Eigenfunction of a spin-orbit coupling Hamiltonian

Tags:
1. Jul 6, 2016

IanBerkman

Dear all,

The Hamiltonian for a spin-orbit coupling is given by:
$$\mathcal{H}_1 = -\frac{\hbar^2\nabla^2}{2m}+\frac{\alpha}{2i}(\boldsymbol \sigma \cdot \nabla + \nabla \cdot \boldsymbol \sigma)$$
Where
$$\boldsymbol \sigma = (\sigma_x, \sigma_y, \sigma_z)$$
are the Pauli-matrices.

I have to find the eigenfunctions of this equation. However, I am not sure how to interpret the part: $$\nabla \cdot \boldsymbol \sigma$$
The Pauli-matrices are 2x2 matrices containing only constants, does this mean this term equals zero?

Ian

2. Jul 6, 2016

QuantumForumUser

I think the nabla symbol ∇ along with σ means derivative of σx with respect to x, derivative of σy with respect to y, and derivative of σz with respect to z.

3. Jul 7, 2016

Fred Wright

Can you see that:
$\nabla \cdot \sigma =\begin{pmatrix} 0 & \frac {\partial} {\partial x}\\ \frac {\partial} {\partial x}&0\end{pmatrix}+\begin{pmatrix} 0 & \frac {-i\partial} {\partial y}\\ \frac{i\partial} {\partial y}&0\end{pmatrix}+\begin{pmatrix} \frac {\partial} {\partial z} & 0\\0 & \frac {-\partial} {\partial z}\end{pmatrix}\\$ ?

4. Jul 7, 2016

QuantumForumUser

It appears mostly correct.

Last edited: Jul 7, 2016
5. Jul 7, 2016

QuantumForumUser

However, the derivative of a constant is equal to 0.

6. Jul 8, 2016

IanBerkman

Yeah that was part I was most confused of:
Does it mean it:
(1) takes the partials of the components (constants) inside the Pauli matrices
or
(2) the partials take the places of the constants in the matrices?
In the first case I would say the term becomes zero, the second case gives indeed the answer of Fred Wright.

The solution I took was using the product rule
$$\boldsymbol\sigma \cdot \nabla \psi + \psi \nabla\cdot\boldsymbol\sigma = \nabla \cdot (\psi \boldsymbol\sigma)$$
And solved the RHS with the plane-wave solution $\psi = e^{i k \cdot r}$

7. Jul 8, 2016

IanBerkman

Remarkable is how
$\boldsymbol \sigma \cdot \nabla e^{ik\cdot r} = \begin{pmatrix} ik_z & ik_x+k_y\\ ik_x-k_y & -ik_z \end{pmatrix}e^{ik\cdot r}$ and
$\nabla \cdot (e^{ik\cdot r} \boldsymbol \sigma) = \begin{pmatrix} ik_z & ik_x+k_y\\ ik_x-k_y & -ik_z \end{pmatrix}e^{ik\cdot r}$
This concludes that the term $e^{ik\cdot r} \nabla \cdot \boldsymbol \sigma = 0\\$
and therefore, if I did nothing wrong
$\nabla \cdot \boldsymbol \sigma = 0$

8. Jul 8, 2016

Fred Wright

Your Hamiltonian appears to be a somewhat abstruse variant of the Rashba effect which is a model of a 2d solid coupling electronic quasi-momentum to spin. In this model symmetry is broken by a transverse time independent E field, $-E_0\hat z$. Due to relativistic corrections an electron moving with velocity v in the electric field will experience an effective magnetic field B(the "orbital" moment). Google Rashba effect for a derivation. The upshot is $H_R = \alpha(\mathbf \sigma \times \mathbf k) \cdot \hat z.$ Where $\mathbf k = \frac {\nabla} {i}$ is the quasi-momentum. Now rewritten with the quasi-momentum, the $\sigma$ term in the Hamiltonian becomes:
$\frac { \mathbf \nabla} {i} \cdot \mathbf\sigma = \mathbf k \cdot \mathbf\sigma= \begin{pmatrix} k_z & k_x - ik_y\\ k_x + ik_y & -k_z\end{pmatrix}$.
The vector inner-product commutes so, $\mathbf k\cdot \mathbf \sigma = \mathbf \sigma \cdot \mathbf k$. I assert that this can only be physical if $k_z=0$ because the symmetry breaking field is along the z-axis. Thus we write the Hamiltionian:
$\frac {ħ^2k^2\sigma_0} {2m}+ \alpha( \mathbf \sigma \cdot \mathbf k)= \begin{pmatrix} \frac {ħ^2k^2} {2m} & \alpha(k_x - ik_y)\\ \alpha(k_x + ik_y) & \frac {ħ^2k^2} {2m}\end{pmatrix}$,
where $\sigma_0$ is the 2x2 unit matrix. You can diagonalize this matrix to find the eigenvectors and eigenvalues which lift the degeneracy of the conduction band due to the spin-orbit interaction.

Last edited: Jul 8, 2016
9. Jul 9, 2016

IanBerkman

Thank you, the Hamiltonian looks exactly the same as the Hamiltonian I have obtained, except for a $\alpha/2$ factor.
I have plotted the energies as function of $k_x$ and it looks like the Rashba effect, which should be no surprise.