Weak Form of the Effective Mass Schrodinger Equation

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 3K views
Morberticus
Messages
82
Reaction score
0
Hi,

I am numerically solving the 2D effective-mass Schrödinger 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}<br /> 1/m^*_x & 1/m^*_{xy} \\<br /> 1/m^*_{yx} & 1/m^*_y \\<br /> \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:
Physics news on Phys.org