}\mathbf{A}$ . Quantizing by the usual rule $\mathbf{p}\to-i\hbar\nabla$ then automatically inserts $\mathbf{A}$ wherever momentum appears. This is why magnetism cannot be added as a potential-energy term: it lives in the momentum, deforming how the particle moves rather than where its energy is low.

Figure

Canonical versus kinetic momentum: minimal coupling deforms the momentum, not the potential

The unsettling feature is that $\mathbf{A}$ is not measurable. Only $\mathbf{B}=\nabla\times\mathbf{A}$ and $\mathbf{E}=-\nabla\phi-\tfrac1c\partial_t\mathbf{A}$ are physical, and these are unchanged by a gauge transformation $\mathbf{A}\to\mathbf{A}+\nabla\chi$ , $\phi\to\phi-\tfrac1c\partial_t\chi$ for any smooth scalar $\chi(\mathbf{r},t)$ . Since the Hamiltonian contains the unphysical $\mathbf{A}$ , one might fear the predicted energies and probabilities would shift when we merely relabel the potentials. They do not — but the reason is subtle and beautiful: the wavefunction is not gauge-invariant either. It rotates by a local phase $e^{iq\chi/\hbar c}$ that tracks $\chi$ point by point, and this rotation is exactly what the covariant derivative $-i\hbar\nabla-\tfrac{q}{c}\mathbf{A}$ needs to stay consistent.

Figure

Gauge freedom in the potentials is matched by local phase freedom in the wavefunction

Once the framework is in place, the physics of orbital magnetism falls out by algebra. Expanding the square $(\mathbf{p}-\tfrac{q}{c}\mathbf{A})^2$ produces a term linear in $\mathbf{A}$ (paramagnetic) and a term quadratic in $\mathbf{A}$ (diamagnetic). In a uniform field the linear term becomes the Zeeman coupling $-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ , and the full problem quantizes into equally spaced Landau levels — the quantum shadow of classical cyclotron orbits.

The core equations

The starting point is the classical Hamiltonian of a charge $q$ in potentials $(\mathbf{A},\phi)$ , promoted to an operator by $\mathbf{p}\to-i\hbar\nabla$ . We work in Gaussian units throughout, where the coupling appears as $q/c$ ; the SI form is recovered by $q/c\to q$ (and $\phi$ in volts).

\hat H = \frac{1}{2m}\Big(\mathbf{p}-\frac{q}{c}\mathbf{A}\Big)^2 + q\phi, \qquad i\hbar\,\partial_t\psi = \hat H\psi \qquad (1)

where: $m$ is the particle mass; $q$ its charge (for an electron $q=-e$ ); $\mathbf{p}=-i\hbar\nabla$ the canonical momentum operator; $\mathbf{A}(\mathbf{r},t)$ the magnetic vector potential with $\mathbf{B}=\nabla\times\mathbf{A}$ ; $\phi(\mathbf{r},t)$ the scalar potential with $\mathbf{E}=-\nabla\phi-\tfrac1c\partial_t\mathbf{A}$ ; and $c$ the speed of light. The kinetic momentum $m\mathbf{v}=\mathbf{p}-\tfrac{q}{c}\mathbf{A}$ is the physical (gauge-covariant) momentum, distinct from the canonical $\mathbf{p}$ . It is convenient to name the covariant derivative that carries all the field dependence. Every $\mathbf{p}$ in $(1)$ is really this operator acting on $\psi$ .

\mathbf{D} \equiv \mathbf{p}-\frac{q}{c}\mathbf{A} = -i\hbar\nabla-\frac{q}{c}\mathbf{A}, \qquad \hat H=\frac{\mathbf{D}\cdot\mathbf{D}}{2m}+q\phi \qquad (2)

where: $\mathbf{D}$ is the gauge-covariant momentum (also called the kinetic-momentum operator). Its defining virtue, proved below, is that under a gauge transformation it transforms homogeneously — it commutes with the local phase rotation up to that same phase — which is the entire reason the equation is gauge-invariant. The gauge transformation itself acts on the potentials and, simultaneously, on the state.

