Minimal coupling

Intuition

A charge in electric and magnetic fields feels the Lorentz force, $q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$ . Its magnetic part depends on the velocity and cannot be written as minus the gradient of a potential energy, so a Hamiltonian needs the potentials instead: the scalar potential $\phi$ and the vector potential $\mathbf{A}$ , with $\mathbf{B}=\nabla\times\mathbf{A}$ and $\mathbf{E}=-\nabla\phi-\partial\mathbf{A}/\partial t$ . Classical mechanics already has the answer, $H=(\mathbf{p}-q\mathbf{A})^{2}/2m+q\phi$ , and quantum mechanics keeps it with $\hat{\mathbf{p}}=-i\hbar\nabla$ . This is minimal coupling: the momentum is shifted by the charge times the vector potential. It splits momentum in two. The canonical momentum $\hat{\mathbf{p}}$ , which generates translations, is no longer $m$ times the velocity; the kinetic momentum $\hat{\mathbf{p}}-q\mathbf{A}$ is, and its components do not commute.

A walker on a moving walkway can be timed by the pace of the legs or by the speed past the walls, and the two differ by the walkway’s speed. Canonical and kinetic momentum differ by $q\mathbf{A}$ in the same way, on a walkway whose speed changes from place to place.

The vector potential $\mathbf{A}=\tfrac12\mathbf{B}\times\mathbf{r}$ of a uniform field $\mathbf{B}$ pointing out of the page, drawn at points on two circles round the origin: it circles the field’s axis and grows in proportion to the distance from it. Its curl is everywhere.

A particle of charge $q$ and mass $m$ in the potentials $\phi(\mathbf{r},t)$ and $\mathbf{A}(\mathbf{r},t)$ , with and , has the Hamiltonian

\begin{aligned}\hat{H}&=\frac{\bigl(\hat{\mathbf{p}}-q\mathbf{A}\bigr)^{2}}{2m}+q\phi\\ \hat{\mathbf{p}}&=-i\hbar\nabla\end{aligned}

Properties

The kinetic momentum $\vec{\Pi}=\hat{\mathbf{p}}-q\mathbf{A}$ is times the velocity, ; the canonical momentum still obeys and generates translations, but it is not .

Velocity in a magnetic field

\hat{\mathbf{v}}=\frac{\vec{\Pi}}{m},\qquad [\hat{\Pi}_{x},\hat{\Pi}_{y}]=i\hbar q\,B_{z}

Heisenberg’s equation gives the velocity as a commutator of the Hamiltonian with the position. Of the kinetic momenta only $\hat{\Pi}_{x}$ fails to commute with $\hat{x}$ , and its square gives $_{}$ . For the second statement, the canonical momenta commute with one another and so do the components of , so only the cross terms survive; each is a derivative of , and together they make its curl.

Proof steps

\hat{v}_{x}=\frac{i}{\hbar}[\hat{H},\hat{x}]=\frac{i}{\hbar}\,\frac{[\vec{\Pi}^{2},\hat{x}]}{2m}

Heisenberg’s equation; $q\phi$ commutes with $\hat{x}$ .

[\hat{\Pi}_{x},\hat{x}]=[\hat{p}_{x},\hat{x}]=-i\hbar\ \Longrightarrow\ [\vec{\Pi}^{2},\hat{x}]=-2i\hbar\,\hat{\Pi}_{x}

$\mathbf{A}$ commutes with $\hat{x}$ , and so do $\hat{\Pi}_{y}$ and .

\hat{v}_{x}=\frac{\hat{\Pi}_{x}}{m}

Put the commutator into the first line.

[\hat{\Pi}_{x},\hat{\Pi}_{y}]=-q[\hat{p}_{x},A_{y}]-q[A_{x},\hat{p}_{y}]

$[\hat{p}_{x},\hat{p}_{y}]=0$ and .

[\hat{\Pi}_{x},\hat{\Pi}_{y}]=i\hbar q\,(\partial_{x}A_{y}-\partial_{y}A_{x})=i\hbar q\,B_{z}

$[\hat{p}_{x},f]=-i\hbar\,\partial_{x}f$ for any function of position.

Applications

Practice

The Hamiltonian

A charge enters the Hamiltonian through the potentials: the momentum is shifted by $q$ times the vector potential, and $q\phi$ is added.

\hat{H}=\frac{(\hat{\mathbf{p}}-q\mathbf{A})^{2}}{2m}+q\phi

Try it

Which Hamiltonian describes a particle of charge $q$ in the potentials $\phi$ and $\mathbf{A}$ ?