Does the Wave Function Approach Zero at Infinity?

[thatguy14]

Homework Statement

If I have a wave function that goes to infinity can I assume that the derivative also goes to 0 at infinity?

Homework Equations

The Attempt at a Solution

The reason I think it does is because the wavefunction and its derivative must be continuous everywhere except at potentials that go to infinity. Is this the correct logic?

 

[TheAustrian]

Well, this is a little too little information. What is the equation for your wavefunction?

 

[thatguy14]

We aren't given any. Some other information which might be helpful though, the wave function is normalized, it is a solution to the time dependent Schrödinger equation and the potential function is real. And again it approaches zero as x goes to +- infinity. Is that enough?

 

[TheAustrian]

Just that in general the most sure way to decide is to calculate. So, what you should do is write down a wavefunction that satisfies the conditions and see what happens if you differentiate it under your stated conditions.

For example, try to solve this one, you will find it interesting:

\psi(x) = \frac{Asin^{2}x}{\sqrt{x^{2}+1}}

 

[thatguy14]

I think I see what you are saying. I will give a little more information into what I am trying to do.

The question is:

Consider two normalizable wave functions Ψ1(x, t) and Ψ2(x, t), both of which are solutions
of the time-dependent Schrödinger equation. Assume that the potential function is real.
The functions are normalized, and the functions both approach zero as x goes to ±∞.

Show that these propeties can be used to prove that

\frac{d}{dt} \int_{-∞}^{∞} Ψ1*(x, t)Ψ2(x, t)dx = 0

We get a hint to conert the temporal derivative to the spatial dertivative using the time-dependent Schrödinger equation and you get:

\frac{i\bar{h}}{2m} \int_{-∞}^{∞} Ψ1*(x, t)\frac{d^2}{dx^2}Ψ2(x, t) - Ψ2(x, t)\frac{d^2}{dx^2}Ψ1*(x, t)

Then using integration by parts you entually get a term (for one part of the integral) that is

Ψ1*(x, t)\frac{d}{dx}Ψ2(x, t) |^{∞}_{-∞}

and I need that to go to 0 or else the rest doesn't really follow through correctly. I justified it by saying that since Ψ2(x, t) goes to 0 and +- infinity then its derivative will go to 0 and +- infinity.

 

[TheAustrian]

Well, as far as I know, in general it is assumed that

Lim(x \rightarrow ∞) \frac{∂^{n}ψ}{∂x^{n}} = 0

where n is some positive integer

but it does not necessarily have to be the case. As far as I know it is only an assumption.

 

[thatguy14]

Okay thank you. I think it is sufficient for my course. I appreciate your time!