# Is the Spin Orbit Hamiltonian really Hermitian?

The regular spin orbit Hamiltonian is

$$H_{SO} = \frac{q\hbar}{4 m^2 c^2}\sigma\cdot(\textbf{E}\times \textbf{p})$$

If I consider a 2D system where E = E(x,y) and p is treated as an operator, i.e. $\hat{p} = \hat{i}p_x + \hat{j}p_y$ then, clearly E and p do not commute, so this doesn't look like a Hermitian operator.

Shouldn't this be added to its Hermitian adjoint (and divided by 2) to get a Hermitian Hamiltonian?

Related Quantum Physics News on Phys.org
Bill_K
The spin-orbit Hamiltonian is usually written ξ(r) L·S where ξ(r) = (1/2m2c2r) dV/dr.

Note that since V(r) is spherically symmetric, ξ(r) and L commute, so the whole thing is Hermitian.

But what if I want to look at the Spin Orbit Coupling in a system due to an external electric field, which may not be spherically symmetrical?

Bill_K
I think E x p is Hermitian as long as E = ∇V, isn't it? That's basically crossing ∇ with itself.

Well, if my position ket is of the form [itex]|x;y\rangle[/tex], and I am representing them on a rectangular grid (discrete representation)

$$\langle x_1;y_1|E_y(x,y) p_x|x_2;y_2\rangle = \sum_{x'',y''}\langle x_1;y_1|E_y(x,y)|x'';y''\rangle \langle x'';y''|p_x|x_2;y_2\rangle$$

Now,

$$\langle x_1;y_1|E_y(x,y)|x'';y''\rangle = E_y(x_1,y_1)\delta_{x_1,x''}\delta_{y_1,y''}$$

and

$$\langle x'';y''|p_x|x_2;y_2\rangle = -i\hbar\langle x'';y''|\left[\frac{|x_2 + a_x;y_2\rangle - |x_2-a_x;y_2\rangle}{2 a_x}\right]$$

The problem is... where you evaluate E_y -- at the location specified by the bra or the ket?

Can you please elaborate on that? I am trying to use this Hamiltonian with an external electric field E, in a quantum transport problem. Is it inapplicable there?

Would the correct prescription be to use 0.5(this + h.c.) where h.c. denotes the Hermitian adjoint of this operator?