Lithium Atom Ground State Radial Wavefunction

[bob012345]

TL;DR Summary

I have been looking a long time for the radial ground state wavefunction for the Lithium atom without success.

I would like to see what the shape of the ground state radial wavefunction for the Lithium atom is. An approximate function that shows the shape would be fine. Thanks.

 

[Vanadium 50]

Is hydrogen close enough? The nucleus and the two inner elections looks sort of like a Q=+1 object. If it's not close enough, this will be difficult, as it is a four-body problem.

 

[PeterDonis]

I would like to see what the shape of the ground state radial wavefunction for the Lithium atom is.

If you consider just the outer electron (i.e., the one in the $2s$ orbital), and treat the rest of the atom as providing a single Coulomb potential, then you can find a ground state radial wave function for that approximation; as @Vanadium 50 says, it's similar to that of a hydrogen atom's single electron, except that it's a $2s$ orbital, not a $1s$ orbital. But this is only an approximation.

If you want to consider all three electrons, with a single Coulomb potential due to the nucleus, then you don't have a "radial" wave function at all, because the Hilbert space for the configuration (even neglecting spin) is not ordinary 3-space but a 9-dimensional space.

 

[bob012345]

If you consider just the outer electron (i.e., the one in the $2s$ orbital), and treat the rest of the atom as providing a single Coulomb potential, then you can find a ground state radial wave function for that approximation; as @Vanadium 50 says, it's similar to that of a hydrogen atom's single electron, except that it's a $2s$ orbital, not a $1s$ orbital. But this is only an approximation.

If you want to consider all three electrons, with a single Coulomb potential due to the nucleus, then you don't have a "radial" wave function at all, because the Hilbert space for the configuration (even neglecting spin) is not ordinary 3-space but a 9-dimensional space.

Thanks. I understand it is a 9 dimensional space but if one looks at approximate Helium two electron models, there are models such as this;

$$\psi(r_1,r_2)≈e^{-α(r_1+r_2)} $$ Or models such as this; $$\psi(r_1,r_2,r_{12})≈e^{-α(r_1+r_2)}׃(r_1,r_2,r_{12})$$

That's what I'm looking to find for Lithium. Thanks.

 

[Vanadium 50]

I think it is unlikely to find such a simple answer.

 

[PeterDonis]

if one looks at approximate Helium two electron models

Can you give a reference for the helium models you are looking at?

 

[bob012345]

Can you give a reference for the helium models you are looking at?

http://www.umich.edu/~chem461/QMChap8.pdf
https://www2.chemistry.msu.edu/faculty/harrison/web_docs_He/Hylleraas_wavefunction.pdf

The first reference is where I got those expressions.

 

[bob012345]

I think it is unlikely to find such a simple answer.

I don't think a general expression such as the following with the function being a polynomial as all that simple! $$\psi(r_1,r_2,r_3,r_{12},r_{13},r_{23})≈e^{-α(r_1+r_2+r_3)}׃(r_1,r_2,r_3,r_{12},r_{13},r_{23})$$
But if an expression with just a few terms could get the basic form of the wavefunction reasonably well, it would still be difficult to calculate and then visualize an overall electron cloud probability density on 3 space from such a wavefunction.

 

[Vanadium 50]

Technically, any function fits that form - put the integral in the denomiantor of your function.

Anyway, you have seen how complicated the Helium atom is, and the trouble you run into by neglecting r12. Here you also neglect r13 and r23, and have a final answer that is the sum of three distributions. It will not be simple.

 

[bob012345]

Technically, any function fits that form - put the integral in the denomiantor of your function.

Anyway, you have seen how complicated the Helium atom is, and the trouble you run into by neglecting r12. Here you also neglect r13 and r23, and have a final answer that is the sum of three distributions. It will not be simple.

