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Tried to Normalize this function but

  1. Aug 17, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is attached in the picture. It is to normalize the wavefunction to find the value of A.



    3. The attempt at a solution

    The solutions used an element volume dV of a very thin shell, then integrated r from 0 to ∞.

    I didn't think of this way, I simply integrated using boundaries of ψ, θ and r.

    The limits of ψ ranges from 0 to 2∏.

    The limits of θ ranges from 0 to 2∏.

    The limits of r ranges from 0 to infinity.


    However, I get a different answer. I've checked my working through a couple of times, not sure where I went wrong..
     

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    Last edited: Aug 17, 2012
  2. jcsd
  3. Aug 17, 2012 #2

    gabbagabbahey

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    You've made two very serious mistakes before even starting to evaluate your integral:

    (1) The volume differential in spherical coordinates is [itex]r^2 \sin\theta dr d\theta d\phi[/itex], not [itex]dr d \theta d \phi[/itex] (which doesn't even have units of volume :eek:)

    (2) Only one of the angles is integrated from [itex]0[/itex] to [itex]2\pi[/itex] (by convention, this is usually the azimuthal angle [itex]\phi[/itex]), this produces a circle (or circular disk if you integrate over [itex]r[/itex] as well), the other angle needs only be integrated from [itex]0[/itex] to [itex]\pi[/itex] (rotating a circle 180 degrees creates a shperical shell/sphere, rotating 360 degrees will give you double the correct result)

    I strongly suggest you pick up a textbook on multivariable/vector calculus and read through how to integrate and differentiate functions in curvilinear coordinates. Most good textbooks will have worked examples, diagrams and sample problems for you to practice with.
     
  4. Aug 17, 2012 #3


    Do you have any recommendations/links that I can use to practice and improve??

    I found a helpful website that explains the derivation of dV : http://www.mas.ncl.ac.uk/~nas13/mas2104/Section511.pdf [Broken]

    I think this makes more sense, by initially including Δθ.

    I first rotated Δθ, then the curved length is Δθ*r.
    Then I rotated Δψ, then won't the 'radius' be r*sin(θ+Δθ) instead of simply r*sin(θ) ?

    ΔV = (Δθ*r)[r*sin(θ+Δθ)*Δψ]*Δr

    Taking limits ΔV → 0, Δr, Δθ and Δψ → 0

    sin(θ+Δθ) → sin (θ)

    dV = r2 sinθ dr dθ dψ
     
    Last edited by a moderator: May 6, 2017
  5. Aug 17, 2012 #4

    chiro

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    Hey unscientific.

    I haven't taken QM but what is the description of the wave-function distribution in this example? How does it vary through space mathematically as a function of the Euler-Angles and the radius (whether implicitly or explicitly)?
     
  6. Aug 17, 2012 #5
    Here's the picture of the full question, hopefully this gives you the answer you're looking for:
     

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  7. Aug 17, 2012 #6

    gabbagabbahey

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    For a quick reference geared more toward physics (and hence lacking some mathematical rigor), my personal favourite is the first chapter of Griffiths' Introduction to Electrodynamics which introduces vector calculus geometrically and contains several examples and problems for practice.

    For a more rigorous text I recommend Mary L. Boas' Mathematical Methods In The Physical Sciences.

    If you are just looking for a free online tutorial, a quick Google search turns up video lectures from MIT here - although I haven't watched them so I can't attest to their usefulness.
     
  8. Aug 17, 2012 #7

    If dV = r2 sinθ dr dθ dψ,

    Then if you ∫dψ ∫ r2 sinθ dθ (from 0 to ∏ first then 0 to 2∏)

    you only end up with 2∏ in your answer when you reach ∫ dr. However, the answer has 4∏2...Am i missing something?

    I worked out everything using V = r2 sinθ dr dθ dψ and got nearly the same answer, just that they had 4∏2 while I only had 2∏...
     
  9. Aug 17, 2012 #8

    ehild

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    There is 4πr2 in the answer. The factor 2 comes in when you substitute the upper and lower bounds π and 0 in the integral with respect to θ: ∫sin(θ)dθ=-cos(π)-(-cos(0))=2.

    ehild
     
  10. Aug 17, 2012 #9
    I see, thanks!
     
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