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Eigenvalues of a spin-orbit Hamiltonian

  • #1
54
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
 

Answers and Replies

  • #2
Paul Colby
Gold Member
1,034
230
The second term in your Hamiltonian doesn't appear to be hermitian? Other than that your steps seem fine.
 
  • #3
54
1
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:
  • #4
54
1
Nevermind the function in my comment above, it is not right.
 
Last edited:
  • #5
vela
Staff Emeritus
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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##?
 
  • #6
Paul Colby
Gold Member
1,034
230
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.
 
  • #7
Paul Colby
Gold Member
1,034
230
Do you mean the αi(σ⋅k) term?
Yes, interesting it appears with an ##i## here but not in your original problem statement.
 
  • #8
54
1
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.
 
  • #9
54
1
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##?
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.
 
  • #10
54
1
The problem was indeed incorrectly stated. Well, sometimes these things happen.

Thank you all for helping.

Ian
 

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