In the presence of a magnetic field $\mathbf{B} = \nabla \times \mathbf{A}$, the Hamiltonian of a classical system is modified to :

$$ H = \frac{(\mathbf{p}-\frac{e}{c}\mathbf{A})^2}{2m}, $$

which is usually justified by minimal coupling, where the kinetic momentum $m\mathbf{\dot{r}}$ goes from $\mathbf{p}$ to $\mathbf{p}-\frac{e}{c}\mathbf{A}$, $\mathbf{p}$ being the canonical momentum.

In quantum mechanics, we quantise $\hat{H}$, $\hat{x}$ and $\hat{p}$ (canonical or kinetic?).

If I quantise the momentum, it means that $\mathbf{\hat{A}}$ has also been promoted to an operator.

What are the eigenstates and eigenvalues of the magnetic vector potential $\mathbf{\hat{A}}$?


1 Answer 1


The components of $\hat{\mathbf A}$ are just multiplication operators $$(\hat A_k \psi)(\mathbf x) = A_k(\mathbf x) \psi(\mathbf x) .$$ Therefore the eigenvectors are the same as those of $\hat{\mathbf x}$, i.e., $| \mathbf x_0 \rangle$ and $$ \hat A_k | \mathbf x_0 \rangle = A_k(\mathbf x_0)\, | \mathbf x_0 \rangle , $$ the eigenvalues are the range of the function $A_k$.

Technically, the eigenvectors are not in $L^2$ and to be understood as generalized eigenvalues, the "eigenvalues" are actually continuous spectrum.
For reference, see e.g. Reed & Simon, Functional Analysis, Proposition VIII.3.1.


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