How is the Wave Function u(r) = Asin(kr) Normalized?

[1v1Dota2RightMeow]

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

Homework Equations

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

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?

 

[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]

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]

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]

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

 

[TSny]

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]

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]

Yes.

facepalm.jpgUgh...makes sense. Sorry for being so dense. Thanks!