1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eigenfunction of a spin-orbit coupling Hamiltonian

  1. Jul 6, 2016 #1
    Dear all,

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

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

    Thanks in advance.

    Ian
     
  2. jcsd
  3. Jul 6, 2016 #2
    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.
     
  4. Jul 7, 2016 #3
    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}\\ ## ?
     
  5. Jul 7, 2016 #4
    It appears mostly correct.
     
    Last edited: Jul 7, 2016
  6. Jul 7, 2016 #5
    However, the derivative of a constant is equal to 0.
     
  7. Jul 8, 2016 #6
    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
    [tex]
    \boldsymbol\sigma \cdot \nabla \psi + \psi \nabla\cdot\boldsymbol\sigma = \nabla \cdot (\psi \boldsymbol\sigma)
    [/tex]
    And solved the RHS with the plane-wave solution ##\psi = e^{i k \cdot r}##
     
  8. Jul 8, 2016 #7
    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
    ##
     
  9. Jul 8, 2016 #8
    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
  10. Jul 9, 2016 #9
    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.

    I think I have enough information to conclude a right answer about this problem, thank you all.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Eigenfunction of a spin-orbit coupling Hamiltonian
  1. Spin-orbit coupling (Replies: 5)

  2. Spin orbit coupling (Replies: 4)

  3. Spin Orbit Coupling. (Replies: 6)

  4. Spin orbit coupling (Replies: 1)

Loading...