Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Weak Form of the Effective Mass Schrodinger Equation

  1. Apr 29, 2014 #1

    I am numerically solving the 2D effective-mass Schrodinger equation

    [itex]\nabla \cdot (\frac{-\hbar^2}{2} c \nabla \psi) + (U - \epsilon) \psi = 0[/itex]

    where [itex]c[/itex] is the effective mass matrix

    [itex]\left( \begin{array}{cc}
    1/m^*_x & 1/m^*_{xy} \\
    1/m^*_{yx} & 1/m^*_y \\
    \end{array} \right)[/itex]

    I know that, when the effective mass is isotropic, the weak form is
    [itex]\int \frac{-\hbar^2}{2m^*}\nabla \psi \cdot \nabla v + U\psi vd\Omega = \int \epsilon \psi vd\Omega[/itex]

    The matrix is giving me trouble however. Is this the correct form?

    [itex]\int \frac{-\hbar^2}{2m^*_x}\frac{\partial u}{\partial x}\frac{ \partial v}{\partial x} + \frac{-\hbar^2}{2m^*_{xy}}\frac{\partial u}{\partial x}\frac{ \partial v}{\partial y} + \frac{-\hbar^2}{2m^*_{yx}}\frac{\partial u}{\partial y}\frac{ \partial v}{\partial x} + \frac{-\hbar^2}{2m^*_y}\frac{\partial u}{\partial y}\frac{ \partial v}{\partial y} + U\psi v d\Omega= \int \epsilon \psi v d\Omega[/itex]
    Last edited: Apr 29, 2014
  2. jcsd
  3. May 6, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook