4.1.2 Electromagnetic Coupling#

Prompts

Starting from $\hat{H} = \frac{(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2}{2m} + q\Phi$, compute $\frac{\mathrm{d}\langle\boldsymbol{x}\rangle}{\mathrm{d}t}$ using Heisenberg equations. What is the kinetic momentum $\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}$? How does it differ from canonical momentum?
Compute $\frac{\mathrm{d}\langle\hat{\boldsymbol{\pi}}\rangle}{\mathrm{d}t}$ and show it gives the Lorentz force $m\boldsymbol{a} = q(\boldsymbol{E} + \boldsymbol{v} \times \boldsymbol{B})$. What does this reveal about the gauge field?
How do $\boldsymbol{E}$ and $\boldsymbol{B}$ arise from the gauge potentials? Write expressions for $\boldsymbol{E}$ and $\boldsymbol{B}$ in terms of $\boldsymbol{A}$ and $\Phi$. What is the field strength tensor $F_{\mu\nu}$?
We derived electromagnetism from the requirement of local U(1) phase invariance. What assumptions went in, and what came out? Why does gauge theory demand the existence of electromagnetism?

Lecture Notes#

Overview#

The gauge-invariant Hamiltonian $\hat{H} = \frac{(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2}{2m} + q\Phi$, derived in §4.1.1 purely from local U(1) phase symmetry, contains a profound discovery: it automatically describes charged particles interacting with electromagnetic fields. By computing equations of motion, we extract the Lorentz force and identify the electric and magnetic fields. The gauge principle transforms a mathematical symmetry requirement into the fundamental interaction that shapes the universe.

Kinetic Momentum#

The Heisenberg equation for position gives:

\[\frac{\mathrm{d}\langle\boldsymbol{x}\rangle}{\mathrm{d}t} = \frac{1}{\mathrm{i}\hbar}\langle[\hat{\boldsymbol{x}}, \hat{H}]\rangle = \frac{\langle\hat{\boldsymbol{p}} - q\boldsymbol{A}\rangle}{m}\]

using $[\hat{x}_i, \hat{p}_j] = \mathrm{i}\hbar\delta_{ij}$ and $[\hat{x}_i, \boldsymbol{A}(\hat{\boldsymbol{x}})] = 0$.

Kinetic Momentum

\[\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}, \qquad \boldsymbol{v} = \frac{\hat{\boldsymbol{\pi}}}{m}\]

The kinetic momentum $\hat{\boldsymbol{\pi}}$ determines the particle’s velocity. The canonical momentum $\hat{\boldsymbol{p}} = -\mathrm{i}\hbar\nabla$ is gauge-dependent; only $\hat{\boldsymbol{\pi}}$ is directly observable.

Caution: Kinetic Momentum Does Not Commute

Unlike canonical momentum ($[\hat{p}_i, \hat{p}_j] = 0$), kinetic momentum components satisfy $[\hat{\pi}_i, \hat{\pi}_j] = \mathrm{i}q\hbar\epsilon_{ijk}B_k$ in a magnetic field. This noncommutativity is central to Landau quantization (§4.3).

The Lorentz Force Law#

Computing $\frac{\mathrm{d}\langle\hat{\boldsymbol{\pi}}\rangle}{\mathrm{d}t}$ requires the commutator $[\hat{\boldsymbol{p}}, \hat{H}]$. Using $[\hat{p}_i, A_j] = -\mathrm{i}\hbar\partial_i A_j$ and the product rule on $(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2$, one extracts gradient and curl terms from the gauge potentials.

Lorentz Force from Gauge Theory

\[m\frac{\mathrm{d}\langle\boldsymbol{v}\rangle}{\mathrm{d}t} = q(\langle\boldsymbol{E}\rangle + \langle\boldsymbol{v} \times \boldsymbol{B}\rangle)\]

where $\boldsymbol{E} = -\nabla\Phi - \frac{\partial\boldsymbol{A}}{\partial t}$ and $\boldsymbol{B} = \nabla \times \boldsymbol{A}$.

This is the quantum Lorentz force law, derived entirely from the gauge-invariant Hamiltonian with no assumption about the existence of electromagnetic forces.

