# Can QT estimate the ionization energy of neutral Helium, He I?

1. Oct 12, 2007

First, I want to understand something. Everything I look at suggests that the ionization energy of an atom or ion is suppose to be the energy needed to remove the 'outermost electron'- which I find troubling conceptually, since they also refer to this as the one with the highest energy. Wouldn't the hardest electron to remove be the groundstate? Why am I told not to think of this electron as a groundstate and certainly not as an n=1 electron, as was the case in Bohr's original theory?

Second, does anyone know how the ionization energy of the atoms and ions are experimentally determined? In a laboratory?

Finally, how does QT estimate or calculate the ionization energy of Helium I or other systems? In retrospect, Bohr was only able to calculate the ionization energy of Helium II... If QT is better, how does it calculate this value- or do these measurements remain experimentally determined?

Anyone have any idea- in whole or in part?

2. Oct 12, 2007

### quetzalcoatl9

you can use perturbation theory to find the ionization energy of He to within <10 %.

alternatively, if you make the Hartree-Fock approximation for He

$$\psi(r_1, r_2) = \phi_1(r_1) \phi_2(r_2)$$

then there is an effective Hamiltonian

$$\hat{H}^{eff}_1 (r_1) = -\frac{\nabla_1^2}{2m} - \frac{2}{r_1} + \int dr_2 \phi(r_1) \frac{1}{r_{12}} \phi(r_2)$$

(and the same for the second electron) defined such that a single-body approximate to the true Schrodinger equation is:

$$\hat{H}^{eff}_1 \phi(r_1) = \epsilon_1 \phi(r_1)$$

where $$\epsilon$$ will be a close approximation to the true ionization energy.

this can be solved to high accuracy on a computer (<1% error). for elements with more electrons, the HF appromixation will be less accurate since it explicitly denies electron correlation. there are, however, systematic ways of recovering the correlation energy to higher order.

3. Oct 12, 2007

So if I am to understand you correctly, perbutation theory can be used to estimate the i.e. of say neutral Helium to within 10%, but that the usefulness of that estimate, therefore rests upon its comparison with experimental results. Wherein, correlation energies -being corrections- may be tabulated and used to reach estimates that are within 1%?

Now you mentioned that computers need to be used to get more accurate results...
Does this mean that it is a difficult calculation, given the sophisticated math involved? If it is hard to make a calculation for Helium, what about BI, Li I, Be I, II, III etc. Can these values also be calculated with the same degree of accuracy? Thanks

Aside I would still like to know how the ionization energy is experimentally determined- in the laboratory- anyone?

4. Oct 13, 2007

### quetzalcoatl9

if you lose the last sentence above, then you are correct. the correlation energy refers to what-you-lose with the Hartree-Fock method.

computers are a useful way of solving equations numerically to arbitrary precision. the other elements can be done (numerically, NOT by perturbation theory analytically), but as i said, the situation starts to become more complex. correlation effects need to be included in some systematic way and furthermore non-relativistic QM breaks down for the heavy elements (i.e. Z>40 or so) since the average momentum of the core electrons approaches $$mc$$. by orthogonality, this affects all of the other states as well (including the valence electrons). for the lanthanides/actinides this effect is non-negligible (it is also one of the reasons for their interesting chemistry).

your questions are good ones, and serve to illustrate what is rarely taught in undergraduate coursework - that nearly all physical problems of interest are either mathematically or numerically intractable. the full weight of this isn't really felt until one is working in this area. we rely upon making suitable approximations to move forward; sometimes even that isn't enough, and so we build very expensive computers.

i have no idea how the laboratory experiments are done.

Last edited: Oct 13, 2007
5. Oct 13, 2007

### Gokul43201

Staff Emeritus
The electron with the highest energy is the one that is easiest to remove, not the other way round.

This is measured by a photoionization experiment conducted at a synchrotron source, using a spectrometer.

UV light of sufficient energy + gaseous atom ---> ion + electron ---> charged species accelerated by a field and detected by an analyser.

6. Oct 14, 2007

non-relativistic QM breaks down

I wonder, this supposed breakdown in the heaveir elements-does it have anything to do with the 'Rydberg atoms' phenomenon?

7. Oct 14, 2007

Okay, for clarity, lets say you have a K atom, it has one outer valence electron. Its oxidation state is K+. This 'outer electron' can make quantum transitions from its ground state to excited states. Wouldn't it be easier to remove an electron from one of its excited states as compared with the same electron's ground state?

Where, as I see it, doesn't an electron in a lower state have more kinetic energy per se, than an outer excited one. Or am I just thinking too 'planetary'? Surely, an electron in an excited state uses up its kinetic energy in making a transition to a greater potential energy.??

Is this reasoning inconsistent with QM doctrine?

8. Oct 14, 2007

x
Thankyou, I looked this up and it is a rather complicated process. I wonder if it has always been done this way. Is it possible that the first ionization energies were done by another method? Any idea? It would appear that it takes a great deal of time and expense to make these empirical measurements?

