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Normalization problem; volume element to use?

  1. Aug 6, 2011 #1
    Hello!

    1. The problem statement, all variables and given/known data

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

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

    [itex] 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 [/itex]
    So:
    [itex] 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
    [/itex]
    Et cetera..

    3. The attempt at a solution

    When I tried to solve it I write the second part of the integral like this:
    [itex]

    \int_0^{2 \pi}\int_0^{\pi} cos^2\theta sin\phi d\theta d\phi

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

    Not that it makes my answer any more like my teachers, but this is how i continue:
    [itex]
    \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 ...
    [/itex]
    Thank you for your time,
    Alfred


    edit: forgot an N^2
     
  2. jcsd
  3. Aug 6, 2011 #2
    Nope. It's r2sinθdrdϕdθ. See for example: http://en.wikipedia.org/wiki/Spherical_coordinate_system under the section "Integration and differentiation in spherical coordinates"
     
  4. Aug 6, 2011 #3

    tiny-tim

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    Hello Alfred! :smile:
    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. :wink:
     
  5. Aug 6, 2011 #4

    SammyS

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    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)
     
  6. Aug 6, 2011 #5

    tiny-tim

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    no, 0 to π :wink:
     
  7. Aug 7, 2011 #6
    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), ( [itex] \int_{a_1}^{a_2}\int_{b_1}^{b_2} function \, db da [/itex]) but apparently that's not the case here :P

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

    [itex] \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\} [/itex]

    [itex]
    = \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 [/itex]

    [itex] = 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)[/itex]

    [itex]
    = 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)
    [/itex]
    [itex]
    =2 + \frac{1}{2} \left( 0 - 2 \right) + \frac{1}{4} \left(\frac{4}{3}\right)

    =\frac{7}{3} \neq \frac{2}{3}
    [/itex]
    In the last step I use:
    [itex]
    \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}
    [/itex]

    Many steps; can you see which one is wrong? :S
     
  8. Aug 7, 2011 #7

    tiny-tim

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    Hello Alfred! :smile:

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

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

    but a u-subsitution would have been more sensible. :redface:
    You're right, it should do so.
     
    Last edited by a moderator: Apr 26, 2017
  9. Aug 7, 2011 #8
    Hi Tim! :)
    That's from [itex] \int cos^2 \theta = \frac{1}{2} \left(\theta + sin \theta cos \theta\right) [/itex]?

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


    [itex] \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)
    [/itex]

    [itex]
    f\left(t\right) = t^2 , t = cos \theta; g'\left(x\right) = -sin \theta
    [/itex]

    [itex]
    \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}
    [/itex] !!! :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!
     
    Last edited by a moderator: Apr 26, 2017
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