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Eigenvalues of a spin-orbit Hamiltonian

  1. May 14, 2016 #1
    Good day everyone,

    The question is as following:


    Consider an electron gas with Hamiltonian:
    [tex] \mathcal{H} = -\frac{\hbar^2 \nabla^2}{2m} + \alpha (\boldsymbol{\sigma} \cdot \nabla)[/tex]

    where α parameterizes a model spin-orbit interaction. Compute the eigenvalues and eigenvectors of wave vector k and plot them in the x-direction. Interpret the results.


    Relevant equations:

    σ is given by the Pauli matrices:
    [tex] \boldsymbol{\sigma} = (\sigma_x, \sigma_y, \sigma_z)[/tex]
    [tex] \sigma_x = \left(\begin{matrix} 0&1\\1&0\end{matrix}\right),[/tex]
    [tex] \sigma_y = \left(\begin{matrix} 0&-i\\i&0\end{matrix}\right),[/tex]
    [tex] \sigma_z = \left(\begin{matrix} 1&0\\0&-1\end{matrix}\right)[/tex]

    Attempt at the solution:

    I started with calling the eigenvector as following:
    [tex]|\Psi> = Ae^{i\textbf{k}\cdot \textbf{r}}|\psi> ,[/tex]
    which need to fulfill the eigenvalue equation:
    [tex]\mathcal{H}|\Psi> = E|\Psi>. [/tex]
    This eigenvector gives the following:
    [tex] \mathcal{H}|\Psi> = Ae^{i\textbf{k}\cdot \textbf{r}}\frac{\hbar^2 k^2}{2m}|\psi> +Ae^{i\textbf{k}\cdot \textbf{r}}\alpha i(\boldsymbol{\sigma} \cdot \textbf{k})|\psi> = E\cdot Ae^{i\textbf{k}\cdot \textbf{r}}|\psi>,[/tex]
    where
    [tex] \alpha i(\boldsymbol{\sigma} \cdot \textbf{k}) = \left(\begin{matrix} \alpha i k_z & \alpha i k_x + \alpha k_y\\ \alpha i k_x - \alpha k_y & -\alpha i k_z \end{matrix}\right).[/tex]
    This can then be reduced to a matrix eigenvalue equation:
    [tex]\begin{align*}E|\psi> &= \left(\frac{\hbar^2 k^2}{2m} +\alpha i (\boldsymbol{\sigma} \cdot \textbf{k})\right)|\psi>\\ E|\psi> &= \left(\begin{matrix} \frac{\hbar^2 k^2}{2m} + \alpha i k_z & \alpha i k_x + \alpha k_y \\ \alpha i k_x - \alpha k_y & \frac{\hbar^2 k^2}{2m} - \alpha i k_z \end{matrix}\right)|\psi>\end{align*}[/tex]

    However, this matrix gives the eigenvalues:
    [tex] E_1 = \frac{\hbar^2k^2}{2m} + \alpha i k \\
    E_2 = \frac{\hbar^2k^2}{2m} - \alpha i k [/tex]

    Which look quite right, except that they contain imaginary parts which suggest some form of energy loss/damping which is not stated in the problem.

    Did anybody see where I went wrong?

    Thanks in advance,

    Ian Berkman
     
  2. jcsd
  3. May 14, 2016 #2

    Paul Colby

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    Gold Member

    The second term in your Hamiltonian doesn't appear to be hermitian? Other than that your steps seem fine.
     
  4. May 14, 2016 #3
    Do you mean the αi(σ⋅k) term?
    I have checked that term couple of times, but I cannot see where it went wrong.

    I could try to solve it with the function
    [tex] |\Psi> = A e^{\textbf{k} \cdot \textbf{r}}|\psi> [/tex]

    Which gives a Hermitian operator, however, as far as I see, it also gives a negative h2k2/2m term in the energy eigenvalue. Furthermore, I cannot remember having ever seen such a function.
     
    Last edited: May 14, 2016
  5. May 14, 2016 #4
    Nevermind the function in my comment above, it is not right.
     
    Last edited: May 14, 2016
  6. May 14, 2016 #5

    vela

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    The spin-orbit term isn't hermitian because of the ##\nabla##. Shouldn't it be written in terms of the momentum, which will supply a factor of ##i##?
     
  7. May 14, 2016 #6

    Paul Colby

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    Well, you have the right eigenvalues, eigenvalues should be real in this case. Choosing ##\alpha## imaginary would do the trick. I suspect your problem has a typo.
     
  8. May 14, 2016 #7

    Paul Colby

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    Yes, interesting it appears with an ##i## here but not in your original problem statement.
     
  9. May 15, 2016 #8
    The i appears in the second term because of the gradient of e^(ikr).
    I am going to mail the professor if alpha could be an imaginary value, however, I would say that that would be stated in the problem if that is the case.
     
  10. May 15, 2016 #9
    The extra factor of ##i## should also appear when ##\alpha## could only take an imaginary value. Also, a momentum operator fits better in a spin-orbit Hamiltonian.

    I think indeed this problem has a typo, I will keep you informed about the professor's reply.
     
  11. May 16, 2016 #10
    The problem was indeed incorrectly stated. Well, sometimes these things happen.

    Thank you all for helping.

    Ian
     
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