# The meaning of Ionization

1. Dec 9, 2006

### Quantum River

What is the definition of ionization? For example, in the case of Hydrogen atom, the Coulomb force is the long-range, so the electron and its wave function is always affected by the left proton. So the QUANTUM description is actually difficult and unclear.

Could anyone explain to the the meaning of Ionization from the quantum perspective?

Forget about the Quantum tunneling, it doesn't solve the problem I have mentioned above.

2. Dec 9, 2006

### chroot

Staff Emeritus
What are you talking about? The left proton?!

- Warren

3. Dec 9, 2006

### Quantum River

The electron has escaped away. So only the proton of the Hydrogen atom is left.

4. Dec 10, 2006

### Parlyne

From any perspective, quantum or otherwise, ionization is when the electron gains enough energy that it's no longer bound to the nucleus. If the electron is a distance r away from the rest of the atom, it's potential energy is $$V(r) = -\frac{1}{4 \pi \epsilon_0}\frac{e^2}{r}$$. Ionization has occurred when the electron's total energy is positive, which will mean that it's free to move anywhere without failing to conserve energy. So, it's kinetic energy must be no smaller than $$-V$$.

5. Dec 10, 2006

### marlon

Well, the ionization energy is defined as the energy necessary to remove an electron from the neutral atom.

So why are you talking about protons that "have left" ?
Besides, you are claiming that QM cannot explain ionization ? If so, you really need to start doing some thorough studying, my dear. You might just wanna start with how electronic energylevels in atoms are described by QM. Leave the "tunneling" for what it is and first start with the basic concepts.

marlon

6. Dec 10, 2006

### Quantum River

I didn't claim QM couldn't explain ionization.

My viewpoint is you are explaining the ionization from a classical (physics) point. Could you remove the electron from the neutral atom? The removed electron (for example the center of the wave function is now 10 nm from the original neutral atom) is still affected by the atom from a quantum point, because the wave function of the electron appears in every point of the space and the Universe. Of course you can approximate that distribution of the wave function is concentrated in a narrow place (for example some small ball with a radius of 1 nm) and so the influence of the original atom will not be taken into account. But it is just an approximation and under many circumstances this approximation will not work any more.

I just want to say because of the long-range nature of the coulomb force. The concept of the Ionization is somewhat not exact as you may assume.

7. Dec 10, 2006

### ZapperZ

Staff Emeritus
So can you show me any effects of the gravitational field from Alpha Centauri in the dynamics of what we observe here on Earth? Do you detect it when you drop a ball, or any other experiment?

Now apply this to a plasma. Can you show what observations of the property of such a plasma that would support your arguments? For example, there were two recent laser-plasma wakefield experiment as an accelerating mechanism[1,2]. In which part of these two experiments would the "long range coulomb force" of the ionized atom with its separated electron played any role in the description of the physics?

Zz.

1. W P Leemans et al. Nature Physics v2, p699 (2006).
2. J Faure, et al, Nature v444, p737 (2006).

Zz.

8. Dec 10, 2006

### Quantum River

I have read Faure's paper and it is interesting. The laser field and wakefield both have a electric field>100GV/m, which is comparable with the Coulomb field when the electron has a distance of one Bohr radius with the atomic nucleus. So I guess the Coulomb force may play a noticeable role in the process of ionization, although plasma physicists will not taken into account the Coulomb force and the subsequent quantum effects at all.

Once the electron has been removed from the atomic nucleus, the Coulomb force will be relatively small compared with the laser electric field. I guess the nucleus may play a role like the nucleus in the Rutherford alpha scattering. So the energy width (in Faure's experiment, the energy spread is 20%-5%) is partly contributed by the existence of atomic nucleus.

Quantum River

Last edited: Dec 11, 2006
9. Dec 11, 2006

### dextercioby

Neglecting spin effects, the Coulomb "force", better call it "interaction" plays the SINGLE POSSIBLE role in the process of ionization. So instead of "may play", i'd rather say "plays".

Daniel.

10. Dec 11, 2006

### ZapperZ

Staff Emeritus
YOu did not read the same paper, nor did you understand it.

The wakefiled is created NOT by the coulombic attraction between the ionized atom and its separated electron. Read the physics carefully. Rather, the accelerating gradient is created by regions of charges where the is an accumulation of FREE charges, which is called the blowout region of the plasma. You do NOT get that kind of a field gradent between an electron an a positively ionized atom when they are already separated! If you think this separated distance is the Bohr radius, then you haven't understood what they were doing.

Also note that the accelerating gradient that is being quoted is the gradient being felt by the electrons being accelerated, NOT the gradient between the ionized atom and the separated electron. No such physics is ever considered in the wakefield experiment. In fact, the physics involved contains a complete separation of the elctrons and the ions (again, the blowout regime). This is the fundamental mechanism of laser-plasma acceleration.

Zz.

