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

[Sean Torrebadel]

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?

 

[quetzalcoatl9]

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?

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.

 

[Sean Torrebadel]

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?

 

[quetzalcoatl9]

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%?

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

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

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.

 

[Gokul43201]

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?

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

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

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.

 

[Sean Torrebadel]

non-relativistic QM breaks down

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).

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

 

[Sean Torrebadel]

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

Okay, for clarity, let's 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?

 

[Sean Torrebadel]

x

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.

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?

 

[quetzalcoatl9]

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

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.

 

[OOO]

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.

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).

Wouldn't the hardest electron to remove be the groundstate?

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.

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?

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.

 

[Sean Torrebadel]

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.

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

My underlying purpose, for this thread, is to gauge 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.

 

[vanesch]

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

 

[jtbell]

you have a wavefunction that is function of the positions of the N electrons

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.

 

[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 vacuum fluctuations <-> quantum field theoretic calculations...

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

 

[Sean Torrebadel]

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

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.

 

[quetzalcoatl9]

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.

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...

 

[OOO]

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.

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

 

[Sean Torrebadel]

The Achilles Heel of QT:

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?

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.

 

[vanesch]

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).

 

[OOO]

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.

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.

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.

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.

 

[quetzalcoatl9]

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 ?).

there is some truth to this. but not just for heavy atoms, getting linewidth is difficult in general because the broadening occurs (largely) due to condensed phase interactions. and it is hard to treat things in the condensed phase. in other words, a gas phase (single molecule) will give you sharp peaks but the bulk substance will have roughly gaussians or lorentzians around those locations.

 

[OOO]

there is some truth to this. but not just for heavy atoms, getting linewidth is difficult in general because the broadening occurs (largely) due to condensed phase interactions. and it is hard to treat things in the condensed phase. in other words, a gas phase (single molecule) will give you sharp peaks but the bulk substance will have roughly gaussians or lorentzians around those locations.

If I remember correctly there are methods to compensate (experimentally) for Doppler and collision broadening. So there should be a way to get natural linewidth from experiment at least for the gas phase.

But then computations of the natural linewidth need not account for the condensed phase. As I have understood my professor, the computation of natrual linewidth from first principles is not possible even in this special case. Maybe I just got him wrong.

 

[Gokul43201]

I think everyone here is going a few steps too far in talking about things like many-body wavefunctions, linewidth calculations, relativistic hamiltonia, etc. There are much more fundamental misconceptions here.

Okay, for clarity, let's say you have a K atom, it has one outer valence electron. Its oxidation state is K+.

First of all, if you have a K-atom, it's oxidation state is 0 (zero). If you ionize it once, it becomes K⁺. If you expose K to say water, then there's a violent reaction resulting in the formation of KOH, in which compound K has an oxidation state of +1. To say that the oxidation state of K is K+ is meaningless.

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?

Ignoring, for the moment, that we should actually be talking about the interaction many-body ground state (and excited states), yes, it would be easier to remove the valence electron from an excited state configuration. This, however, does not change the ionization potential for the atom, which is defined as the energy required to ionize the atom in its ground state configuration. Furthermore, it would have taken some energy in the first place to excite the atom into an excited state. If you include this in your accounting of energy spent, you will still end up with the same number.

Where, as I see it, doesn't an electron in a lower state have more kinetic energy per se, than an outer excited one.

Yes, an electron in a lower state has more kinetic energy than an electron in a higher state. It also has a more negative potential energy (recall or look up the virial theorem), giving it a smaller total energy.

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.??

This is incorrect thinking: transitions between states do not conserve energy (for the atom). So you can not use this to reason that a transition to a higher state must involve a gain of PE at an equal cost to KE. What the electron actually "uses up" in making a transition to a higher state is the energy of the exciting photon.

 

[Gokul43201]

Oops! I didn't notice that there was a second page of posts!

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.

... 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.

I think this has gone beyond asking questions, into denying QM.

 

[Count Iblis]

Why not do some computations yourself? Todays PCs are more powerful than the computers used by many physicsts even ten years ago. Take some atom, write down a trial wavefunction and minimize $$\newcommand{\lhk}[1]{\left | #1\right |} \newcommand{\gem}[1]{\left\langle #1\right\rangle}\newcommand{\brak}[2]{\left\langle #1\vphantom{#2}\right | \!\left.\vphantom{#1}#2 \right\rangle}\newcommand{\braket}[3]{\gem{#1\lhk{\vphantom{#1}#2\vphantom{#3}}#3}}\frac{\braket{\psi}{H}{\psi}}{\brak{\psi}{\psi}}$$

 

[Sean Torrebadel]

I just spent $180.00 on a quantum chemistry text, I'm not trying to deny it- only to come to terms with what it represents. Why? Novice.