\mathbf{A}\to\mathbf{A}+\nabla\chi,\quad \phi\to\phi-\frac1c\partial_t\chi,\quad \psi\to\psi'=e^{iq\chi/\hbar c}\,\psi \qquad (3)

where: $\chi(\mathbf{r},t)$ is an arbitrary smooth gauge function and $e^{iq\chi/\hbar c}$ is a local (position- and time-dependent) $U(1)$ phase. The physical fields $\mathbf{E},\mathbf{B}$ are invariant under $(3)$ by construction; the claim is that $\psi'$ solves the transformed equation $(1)$ exactly when $\psi$ solves the original, so all measurable predictions agree. The probability current inherits a field-dependent piece. It follows from $(1)$ exactly as the free current follows from the free equation, but with $\nabla\to$ the covariant $\mathbf{D}/(-i\hbar)$ .

\rho=|\psi|^2,\qquad \mathbf{j}=\frac{\hbar}{m}\,\mathrm{Im}\!\big(\psi^*\nabla\psi\big)-\frac{q}{mc}\mathbf{A}\,|\psi|^2=\frac{1}{m}\,\mathrm{Re}\!\big(\psi^*\mathbf{D}\psi\big),\qquad \partial_t\rho+\nabla\cdot\mathbf{j}=0 \qquad (4)

where: $\rho$ is the probability density and $\mathbf{j}$ the probability current. The added $-\tfrac{q}{mc}\mathbf{A}|\psi|^2$ term is exactly what makes $\mathbf{j}$ gauge-invariant: under $(3)$ the phase gradient added to $\mathrm{Im}(\psi^*\nabla\psi)$ is cancelled by the shift $\mathbf{A}\to\mathbf{A}+\nabla\chi$ . Without it, the current would depend on the arbitrary gauge. Expanding the kinetic term exposes the magnetic physics. With the Coulomb gauge $\nabla\cdot\mathbf{A}=0$ (so $\mathbf{p}\cdot\mathbf{A}=\mathbf{A}\cdot\mathbf{p}$ as operators) and a uniform field $\mathbf{B}=B\hat{\mathbf z}$ in the symmetric gauge $\mathbf{A}=\tfrac12\mathbf{B}\times\mathbf{r}$ :

\hat H=\frac{p^2}{2m}+q\phi-\frac{q}{2mc}\,\mathbf{L}\cdot\mathbf{B}+\frac{q^2}{8mc^2}\,(\mathbf{B}\times\mathbf{r})^2 \qquad (5)

where: $\tfrac{p^2}{2m}+q\phi$ is the field-free part; $-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ is the paramagnetic (Zeeman) term with $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ the orbital angular momentum; and $\tfrac{q^2}{8mc^2}(\mathbf{B}\times\mathbf{r})^2$ is the diamagnetic term, quadratic in $B$ . For a uniform field these terms produce the Landau spectrum $E_n=\hbar\omega_c(n+\tfrac12)$ with the cyclotron frequency $\omega_c=|q|B/mc$ , developed for the density of states in the solid-state track.

Derivation AGauge invariance from the covariant derivative

The goal is to prove the central claim of the whole subject: that the minimally coupled Schrödinger equation is form-invariant under the joint transformation $(3)$ , so no observable can depend on the gauge choice. The route is to show that the covariant derivative $\mathbf{D}=-i\hbar\nabla-\tfrac{q}{c}\mathbf{A}$ transforms homogeneously — it simply carries the local phase through itself — and that the scalar-potential change is exactly cancelled by the time derivative of the phase. The one thing to watch is that the phase is local: because $\chi$ depends on position and time, differentiating $e^{iq\chi/\hbar c}\psi$ produces an extra gradient term, and that extra term is precisely what the shift $\mathbf{A}\to\mathbf{A}+\nabla\chi$ is designed to absorb.

