# Special cases of the Schroedinger Eq?

1. Apr 20, 2010

### ManDay

According to the lecture I'm hearing the SE can be simplified for the cases of a timeinvariant potential and a constant potential:

Time invariant

$\frac{\partial^2\psi}{\partial x^2} = 2\frac m{\hbar^2}(V(x)-E)\psi$.

Then, the lecture states that for the case of the potential being not just time but also spatially invariant it can be simplified to

$\frac{\partial^2\psi}{\partial x^2} = -k^2\psi$

My question is, why this is said to be possible, only if the potential is constant. Given the term $V(x) - E$ we can conclude that $V(x) - E = E_{kin}$ and hence $\frac12\hbar^2\frac{k^2}m$

Thanks

2. Apr 20, 2010

### LostConjugate

If the potential is non-constant then the kinetic energy V-E is once again a function of position. Your last result is not a function of position.

3. Apr 21, 2010

### ManDay

Call it k(x) Then it is a function of position.

nvm replying, I got the "idea" and I've to blame it on the lecture that this wasn't clear.

Last edited: Apr 21, 2010