11. Dec 11, 2006

### Quantum River

The wakefiled is created NOT by the coulombic attraction between the ionized atom and its separated electron.
Did I say the wakefield was created by the coulombic attraction.
Read the physics carefully. Rather, the accelerating gradient is created by regions of charges where the is an accumulation of FREE charges, which is called the blowout region of the plasma. You do NOT get that kind of a field gradent between an electron an a positively ionized atom when they are already separated! If you think this separated distance is the Bohr radius, then you haven't understood what they were doing.
I am so luck to think that their separated distance is not the Bohr radius.
Also note that the accelerating gradient that is being quoted is the gradient being felt by the electrons being accelerated, NOT the gradient between the ionized atom and the separated electron. No such physics is ever considered in the wakefield experiment. In fact, the physics involved contains a complete separation of the elctrons and the ions (again, the blowout regime). This is the fundamental mechanism of laser-plasma acceleration.
Actually I have said the Coulomb force plays a noticeable role in the process of ionization and it may still play a role after the ionization because the electron still could collide with the atomic nucleus after the ionization.

12. Dec 11, 2006

### ZapperZ

Staff Emeritus
You implied! You quoted the wakefield gradient and said that it is the field at a distance of Bohr radius. This association is misleading at best, and wrong at worst. The high gradient is created by the large amount of charges that have been segregated in regions that are separated by at least equal to the wavelength of the laser. This is many, many times larger than the Bohr radius.

But this is what TYPE of coulomb force, and why would this be relevant to what you insisted in the OP? The existence of this accelerating gradient is simply due to the segregation of those charges. The PHYSICS that describes them have nothing in it that contains a coupling, even weakly, between the electrons and the ions, which you insisted must still be there. Yet, you latched on to this example which I brought up as if it justifies what you are arguing. This is incorrect.

I brought up this example because this is exactly the type of area that I work in, except that I generate wakefields in dielectric structures. So I deal with the physics and experimental measurements of such things daily. When I hear people make claims of something that simply do not jive with a whole field of study, then something has been seriously overlooked. You cannot make unjustified claim without showing valid experimental evidence, which I have requested that you do and haven't provide. I, on the other hand, can provide a few more (photoionization, secondary electron emission, etc.) to counter your claim.

Zz.

13. Dec 11, 2006

### Jheriko

Please calm down guys, all the red text and quotes hurt my eyes.

Has anybody answered the original question? I don't see any difficulty in interpreting the original post, it seems like a simple enough question to me. I don't see that he is claiming that quantum physics fails to explain ionisation, or that there are any significant effects...

I am interested if by seperating a proton and electron it is possible to stop their wave functions from overlapping... if they must always overlap then, presumably, there must always be some non-zero probability that an ionised atom is actually neutral, or is going to become neutral very soon.

So, how do we define ionisation in quantum terms? Must it necessarily be fuzzy like so many other quantities? Can we decide exactly when an electron is free and when it is not? Can we label atoms as exactly ionised without introducing some error (maybe really small)?

14. Dec 11, 2006

### ZapperZ

Staff Emeritus
But I can already verify this by assuming the negative. IF we can't remove the overlap, then we will never have such thing as free charges. Then all of what we have assumed to be free charges are really not free. If that is the case, then there will be some "new" physics that we should be seeing that isn't part of physics that we're using to describe the system. Where are they?

That is what I have been asking when I cited the plasma wakefields. The physics describing such a system explicitly assumes that the electrons and the ions are not coupled other than "globs of electrons" and "globs of ions" producing E-field. Another such example would be simply the photoelectric effect. When the electron IS coupled to the ions, you get the work function. But when it overcomes that, there are no interaction of any kind with the original solid and you get free electrons. To argue if we still any "overlap", to me, is rather puzzling when the description that we have of the photoelectron as free particles unthetered to the original atom/solid works! The same can be said with photoionization phenomenon.

This is what I've been asking. Is there any observation to indicate otherwise?

Zz.

15. Dec 11, 2006

### reilly

Quantum River -- First, ionization has been around for a very long time. In other words it is a well accepted phenomena, and, indeed, a well understood phenomena. check out a chemistry book or two, check out biological work -- free radicals and all that -- ions are big trouble in air pollution.

Second -- review your knowledge of the H atom. There are two types of solutions. One type describes the well known bound states. The other describes scattering states. So when something disturbs a normal H atom enough for a transition from bound to scattering, then, voila, you've got your basic ionization process. Check out photoelectric and photo-ionization phenomena. Further note that the a basic phenomena in the details of human vision involves some ionization -- photoinonization of rhodopsin by a photon --. Don't forget the ozone layer.

Read up on atomic and molecular physics -- these ionization issues have been settled for a very long time. Again, in QM, ionization is directly associated with transitions from bound to continuous states. If you are worried about distance, study the problem with wave packets.
Regards,
Reilly Atkinson

16. Dec 12, 2006

### lightarrow

If the removed electrons couldn't have the possibility to interact with something else outside the atom, your consideration could have some meaning, but of course, it's not this the case.