Step 1. Write the transformed state $\psi'=e^{i\lambda}\psi$ with $\lambda\equiv q\chi/\hbar c$ , and the transformed potentials $\mathbf{A}'=\mathbf{A}+\nabla\chi$ , $\phi'=\phi-\tfrac1c\partial_t\chi$ . Apply the transformed covariant derivative $\mathbf{D}'=-i\hbar\nabla-\tfrac{q}{c}\mathbf{A}'$ to $\psi'$ .

Step 2. Expand the gradient: $\nabla\psi'=e^{i\lambda}(\nabla\psi+i(\nabla\lambda)\psi)$ . Since $\nabla\lambda=\tfrac{q}{\hbar c}\nabla\chi$ , the extra piece is $-i\hbar\cdot i\tfrac{q}{\hbar c}(\nabla\chi)\psi\,e^{i\lambda}=+\tfrac{q}{c}(\nabla\chi)\psi\,e^{i\lambda}$ .

Step 3. Subtract the shifted vector potential: $-\tfrac{q}{c}\mathbf{A}'\psi'=-\tfrac{q}{c}(\mathbf{A}+\nabla\chi)e^{i\lambda}\psi$ . The $-\tfrac{q}{c}\nabla\chi$ term cancels the $+\tfrac{q}{c}\nabla\chi$ generated in Step 2, leaving $\mathbf{D}'\psi'=e^{i\lambda}\big(-i\hbar\nabla-\tfrac{q}{c}\mathbf{A}\big)\psi=e^{i\lambda}\,\mathbf{D}\psi$ . The covariant derivative transforms homogeneously — that is the whole point. Why this is the crux: an ordinary derivative would leave the stray $\nabla\chi$ term uncancelled and spoil form-invariance. The vector potential is exactly the compensating field that makes the derivative covariant.

Step 4. Because $\mathbf{D}'\psi'=e^{i\lambda}\mathbf{D}\psi$ , applying $\mathbf{D}'$ twice gives $\mathbf{D}'^2\psi'=e^{i\lambda}\mathbf{D}^2\psi$ , so the kinetic operator $\tfrac{1}{2m}\mathbf{D}^2$ carries the phase cleanly through.

Step 5. Handle the time side. The transformed left member is $i\hbar\partial_t\psi'=i\hbar\,e^{i\lambda}(\partial_t\psi+i(\partial_t\lambda)\psi)=e^{i\lambda}\big(i\hbar\partial_t\psi-\hbar(\partial_t\lambda)\psi\big)$ . Now $\hbar\partial_t\lambda=\tfrac{q}{c}\partial_t\chi$ , and $q\phi'=q\phi-\tfrac{q}{c}\partial_t\chi$ , so the $-\tfrac{q}{c}\partial_t\chi$ from the phase exactly cancels the $-\tfrac{q}{c}\partial_t\chi$ hidden in $q\phi'$ . Assembling: $\big(i\hbar\partial_t-\hat H'\big)\psi'=e^{i\lambda}\big(i\hbar\partial_t-\hat H\big)\psi$ . Hence $\psi'$ solves the primed equation iff $\psi$ solves the original. Verified: the full symbolic transformation $S'[\psi']=e^{i\lambda}S[\psi]$ , including the exact cancellation of the $\partial_t\chi$ term against the phase's time derivative, was reproduced identically in the companion verifier for an arbitrary $\chi(x,t)$ , along with the homogeneous transformation of $\mathbf{D}$ and $\mathbf{D}^2$ .

Go deeper · Derivation B · 1 minThe gauge-invariant probability current

Derivation BThe gauge-invariant probability current

Here the aim is to derive the probability current for the coupled equation and show that the extra $-\tfrac{q}{mc}\mathbf{A}|\psi|^2$ term is not optional decoration but exactly what is required for gauge invariance and for the continuity equation to hold. The route is to build $\mathbf{j}$ from the covariant derivative rather than the ordinary one, then verify both that it reduces to the familiar expression and that it is invariant under $(3)$ . The subtlety is that the naive current $\tfrac{\hbar}{m}\mathrm{Im}(\psi^*\nabla\psi)$ alone is not gauge-invariant — it changes by a term proportional to $\nabla\chi$ — so it cannot be the physical current on its own.

