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Plotting the Probability Density of the Coulomb Wave Function

  1. Nov 19, 2015 #1
    Hey there - I think I have an issue with my 3D density plots of the probability density of the Coulomb wave function. The reason I think something is going wrong is because my plots of |ψ(n=2, l=1, m=-1)|² and |ψ(2, 1, 1)|² are identical, while I would expect them to have the same shape but be rotationally symmetric along different orthogonal axes.

    cph1joM.png oF3Tp6b.png DXDLHxW.png

    The above images are Mathematica's plots of |ψ(2, 1, -1)|², |ψ(2, 1, 0)|², |ψ(2, 1, 1)|², respectively. As you can see, the first and third are identical, and not the shape of 2p orbitals, while the second plot actually looks like what I would expect - one of the three 2p orbitals.

    Here is my wave function - it's possible that the conversion from spherical to Cartesian coordinates is a problem, but I'm not sure:

    AfffBBP.png

    If the above needs clarifying, do ask. Thanks for any help in advance ;)
     
  2. jcsd
  3. Nov 19, 2015 #2

    fzero

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  4. Nov 19, 2015 #3

    jtbell

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    If you take suitable linear combinations of ##\psi(2,1,-1)## and ##\psi(2,1,+1)##, and then find the probability densities, you do get the ##p_x## and ##p_y## orbitals.
     
  5. Nov 20, 2015 #4
    Ah, I've managed to make linear combinations which give the x and y-direction p orbitals, but I am still confused - does that mean some orbital wave functions aren't energy energy eigenfunctions? How/why would that be true?
     
  6. Nov 20, 2015 #5

    DrClaude

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    They are still energy eigenfunctions. However, ##p_x## and ##p_y## are not eigenfunctions of ##\hat{L}_z##.
     
  7. Nov 20, 2015 #6

    jtbell

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  8. Nov 20, 2015 #7
    Right, I understand that, but I've now got more questions - what are those donut-shaped probability densities? Why aren't they valid orbitals? Also, is there a systematic way for me to combine these 'base' wave functions to create the correct electron orbitals? I managed to make an intuitive guess with the other two p orbitals, but is there a general set of linear combinations which give all of the orbitals?
     
    Last edited: Nov 20, 2015
  9. Nov 20, 2015 #8

    DrClaude

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    They are completely valid orbitals. But they are complex functions, which can make them more difficult to work with in certain situations. The ##p_x## and ##p_y## orbitals are completely real functions, and their orientation along the Cartesian coordinates makes them useful to understand things like chemical bonding.

    See https://en.wikipedia.org/wiki/Atomic_orbital#Real_orbitals
     
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