# Eigenvalues of a spin-orbit Hamiltonian

• IanBerkman
BerkmanIn summary, the eigenvalues and eigenvectors of wave vector k in an electron gas with Hamiltonian:- are given by the Pauli matrices- are: 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>- can be reduced to a matrix eigenvalue equation- has E_1 = \frac{\hbar^2k^2}{

#### IanBerkman

Good day everyone,

The question is as following:

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

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:
$$\boldsymbol{\sigma} = (\sigma_x, \sigma_y, \sigma_z)$$
$$\sigma_x = \left(\begin{matrix} 0&1\\1&0\end{matrix}\right),$$
$$\sigma_y = \left(\begin{matrix} 0&-i\\i&0\end{matrix}\right),$$
$$\sigma_z = \left(\begin{matrix} 1&0\\0&-1\end{matrix}\right)$$

Attempt at the solution:

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

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

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?

Ian Berkman

The second term in your Hamiltonian doesn't appear to be hermitian? Other than that your steps seem fine.

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
$$|\Psi> = A e^{\textbf{k} \cdot \textbf{r}}|\psi>$$

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.

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Nevermind the function in my comment above, it is not right.

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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##?

blue_leaf77
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.

IanBerkman said:
Do you mean the αi(σ⋅k) term?

Yes, interesting it appears with an ##i## here but not in your original problem statement.

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.

vela said:
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.

The problem was indeed incorrectly stated. Well, sometimes these things happen.

Thank you all for helping.

Ian