# Eigenfunction of a spin-orbit coupling Hamiltonian

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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.