Is the Spin Orbit Hamiltonian really Hermitian?

In summary, the spin-orbit Hamiltonian is usually written ξ(r) L·S where ξ(r) = (1/2m2c2r) dV/dr. 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. However, if my position ket is of the form [itex]|x;y\rangle[/tex], and I am representing them on a rectangular grid
  • #1
maverick280857
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The regular spin orbit Hamiltonian is

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

If I consider a 2D system where E = E(x,y) and p is treated as an operator, i.e. [itex]\hat{p} = \hat{i}p_x + \hat{j}p_y[/itex] 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?
 
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  • #2
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.
 
  • #3
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?
 
  • #4
I think E x p is Hermitian as long as E = ∇V, isn't it? That's basically crossing ∇ with itself.
 
  • #5
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)

[tex]\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[/tex]

Now,

[tex]\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''}[/tex]

and

[tex]\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][/tex]

The problem is... where you evaluate E_y -- at the location specified by the bra or the ket?
 
  • #7
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?
 

1. What is a Spin Orbit Hamiltonian?

A Spin Orbit Hamiltonian is a mathematical representation of the interaction between an electron's spin and its orbital motion in an atom or molecule. It describes the coupling between the spin and orbital angular momentum, and is essential for understanding many physical phenomena.

2. Is the Spin Orbit Hamiltonian a Hermitian operator?

Yes, the Spin Orbit Hamiltonian is a Hermitian operator. This means that it is self-adjoint, and its eigenvalues are real. This is important for quantum mechanical calculations as it ensures that the results obtained are physically meaningful.

3. Why is it important for the Spin Orbit Hamiltonian to be Hermitian?

The Hermiticity of the Spin Orbit Hamiltonian ensures that it obeys the laws of quantum mechanics and accurately describes the physical system. It also guarantees that the energy levels and other properties calculated using this operator are physically meaningful and can be compared to experimental results.

4. How is the Hermiticity of the Spin Orbit Hamiltonian verified?

The Hermiticity of the Spin Orbit Hamiltonian can be verified by checking if it satisfies the condition of self-adjointness, which states that the operator is equal to its own adjoint. This can be done by taking the complex conjugate of the operator and comparing it to the original operator.

5. Are there any exceptions to the Hermiticity of the Spin Orbit Hamiltonian?

In some cases, the Spin Orbit Hamiltonian may not be strictly Hermitian. This can happen when there is a strong external magnetic field or in systems with strong spin-orbit coupling. In these cases, the operator may still be approximately Hermitian, and the deviations can be accounted for in the calculations.

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