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Showing wavefunctions are orthonormal.

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data

    I've attached the question as a jpeg.

    I'm having trouble showing that the wavefunctions are orthonormal (i.e. Orthogonal and normal)

    When I try to show U_210 is normal I dont get 1:

    My working:


    ∫ |U_210|² = 1

    I set the limits from 0 to infinity for the radial part and 0 to π for cos²θ , so:

    1/(16πa^5) . ∫ (r² . exp(-r/a)) dr . ∫ cos²θ dθ = 1


    I used the identity given in the question to integrate the radial part and I get 2a³

    Then I integrate the cos²θ and I get π/2

    Therefore,

    1/(16πa^5) . 2a³ . π/2 = 1 (But of course this does not equal to 1, and I dont know what I'm doing wrong)

    Thanks very much.
     

    Attached Files:

  2. jcsd
  3. May 9, 2010 #2

    Cyosis

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    You forgot the Jacobian for spherical coordinates.
     
  4. May 9, 2010 #3
    Aha! So I'm integrating cos²θ w.r.t to dr and not dθ?

    If so how can I convert the integral from dr to dθ..
     
  5. May 9, 2010 #4

    Cyosis

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    No you will need to integrate cos²θ with respect to θ of course. You are integrating over a volume, in Cartesian coordinates the volume element would be dxdydz. However we're not using Cartesian coordinates, but spherical coordinates. Therefore you need to integrate over the volume element in spherical coordinates.
     
  6. May 9, 2010 #5

    jtbell

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    To put what Cyosis said (in post #2) in other words, the normalization integral is a volume integral:

    [tex]\int {\int {\int {|\psi_{210}|^2 dV}}} = 1[/tex]

    where dV is a volume element. If you were doing the integral in rectangular coordinates, dV would be dx dy dz. But you're actually using spherical polar coordinates. What's dV in spherical polar coordinates?

    (Cyosis beat me to it while I was proofreading my LaTeX!)
     
  7. May 9, 2010 #6
    Oh right yeap, I completely forgot I was integrating w.r.t to the volume. Thanks!

    There is something else im not exactly sure on. The very first part of the question asks me to prove that the degeneracy is n² .

    I know that the number of values of m for a particular value of l is (2l + 1) and that there are n values of l. When you add all the combinations together it is actually n² but how can I show this.
     
  8. May 9, 2010 #7

    vela

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    Just add them up for a given n. For each value of l, you have 2l+1 states, so the total number of states is

    [tex]\sum_l (2l+1)[/tex]

    You just need to fill in the correct limits for the summation and evaluate the sum.
     
  9. May 10, 2010 #8
    So there is no neat way of proving it is equal to n² . You just have to show that the summation of all the states is the same as n²?
     
  10. May 10, 2010 #9

    vela

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    Yeah, there's no clever way I'm aware of, but you don't need one. Showing the sum is equal to n2 is straightforward.
     
  11. May 10, 2010 #10
    One last thing that is bothering me, is when I try to normalise the wavefunction U_211

    I get part of the integral:

    ∫ exp(2iφ) dφ

    limits between 2Pi to 0. How can I evaluate this, to show that the wavefunction is normalised.
     
  12. May 10, 2010 #11

    Cyosis

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    I am sure you can find the primitive of a simple exponential function. Once you do this you will notice that something is very wrong!
     
  13. May 10, 2010 #12

    vela

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    Is it bothering you because what you tried gave you 0 as the answer? If so, you integrated correctly. That factor shouldn't be there.
     
  14. May 10, 2010 #13
    Of course I forgot about the complex conjugate when normalising the wavefunctions, so essentially this will get rid of the exponential of the wavefunction im taking the complex conjugate of, but there will still be an exponential term in the integral. Im not sure where this is going..
     
  15. May 11, 2010 #14

    Cyosis

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    U_210 had an exponential in the integral as well. What's the problem? The complex exponential disappears, but the real exponential, just as for u210, stays.
     
  16. May 11, 2010 #15
    What I'm trying to do is integrate the following:

    ∫exp(φ).exp(iφ) dφ

    Before, I could easily integrate the exponential because I had the identity, but when I try integrating this I get a factor of 'i'. My maths is terrible, this is partly the reason why I cant see the problem very well.
     
  17. May 11, 2010 #16

    vela

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    You need to show us what you're doing from the beginning because you shouldn't have a term like that.
     
  18. May 12, 2010 #17

    Cyosis

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    As Vela said that term shouldn't be there. Remember [itex]|\psi|^2=\psi^* \psi[/itex].

    That said I find it a little odd that you aren't able to compute a simple exponential function and I suggest you brush up your calculus immediately or you will run into a lot of difficulties during your QM course.

    If you integrate that expression you will indeed get a factor 'i'. However to not confuse you I will say it again, those exponentials shouldn't be there in the first place!
     
  19. May 12, 2010 #18
    Wow! that was very stupid of me! The mix up was with the complex conjugate, very often examples in QM from my course that involve normalisation of a wavefunction dont have complex numbers. After fixing that I got a factor of Pi, hence normalising the wavefunction. I assume the limits are from 0 to Pi
     
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