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Visualizing legendre polynomials in the hydrogen atom.

  1. Jan 21, 2015 #1
    1. The way we solved this problem was proposing that the wave function has to form of ##\Psi=\Theta\Phi R## where the three latter variables represent the anlge and radius function which are independent. The legendre polynomials were the solution to the ##\Theta## part. I am having some trouble understanding the commonly used plot of these equations that looks something like this:

    http://www.physics.umd.edu/courses/Phys402/AnlageSpring09/spherical_harmonics.gif

    What am I looking at? Well first of all, what does the ##\Theta## function represent? It represents a variation in the wave function as ##\Theta## changes. Assuming that ##R## and ##\Phi## are constant I'm looking at values of the wave function located on a circle. I seem to not be able to reconcile that notion with these kinds of 3D plots.

    Edit: Before anyone points out that the plots I posted are function of two angles, in my class I saw planar plots of solely the ##\Theta## part, which looks like the plots I linked only in a plane. So that wouldn't really lift the confusion.
     
  2. jcsd
  3. Jan 21, 2015 #2

    Quantum Defect

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    The 3-d plots are showing surfaces at some arbitrary "probablility" of the particle. Typically, I think that they are drawn at the 90% level. I.e. 90% of the time the electron would be found within this surface...
     
  4. Jan 21, 2015 #3
    Hmm, alright so I opened up Matlab and played around with the equations and I seem to be finding EXACTLY the shapes like in my book when I'm doing the following.

    So I take the Legendre polynomials where ''m=0'' and m represents the whatever quantum number. I have an infinite countable set of polynomials right now, depending on the value of l(l+1). Anyway this doesn't really matter. The first few polynomials they look like this:

    ##\Theta_{1}=1##

    ##\Theta_{2}=cos(\theta)##

    ##\Theta_{3}=0.5(3cos^{2}(\theta)-1)##

    Now I plot them in matlab in polar coordinates where I put ##r=\Theta_{1}## , ##r=\Theta_{2}##, ##r=\Theta_{3}##, and so on...

    These exactly the same figures as in my coursenotes. An example in the attachment.

    So I'm very confused at what I'm looking at. Here I have a variable ''r'' on the figures, while the ''r'' of the wavefunction is totally independent of the angle.
     

    Attached Files:

  5. Jan 21, 2015 #4

    Quantum Defect

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    For the lowest one, you are looking at what a chemist would call an s orbital. Spherical symmetry. No angular dependence. If you draw a surface at 90% (contianing 90% of the probability) you will get a ball.

    For L=1, mL=0 you will get the dumbell shape that chemists say is the p_z orbital shape. p_x and p_y shapes can be obtained from linear combinations of the L=1, mL = +/- 1 wavefunctions.

    For L=2, mL=0 you will get the dz^2 orbital -- like a p-orbital with a donut at the waist. Various linear combinations of the others will give you the other d-orbital shapes.
     
  6. Jan 21, 2015 #5
    Thanks for taking the time. I probably messed up my wording. I know the shapes and I have seen them in chemisty class and such. The problem is that I'm not sure how to connect the position coordinates as seen on the image to the position coordinates of the wave function input.

    Why do we get these shapes by plugging the angle dependence into spherical coordinates. For this we make the radius a function of the angle. The way you explain the first one seems like the image indeed does represent the wave function coordinates.

    Basically I am looking for a reasoning to connect the fact that we solve the ##\Theta## function independently and find some polynomials, to plotting a radius in terms of these functions. Nothing in the wave function implies any connection between the radius and the angle.
     
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