Step 1. Define the covariant current $\mathbf{j}=\tfrac1m\mathrm{Re}(\psi^*\mathbf{D}\psi)$ with $\mathbf{D}=-i\hbar\nabla-\tfrac{q}{c}\mathbf{A}$ . Writing $\psi=Re^{iS}$ with $R,S$ real, compute $\mathbf{D}\psi=e^{iS}(\hbar R\nabla S-i\hbar\nabla R-\tfrac{q}{c}\mathbf{A}R)$ .

Step 2. Form $\psi^*\mathbf{D}\psi=R^2\hbar\nabla S-i\hbar R\nabla R-\tfrac{q}{c}\mathbf{A}R^2$ . The real part drops the pure-imaginary middle term, giving $\mathbf{j}=\tfrac1m\big(\hbar R^2\nabla S-\tfrac{q}{c}\mathbf{A}R^2\big)=\tfrac{\hbar}{m}|\psi|^2\nabla S-\tfrac{q}{mc}\mathbf{A}|\psi|^2$ , which is exactly $(4)$ since $\tfrac{\hbar}{m}\mathrm{Im}(\psi^*\nabla\psi)=\tfrac{\hbar}{m}R^2\nabla S$ .

Step 3. Test gauge invariance. Under $(3)$ , $R$ is unchanged, the phase shifts $S\to S+\tfrac{q}{\hbar c}\chi$ , and $\mathbf{A}\to\mathbf{A}+\nabla\chi$ . Then $\tfrac{\hbar}{m}R^2\nabla S\to\tfrac{\hbar}{m}R^2\nabla S+\tfrac{q}{mc}R^2\nabla\chi$ , while $-\tfrac{q}{mc}\mathbf{A}R^2\to-\tfrac{q}{mc}\mathbf{A}R^2-\tfrac{q}{mc}R^2\nabla\chi$ . The two $\nabla\chi$ pieces cancel: $\mathbf{j}$ is gauge-invariant.

Step 4. Continuity follows from the equation itself: taking $\partial_t|\psi|^2$ and using $(1)$ for $\partial_t\psi$ and its conjugate, the terms organize into $\partial_t\rho=-\nabla\cdot\mathbf{j}$ with exactly the $\mathbf{j}$ above; the $\mathbf{A}$ -dependent pieces are needed for the cancellation, since the Hamiltonian is $\mathbf{D}^2/2m$ , not $p^2/2m$ . Verified: the companion verifier confirms symbolically that $\tfrac1m\mathrm{Re}(\psi^*\mathbf{D}\psi)$ equals $\tfrac{\hbar}{m}|\psi|^2\nabla S-\tfrac{q}{mc}\mathbf{A}|\psi|^2$ for $\psi=Re^{iS}$ , and that the gauge shifts of $S$ and $\mathbf{A}$ cancel to leave $\mathbf{j}$ invariant.

Limiting cases

Each limit is a consistency check on the coupled equation.

Case	Result	Meaning
No field, $\mathbf{A}=0,\ \phi=0$	$\hat H=p^2/2m$	Recovers the free Schrödinger equation; canonical and kinetic momenta coincide.
Pure electrostatics, $\mathbf{A}=0$	$\hat H=p^2/2m+q\phi$	The familiar potential-energy problem; the phase freedom reduces to a global constant.
Weak uniform $\mathbf{B}$	$\hat H\approx H_0-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$	Linear (paramagnetic) Zeeman splitting; the $A^2$ term is negligible at laboratory fields.
Strong uniform $\mathbf{B}$ , free particle	$E_n=\hbar\omega_c(n+\tfrac12)$ , $\omega_c=\|q\|B/mc$	Motion quantizes into Landau levels; the classical cyclotron orbit becomes a ladder.
Time-independent gauge, $\chi=\chi(\mathbf{r})$	$\psi\to e^{iq\chi/\hbar c}\psi$	A static gauge choice is a position-dependent phase; the Aharonov–Bohm phase is the line integral $\tfrac{q}{\hbar c}\oint\mathbf{A}\cdot d\mathbf{r}$ .

