Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
I have encountered this situation once before whilst studying the quantum Hall effect, check out these lecture notes arxiv.org/abs/0909.1998 Chapter 2 and especially page 35. –  Funzies Jun 20 '13 at 21:48
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.