Do you know if anyone has ever used atomic spectral data to accomplish the same goal? Specifically, the ability to determine the ionization energy of an atom or ion from the empirical wavelengths of its spectra?

Surely, there has to be more than one way to empirically measure the ionization energy of an atom or ion... It would be ideal, would it not, to have two methods that could be cross referenced?

Last edited: Oct 14, 2007
9. Oct 14, 2007

### quetzalcoatl9

no, it does not.

the average velocity of the core electrons becomes a sizeable fraction of the speed of light for large Z. therefore, relativistic effects must be accounted for in some way when describing the Hamiltonian for such an atom.

10. Oct 15, 2007

### OOO

It's the definition of the term ionization energy. But if you like you can also ionize an atom by removing one of its inner electrons. That's what is done in X-ray absorption spectroscopy. For that you need more energy (thus X-ray).

Depends what you call the ground state. The true ground state of a multi electron system is a multielectron wave function. From this point of view you can't just "remove the ground state of one electron". But if you are doing Hartree-Fock then you consider the movement of one electron in the effective field of the others + the nucleus. If this electron is the outermost electron then its ground state is indeed the one you are interested in for ionization. Likewise you could consider the "ground state" of inner Hartree-Fock electrons, but I don't know if this is directly connected to measured X-ray energies.

What you label by "n=1" is a matter of taste in a sense similar to "ground state". If you think of one electron moving around a cloud of other electrons + nucleus then you are essentially solving a modified hydrogen problem. So it would be consequent to call the ground state of this one-electron system "n=1". But if you think of the cloud of inner electrons as building up "shells", especially when obeying Pauli's principle, then it's consequent to enumerate this state as "n>1".

In the end you must not forget that the partitioning into single-electron wave functions is a consequence of an approximation. In reality electrons are indistinguishable and have a common wave function where you can't say which is outermost and which is innermost.

11. Oct 15, 2007

'In reality'-whose reality? Yours, mine or quantum mechanics? No offense...

My underlying purpose, for this thread, is to guage the limits of quantum theory. I actually follow a more classical route. Since I have been entrenched in that realm so long, it is difficult to dissociate from it. What I want to know is how good quantum theory is at doing what it does. What it actually accomplishes, and what are its limitations. This is difficult because QT presents itself in a premtory manner.

From my perspective, Quantum mechanics is and will always be a sophisticated, time consuming probability theory. I'm not saying that it is not valid, I just want to know its limitations.

I want to thank all of you for engaging in this thread. You have given me several points of reference from which I can continue my investigation. Thanks.

12. Oct 15, 2007

### vanesch

Staff Emeritus
It's difficult to gauge the limits of a theory when one refuses to consider how the theory works. Think about an Aristotelian gauging the limits of Newtonian gravity, and refuses to consider such concepts as inertia and force...

What people here have been trying to tell, is that the way quantum mechanics deals with an atom, is as follows:
you have a wavefunction that is function of the positions of the N electrons (and their spins), and that wavefunction is going to describe the quantum state of the atom. As such there is ONE state describing the entire CLOUD of electrons, and there are a priori no different individual states for each individual electron. But one can make an approximation, and one can suppose that this total wavefunction is built up by individual states of individual electrons. That's not what quantum mechanics tells us, that's an extra constraint that we put in by hand. There's a mathematical rule that helps us go from N individual states to one global N-state. In fact there are two: a simplistic one (just make the product), and a more correct one, the so-called Slater determinant. You feed in the N different states, and out comes a single N-state.
THIS is what is implicitly assumed when you talk about the "outer electron" and the "electron orbitals" and all that. Most "intuitive" chemistry is based on that. And when you do that, the problem becomes numerically tractable, but makes some errors. One calls this approach: the Hartree-Fock approximation (the simplistic approximation is the Hartree approximation). It gives moderately good results, but with some observable deviations from experiment. It is most of the time good enough to explain the periodic table for instance.
But as we said, we put this specific form of the wavefunction in by hand. Quantum mechanics doesn't impose this, we do, for sake of computational comfort (and probably also because of the link to intuitive chemistry). But we can take this result now as a starting point to do better, full quantum calculations. The problem is that these are numerically very difficult, but for simple atoms, they give very good results.
You can find this in a few books on quantum chemistry. The main calculational trick remains on how to pick the "parametrized function space" on which to let the computer run the quantum mechanical computation.

Googling some stuff came up with this:
http://arxiv.org/ftp/physics/papers/0509/0509135.pdf

13. Oct 16, 2007

### Staff: Mentor

Just to make this more explicit, with helium for example the two electrons are described, strictly speaking, by a single wave function $\Psi(x_1, y_1, z_1, x_2, y_2, z_2, t)$, which gives the joint probability amplitude for finding one electron at $(x_1, y_1, z_1)$ and the other at $(x_2, y_2, z_2)$ at time t. For atoms with more electrons we still have a single wave function, but with more position variables.

To make the solution tractable, we assume that this can be approximated by separating it into individual wave functions for each electron.

