Schroedinger equation Hydrogenic atom?

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In summary: If you want to know how to do this, you'll have to find the book or find someone who knows how to do it. Thank you for your question.
  • #1
shaun2985
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Hi

I'm second year undergrad in Physics and I've been studying the time-independent Schroedinger Equation (TISE), QM operators, etc and now application of this to a hydrogenic atom.

I've come to a bit of a dead-end though - I've got to the stage where I've separated the TISE into Radial and Angular parts. With the angular part, I've deduced that the corresponding eigenfunctions are spherical harmonics with corresponding eigenvalue [tex]\hbar l(l+1)[/tex] which I've put equal to [tex]2m(\lambda)[/tex].

Now, in solving the radial part of the TISE for the hydrogenic atom, the lecturer has considered the radial equation at very large distances from the nucleus, when the effective potential [tex]/V_eff[/tex] can be neglected. i.e. considering bound states of the atom. He has put [tex]\E = -\frac{\kappa^2}{2}[/tex], substituted into TISE (using the relation R = chi/r) and has obtained the solution in term of an unknown function, F(r), [tex]\chi(r) = F(r) \exp(-\kappa r)[/tex]. He has obtained a corresponding differential equation for F, resulting in [tex]\frac{d^2 F}{dr^2} = -frac{l(l+1)}{r^2} F = 2k \frac{dF}{dr} - \frac{2z}{r}F[/tex]. Solution of this is by series solution with condition that r is well behaved as it tends to 0. Now the problem is that I've lost notes on this particular derivation and am unsure how to find the energy for hydrogenic atom in the parameters I've described. I've looked at books, but they all show a different method, one that won't be examined in my course.

I was wondering if one of you could perhaps fill in the gaps for me, by using a series solution method.

Thanks




I'll have a go showing a rough derivation of where I've got to..

Hamiltonian in 3D in spherical polar coordinates with coulomb potential:
[tex]\hat{H} = -\frac{\hbar^2}{2m_e} [\frac{1}{r^2} \pd{}{r}{} (r^2 \pd{}{r}{}) - \frac{\hat{L^2}}{\hbar^2 f^2}] - \frac{Ze^2}{4 \pi \epsilon_0 r}[/tex]

Therefore TISE:
 
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  • #2
lol, the latex tags didn't quite work out.

it's much easier if I show you what I've done...

http://img241.echo.cx/img241/5253/phy15mn.jpg

http://img260.echo.cx/img260/44/phy10013mj.jpg

http://img260.echo.cx/img260/1706/phy10042wn.jpg

http://img260.echo.cx/img260/1198/phy10051ko.jpg

http://img260.echo.cx/img260/2923/phy10060wh.jpg

http://img260.echo.cx/img260/4489/phy10072kz.jpg

http://img260.echo.cx/img260/191/phy10083fn.jpg

http://img260.echo.cx/img260/523/phy10090xx.jpg

http://img229.echo.cx/img229/2540/phy10105fp.jpg


Ok - the above images show the separation of the TISE under electric potential into radial and angular parts. Angular parts have thus yielded spherical harmonic eigenfunctions and the respective eigenvalues.

The solution of the radial part has required substitution of [tex]\chi = rR(r)[/tex] into the TISE. Then it was considered as r -> infinity. The V term disappears and left with linear second-order DE. Solutions give [tex]\chi = B \exp(-\kappa r)[/tex]. Rewriting the solution in terms of an unknown function yields [tex]\chi = F(r) \exp(-\kappa r)[/tex].

This was differentiated twice and substituted into the TISE in atomic units,
[tex]-\frac{1}{2} \frac{d^2 \chi}{dr^2} + \left[ \frac{l(l+1)}{2r^2} - \frac{z}{r} \right] \chi = E\chi[/tex]

This gave a 2nd order DE in terms of F:
[tex]\frac{d^2 F}{dr^2} - \frac{l(l+1)}{r^2} F = 2k \frac{dF}{dr} - \frac{2z}{r} F[/tex]

I did the first stages of Frobenius' Method but am unsure where to go from slide "Series Solution 2". I'd be grateful if anyone could fill in the rest for me, in finding the energy as shown in the slides below...

http://img229.echo.cx/img229/8739/phy10110wo.jpg
 
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  • #3
edit the .th parts out of your links. and try again with the latex, the text in the pics is really really hard to read.
 
  • #4
hi, I've edited out the .th parts - they should be much bigger to see now. i'd try again with the latex tags, but it's my first time using them and could take hours.
 
  • #5
I really didn't understand what u were looking for,but i can tell that the most rigurous treatment for the H atom using SE is made in A.Messiah's book.

In the most rigurous approach,th solution of SE for the H atom is expressible as a product between a special case of degenerate/confluent hypergeometric function (the Laguerre polynomial) and a special case of a Gauss hypergeometric function (the spherical harmonic).This discussion is made in Messiah's book.

Daniel.
 

Related to Schroedinger equation Hydrogenic atom?

1. What is the Schroedinger equation for a Hydrogenic atom?

The Schroedinger equation for a Hydrogenic atom is a mathematical equation that describes the behavior of an electron in the atom's quantum mechanical state. It takes into account the attractive force between the positively charged nucleus and the negatively charged electron, as well as the electron's wave-like behavior.

2. How is the Schroedinger equation derived for a Hydrogenic atom?

The Schroedinger equation for a Hydrogenic atom is derived using the principles of quantum mechanics and the concept of energy quantization. It is based on the idea that the electron's motion can be described by a wave function, and the equation determines the allowed energy levels and corresponding wave functions for the electron.

3. What is the significance of the Hydrogenic atom in quantum mechanics?

The Hydrogenic atom is significant because it is the simplest model of an atom that can be described using the Schroedinger equation. It serves as a starting point for understanding more complex atoms and molecules, and it allows for the prediction of the atom's energy levels and behavior.

4. How does the Schroedinger equation explain the spectral lines of a Hydrogenic atom?

The Schroedinger equation predicts the energy levels of a Hydrogenic atom, and these energy levels correspond to the different spectral lines observed in the atom's emission spectrum. The energy difference between these levels determines the frequency of the emitted light, which creates the distinctive spectral lines.

5. Can the Schroedinger equation be applied to other atoms besides Hydrogen?

Yes, the Schroedinger equation can be applied to other atoms besides Hydrogen. However, for more complex atoms, the equation becomes more complicated and difficult to solve exactly. Approximations and numerical methods are often used to solve the equation for these atoms.

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