My understanding is that the solutions I seek are forms of trial solutions to the full Hamiltonian which contain the interaction terms. The trial solutions themselves may or may not contain the interaction variables in every part. I believe this was the case for Helium solutions mentioned. So, for Lithium;
$$ (-\frac 1 2 ∇_1^2 -\frac 1 2 ∇_2^2 -\frac 1 2 ∇^2_3 -\frac Z {r_1 }-\frac Z {r_2 }-\frac Z {r_3 } + \frac 1 {r_{12} } + \frac 1 {r_{13} } + \frac 1 {r_{23} })\psi=E\psi $$

where $$\psi=\psi(r_1,r_2,r_3,r_{12},r_{13},r_{23})≈e^{-α(r_1+r_2+r_3)}׃(r_1,r_2,r_3,r_{12},r_{13},r_{23})$$
or some other trial function where ƒ is a polynomial. This is the form Hylleraas used for Helium in 1929. But some graduate student probably did this calculation for a thesis decades ago. I just want to find it somewhere.

 

[Dr_Nate]

Is this what you are looking for?

 

[Vanadium 50]

Is this what you are looking for?

My favorite phrase: "...using the 9576-term wavefunction..."

 

[bob012345]

Is this what you are looking for?

Yes, thank you very very much!

I find it very interesting that the authors suggests that a more complete trial function pre-factor such as below;

$$f(r_1, r_2, r_3, r_{12}, r_{13}, r_{23}) = (1 + β_1 r_1 + β_2 r_2 + β_3 r_3 + γ_1 r_{23} + γ_2 r_{13} + γ_3 r_{12})$$

"..seems beyond the computer resources presently available to us." because "It is well-known that the standard quantum chemistry approaches to calculation of the energies of the low-lying states of few-electron atoms are characterized by slow convergence."

This is why I am interested in the potential use of analog computation for multi-electron atoms. One would not pick a trial function at all but the analog model would converge to the correct shape of the wavefunction for the correct energy... if it worked. Results like this paper would allow a comparison. THANKS AGAIN!

 

[Dr_Nate]

My favorite phrase: "...using the 9576-term wavefunction..."

Then they get to go on and say the results are 'Exact' in the table.

 

[Dr_Nate]

Yes, thank you very very much!

I find it very interesting that the authors suggests that a more complete trial function pre-factor such as below;

$$f(r1, r2, r3, r12, r13, r23) = (1 + β1 r1 + β2 r2 + β3 r3 + γ1 r23 + γ2 r13 + γ3 r12),$$

"..seems beyond the computer resources presently available to us." because "It is well-known that the standard quantum chemistry approaches to calculation of the energies of the low-lying states of few-electron atoms are characterized by slow convergence."

This is why I am interested in the potential use of analog computation for multi-electron atoms. One would not pick a trial function at all but the analog model would converge to the correct shape of the wavefunction for the correct energy... if it worked. Results like this paper would allow a comparison. THANKS AGAIN!

You're welcome

Your analog idea sounds like quantum dots as artificial atoms. You might want to look into that research area.

 

[bob012345]

You're welcome

Your analog idea sounds like quantum dots as artificial atoms. You might want to look into that research area.

Thanks I will. The analog model I have been using for solving the Schrodinger equation for simple systems and Hydrogenic atoms is based on circuit elements. Actually, I'm using a circuit solver to simulate the analog computer (yes, on a digital computer) because such solvers are very robust for solving systems of 2nd order differential equations, and they are freely available and convenient.

 

[bob012345]

I just wanted to add a comment that looking at the paper referenced above it seems that even with only a 'few' parameters, there seems to be a few dozen terms in the wavefunction counting permutations and spin. Ultimately it would be interesting to reduce it to a function of total electron cloud probability density as a function of distance from the nucleus such as in the image below however I suspect this plot was constructed from one-electron wavefunctions.

 

[PeterDonis]

I suspect this plot was constructed from one-electron wavefunctions.

We have no way of assessing that unless you give us a reference for where the image came from (hint, hint...).

 

[bob012345]

We have no way of assessing that unless you give us a reference for where the image came from (hint, hint...).

Sorry. Here it is. I may have been hasty in concluding it was based on single electron wavefunctions. I can say I plotted the sums of probability densities from such hydrogenic wavefunctions and got a very similar plot.

https://chem.libretexts.org/Courses/Saint_Marys_College%2C_Notre_Dame%2C_IN/CHEM_342%3A_Bio-inorganic_Chemistry/Readings/Purgatory/D._Sizes_of_Atoms_and_Ions

 

[PeterDonis]

Here it is.