Common trapThinking the vector potential is unphysical because it is gauge-dependent

It is tempting to conclude that because $\mathbf{A}$ is not measurable and can be changed by a gauge transformation, it plays no real role — that only $\mathbf{B}$ matters. Quantum mechanics disproves this. The Aharonov–Bohm effect shows that a charged particle acquires a measurable phase $\tfrac{q}{\hbar c}\oint\mathbf{A}\cdot d\mathbf{r}$ around a loop even in a region where $\mathbf{B}=0$ everywhere on its path. The gauge-invariant loop integral of $\mathbf{A}$ — equal to the enclosed flux — is physical, even though the pointwise value of $\mathbf{A}$ is not. So $\mathbf{A}$ carries genuine physics that $\mathbf{B}$ alone (evaluated only where the particle is) cannot capture. The correct statement is not " $\mathbf{A}$ is unphysical" but "only gauge-invariant functionals of $\mathbf{A}$ are physical." The full argument lives in the Aharonov–Bohm article.

Worked examples

Example 1 — Cyclotron frequency and Landau spacing for an electron in a 1 T field. Compute the cyclotron frequency, the Landau-level spacing, and compare it to the Bohr-magneton Zeeman scale for an electron ( $q=-e$ , $m=m_e$ ) in $B=1\ \mathrm{T}$ .

Step 1. Cyclotron angular frequency. In SI, $\omega_c=eB/m_e=(1.602\times10^{-19}\,\mathrm{C})(1\,\mathrm{T})/(9.109\times10^{-31}\,\mathrm{kg})=1.759\times10^{11}\ \mathrm{rad/s}$ . As an ordinary frequency, $f_c=\omega_c/2\pi=\mathbf{28.0\ GHz}$ — squarely in the microwave band, which is why cyclotron resonance is a microwave experiment.

Step 2. Landau-level spacing. Adjacent levels are separated by $\hbar\omega_c=(1.055\times10^{-34})(1.759\times10^{11})\,\mathrm{J}=1.855\times10^{-23}\,\mathrm{J}$ . Dividing by $1.602\times10^{-19}\,\mathrm{J/eV}$ gives $\hbar\omega_c=\mathbf{1.16\times10^{-4}\ eV}$ . This tiny spacing is why quantum-Hall physics requires low temperatures ( $k_BT\lesssim\hbar\omega_c$ means $T\lesssim1\,\mathrm{K}$ at 1 T).

Step 3. Zeeman consistency check. The Bohr-magneton scale is $\mu_B B=(9.274\times10^{-24}\,\mathrm{J/T})(1\,\mathrm{T})=9.274\times10^{-24}\,\mathrm{J}=\mathbf{5.79\times10^{-5}\ eV}$ . Note $\hbar\omega_c=2\mu_B B$ exactly, because $\mu_B=e\hbar/2m_e$ while $\hbar\omega_c=e\hbar B/m_e$ ; the factor of two is the orbital-versus-spin bookkeeping and is a useful sanity check. These numbers are reproduced in the verifier; the takeaways are a 28.0 GHz cyclotron frequency, a $1.16\times10^{-4}$ eV Landau spacing, and $\hbar\omega_c=2\mu_B B$ .

Example 2 — The paramagnetic term is the orbital Zeeman coupling. Show that in the symmetric gauge $\mathbf{A}=\tfrac12\mathbf{B}\times\mathbf{r}$ with $\mathbf{B}=B\hat{\mathbf z}$ , the linear term $-\tfrac{q}{mc}\mathbf{A}\cdot\mathbf{p}$ equals $-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ .

