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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Big-Daddy on February 20, 2013, 08:43:15 PM

Title: Schrodinger Equation
Post by: Big-Daddy on February 20, 2013, 08:43:15 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.
Title: Re: Schrodinger Equation
Post by: Borek on February 21, 2013, 04:22:11 AM
Normalization factor?
Title: Re: Schrodinger Equation
Post by: Big-Daddy on February 21, 2013, 07:50:19 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.
Title: Re: Schrodinger Equation
Post by: Borek on February 21, 2013, 08:05:07 AM
[tex]N'_{nl}=-\sqrt{\left(\frac{2Z}{na_0}\right)^3\frac{(n-l-1)!}{2n((n+l)!)^3}}[/tex]

Unless I am wrong. That's what I found in my copy of Quantum Chemistry by Kołos (Polish book).
Title: Re: Schrodinger Equation
Post by: Big-Daddy on February 21, 2013, 08:55:22 AM
[tex]N'_{nl}=-\sqrt{\left(\frac{2Z}{na_0}\right)^3\frac{(n-l-1)!}{2n((n+l)!)^3}}[/tex]

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.
Title: Re: Schrodinger Equation
Post by: Big-Daddy on February 21, 2013, 10:45:28 AM
[tex]N'_{nl}=-\sqrt{\left(\frac{2Z}{na_0}\right)^3\frac{(n-l-1)!}{2n((n+l)!)^3}}[/tex]

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.
Title: Re: Schrodinger Equation
Post by: Borek on February 22, 2013, 07:03:12 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.
Title: Re: Schrodinger Equation
Post by: Big-Daddy on February 22, 2013, 07:51:31 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)?
Title: Re: Schrodinger Equation
Post by: Borek on February 22, 2013, 11:25:45 AM
d orbitals are sometimes listed as dz2-r2, dxy, dxz, dyz, dx2-y2 - guess why ;)
Title: Re: Schrodinger Equation
Post by: Big-Daddy on February 22, 2013, 11:37:09 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?