Help normalizing a wave function

1v1Dota2RightMeow

1. Homework Statement
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. Homework 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?

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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 $A$

TSny

Homework Helper
Gold Member
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).

1v1Dota2RightMeow

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

1v1Dota2RightMeow

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 $A$
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.

TSny

Homework Helper
Gold Member
Did you take account of the fact that the wavefunction vanishes for r > a?

1v1Dota2RightMeow

Did you take account of the fact that the wavefunction vanishes for r > a?
But I thought whenever you normalize, you integrate over all space?

TSny

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

TSny

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

1v1Dota2RightMeow

Yes, you integrate over all space. But you have to use the correct wavefunction for all of space.
So you mean the integral should technically be $\int_0^a |A|^2 sin^2 (n\pi r/a)dr + \int_a^{\infty} 0 dr$ ??

TSny

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

1v1Dota2RightMeow

facepalm.jpg

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

"Help normalizing a wave function"

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