Step 1. Coulomb gauge check. Compute $\nabla\cdot\mathbf{A}=\nabla\cdot\tfrac12(\mathbf{B}\times\mathbf{r})=0$ for uniform $\mathbf{B}$ , so $\mathbf{p}\cdot\mathbf{A}=\mathbf{A}\cdot\mathbf{p}$ as operators — no ordering correction. The cross term in $(\mathbf{p}-\tfrac{q}{c}\mathbf{A})^2$ is therefore $-\tfrac{2q}{c}\mathbf{A}\cdot\mathbf{p}$ , giving $-\tfrac{q}{mc}\mathbf{A}\cdot\mathbf{p}$ in $\hat H$ .

Step 2. Substitute $\mathbf{A}=\tfrac12\mathbf{B}\times\mathbf{r}$ : $\mathbf{A}\cdot\mathbf{p}=\tfrac12(\mathbf{B}\times\mathbf{r})\cdot\mathbf{p}$ . Use the scalar triple-product identity $(\mathbf{B}\times\mathbf{r})\cdot\mathbf{p}=\mathbf{B}\cdot(\mathbf{r}\times\mathbf{p})=\mathbf{B}\cdot\mathbf{L}$ , so $\mathbf{A}\cdot\mathbf{p}=\tfrac12\mathbf{B}\cdot\mathbf{L}$ .

Step 3. Assemble: $-\tfrac{q}{mc}\mathbf{A}\cdot\mathbf{p}=-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ . Writing $q=-e$ for an electron, this is $+\tfrac{e}{2m_ec}\mathbf{L}\cdot\mathbf{B}=\tfrac{\mu_B}{\hbar}\mathbf{L}\cdot\mathbf{B}$ (Gaussian $\mu_B=e\hbar/2m_ec$ ), the orbital Zeeman coupling with $g_\ell=1$ . The symbolic identity $\tfrac12(\mathbf{B}\times\mathbf{r})\cdot\mathbf{p}=\tfrac12\mathbf{B}\cdot\mathbf{L}$ and $\nabla\cdot\mathbf{A}=0$ are both checked in the verifier; the result is that the paramagnetic term is exactly the orbital Zeeman coupling $-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ .

Origins · 1 mingauge invariance and how this was discovered

gauge invariance

principle, 19th–20th c. · the symmetry that fixes how matter couples to force fields

Gauge invariance began as a bookkeeping curiosity of the electromagnetic potentials — the freedom to add $\nabla\chi$ to $\mathbf{A}$ without changing the fields — and grew into the organizing principle of modern physics. Hermann Weyl introduced the term Eichinvarianz (gauge invariance) in 1918–1929, first in a failed attempt to unify gravity and electromagnetism by rescaling lengths, then correctly reinterpreted as a rescaling of the quantum phase. Vladimir Fock and Fritz London recognized (1927) that minimal coupling in the Schrödinger equation is exactly a local phase symmetry. The insight — that demanding local phase invariance forces the existence of the gauge field $\mathbf{A}$ — was generalized by Yang and Mills (1954) to non-abelian groups and now underlies the entire Standard Model. In this article gauge invariance appears in its original home: the electromagnetic coupling of a single charged particle.

History

1918

— Weyl proposes a "gauge" (scale) invariance in an attempt to unify gravity and electromagnetism

1926

— Schrödinger, Klein, Gordon, and Fock write the wave equation for a charged particle with the minimal-coupling substitution

1927

— Fock and London identify minimal coupling as a local phase (gauge) symmetry of the quantum wavefunction

1929

— Weyl gives the modern reinterpretation: gauge invariance is the invariance of the quantum phase, not of length

1930

— Landau solves the uniform-field problem, obtaining the quantized levels

E_n=\hbar\omega_c(n+\tfrac12)

1959

— Aharonov and Bohm predict a measurable phase from

\mathbf{A}

in a field-free region, proving the potential carries physics

Going further

References. Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.7 (Potentials and Gauge Transformations) · Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.3 and Problem 4.61 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. I, Ch. III complement · Landau, "Diamagnetismus der Metalle," Z. Phys. 64, 629 (1930) · Aharonov & Bohm, Phys. Rev. 115, 485 (1959).