14. Oct 16, 2007

### OOO

Since Sean Torrebadel wants to explore the limits of quantum theory: One of my professors once told me (it was ages ago so forgive me if the following is not an issue anymore) that there is no method to calculate natural linewidths of electronic transitions in heavier atoms (how heavy ?).

Is this still a problem at present ? Do the computations grow ever more precise ? Can they even be computed with satisfactory precision ? How does one tackle the computation of natural linewidths at all ? I guess it must be something like: natural linewidth <-> spontaneous emission <-> interaction with the vaccuum fluctuations <-> quantum field theoretic calculations...

Sorry if this question sounds a bit stupid, but I used to avoid atomic spectroscopy wherever I could ...

Last edited: Oct 16, 2007
15. Oct 16, 2007

I'm not refusing to consider this theory, I am considering this theory. There is a difference between reading a text and gathering an overall image like the one you have just provided. And I appreciate that. I also appreciate the paper you reference, wherein the introduction is quite illuminating.

Look I'm not here to question the validity of quantum theory. I am here to see how well it does what it does- given that it is a probability/statistically structured doctrine... I was particularly interested in how and to what accuracy it was able to calculate the ionization energy of multi electron systems. I am also interesting in how long it takes to make such determinations, how many components there are in such a calculation and how many empirical correlations are required to adjust for accuracy.

The underlying reason for this line of questioning is simply a new theory which makes these determinations from the spectra of each atom and ion within minutes and with considerable accuracy- within .02% accuracy It is a personal theory under peer review and I cannot include it here/forum rules.

16. Oct 16, 2007

### quetzalcoatl9

are you telling me that you are proposing a "new theory" of electronic spectroscopy and at the same time that you are unfamiliar with quantum mechanics (as is also apparent from your other post https://www.physicsforums.com/showthread.php?p=1463172), and that such a theory is under peer review?

in the other thread you claim that you don't accept that a photon has no rest mass? i'm getting worried here...

17. Oct 16, 2007

### OOO

Yeah, I've got my personal theory too. It keeps you economically independent... :rofl:

18. Oct 16, 2007

The Achilles Heel of QT:

First, I am not entirely unfamiliar with quantum mechanics. Do I need to look at it in more depth-yes. What I have attempted to do here is to zero in on some key areas.
Second, every crack-pot has a theory... I know that, I've seen the same things on-line as most people have. We, however, are discussing quantum theory in its current state.
Third, I believe that peer review is the best way to have an idea or theory judged-is it not?
Lastly, and this is for OOO, I should think that our respective pursuits are humble, rather than self indulgent attempts at vanity. I seek what most people who come here seek, which is help.

Look, I'm outside the box, looking in. I don't like the idea of having to include some 1600 terms in order to make an accurate calculation of this or that atom's (valence electron's) groundstate... When I make an attempt to calculate the ionization energy of a multi electron atom in any state, using my theory, it is from the wavelengths themselves, rather than from a fundamental route. Those wavelengths are precise, ordered, and bound by a quantum logic. I believe that these spectra can be used as evidence for the structure of the atoms. I find it difficult, therefore, to accept that the picture of the atom, in Heisenberg vision, is statistically structured. The precise nature of the wavelengths suggest, that there is an underlying structural logic to the atoms that we have chosen to blur with electron clouds. While I am forced to accept that quantum theory is a successful method of approximating an electron's behaviour, I do not accept the atomic picture it presents.

Let's go back to 1913. Bohr had just reproduced Balmer's, or is it the Rydberg equation using an electrostatic model of the Hydrogen atom... Bohr only found success for single electron systems. I ask this question. What would have happened, which direction would science have turned if someone-at that time- found a way to reproduce the spectra of a multi electron system using a modified Bohr theory? A different route, a different direction, a different science? The intellectual gap between you and I...

I'll accept that probability theories have their place, but I refuse to accept that the foundation of my structural being is structured with such uncertainty.

19. Oct 17, 2007

### vanesch

Staff Emeritus
Well, first of all, if you have a personal theory, there's a place for exactly that on PF, it is called the Independent Research forum, where we tried to set up a discussion forum for personal theories. All the rules of scientific research have to be respected (which eliminates lunatic crackpot theories) but you can be outside of the mainstream as much as you like, and nothing has to have appeared in peer review (which is the rule elsewhere on PF).

So if you really have a formalism that can spit out the spectral lines of say, a hydrogen molecule, or a Lithium atom or the like with high accuracy, from first principles, then that is very welcome there (there's a cue "waiting for validation", and the cue is long because of the huge amount of totally bezerk crackpot theories we have to wade through, so it can take some time before validation).

20. Oct 17, 2007

### OOO

Sometimes a little fun can be helpful too. But since neither I nor anybody else here has seen something of your work, we cannot really appreciate your achievements.

In my opinion what you say is reasonable in principle. If one gets stuck it's sometimes good to go back a little. But in the case of quantum mechanics this is extraordinarily difficult since so much has been done since 1913. If you fiddle with the basement then in most cases the whole building will collapse.

Thus, I feel it is rather obvious that all serious questions about the foundations of QM/QFT today are about the right interpretation and not the right math.