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Topic: Schrodinger Equation orbital solutions  (Read 6088 times)

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Offline Big-Daddy

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Schrodinger Equation orbital solutions
« on: May 20, 2013, 02:04:39 PM »
On this Wikipedia page (http://en.wikipedia.org/wiki/Hydrogen-like_atom) under the section "Non-relativistic wavefunction and energy", the general solution of the Schrodinger equation for a system with a Z-proton nucleus and 1 electron is written.

How is this equation arrived at from the usual time-independent 3D Schrodinger equation? Do we arrive at the general solution by deriving it directly from the Schrodinger, or derive specific cases and then classify the general trends into a general solution?

Offline Corribus

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Re: Schrodinger Equation orbital solutions
« Reply #1 on: May 20, 2013, 02:18:44 PM »
You have to solve the SE directly using the hydrogenic atom Hamiltonian.
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Offline curiouscat

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Re: Schrodinger Equation orbital solutions
« Reply #2 on: May 20, 2013, 02:21:48 PM »
Directly AFAIK. (And it wasn't easy....)

Offline Borek

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Re: Schrodinger Equation orbital solutions
« Reply #3 on: May 20, 2013, 03:03:04 PM »
From what I remember it is not that difficult, as variables are separable.

Not that I would know how to solve it now.
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Offline curiouscat

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Re: Schrodinger Equation orbital solutions
« Reply #4 on: May 20, 2013, 03:21:09 PM »
From what I remember it is not that difficult, as variables are separable.


You may be right.

It's all relative I guess. As an undergrad it seemed pretty hard to me. But then again I was never too good at math.


Offline Corribus

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Re: Schrodinger Equation orbital solutions
« Reply #5 on: May 20, 2013, 10:55:38 PM »
@BD

If you need help solving this equation, let me know.  I will try to knock my rust off. :) But there are probably solutions online already existing. 
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Big-Daddy

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Re: Schrodinger Equation orbital solutions
« Reply #6 on: May 21, 2013, 06:12:59 AM »
Thanks for the responses. I found a good derivation on this website: http://users.aber.ac.uk/ruw/teach/237/hatom.php

I have one question. The radial solution Rn,l seems completely different in the webpage (see the section titled "Solving the radial equation") to the Wikipedia form on this page http://en.wikipedia.org/wiki/Hydrogenlike_atom under "Non-relativistic wavefunction and energy". How do we get from the website's solution to Wikipedia's?

Offline Borek

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Re: Schrodinger Equation orbital solutions
« Reply #7 on: May 21, 2013, 08:26:31 AM »
Ignore wikipedia. The other page solves the equation the way I remember it (not that I checked all the details, I mostly skimmed).
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Offline Corribus

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Re: Schrodinger Equation orbital solutions
« Reply #8 on: May 21, 2013, 10:12:30 AM »
On the contrary, it would be my advice to ignore the other website.  The form of the solution is important, and the one shown in the wikipedia article is the most useful form, which includes Laguerre polynomials that depend clearly on the typical quantum numbers.  If you want to be able to clearly specify what the wavefunction is for a given set of quantum numbers, this form is what you should use.  The form specified on the other website is useless and incomplete, because it doesn't explicitly include the l-quantum number dependence. (It leaves it as an undefined coefficient.) The wikipedia form is also the one used in most physical chemistry textbooks.

To be honest, I'm not sure why you are interested in working through the mathematical solution.  I suppose it might give you some satisfaction to follow in the footsteps of Niels Bohr and his contemporaries, but it's strictly a mathematical exercise and I feel there are probably better ways to spend your time.  Looking over the website you linked and some others like it, I admit it would take me the better part of a day (maybe a week - I haven't solved any real differential equations since college and the rust doesn't come off so easily!) to try to remember how to go through and derive the form on the wikipedia page.  And to what end?  To get something that is already well known to everyone?  My advice would be to skip the derivation and focus on fully understanding what the solutions mean.  This is what is really important to chemists.

That said, if you are really interested in understanding how to derive the common form shown on the wikipedia page, you are probably better off asking at a physics or mathematics forum, which will be populated by people who do these kinds of complex mathematical manipulations on a more frequent basis.

EDIT: This is the type of derivation I remember from college which gives the formula shown in the wikipedia article:

http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/HydrogenAtom.htm
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Offline curiouscat

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Re: Schrodinger Equation orbital solutions
« Reply #9 on: May 21, 2013, 10:33:53 AM »

To be honest, I'm not sure why you are interested in working through the mathematical solution.  I suppose it might give you some satisfaction to follow in the footsteps of Niels Bohr and his contemporaries, but it's strictly a mathematical exercise and I feel there are probably better ways to spend your time.  Looking over the website you linked and some others like it, I admit it would take me the better part of a day (maybe a week - I haven't solved any real differential equations since college and the rust doesn't come off so easily!) to try to remember how to go through and derive the form on the wikipedia page.  And to what end?  To get something that is already well known to everyone?  My advice would be to skip the derivation and focus on fully understanding what the solutions mean.  This is what is really important to chemists.


I agree with @corribus. The first time I ever did this derivation was in my Quantum Physics Class (an advanced undergrad class). To try and attempt this at High School Level is fairly ambitious.

@BigDaddy: Do you have any exposure to solving PDE's before this? Even as PDE's go this isn't a simple one.

Offline Big-Daddy

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Re: Schrodinger Equation orbital solutions
« Reply #10 on: May 21, 2013, 02:22:34 PM »
To be honest, I'm not sure why you are interested in working through the mathematical solution.  I suppose it might give you some satisfaction to follow in the footsteps of Niels Bohr and his contemporaries, but it's strictly a mathematical exercise and I feel there are probably better ways to spend your time.  Looking over the website you linked and some others like it, I admit it would take me the better part of a day (maybe a week - I haven't solved any real differential equations since college and the rust doesn't come off so easily!) to try to remember how to go through and derive the form on the wikipedia page.  And to what end?  To get something that is already well known to everyone?  My advice would be to skip the derivation and focus on fully understanding what the solutions mean.  This is what is really important to chemists.

I only wanted to have a look at the derivation. I'm not planning to look through all the stages very carefully or repeat it by hand.


[quote author=Corribus link=topic=68328.msg246329#msg246329 date=1369145550EDIT: This is the type of derivation I remember from college which gives the formula shown in the wikipedia article:

http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/HydrogenAtom.htm
[/quote]

Thanks, this was a helpful website, which took it all the way up to the form shown in the Wikipedia article.

I agree with @corribus. The first time I ever did this derivation was in my Quantum Physics Class (an advanced undergrad class). To try and attempt this at High School Level is fairly ambitious.

I'm reading it, just for the maths pretty much. And there are many steps where there is Maths I haven't encountered. But on the whole I can follow most parts.

@BigDaddy: Do you have any exposure to solving PDE's before this? Even as PDE's go this isn't a simple one.

No. But I read a little when I saw that the solution involved partial differentials. Not much of what I learnt in that reading seemed to actually be used in the derivation  ::) That is ok, I was concerned with the overrall process. If each step isn't clear right now it can wait until my advanced undergrad years!  ;)

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