Born Conditions on Wavefunctions

[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 continuous at x=0?

 

[Meir Achuz]

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

 

[jtbell]

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.

 

[scarecrow]

F(x) is continuous and has a "cusp" at x = 0, hence the first derivative of F(x) is discontinous at x=0 and is only piecewise continuous. But this doesn't prevent F(x) from being an acceptable "well-behaved" wavefunction.

F(x) must be continuous, but F' can be piecewise continuous for the wavefunction to be an acceptable Born function.

This is a paraphrase from one of my textbooks.

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

 

[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.

 

[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 canceled by the potential term, namely the coulomb term. This is an exact wave-function and there are no problems here.