 

[reilly]

Sean-- Do not forget that QM is by far the most checked theory in physics; it has yet to come up short. Second; do not forget that the n-body problem (n>3)in classical mechanics cannot be solved analytically; or, why not try to compute the volume of a balloon, initially full of helium, after a pinprick hole is created -- do the trajectory as well. As I'm sure that you really know, almost any field of physics can be brought to a stop by highly difficult and complex problems.

OM, for most of us, is very difficult, and takes a long time of very hard work to get a basic understanding -- like several years . I say this after 2 QM courses as an undergraduate, 4 in graduate school; and after teaching QM 4 or 5 time as a professor, and after using it in my research. Many professional physicists had to swallow their doubts and classical perspective to make any progress in learning QM. You will never be able to understand the guts and limitations of QM without substantial and continuous effort over a least a few years.

I swallowed my classical pride, as an undergraduate, because I knew that QM was the best game in town; still is -- no one has come even close to developing an equal and better alternative. That does not mean that such an alternative does not exist, rather it suggests to me that it will take new, unexpected phenomena to generate progress outside of the "QM box". You can get a sense of how sophisticated atomic physics and how powerful QM have become by doing a Google on "ionization atoms."Finally, in my opinion, you need numerous books on QM to get a solid understanding of QM; should include Dirac's QM book as well as Landau and Lifschitz. A course or two would undoubtedly help.And, yes, in spite of Feynman's dictum, I think that QM can be very understandable -=- as the Nike folks say, "Just do it." Perhaps my next notion is controversial. I'm quite convinced, from experience, that any successful student of QM and classical E&M needs to be highly motivated and driven -- these subjects are just plain hard.
Regards,
Reilly Atkinson

 

[Count Iblis]

Reilly, it's hard only because people have the wrong attitude. Physics and maths in general is seen to be very difficult only because science education is not taken seriously in schools right from primary school onwards.

Compare science education with language education. If we were to teach English the way we teach science, then we would spend a year on the alphabet, spelling etc. By the time students would actually learn to read and write, most of the students would already have left school.

If you don't learn things at an early age it only get's more difficult to laern it. Just imagine being lost somewhere in China and trying to make sense of the Chinese language.

 

[Sean Torrebadel]

Finally, in my opinion, you need numerous books on QM to get a solid understanding of QM; should include Dirac's QM book as well as Landau and Lifschitz. A course or two would undoubtedly help.And, yes, in spite of Feynman's dictum, I think that QM can be very understandable -=- as the Nike folks say, "Just do it."
Regards,
Reilly Atkinson

Would these books still be relevant, ie, has quantum theory changed at all during say the last 30 years or has it only grown? thanks

I'm looking at a book by Frank J. Brockhoff 1976 and an MIT introduction to ... by French and Taylor 1978- and I was worried that they might be out of date.

 

[reilly]

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.

The best of the best tried to explain the structure and spectra of multi-electron atoms in Bohrian terms to no avail. The research along that Bohr-like line pretty much stopped when it became clear that a new and better theory, QM, was being developed. And not too many years after the advent of modern QM, Hartree, Fock, Fermi and Thomas developed a framework to deal with heavy atoms -- see Condon and Shortley, Theory of Atomic Spectra(1935), which was for many years one of the preferred QM texts. And, it is still a superb text on QM as well as on atomic physics. The only real change needed to make the book modern are some change in notation and matters of style -- eg they use dyadics as the basis of multipole expansions rather than the more modern angular momentum --Clebsch Gordon approach. That's really no big deal.

So, yes to a later question, the Dirac and Landau and Lifschitz books give as good discussions of QM as you will ever find -- it certainly doesn't hurt that Dirac was one of the inventors of QM, Landau was a major player from the early days of QM to the beginnings of quarks and all that. Both Dirac and Landau won Nobel Prizes.

Why should Nature not be probabilistic? Nature pays little heed to personal prejudices.

Regards, Reilly Atkinson

 

[lalbatros]

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.

Still you are born because of a totally random egg-activation,
and you had about 50% probability of being structurally a male or a female.

From the rest of you comments on QM, I can also conclude that your training in this field was a bit chaotic ...
... and this influenced considerably (structurally) your way of thinking.

If you just considered all experimental data, your conclusion would be certainly different.
Experimental data would reduce any uncertainty in your understanding of physics.