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Weak Form of the Effective Mass Schrodinger Equation

by Morberticus
Tags: effective, form, mass, schrodinger, weak
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Morberticus
#1
Apr29-14, 04:09 PM
P: 81
Hi,

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]
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Greg Bernhardt
#2
May6-14, 11:48 PM
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P: 9,336
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?


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