Born Conditions on Wavefunctions

1. Sep 23, 2006

scarecrow

Born's conditions for an acceptable "well-behaved" wavefunction F(x):
1. it must be finite everywhere, i.e. converge to 0 as x -> infinity
2. it must be single-valued
3. it must be a continuous function
4. and dF/dx must be continuous.

I'm having difficulty understanding the last condition for a specific example. I have a wavefunction, F(x) = exp[-|x|], and the derivative at x = 0 does not exist. Is dF/dx still continous at x=0?

2. Sep 23, 2006

Meir Achuz

"Is dF/dx still continous at x=0?"
It is not. If F' is not continuous, then F"-->infinity, which corresponds to an infinite energy in the Schrodinger eq. This can only happen at an infinite potential step, such as in what is called an "infinite square well".

3. Sep 23, 2006

Staff: Mentor

And in such cases, the discontinuous step in the potential and in F' are idealizations which are not possible in reality, although they are useful as approximations to make the solutions simpler. In reality, the potential always varies continuously (although very rapidly in this case) and F' also varies continuously but rapidly.

Instead of a perfectly "vertical" potential step, V(x) might have a steep slope that is almost (but not quite) vertical, and F(x) has a short rounded section instead of a sharp kink.

4. Sep 23, 2006

scarecrow

This is a paraphrase from one of my textbooks.

So exp[-|x|] is an acceptable "well-behaved" wavefunction according to this?

5. Sep 25, 2006

Severian

The correct answer to your question is 'no it is not'. It cannot represent reality perfectly. However, it may be a good approximation of the true wavefunction away from 0.

6. Sep 25, 2006

Epicurus

By definition, an exact wave-function will have a constant local energy as a function of position. The cusp in the wave function at x = 0 poses no problems as the divergence in the kinetic component is exactly cancelled by the potential term, namely the coulomb term. This is an exact wave-function and there are no problems here.