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Electromagnetism Checklist

Electromagnetism enters quantum mechanics through charged particles, potentials, magnetic moments, radiation, spectroscopy, condensed matter, and the bridge to relativistic field theory. A reader does not need all of advanced electrodynamics before studying quantum mechanics, but the basic field and potential language should be solid.

This checklist uses SI-style equations for orientation. Later pages may choose different unit conventions, so always check the local convention for factors of cc, ϵ0\epsilon_0, and μ0\mu_0.

  • Distinguish electric fields, magnetic fields, scalar potentials, and vector potentials.
  • State the Lorentz force law.
  • Recognize the physical meaning of electric potential energy.
  • Explain why a magnetic field changes the motion of a charged particle.
  • Identify electromagnetic waves, propagation direction, polarization, wavelength, and frequency.
  • Use the relation between E\mathbf E, B\mathbf B, and propagation direction for a plane wave.
  • Explain gauge freedom at an introductory level.
  • Know that potentials can affect quantum phases.
  • Recognize magnetic dipoles and magnetic moments.
  • State qualitatively how spin and magnetic fields couple.
  • Distinguish radiation fields from static fields.
  • Recognize when a nonrelativistic treatment of light-matter coupling is an approximation.

In SI notation, Maxwell’s equations are

∇⋅E=ρϵ0,∇⋅B=0,\nabla\cdot\mathbf E=\frac{\rho}{\epsilon_0}, \qquad \nabla\cdot\mathbf B=0, ∇×E=−∂B∂t,∇×B=μ0J+μ0ϵ0∂E∂t.\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}, \qquad \nabla\times\mathbf B =\mu_0\mathbf J+\mu_0\epsilon_0 \frac{\partial\mathbf E}{\partial t}.

The Lorentz force is

F=q(E+v×B).\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B).

Potentials are defined by

B=∇×A,E=−∇Φ−∂A∂t.\mathbf B=\nabla\times\mathbf A, \qquad \mathbf E=-\nabla\Phi-\frac{\partial\mathbf A}{\partial t}.

A gauge transformation changes the potentials by

A′=A+∇χ,Φ′=Φ−∂χ∂t,\mathbf A'=\mathbf A+\nabla\chi, \qquad \Phi'=\Phi-\frac{\partial\chi}{\partial t},

while leaving E\mathbf E and B\mathbf B unchanged.

For a nonrelativistic particle of charge qq in prescribed electromagnetic potentials, a common Hamiltonian is

H=12m(−iℏ∇−qA)2+qΦ.H=\frac{1}{2m} \left(-i\hbar\nabla-q\mathbf A\right)^2 +q\Phi.

This formula explains why the vector potential is not merely bookkeeping in quantum mechanics. The field strengths remain gauge-invariant observables, but the potential appears directly in the operator acting on wavefunctions. Gauge changes must therefore be accompanied by the corresponding change of phase convention for the quantum state.

Magnetic fields also couple to magnetic moments. For a spin degree of freedom, a basic Zeeman form is

HZ=−μ⋅B,μ=γS,H_Z=-\boldsymbol\mu\cdot\mathbf B, \qquad \boldsymbol\mu=\gamma\mathbf S,

with the gyromagnetic factor depending on the particle and convention. This is one reason electromagnetism is inseparable from spin, spectroscopy, resonance, and many quantum technologies.

  1. A plane electromagnetic wave travels in the +z+z direction. Its electric field points in the +x+x direction. What direction should its magnetic field point in vacuum?
Solution

For a wave traveling in direction k^\hat{\mathbf k}, the fields satisfy

B=1ck^×E\mathbf B=\frac{1}{c}\hat{\mathbf k}\times\mathbf E

in vacuum. With k^=z^\hat{\mathbf k}=\hat{\mathbf z} and E\mathbf E along x^\hat{\mathbf x},

z^×x^=y^.\hat{\mathbf z}\times\hat{\mathbf x}=\hat{\mathbf y}.

So B\mathbf B points in the +y+y direction, with magnitude E/cE/c.

  1. Show that the vector potential
A=12B(−y,x,0)\mathbf A=\frac12 B(-y,x,0)

gives a uniform magnetic field Bz^B\hat{\mathbf z}.

Solution

The curl has components

(∇×A)x=0,(∇×A)y=0,(\nabla\times\mathbf A)_x=0, \qquad (\nabla\times\mathbf A)_y=0,

and

(∇×A)z=∂Ay∂x−∂Ax∂y=B2−(−B2)=B.(\nabla\times\mathbf A)_z =\frac{\partial A_y}{\partial x} -\frac{\partial A_x}{\partial y} =\frac{B}{2}-\left(-\frac{B}{2}\right) =B.

Therefore ∇×A=Bz^\nabla\times\mathbf A=B\hat{\mathbf z}.

  1. Verify that the gauge transformation above leaves B\mathbf B unchanged.
Solution

Using B′=∇×A′\mathbf B'=\nabla\times\mathbf A',

B′=∇×(A+∇χ)=∇×A+∇×∇χ.\mathbf B' =\nabla\times(\mathbf A+\nabla\chi) =\nabla\times\mathbf A+\nabla\times\nabla\chi.

The curl of a gradient is zero for smooth χ\chi, so

B′=∇×A=B.\mathbf B'=\nabla\times\mathbf A=\mathbf B.
  1. Why is it misleading to say that only E\mathbf E and B\mathbf B matter in quantum mechanics?
Solution

Observable predictions are gauge invariant, but the wavefunction representation of a charged particle can depend directly on the potentials. In the Hamiltonian, A\mathbf A appears inside the kinetic momentum operator. A gauge change can alter the phase convention of the wavefunction without changing fields or measured probabilities. This is one reason potential language is essential rather than optional.

  • Confusing electric potential Φ\Phi with electric potential energy qΦq\Phi.
  • Treating A\mathbf A as unique. Different vector potentials can describe the same magnetic field.
  • Forgetting that magnetic forces do no work on a classical point charge because v⋅(v×B)=0\mathbf v\cdot(\mathbf v\times\mathbf B)=0.
  • Assuming polarization is a property of photons only. Classical waves already have polarization; quantum theory changes the state description.
  • Ignoring unit conventions when comparing formulas from different sources.
  • Treating light-matter coupling as exact without specifying whether the electromagnetic field is classical, quantized, prescribed, or dynamical.

Use these pages when a checklist item is weak:

  • D. J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017.
  • J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley, 1998.
  • A. Zangwill, Modern Electrodynamics, Cambridge University Press, 2013.
  • C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 1, Wiley, 1977.