Derivation: Equation of Motion for Kinetic Momentum

Expand $\hat{H} = \frac{1}{2m}(\hat{\boldsymbol{p}}^2 - q\boldsymbol{A}\cdot\hat{\boldsymbol{p}} - q\hat{\boldsymbol{p}}\cdot\boldsymbol{A} + q^2\boldsymbol{A}^2) + q\Phi$. The relevant commutators are:

\[[\hat{p}_i, A_j\hat{p}_j] = -\mathrm{i}\hbar(\partial_i A_j)\hat{p}_j, \quad [\hat{p}_i, A_j^2] = -2\mathrm{i}\hbar A_j(\partial_i A_j), \quad [\hat{p}_i, q\Phi] = -\mathrm{i}\hbar q\,\partial_i\Phi\]

Collecting all terms in $\frac{\mathrm{d}\langle\hat{\boldsymbol{p}}\rangle}{\mathrm{d}t} = \frac{1}{\mathrm{i}\hbar}\langle[\hat{\boldsymbol{p}}, \hat{H}]\rangle$ and using the total time derivative $\frac{\mathrm{d}\hat{\boldsymbol{\pi}}}{\mathrm{d}t} = \frac{\mathrm{d}\hat{\boldsymbol{p}}}{\mathrm{d}t} - q\frac{\partial\boldsymbol{A}}{\partial t}$:

\[m\frac{\mathrm{d}\langle\boldsymbol{v}\rangle}{\mathrm{d}t} = q\left(-\langle\nabla\Phi\rangle - \left\langle\frac{\partial\boldsymbol{A}}{\partial t}\right\rangle\right) + \frac{q}{m}\langle(\nabla\boldsymbol{A} - (\nabla\boldsymbol{A})^T)\cdot\hat{\boldsymbol{\pi}}\rangle\]

Recognizing $\boldsymbol{E} = -\nabla\Phi - \partial_t\boldsymbol{A}$ and $(\nabla\boldsymbol{A} - (\nabla\boldsymbol{A})^T)_{ij} = \partial_i A_j - \partial_j A_i = \epsilon_{ijk}B_k$ yields the Lorentz force.

Discussion

In classical mechanics, the Lorentz force $m\boldsymbol{a} = q(\boldsymbol{E} + \boldsymbol{v} \times \boldsymbol{B})$ is an empirical law. In quantum mechanics, we derived it from the gauge principle alone. Does this mean quantum mechanics is “deeper” than classical mechanics? Could a different gauge group produce a different force law?

Electric and Magnetic Fields from Potentials#

The gauge potentials $(\Phi, \boldsymbol{A})$ are gauge-dependent and not directly observable. The physical fields are:

\[\boldsymbol{E} = -\nabla\Phi - \frac{\partial\boldsymbol{A}}{\partial t}, \qquad \boldsymbol{B} = \nabla \times \boldsymbol{A}\]

In spacetime notation with $A_\mu = (\Phi, \boldsymbol{A})$, both fields are encoded in the field strength tensor:

Field Strength Tensor

\[F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\]

$F_{\mu\nu}$ is gauge-invariant: under $A_\mu \to A_\mu + \partial_\mu\chi$, the antisymmetry gives $F'_{\mu\nu} = F_{\mu\nu}$ because $\partial_\mu\partial_\nu\chi = \partial_\nu\partial_\mu\chi$.

The Gauge Principle Requires Electromagnetism#

The Gauge Principle

Starting from only two inputs:

Local U(1) phase invariance: $\psi(\boldsymbol{x},t) \to \mathrm{e}^{\mathrm{i}q\chi(\boldsymbol{x},t)}\psi(\boldsymbol{x},t)$
The Schrödinger equation retains its form

we were forced to introduce a gauge field $\boldsymbol{A}$, replace $\nabla \to \nabla - \mathrm{i}(q/\hbar)\boldsymbol{A}$ (covariant derivative), and adopt $\hat{H} = \frac{(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2}{2m} + q\Phi$. From this, the Lorentz force and the electric/magnetic fields emerged automatically.

Symmetry $\to$ Interaction: The gauge field is not freely chosen—it is required by quantum consistency. Every charged particle couples to the same universal gauge field.

Summary#

Kinetic momentum $\hat{\boldsymbol{\pi}} = \hat{\boldsymbol{p}} - q\boldsymbol{A}$ determines velocity; its components do not commute in a magnetic field.
The Lorentz force $m\boldsymbol{a} = q(\boldsymbol{E} + \boldsymbol{v}\times\boldsymbol{B})$ is derived from the gauge-invariant Hamiltonian, not assumed.
Electric and magnetic fields arise from gauge potentials: $\boldsymbol{E} = -\nabla\Phi - \partial_t\boldsymbol{A}$, $\boldsymbol{B} = \nabla\times\boldsymbol{A}$.
The field strength tensor $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is gauge-invariant.
The gauge principle transforms symmetry into interaction: local U(1) invariance requires electromagnetism.

