Normalization problem; volume element to use?

[adh2]

Hello!

Homework Statement

I'm revising my quantum mechanics course, but I don't get this normalizing-problem. $\psi = N r cos \theta e^{-r/a_0}$

To begin with, this is how my teacher solves it in the solutions manual:

$1=N^2\int_0^\infty r^2 e^{-r/a_0} r^2 dr \int_0^{\pi} \int_0^{2 \pi} cos^2\theta sin\theta d\theta d\phi$
So:
$N^2 = \int_0^\infty r^4 e^{-r/a_0} dr \cdot 2 \pi \cdot \frac{2}{3} = \frac{4 \pi}{3} 4! a_0^5$
Et cetera..

The Attempt at a Solution

When I tried to solve it I write the second part of the integral like this:
$\int_0^{2 \pi}\int_0^{\pi} cos^2\theta sin\phi d\theta d\phi$
Here the problem is obvious. I say $sin\phi$, but my teacher says $sin\theta$. As I assume my teacher is the one who's correct, why isn't it $sin\theta$? I mean, the volume element when going from cartesian coordinates to spherical is $dV= r^2 sin\phi drd\phi d\theta$, right?

Not that it makes my answer any more like my teachers, but this is how i continue:
$\int_0^{2 \pi} cos^2\theta d\theta \int_0^{\pi} sin\phi = \frac{1}{2} \left[\theta + sin\theta cos\theta\right]_0^{2 pi} \left[-cos\phi \right]_0^{\pi} = \left\{\frac{1}{2}\left(2 \pi + o\right) - \frac{1}{2}\left(0 + o\right)\right\} \left\{1-0\right\} = \pi ...$
Thank you for your time,
Alfrededit: forgot an N^2

 

[Mike Pemulis]

I mean, the volume element when going from cartesian coordinates to spherical is dV=r²sinϕdrdϕdθ, right?

Nope. It's r²sinθdrdϕdθ. See for example: http://en.wikipedia.org/wiki/Spherical_coordinate_system under the section "Integration and differentiation in spherical coordinates"

 

[tiny-tim]

Hello Alfred!

… Here the problem is obvious. I say $sin\phi$, but my teacher says $sin\theta$. As I assume my teacher is the one who's correct, why isn't it $sin\theta$? I mean, the volume element when going from cartesian coordinates to spherical is $dV= r^2 sin\phi drd\phi d\theta$, right?

I think your teacher is using spherical coordinates with θ and φ the other way round from you.

(mathematicians do it one way, physicists do it the other)

The θ in cos²θ is measured from pole to pole isn't it? Well, the sin in the volume element has to be measured the same way.

 

[SammyS]

It's the sine of whichever variable has integration limits from 0 to 2π.

I your teacher's solution it should be (and is) sin(θ) .

It is not used consistently in your expressions. (in your attempt ta a solution)

 

[tiny-tim]

It's the sine of whichever variable has integration limits from 0 to 2π.

no, 0 to π

 

[adh2]

Thank you, that makes more sense! I just assumed there was a standard. In my calculus textbook by R A Adams the integration element used is with sin phi. But indeed, a closer look in my physical chemistry book (Atkin's Physical Chemistry) and I see he uses sin theta! :D

I also thought the limits on the first integral sign always referred to the last d(variable), ( $\int_{a_1}^{a_2}\int_{b_1}^{b_2} function \, db da$) but apparently that's not the case here :P

However, I still can't solve it. $\int_0^{2\pi} d\phi$ gives the $2\pi$. But the rest doesn't work out to $\frac{2}{3}$:

$\int_0^{\pi} cos^2 \theta sin \theta d \theta = \left\{ as \, \int_a^b U(x) dV(x) dx = \left[U(x)V(x)\right] - \int_a^b V(x)dU(x) dx \right\}$

$= \left[-cos^2\theta cos\theta\right]_0^{\pi} - \int_0^\pi \left(\frac{1}{2} \theta + \frac{1}{2} cos\theta sin\theta\right) \left(-cos\theta\right) d\theta$

$= 2 + \left( \frac{1}{2} \int_0^\pi \theta cos\theta d\theta + \frac{1}{4} \int_0^\pi sin 2\theta cos\theta d\theta \right)$

$= 2 + \frac{1}{2}\left( \left[ sin\theta \theta^2 1/2 \right]_0^\pi - \int_0^\pi 1 sin\theta d\theta \right) +\left( \frac{1}{4} \int_0^\pi sin 2\theta cos \theta d\theta \right)$
$=2 + \frac{1}{2} \left( 0 - 2 \right) + \frac{1}{4} \left(\frac{4}{3}\right) =\frac{7}{3} \neq \frac{2}{3}$
In the last step I use:
$\int_0^\pi sin mx \, cos nx \, dx = \stackrel{0 \,if\, m, n \, integers; m+m \, even}{\frac{2m}{ m^2+n^2} \,if\, m, n \, integers; m+n \, odd}$

Many steps; can you see which one is wrong? :S

 

[tiny-tim]

Hello Alfred!

I don't understand where you got 1/2 θ from in your https://www.physicsforums.com/library.php?do=view_item&itemid=199" …

it should have gone ∫ cos²θsinθ dθ = -cos²θcosθ - ∫ 2cos²θsinθ dθ …

but a u-subsitution would have been more sensible. ​

I also thought the limits on the first integral sign always referred to the last d(variable), ( $\int_{a_1}^{a_2}\int_{b_1}^{b_2} function \, db da$) but apparently that's not the case here :P

You're right, it should do so.

 

[adh2]

Hi Tim! :)
That's from $\int cos^2 \theta = \frac{1}{2} \left(\theta + sin \theta cos \theta\right)$?

U-substitution. My calculus-skills are somewhat rusty, so that didn't seem like a possibility. But of course you're right!


$\int_a^b f\left(g\left(x\right)\right)g'\left(x\right) dx = \int_{g\left(a\right)}^{g\left(b\right)} f\left(t\right) dt, t=g\left(x\right)$

$f\left(t\right) = t^2 , t = cos \theta; g'\left(x\right) = -sin \theta$

$\int_0^\pi cos^2 \theta sin \theta d \theta = - \int_{cos 0}^{cos \pi} u^2 du = \left[ \frac{1}{3} u^3 \right]_{1}^{-1} = -\left( \frac{-1}{3} - \frac{1}{3} \right) = \frac{2}{3}$ ! :D

Wow. Much simpler. I always get exactly this http://www.youtube.com/watch?v=fXW02XmBGQw" (- the last few seconds..) when I solved a difficult math-problem.

Thank you very, very much Tim!