Provenance. A verifier should reproduce: (1) the homogeneous transformation $\mathbf{D}'\psi'=e^{iq\chi/\hbar c}\mathbf{D}\psi$ and the full form-invariance $S'[\psi']=e^{iq\chi/\hbar c}S[\psi]$ including the $\partial_t\chi$ cancellation; (2) the gauge current $\mathbf{j}=\tfrac1m\mathrm{Re}(\psi^*\mathbf{D}\psi)$ equals $\tfrac{\hbar}{m}|\psi|^2\nabla S-\tfrac{q}{mc}\mathbf{A}|\psi|^2$ and is gauge-invariant; (3) the kinetic expansion $(\mathbf{p}-\tfrac{q}{c}\mathbf{A})^2=p^2-\tfrac{q}{c}(\mathbf{p}\cdot\mathbf{A}+\mathbf{A}\cdot\mathbf{p})+\tfrac{q^2}{c^2}A^2$ ; (4) the symmetric-gauge identity $\tfrac12(\mathbf{B}\times\mathbf{r})\cdot\mathbf{p}=\tfrac12\mathbf{B}\cdot\mathbf{L}$ with $\nabla\cdot\mathbf{A}=0$ ; (5) the numerics $f_c=28.0$ GHz, $\hbar\omega_c=1.16\times10^{-4}$ eV, $\mu_B B=5.79\times10^{-5}$ eV, and $\hbar\omega_c=2\mu_B B$ at $B=1$ T.

Builds on: Time-Dependent Schrödinger Equation · Probability Current and the Continuity Equation · Hamiltonian mechanics of a charged particle. Leads to: Landau Levels and the Quantum Hall Effect (solid-state track) · The Aharonov–Bohm Effect · The Zeeman Effect · Lorentz Covariance of the Dirac Equation.

Why is it called "minimal" coupling?

Because it introduces the electromagnetic interaction with the fewest possible new terms — just the single replacement $\mathbf{p}\to\mathbf{p}-\tfrac{q}{c}\mathbf{A}$ (plus $+q\phi$ ) — and no extra, independent coupling constants. It is the unique substitution that (a) reproduces the classical Lorentz force in the correspondence limit, (b) preserves gauge invariance, and (c) adds no new adjustable parameters beyond the charge $q$ . One could add a "non-minimal" term like a Pauli moment $\propto\boldsymbol{\sigma}\cdot\mathbf{B}$ by hand, but minimal coupling deliberately does not; for the electron that spin term instead falls out automatically from the Dirac equation, which is one reason the Dirac construction is considered so deep.

Does the electron's spin appear in this Hamiltonian?

No. The Hamiltonian $(1)$ is the scalar Schrödinger equation, which describes a spinless charged particle or the orbital motion of a particle whose spin we ignore. Its magnetism is entirely orbital: the $-\tfrac{q}{2mc}\mathbf{L}\cdot\mathbf{B}$ term has $g_\ell=1$ . Spin magnetism — the $-\tfrac{gq}{2mc}\mathbf{S}\cdot\mathbf{B}$ coupling with $g=2$ — must be added by hand in the Pauli equation, or derived from first principles in the Dirac equation, where minimal coupling of the relativistic equation produces $g=2$ automatically. So this article captures the orbital half of atomic magnetism; the spin half is signposted to the Pauli and Dirac articles.

Why does the wavefunction phase depend on

q/\hbar c

specifically?

The exponent $q\chi/\hbar c$ is fixed by requiring the covariant derivative to transform homogeneously. When you differentiate $e^{i\lambda}\psi$ , the gradient of the phase produces a term $-i\hbar(i\nabla\lambda)=\hbar\nabla\lambda$ ; to cancel the $\tfrac{q}{c}\nabla\chi$ from the shifted potential you need $\hbar\nabla\lambda=\tfrac{q}{c}\nabla\chi$ , i.e. $\lambda=\tfrac{q}{\hbar c}\chi$ . The $\hbar$ makes it a quantum phase (classically there is no wavefunction to rotate), the $q$ makes the coupling proportional to charge, and the $c$ is the Gaussian-unit bookkeeping (absent in SI). It is not a free choice — any other coefficient would leave an uncancelled $\nabla\chi$ term and break gauge invariance.

