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Is the Spin Orbit Hamiltonian really Hermitian?

  1. Jun 5, 2013 #1
    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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  3. Jun 5, 2013 #2


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    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.
  4. Jun 5, 2013 #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?
  5. Jun 5, 2013 #4


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    I think E x p is Hermitian as long as E = ∇V, isn't it? That's basically crossing ∇ with itself.
  6. Jun 5, 2013 #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]


    [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]


    [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. Jun 6, 2013 #6


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  8. Jun 6, 2013 #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?
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