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Minimal coupling

For a particle of charge $q$ in a vector potential $\mathbf A$, the canonical momentum in the free Hamiltonian is replaced by $\mathbf p-q\mathbf A$, giving
$$
H=\frac1{2m}(-i\hbar\nabla-q\mathbf A)^2.
$$


Magnetic minimal coupling

Aharonov-Bohm effect

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 34D / a / Solution

Magnetic minimal coupling

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Magnetic minimal coupling

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