# I Kinetic energy with effective mass

1. Mar 20, 2016

### aaaa202

I want to write the kinetic energy operator as a matrix within a finite element approach for electrons moving in a crystal with some effective mass that is a function of position.
Now usually we have:
K = -ħ2/2m d2/dx2
such that the second order derivative of a wavefunction maybe written as:
d2/dx2 = 2/(xi+1-xi-1)* (ψi+1i)/(xi+1-xi) - (ψii-1)/(xi-xi-1))
But for electrons moving in a crystal where the effective mass depends on the spatial coordinate, then the kinetic energy operator is:
K = -ħ2/2 d/dx(1/m*(x) d/dx)
How can I write this in a finite element approach? Do I just put 1/mi in front of ψi etc.? I tried that but it gives some funny results that do not seem physical. In the problem I am solving the effective mass makes a large jump from one grid point to the next, so maybe this could cause some problems?

2. Mar 22, 2016

### DrDu

Write it as the sum of a second derivative and a first derivative of psi.

3. Mar 22, 2016

### DrDu

You could also try $\frac{d}{dx}\frac{1}{m}\frac{d \psi}{dx}\approx \{\frac{1}{m_{i+1}}\frac{\psi_{i+2}-\psi_{i}}{x_{i+2}-x_i}-\frac{1}{m_{i-1}}\frac{\psi_i-\psi_{i-2}}{x_i-x_{i-2}}\}\frac{1}{x_{i+1}-x_{i-1}}$

4. Mar 22, 2016

### aaaa202

okay but why is it necessary to have ψi+2 and not just ψi+1? It seems you want the effective mass point in the middle between the two wavefunction points, why is that? (I looked up an article that do the exact same as you, so I guess it is correct).

5. Mar 22, 2016

### DrDu

There are different ways to approximate the first derivative; e.g. you could use $\psi'(x_i)\approx (\psi(x_{i+1})-\psi(x_i))/(x_{i+1}-x_i)$, but, it should be intuitively clear that a better approximation is $\psi'(x_i)\approx (\psi(x_{i+1})-\psi(x_{i-1}))/(x_{i+1}-x_{i-1})$. A more formal way would be to use the Taylor expansion on the RHS and show that for the second definition, second order corrections are absent, but not for the first one (assuming that the x_i are equidistant). So you are approximating
$1/m(x_i)\psi'(x_i)$ and use the central difference formula.

Last edited: Mar 22, 2016