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##### Schrodinger Equation
« on: February 20, 2013, 02:43:15 PM »

The associated Laguerre polynomial required for the Schrodinger equation for orbitals is L(2l+1,n+l-1), where the polynomial is a function of p. Would you agree? p=2*Z*r/(n*a0), where Z is the atomic number, r is the distance from nucleus to electron, and n is the first quantum number. Let us take l as 0 (so we are looking at the ns orbital case).

The problem is then L(1,2), which should pertain to n=3. According to Wikipedia's sum for n=3 (in terms of a, which we know to be 1), which is L(1,2)=(1/2)p2-3p+3, I am getting, after I substitute in p=2*Z*r/(n*a0), which is now p=2*Z*r/(3*a0),  ((2/9)*Z2*(r/a0)2)-(2*Z*(r/a0))+3. When I put Z=1 then I get ((2/9)*(r/a0)2)-(2*(r/a0))+3. Whereas I should be finding (1-(2/3)*(r/a0)+(2/27)*(r/a0)^2) acording to this website (http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html). Anyone know the problem?

Strangely enough, I can reach the right answer by dividing by 3. But why is this that I have to? I can't see what I did wrong so that diving by 3 should rectify it. Because I did use the correct expression for p and I've checked the polynomial simplification with WolframAlpha. So something must be going on outside the polynomial which reduces it down by 1/3, but I can't see what.
« Last Edit: February 20, 2013, 02:58:05 PM by Big-Daddy »
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#### Borek

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##### Re: Schrodinger Equation
« Reply #1 on: February 20, 2013, 10:22:11 PM »

Normalization factor?
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##### Re: Schrodinger Equation
« Reply #2 on: February 21, 2013, 01:50:19 AM »

Normalization factor?

Yes that has been in my thoughts overnight.
It seems to be 2/n (where n is the first quantum number) so long as l=0. This would explain the overall multiplication of 2 for n=1, no change for n=2 (2/2), multiplication of 2 and division of the polynomial by 3 for n=3 (2/3).

Can you link me to somewhere showing how to work out normalization factor for general n and l? (So far is only for general n where l=0)

Edit: I mean general method for n and l quantum numbers. I'm aware that normalization of the "Schrodinger equation in general" is a vastly complicated field and I'm only looking at the orbital special case for now.
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#### Borek

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##### Re: Schrodinger Equation
« Reply #3 on: February 21, 2013, 02:05:07 AM »

$$N'_{nl}=-\sqrt{\left(\frac{2Z}{na_0}\right)^3\frac{(n-l-1)!}{2n((n+l)!)^3}}$$

Unless I am wrong. That's what I found in my copy of Quantum Chemistry by Kołos (Polish book).
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##### Re: Schrodinger Equation
« Reply #4 on: February 21, 2013, 02:55:22 AM »

$$N'_{nl}=-\sqrt{\left(\frac{2Z}{na_0}\right)^3\frac{(n-l-1)!}{2n((n+l)!)^3}}$$

Unless I am wrong. That's what I found in my copy of Quantum Chemistry by Kołos (Polish book).

I'm not getting the right answers from that equation. This website - http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html - I think does not account for the fact that 2r/(n*a0) is your usual term (it treats it just as r/a0). But I would guess that the constants are still trustworthy (and have not accounted for the n in 2r/(n*a0)).

If we try n=3, l=1, the website suggests we should get (4*root 2)/3 whereas this equation gives -1.89*10-3 (Z=1 for all cases as the website is recording for hydrogen). If we try n=3, l=2, the website gives (2*root 2)/(27*root 5) but the equation -1.69*10-4. (I've excluded a0, i.e took it as =1, because I can't see how to include it and it doesn't appear in the normalizing constant I can see for any of the equations given). Also surely the constant will not be negative if the rest of the operation is positive because the function is finally positive.
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##### Re: Schrodinger Equation
« Reply #5 on: February 21, 2013, 04:45:28 AM »

$$N'_{nl}=-\sqrt{\left(\frac{2Z}{na_0}\right)^3\frac{(n-l-1)!}{2n((n+l)!)^3}}$$

Unless I am wrong. That's what I found in my copy of Quantum Chemistry by Kołos (Polish book).

I've found a modified version of this equation on Wikipedia which seems to relate closely to the values found on the website.

http://en.wikipedia.org/wiki/Hydrogen-like_atom "Non-relativistic wavefunction"

There is just one problem now: the normalization listed on the website for n=3, l=1 is (exactly) 8 times greater than that found by Wikipedia, whereas by my calculations it should be 8/3 times greater. Am I just to take this as an error on the part of the author of the website (because if he had taken simply 2r/a0 rather than 2r/(n*a0) then it would make sense that his factor is 8 rather than 8/3, as we are looking at n=3)?

For the other normalizations I have tested - n=2,l=0, n=3,l=0 and n=3,l=2 - I have been able to figure out why both sources show correct values.
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#### Borek

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##### Re: Schrodinger Equation
« Reply #6 on: February 22, 2013, 01:03:12 AM »

On the second read normalization factor I have listed is just for the radial function (full wave function can be expressed as a product of a radial and angular functions).

No idea why it is listed as negative, but it doesn't matter much - norm is calculated for the squared function, so it will be positive no matter what.
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##### Re: Schrodinger Equation
« Reply #7 on: February 22, 2013, 01:51:31 AM »

On the second read normalization factor I have listed is just for the radial function (full wave function can be expressed as a product of a radial and angular functions).

No idea why it is listed as negative, but it doesn't matter much - norm is calculated for the squared function, so it will be positive no matter what.

Ok thank you, I got it.

I have heard that for p, d, f orbitals you have to combine wave functions to get a real result by simply adding them (because the original wavefunctions are complex, with i). e.g. px or py means combining the l=1,ml=1 and l=1,ml=-1 wavefunctions. (l=1,ml=0 is real and corresponds to pz) Can you link to somewhere showing how to do this (i.e. which wavefunctions to combine to find which orbitals) for d-orbitals (and even f if you have it)?
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#### Borek

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##### Re: Schrodinger Equation
« Reply #8 on: February 22, 2013, 05:25:45 AM »

d orbitals are sometimes listed as dz2-r2, dxy, dxz, dyz, dx2-y2 - guess why
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##### Re: Schrodinger Equation
« Reply #9 on: February 22, 2013, 05:37:09 AM »

d orbitals are sometimes listed as dz2-r2, dxy, dxz, dyz, dx2-y2 - guess why

Is the first one not simply dz2? So this is the orbital corresponding to uncombined wavefunction l=2,ml=0; dxy and dyz correspond to a combination of the l=2,ml=1 and l=2,ml=-1 wavefunctions; and dxy and dx2-y2 correspond to a combination of the l=2,ml=2 and l=2,ml=-2 wavefunctions.

But how are they combined to reach each orbital? Wikipedia shows the two combinations needed for the px and py orbitals (see http://en.wikipedia.org/wiki/Atomic_orbital "Real orbitals") but doesn't show how the wavefunctions are combined for d or f orbitals?
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