Unfortunately, Firefox appears to insist on removing the apostrophe in Saint Mary's from the URL, which understandably gives a page not found error. (Technically the apostrophe should be URL encoded, so the URL is not correct according to web standards, but in the real world there are far too many URLs out there that break those standards for insisting on strict compliance to be realistic.)

I was able to get to the page from this search:

https://chem.libretexts.org/Special...fpth=&query=sizes+of+atoms+and+ions&type=wiki

Clicking on the second link works for me.

 

[PeterDonis]

I may have been hasty in concluding it was based on single electron wavefunctions.

It says "total electron density for all occupied orbitals", which looks to me like it is doing the following:

(1) Start with the wave function for each orbital for the given atom that is occupied in the ground state. This will not be precisely the wave function for that orbital for hydrogen, because of the different nuclear charge and shielding from electrons in other orbitals, but it will generally look reasonably "hydrogenic" in qualitative form.

(2) For orbitals that have nonzero orbital angular momentum (i.e., all except $s$), average over all angles to remove the angular dependence and obtain a function of radius alone.

(3) Multiply each orbital radial function by the number of electrons in that orbital in the ground state (which will be either 1 or 2--note that this particular graph only includes noble gases, which have completely filled shells so all orbitals have 2 electrons).

(4) Add the radial functions for all the orbitals together to get the total electron density as a function of radius.

There is no math on the page so I can't be sure of the above, but that's what makes sense to me.

 

[pbuk]

Unfortunately, Firefox appears to insist on removing the apostrophe in Saint Mary's from the URL...

No its not FireFox, look below and you can see it's missing from the code in the post - probably stripped by the JavaScript WYSIWYG editor:

Sorry. Here it is. I may have been hasty...
https://chem.libretexts.org/Courses/Saint_Marys_College%2C_Notre_Dame%2C_IN/CHEM_342%3A_Bio-inorganic_Chemistry/Readings/Purgatory/D._Sizes_of_Atoms_and_Ions

If you add the URL manually it works (as long as you delimit it with "s of course surprisingly you can even delimit it with 's).

If you add the URL manually [url='https://chem.libretexts.org/Courses/Saint_Mary's_College%2C_Notre_Dame%2C_IN/CHEM_342%3A_Bio-inorganic_Chemistry/Readings/Purgatory/D._Sizes_of_Atoms_and_Ions']it works[/url] 
([S]as long as you delimit it with "s of course[/S] surprisingly you can even delimit it with 's).

Edit: I'll follow this up elsewhere.

 

[bob012345]

It says "total electron density for all occupied orbitals", which looks to me like it is doing the following:

(1) Start with the wave function for each orbital for the given atom that is occupied in the ground state. This will not be precisely the wave function for that orbital for hydrogen, because of the different nuclear charge and shielding from electrons in other orbitals, but it will generally look reasonably "hydrogenic" in qualitative form.

(2) For orbitals that have nonzero orbital angular momentum (i.e., all except $s$), average over all angles to remove the angular dependence and obtain a function of radius alone.

(3) Multiply each orbital radial function by the number of electrons in that orbital in the ground state (which will be either 1 or 2--note that this particular graph only includes noble gases, which have completely filled shells so all orbitals have 2 electrons).

(4) Add the radial functions for all the orbitals together to get the total electron density as a function of radius.

There is no math on the page so I can't be sure of the above, but that's what makes sense to me.

Thanks! That seems very reasonable. I didn't think it was the actual multi-electron wave-functions since it seems extremely difficult to solve much past Lithium directly. I'm going to try it and hopefully, return a plot to compare the other one to. Obviously, wave-functions for multi-electron atoms aren't exactly the same for single electron hydrogenic atoms. Helium, for example, with two electrons has a very slightly larger radius (peak probability as one measure) than with one electron. One technical question in regards to the overall probability density is whether to add individual densities or add all the wave-functions first and then square? Thanks again.

 

[PeterDonis]

Helium, for example, with two electrons has a very slightly larger radius (peak probability as one measure) than with one electron.

Do you mean a Helium+ ion (one electron) as compared to a neutral Helium atom (two electrons)?

 

[bob012345]

Do you mean a Helium+ ion (one electron) as compared to a neutral Helium atom (two electrons)?