What is the difference between canonical and kinetic momentum, and which is conserved?

The canonical momentum $\mathbf{p}=-i\hbar\nabla$ is conjugate to position and is what appears in commutation relations $[x_i,p_j]=i\hbar\delta_{ij}$ . The kinetic momentum $\mathbf{\Pi}=\mathbf{p}-\tfrac{q}{c}\mathbf{A}=m\mathbf{v}$ is the physical momentum of motion. They differ by $\tfrac{q}{c}\mathbf{A}$ . Crucially, the canonical momentum is gauge-dependent (it changes when $\mathbf{A}$ changes), while the kinetic momentum is gauge-covariant (physical). If a coordinate is cyclic — say the potentials do not depend on $x$ — then the canonical $p_x$ is conserved, even though the kinetic $\Pi_x$ is not; this is a frequent source of confusion in Landau-level problems, where $p_x$ (canonical) labels the states but $\Pi_x$ (kinetic) does the physics.

Is the diamagnetic

A^2

term ever important?

For ordinary atoms in laboratory fields it is tiny — smaller than the paramagnetic term by a factor of order $\hbar\omega_c/(\text{atomic energy})\sim10^{-5}$ per tesla — which is why the linear Zeeman effect dominates atomic spectra. But it is not always negligible. In neutron stars and white dwarfs, fields reach $10^8$ – $10^{12}$ T, and the $A^2$ term reshapes atomic structure entirely. In superconductors the collective diamagnetic response produces the Meissner effect (perfect field expulsion) and flux quantization. And in strong-field Landau physics the $A^2$ term is not a perturbation at all — it combines with the paramagnetic term to give the exact $E_n=\hbar\omega_c(n+\tfrac12)$ spectrum, which is why Landau levels cannot be obtained by keeping only the linear term.

How does this connect to the Aharonov–Bohm effect?

Directly, through the local phase. A time-independent gauge transformation $\chi(\mathbf{r})$ multiplies the wavefunction by $e^{iq\chi/\hbar c}$ . If a particle travels a path and $\mathbf{A}=\nabla\chi$ along it (so $\mathbf{B}=0$ locally), the accumulated phase is $\tfrac{q}{\hbar c}\int\mathbf{A}\cdot d\mathbf{r}=\tfrac{q}{\hbar c}[\chi(\text{end})-\chi(\text{start})]$ . For a closed loop enclosing magnetic flux, $\chi$ is not single-valued and the phase is $\tfrac{q}{\hbar c}\oint\mathbf{A}\cdot d\mathbf{r}=\tfrac{q}{\hbar c}\Phi$ , the enclosed flux — an observable interference shift even where the field vanishes on the path. Minimal coupling is thus the exact framework in which the Aharonov–Bohm phase is defined; the effect is developed fully in its own article.

Does gauge invariance change the energy eigenvalues?

No — energy eigenvalues are gauge-invariant, but you have to be careful about what "the energy" means when the potentials are time-dependent. For a static gauge transformation $\chi(\mathbf{r})$ , the Hamiltonian is unitarily conjugated $\hat H'=U\hat H U^\dagger$ with $U=e^{iq\chi/\hbar c}$ , so the spectrum is identical and eigenstates just pick up the phase. For a time-dependent $\chi(\mathbf{r},t)$ , the scalar potential changes, and the naive "expectation of $\hat H$ " is not gauge-invariant because $\hat H$ itself is not — but all physically measurable transition energies, currents, and probabilities are. The lesson is that gauge-dependent intermediate quantities (like the canonical momentum or the instantaneous $\langle\hat H\rangle$ in a time-dependent gauge) can differ between gauges, while every measurable output agrees.