17. Dec 12, 2006

### Jheriko

I don't really undeerstand this. For me it just means that the contributions from overlapping are either unobservably small or that overlapping is forbidden. I think I must be misunderstanding you somehow...

If I apply what I think is your logic then there should be new physics on Earth. We ignore the gravitation of distant astronomical objects when making calculations, but if this gravitation really effects us there should be some new physics? We don't remove the 'overlap' in the case of gravitation... it is just neglected since it makes no observable difference to the results of our calculations.

So, my original question still stands: is the overlapping forbidden?

18. Dec 12, 2006

### ZapperZ

Staff Emeritus
At some point, the inclusion of every single possible effects on the description of a system becomes absurd. I'm sure even you would understand that.

Let's say I only include the strongest possible forces in a system. If what I do not include actually does make an effect, then my descrption of the system using what I started with will reveal "new" stuff that I could not account for. This is what I would call "new physics", something that I never accounted for with my description.

For example, if I drop a ball from rest and only describe this using gravitational forces, then the dynamics of the ball should be completely described by the relevant forces. However, if I start doing this very carefully, and my ball has flaps, then I will notice that my description of the system isn't quite right. It didn't fall at the time that I expect it to. Why? Because the formulation that I used didn't take into account air friction in which, for this sytem and for the accuracy that I want, will now rear its effects that I can no longer ignore, which is not part of the description that I had. If I were ignorant about air friction in the first place, then this is "new physics" that I never realized before.

That is what I asked for. If there truly are effects due to such overlap in system where the description of it completely ignored such effects, were are they? In a photoemission experiment, for example, where the resolution of the electron analyzers are getting to be extremely fine, where would the ignoring of any kind of overlap of the photoelectrons with the atoms in the cathode manifest it effects? Where would such effects reveal themselves in plasma physics?

Note that it is not as if we don't know, or can't tell, when the approximation we make isn't adequate. The "free electron gas" model for a conductor, for example, is quite adequate to explain most of the behavior of an ordinary conductor, such as Ohm's Law. However, we also start seeing breakdown of this model under certain conditions, and for strongly-correlated electron systems, the "overlap" not only between electrons and ions of the solid, but also between electron-electron can manifest itself as a noticeable effects. This forces us to reformulate how we describe such a system, and take into account the necessary interactions.

So yes, we do have experience in dealing with situation where our description is inadequate. In fact, that is what physicists do for a living. That is why I asked for where such a thing would occur and be observable already. Simply by pointing out that the mathematical description of coulomb's law is infinite in range is pointless when its presence cannot be detected or no longer relevant beyond a certain range. I asked for examples where such omissions DO make a difference.

Zz.

19. Dec 12, 2006

### reilly

Here we go again. For physics, ionization is not, repeat not, rocket science. I mentioned above, ionization`is and has been understood for a very long time.(How about second semester of a first graduate level course in QM) And, surprise, classical physics does not play a role -- any more than it does in radioactive decay(which can produce ions)

For example, see Chapter 11, section 2.2, p 232 and on, in Mott and Massey's classsic, The Theory of Atomic Collisions, Oxford U. Press, 1933, Yes, 1933. They give a quite thorough discussion of ionization. Also see Linus Pauling's classic General Chemistry, 1947. If I was still teaching, I'd assign each student to find three references on ionization, and write a short paper thereon.

I'd say end of story, except that ionization continued to be an important phenomena in cosmic ray physics, and particle physics prior to WWII, and even through the early 1950s. Why? (There are famous pieces of equipment used then and now that might be related to ionization......)

Those who neglect history are, in physics, very likely to get things wrong.
If there's one thing that bothers me about many posts in this Forum, it is a remarkable ignorance of science history, with elaborate discussions about things that are clearly discussed in the literature. With all due respect, this thread is a good example of what I'm talking about.

Sorry 'bout that.

Regards,
Reilly Atkinson

By the way, just for completeness, an ion is an atom with net non-zero electric charge. Why, every time you breath you take in a few ions -- what kind?

20. Dec 12, 2006

### Jheriko

Sorry about all of the confusion, but this is where I misinterpreted you. By new physics I assumed you meant some new physical law(s), rather than having more accurate starting/boundary conditions, but using the same laws. Personally, I would say both systems have the same physics, since they are described by the same laws and would instead use a less ambiguous term like "new phenomena" to describe the effects of having more detailed starting/boundary conditions.

So I guess that the wave-functions overlap, but the contribution towards any effect are neglible in practice?

I apologise if I am ignorant of the history of Physics, I am mostly self-taught and most of the literature I have access to is online... so you are most probably correct that I am ignorant in that respect.

Still, I felt that the original question from the original post was perfectly valid... regardless as to what the history of a subject is or how well documented it is, that doesn't change the validity of a question, nor the fact that nobody gave a simple answer.

Last edited: Dec 12, 2006