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Consider a charged quantum particle in a magnetic field. The Hamiltonian can be written using minimal coupling: $$ H = \frac{1}{2m} \left( \mathbf{p} - \frac{e}{c} \mathbf{A}(\mathbf{x}) \right)^2 $$

where $\mathbf{A}(\mathbf{x})$ is the vector potential.

If one expands this, one gets: $$ H = \frac{1}{2m} \left\{ \mathbf{p}^2 + \left(\frac{e}{c}\mathbf{A}(\mathbf{x})\right)^2 - \frac{2 e}{c} \mathbf{A}(\mathbf{x}) ~\mathbf{p} - \frac{e}{c} \underbrace{\left[ \mathbf{p}, \mathbf{A}(\mathbf{x}) \right]}_{\textrm{commutator}} \right\} $$

Now, the commutator becomes $$[\mathbf{p}, \mathbf{A}(\mathbf{x})] = (\nabla \cdot \mathbf{A}(\mathbf{x}))~~ [\mathbf{p}, \mathbf{x}] = -i (\nabla \cdot \mathbf{A}(\mathbf{x}))$$ which is imaginary and independent of momentum.

Now, this term drops out when one is in the Coulomb gauge. But if one picks a different gauge, say $\mathbf{A} = ( xy, 0, 0)$ for a gradient magnetic field, doesn't such a term make the Hamiltonian non-Hermitian? Which would in turn lead to complex energy eigenvalues?

Or, else, do we have to insist on using the Coulomb gauge for non-relativistic problems?

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I have encountered this situation once before whilst studying the quantum Hall effect, check out these lecture notes Chapter 2 and especially page 35. – Funzies Jun 20 '13 at 21:48

Since the Hamiltonian $2mH$ is self-adjoint$^1$ (the square of a Hermitian operator is again Hermitian), there is formally no problem. If one of the terms $$- \frac{e}{c} \left[ \mathbf{p}, \mathbf{A}(\mathbf{x}) \right]$$

in the Hamiltonian $2mH$ is non-Hermitian, there must be other non-Hermitian terms that cancel any non-Hermitian effect in the full Hamiltonian $2mH$. In this case the culprit is

$$ - \frac{2 e}{c} \mathbf{A}(\mathbf{x}) ~\mathbf{p}, $$

which is non-Hermitian as well.


$^1$ Here we ignore potential problems related to domains of unbounded operators, i.e. we do not properly distinguish between symmetric, Hermitian, and self-adjoint operators.

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