Yes. Here is a reference which contains the following chart. Notice the relative size of He+ to He which is about 17% larger. This would mean a full 1s shell would have a slightly different peak than a hydrogenic orbital function derived from a one-electron Schrodinger equation and so forth up the chain. But still maybe it's close enough.

https://www.pnas.org/content/115/50/E11578

 

[PeterDonis]

Here is a reference which contains the following chart. Notice the relative size of He+ to He which is about 17% larger.

This is a chart of "non-relativistic classical turning radii", which is a different concept of "size" than your previous reference was using. For one-electron atoms/ions, this is simply twice the radius of the peak of the 1s orbital wave function (you can see this from the value for hydrogen, which is twice the Bohr radius of 0.53 Angstroms--then He+ and Li++ are one-half and one-third of the hydrogen value, as expected from the nuclear charges of 2 and 3 vs. 1), which is twice the value of the previous "size" concept. But for atoms/ions with multiple electrons, I don't know if the same simple relationship holds.

 

[PeterDonis]

This would mean a full 1s shell would have a slightly different peak than a hydrogenic orbital function derived from a one-electron Schrodinger equation and so forth up the chain.

Just on general principles, I would expect removing one electron from any neutral atom to leave a positive ion that is smaller. Once the first electron is removed, the binding energy of the most loosely bound one that remains will be larger than the binding energy of the one that got removed (i.e., the energy it took to singly ionize the atom), and "larger binding energy", heuristically, means "closer average distance of electron from nucleus", i.e., a peak of the electron radial wave function at a smaller radius.

Similarly, I would expect adding an electron to any neutral atom to give a negative ion that is larger, since the new electron will have to be in a more loosely bound state than the ones that were there in the neutral atom. (Of course this assumes that the neutral atom started off in its ground state.)

How much of an energy difference there is in these cases will of course depend on the details of which atom, which shells are filled, partially filled, etc.

 

[bob012345]

This is a chart of "non-relativistic classical turning radii", which is a different concept of "size" than your previous reference was using. For one-electron atoms/ions, this is simply twice the radius of the peak of the 1s orbital wave function (you can see this from the value for hydrogen, which is twice the Bohr radius of 0.53 Angstroms--then He+ and Li++ are one-half and one-third of the hydrogen value, as expected from the nuclear charges of 2 and 3 vs. 1), which is twice the value of the previous "size" concept. But for atoms/ions with multiple electrons, I don't know if the same simple relationship holds.

Thanks. Every chart seems to have a different way of calculating the effective size of atoms and ions but the relative differences are useful. From the chart above, the size ratios for Li+ vs. Helium is not quite the same as for Li++ and He+. $$ He^+/Li^{++} ≈ 1.51 ~~vs.~~ He/Li^+ ≈ 1.59$$

 

[bob012345]

It says "total electron density for all occupied orbitals", which looks to me like it is doing the following:

(1) Start with the wave function for each orbital for the given atom that is occupied in the ground state. This will not be precisely the wave function for that orbital for hydrogen, because of the different nuclear charge and shielding from electrons in other orbitals, but it will generally look reasonably "hydrogenic" in qualitative form.

(2) For orbitals that have nonzero orbital angular momentum (i.e., all except $s$), average over all angles to remove the angular dependence and obtain a function of radius alone.

(3) Multiply each orbital radial function by the number of electrons in that orbital in the ground state (which will be either 1 or 2--note that this particular graph only includes noble gases, which have completely filled shells so all orbitals have 2 electrons).

(4) Add the radial functions for all the orbitals together to get the total electron density as a function of radius.

There is no math on the page so I can't be sure of the above, but that's what makes sense to me.

Here is my attempt to do your suggested procedure. The original plot is included for reference. Since the original plot looked to me that it might be a drawing for a textbook, I wasn't sure how accurate it was. The main conclusion I can make is that the greatest factor is an estimation of the screening by changing the nuclear charge to an effective nuclear charge. I started with my old QM text's Zeff for Helium (1.6875). In the analog computer simulation, time is distance. The horizontal scales are 0 to 100pm on both and are distance from the nucleus. The vertical scale is radial probability in arbitrary units. I did have to tweak the relative strengths of the wavefunctions a bit to get the right relative amplitudes but the normalized wave-functions were in the ballpark. I used Zeff to get the right shapes as a function of distance.