Does derivative of wave function equal zero at infinity?

[Sturk200]

I understand that ψ goes to zero as x goes to infinity. Is it also true that dψ/dx must go to zero as x goes to infinity?

 

[Dr. Courtney]

I can't think of any counter examples.

 

[Nugatory]

It is true as long as $\psi$ and its first deriviative are well-behaved (differentiable, both limits exist, ...). This is a fun old calculus problem - you can prove it by contradiction from the mean value theorem.

 

[Heinera]

It is true as long as $\psi$ and its first deriviative are well-behaved (differentiable, both limits exist, ...). This is a fun old calculus problem - you can prove it by contradiction from the mean value theorem.

Yes, so a counterexample would e.g. be let amplitude go to zero but frequency to infinity. This would be possible if potentials were introduced.

 

[bhobba]

Yes. And that it goes to zero at infinity is a general assumption of physically realisable wavefunctions.

The correct formalism for QM is what's known as Rigged Hilbert Spaces and the restriction that's often imposed is somewhat stronger in the sense of being continuously differentiable and goes to zero fast enough. They are called good functions:
http://euclid.ucc.ie/pages/staff/thomas/AM2071/Notes/S3notes2011.pdf

It makes many things a lot simpler such as being able to rigorously define the Dirac Delta function and Fourier transforms become a snap with the usual issues of convergence etc a piece of cake.

Knowledge of this stuff really belongs in the toolkit of any applied mathematician in just about any area, not just QM. I stronly recommend the following book:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill

 

[PeroK]

A function based on $\frac{sin(x^2)}{x}$ would be a counter-example for the general case.