Why doesn't the spin-orbit coupling term in the Hamiltonian have to be Hermitian?

Question

I am currently looking at a paper (C. Ortix, "Quantum mechanics of a spin-orbit coupled electron constrained to a space curve", Phys. Rev. B 91 (2015) 245412, arXiv:1504.00840) where spin-orbit coupling (SOC) in quasi-one-dimensional quantum wires is studied. The starting point is a SOC Hamiltonian

$$ H= \frac{\mathbf p^2}{2m} + \mathbf\alpha\cdot(\mathbf\sigma\times\mathbf p). \tag{1} $$

Now, since $\alpha$ should in general be position dependent (the authors, for example, eventually turn to a helix), I don't see how $(1)$ would be Hermitian. In the paper, the authors go to a co-moving coordinate system and translate the Hamiltonian to general curved-coordinate form: $$ H = -\frac{\hbar^2}{2m}G^{\mu\nu}D_\mu D_\nu-i\hbar \frac{\varepsilon^{\mu\nu\lambda}}{\|G\|}\alpha_\mu\gamma_\nu\partial_\lambda, \tag{2} $$

where $\gamma_\mu$ generate the curved space Clifford algebra, $\{\gamma_\mu,\gamma_\nu\}=2G_{\mu\nu}$. Here, the components $\alpha_\mu$ are constant while the matrices $\gamma_\mu$ clearly depend on the coordinates. Hence, again, I do not see how $(2)$ is Hermitian.

At the same time, the effective 1-dim. Hamiltonian that is derived from this by adding strong confining potentials, expanding the Hamiltonian to second order in the directions normal and binormal to the path and averaging out these degrees of freedom, is Hermitian.

My question is: Why does the general 3d SOC Hamiltonian $(1)$ or $(2)$ not have to be Hermitian? Ok, the 1d derived from it somehow is (I'm not quoting the latter because I don't think its precise form matters here) but why would we even start with a Hamiltonian that's not physical?

PS: I found one related question (Spin-orbit model; Hamiltonian seems to be non-Hermitian) but there, the issue is only mentioned, not discussed further.

Answer 1

In the systems that I know about that have the Spin orbit coupling term that you have (the Rashba effect in solids) it is of the form ${\bf E}\cdot ({\boldsymbol \sigma}\times {\bf p} )$ where ${\bf E}$ is an electric field. In that situation the field is usually constant. I conjecture, however, that the Rashba term is hermitian provided that $\nabla \times {\bf E}=0$, which is usually the case.

Your cited paper is pretty vague about whether $\alpha$ is to be a constant or not. They do write $\alpha_T$, $\alpha_B$, $\alpha_N$ for the componets along the Serret-Frenet triad, and these will be position dependent even if the original vector $\alpha$ is constant. I see no derivatives of them though, so maybe the author was just careless. You could write to the author and ask (politely) whether the $\alpha$'s are assumed constant.

Answer 2

Usually there are more than one quantum mechanical counterparts of a classical operator. The true corresponding quantum operator should be a linear combination of them determined by basic principles (such as Hermicity) and experiments. In your case, I think you should hermitize the SOC term since alpha and p do not commute.

A similar example, if you want to write down the term due to the coupling between angular momentum L and the magnetic field in a Hamiltonian, the correct way of writing down L is 1/2(r^p-p^r) rather than r^p.