# Homework Help: Quantum Mechanics: Choose an acceptable bound state function

1. Jul 4, 2011

### 511mev

1. Which of the following is an allowed wave function for a particle in a bound state? N is
a constant and α, β>0.
1) Ψ=N e-α r
2) Ψ=N(1-e-α r)
3) Ψ=Ne-α x e-β(x2+y2+z2)
4) Ψ=Non-zero constant if r<R , Ψ=0 if r>R

Only one is correct.

2. What are the criteria for acceptable bound state wave functions?

3. I did assume that one of the criteria is that the wave function must go to zero at infinity. To show this, I took the limits as r goes to infinity and, for the function given in 3, as x,y,z, go to positive and negative infinty. I got that all are zero at infinity except 2.
Another requirement is that the function be smooth and continuous. Since 4 has an abrupt change, i.e., its derivative is infinite at R, then it is not smooth.
That leaves 1 and 3. What additional property am I forgetting?

2. Jul 4, 2011

### vela

Staff Emeritus
It seems to me the problem is wrong in claiming only one answer is correct. The wave functions in (1) and (3) are solutions to, respectively, the hydrogen atom and the simple harmonic oscillator, for the appropriate choices of α and β.

3. Jul 4, 2011

### dextercioby

This is wrong. The requirement is being a member of a properly chosen L^2.

4. Jul 4, 2011

### 511mev

Are you saying that my assertion that a bound state wave function go to zero at infinity is incorrect? Would not a member of a well chosen L^2 have this property if it were to represent a bound state?

5. Jul 4, 2011

### 511mev

That is what I thought. I am attempting to help a student with this problem on Cramster. He claims that there is only one answer. I cannot formulate a proof, though.

6. Jul 5, 2011

### dextercioby

I'm saying that there are wavefunctions which don't go to 0 when approaching infinity, but are still square integrable. But such <oddities> can be the limits of convergent sequences of <good> wavefunctions (=those who have the particular infinity behavior). In the language of functional analysis, S(R) is a proper subset of L^2(R) and dense everywhere in it.

Anyways, regarding the problem itself, as unveiled above, both (1) & (3) are valid solutions to your problem.