Homework#

1. Compute the commutator $[\hat{x}_i, (\hat{\boldsymbol{p}} - q\boldsymbol{A})^2]$ using $[\hat{x}_i, \hat{p}_j] = \mathrm{i}\hbar\delta_{ij}$ and $[\hat{x}_i, \boldsymbol{A}(\hat{\boldsymbol{x}})] = 0$. Show that: $$ \frac{\mathrm{d}\langle\boldsymbol{x}\rangle}{\mathrm{d}t} = \frac{\langle\hat{\boldsymbol{p}} - q\boldsymbol{A}\rangle}{m} = \frac{\langle\hat{\boldsymbol{\pi}}\rangle}{m} $$ Explain why this result identifies the kinetic momentum with the “true” momentum that determines velocity.

2. Show that the kinetic momentum $\hat{\boldsymbol{\pi}}$ commutators differ from canonical momentum commutators: $$ [\hat{\pi}_i, \hat{\pi}_j] = [\hat{p}_i, \hat{p}_j] - q([\hat{p}_i, A_j] + [A_i, \hat{p}_j]) + q^2[A_i, A_j] $$ Simplify this using $[\hat{p}_i, A_j] = -\mathrm{i}\hbar\partial_i A_j$ and evaluate the commutator in the Coulomb gauge. Show that $[\hat{\pi}_i, \hat{\pi}_j] = \mathrm{i}q\hbar B_k$ for a uniform magnetic field in the $z$-direction.

3. Derive the Lorentz force law from first principles. Starting with: $$ \frac{\mathrm{d}\hat{\boldsymbol{p}}}{\mathrm{d}t} = \frac{1}{\mathrm{i}\hbar}[\hat{\boldsymbol{p}}, \hat{H}] $$ and $\hat{H} = \frac{(\hat{\boldsymbol{p}} - q\boldsymbol{A})^2}{2m} + q\Phi$, compute the commutators step by step. Show that: $$ \frac{\mathrm{d}\langle\hat{\boldsymbol{p}}\rangle}{\mathrm{d}t} = -q\nabla\Phi + q\langle(\nabla \times \boldsymbol{A}) \times \boldsymbol{v}\rangle + q\frac{\partial\boldsymbol{A}}{\partial t} $$ Then use $\boldsymbol{E} = -\nabla\Phi - \partial_t\boldsymbol{A}$ and $\boldsymbol{B} = \nabla \times \boldsymbol{A}$ to recover the Lorentz force.

4. In a uniform magnetic field $\boldsymbol{B} = B\hat{z}$, compute $[\hat{\pi}_x, \hat{\pi}_y]$ explicitly. Use the symmetric gauge $\boldsymbol{A} = \frac{B}{2}(-y, x, 0)$ to verify that $\boldsymbol{B} = \nabla \times \boldsymbol{A}$. Then show that: $$ [\hat{\pi}_x, \hat{\pi}_y] = \mathrm{i}\hbar q B $$ Interpret this as a measure of the field strength.

5. Show that the electric and magnetic fields are gauge-invariant. Under the gauge transformation $\boldsymbol{A} \to \boldsymbol{A}' = \boldsymbol{A} - \nabla\chi$ and $\Phi \to \Phi' = \Phi + \partial_t\chi$, prove: $$ \boldsymbol{E}' = \boldsymbol{E} \quad \text{and} \quad \boldsymbol{B}' = \boldsymbol{B} $$ Explain why this guarantees that all observable physics is independent of gauge choice.

6. Compute $\boldsymbol{E}$ and $\boldsymbol{B}$ from the potentials in the following gauge choices for a uniform field $\boldsymbol{B} = B\hat{z}$:

(a) Coulomb gauge: $\boldsymbol{A} = (0, Bx, 0)$ with $\Phi = 0$
(b) Symmetric gauge: $\boldsymbol{A} = \frac{B}{2}(-y, x, 0)$ with $\Phi = 0$
(c) Landau gauge: $\boldsymbol{A} = (-By, 0, 0)$ with $\Phi = 0$

Verify that all three gauges produce the same $\boldsymbol{B}$ but have different $\boldsymbol{A}$.

7. The field strength tensor in spacetime is $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. Write out all components of $F_{\mu\nu}$ (in matrix form, with $\mu, \nu \in \{0, x, y, z\}$) for a uniform electric field $\boldsymbol{E} = E\hat{x}$ and magnetic field $\boldsymbol{B} = B\hat{z}$. (Assume $\Phi$ and $\boldsymbol{A}$ are time-independent and spatial potentials are appropriately chosen.) What patterns do you notice in the structure of the tensor?

8. Explain why the gauge principle (local U(1) invariance) forces the existence of electromagnetism. What would change if:

(a) We demanded only global phase invariance instead?
(b) We used a different gauge group (say, SU(2)) instead of U(1)?
(c) We had no gauge symmetry requirement at all?

Discuss how this principle generalizes beyond electromagnetism to strong and weak nuclear forces.