Is the Spin Orbit Hamiltonian really Hermitian?

[maverick280857]

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?

 

[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.

 

[maverick280857]

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.

 

[maverick280857]

Well, if my position ket is of the form |x;y\rangle[/tex], and I am representing them on a rectangular grid (discrete representation)<br /> <br /> \langle x_1;y_1|E_y(x,y) p_x|x_2;y_2\rangle = \sum_{x&amp;#039;&amp;#039;,y&amp;#039;&amp;#039;}\langle x_1;y_1|E_y(x,y)|x&amp;#039;&amp;#039;;y&amp;#039;&amp;#039;\rangle \langle x&amp;#039;&amp;#039;;y&amp;#039;&amp;#039;|p_x|x_2;y_2\rangle<br /> <br /> Now,<br /> <br /> \langle x_1;y_1|E_y(x,y)|x&amp;#039;&amp;#039;;y&amp;#039;&amp;#039;\rangle = E_y(x_1,y_1)\delta_{x_1,x&amp;#039;&amp;#039;}\delta_{y_1,y&amp;#039;&amp;#039;}<br /> <br /> and<br /> <br /> \langle x&amp;#039;&amp;#039;;y&amp;#039;&amp;#039;|p_x|x_2;y_2\rangle = -i\hbar\langle x&amp;#039;&amp;#039;;y&amp;#039;&amp;#039;|\left[\frac{|x_2 + a_x;y_2\rangle - |x_2-a_x;y_2\rangle}{2 a_x}\right]<br /> <br /> The problem is... where you evaluate E_y -- at the location specified by the bra or the ket?

 

[DrDu]

We had a discussion just recently about the SO Hamiltonian.
https://www.physicsforums.com/showthread.php?t=670912&highlight=spin+orbit
The form you cited (which is only part of the hamiltonian) does not yield a self-adjoint hamiltonian and you can use it at best to obtain some low-order perturbational corrections.

 

[maverick280857]

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?