# Help normalizing a wave function

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1. Nov 8, 2016

### 1v1Dota2RightMeow

1. The problem statement, all variables and given/known data
I don't see how the author normalizes $u(r)=Asin(kr)$. From Griffiths, Introduction to Quantum Mechanics, 2nd edition, page 141-142:

http://imgur.com/a/bo8v6

2. Relevant equations
$\int_0^{\infty} \int_0^{\pi} \int_0^{2\pi}|A|^2 \sin^2(\frac{n\pi r}{a})r^2 \sin \theta drd\theta d\phi=1$

3. The attempt at a solution
My integral was
$\int_0^{\infty} \int_0^{\pi} \int_0^{2\pi}|A|^2 \sin^2(\frac{n\pi r}{a})r^2 \sin \theta drd\theta d\phi=1$

Mathematica simplifies the integral (without the $A$ for simplicity) to
$=\int_0^{\infty}4\pi r^2 \sin^2(\frac{n\pi r}{a})dr$

but it stops there. I don't think this integral converges. Did I make a mistake somewhere?

2. Nov 8, 2016

### Fightfish

As Griffiths himself makes clear - the radial part of the wavefunction is $R(r) = u(r) /r$, so that is what you should be putting in the integral instead. When he says normalising $u(r)$, what he really means is finding the coefficient $A$

3. Nov 8, 2016

### TSny

What is the value of the wavefunction for r > a?

The normalization condition for the radial part of the wavefunction R(r) is given in Griffith's equation [4.31] (not shown in your picture).

4. Nov 8, 2016

### 1v1Dota2RightMeow

The wavefunction is $0$ for $r>a$.

5. Nov 8, 2016

### 1v1Dota2RightMeow

Ah I see. Ok then but normalizing according to that equation yields $\int_0^{\infty} |A|^2 sin^2 (n*\pi *r/a)dr$, which does not converge.

6. Nov 8, 2016

### TSny

Did you take account of the fact that the wavefunction vanishes for r > a?

7. Nov 8, 2016

### 1v1Dota2RightMeow

But I thought whenever you normalize, you integrate over all space?

8. Nov 8, 2016

### TSny

Yes, you integrate over all space. But you have to use the correct wavefunction for all of space.

9. Nov 8, 2016

### TSny

To put it another way, is there a need to integrate over regions of space where the wavefunction is zero?

10. Nov 8, 2016

### 1v1Dota2RightMeow

So you mean the integral should technically be $\int_0^a |A|^2 sin^2 (n\pi r/a)dr + \int_a^{\infty} 0 dr$ ??

11. Nov 8, 2016

### TSny

Yes.

12. Nov 8, 2016

### 1v1Dota2RightMeow

facepalm.jpg

Ugh....makes sense. Sorry for being so dense. Thanks!