So you want to go the electron-ion system go into the lowest energy level at once (i.e. without any quantum leaps in between)?

I don't think that this is likely or possible at all. Imagine for a start only one electron and one ion. If it's truly an ion, you have to "catch" the electron from infinite distance - in order to do this you need exactly the ionisation energy, which is provided by the Coulomb force between the two particles. In this case you will have a neutral atom at the ground state as a result only if both particles were at rest.

As you cannot have particles with a relative velocity below zero and you will gain additional energy from the electromagnetic attraction the direct transition

I* + e- --> Aneutral

seems very unlikely to me. I guess you will have some intermittend quantum states in between, but if you wait some time the excited atom will reach the ground state for sure, according to the minimum energy principle.

You will have a number of possible channels for recombination. But the probability of going directly to the ground state is very low close to zero in most cases. But you will get radiative transitions to lower lying states and provided that there does not exist any metastabile states you will eventually get most in the groundstate. So the time is important too. But the general answer to your question will be no.

the electron-ion recombinations are  dominant events during various types of plasma formation ( originated by electric discharge or induced by laser) that accompany background radiation in the form of continum (due to Brestrulung radiation) and after a time delay, the characteristic emissions of  species in the plasma. In fact, the electron kinetic energy finally converts to radiation. Conersely, the plasma tempreature may be assessed  using Boltsmann relation based on the relative amplitudes of characteristic emissions.

On the other hand,  the species in their exited stetes  finally  decay to the Ground state via collisional or radiative deactivation processes. According to uncertainty principle the excited state with lifetime T deactivates to the lower state. There is no residual excited energy in atom/molecule or ion after its corresponding lifetime.

However, during superelastic reactions, there is a likelihood to promote atom/molecule to higher energy due to electron collissions while the excited energy will be released again as discussed above.

You may view our relevant papers in RG too.

There are exist 3 elementary processes, resulting in electron-ion recombination.

1. 3-body recombination, two electrons collide in the electrical field of the ion, one of the electrons lose its energy and remains attached to the ion, another one leaves the region of the interaction. The rate of 3-body recombination is proportional to ne2so it is effective in dense plasmas. The upper states of the recombined system are most probably populated.

2. Photo recombination. The electron loses its energy in the form of the emitted photon. The most probable channel is to the ground state of the recombined ion.

3. Dielectronic recombination, Free electron loses its energy by the exitation of the ion, the result is doubly excited state. The process is effective in the plasma , which contains multiply charged ions, because high radiative rate is nesessary for the recombined ion stabilization.

The information about the rates of the  particular process is avalable in the internet, there are a lot of literature dealing with electron-ion collisions.

Hi and thanks all very much.

What I am aksing for is exactly about a way to recombine a single ion and a single electron directly into the ground state (no any other quantum levels included). Of course the ionization energy must be released in the process as a photon. Apparently there are many other channels for the other excited levels but I want to escape them.

The velocity of the electron and its trajectory may be well known so the velocity and the trajectory of the ion may be choosen so that they come at almost parallel paths and the ion has the same velocity. So they will be almost at rest to each other.

Afterwards I imagine a few schemes in order to escape formation of the atom in any exited state:

1. To make the recombination in a cavity which allows only energy release greater than ionization energy. I suppose the other levels would than be prohibited.

2. To switch on an appropriete electric field perpendicular to the trajectory so the recombination is achieved at zero velocity (in order to compensate the Coulomb attraction between the ion and the electron)? It seems although that I am here at odds with quantum rule dt.dE>h because dt will be great and energy jump will be in many small steps but not one as I wish.

3. The ion may have a velocity of escaping (moving away) from the electron and so its kinetic energy is equal to the ionization energy.

4. To slow down the electron and attach the ion to some complex and have this way compensation for the ionization energy.

5. To bring two ions with velocities equal with the electron's velocity at the same time in order to compensate the ionization energy

6. In analogy with the laser  principle to put the ion and the electron in a photon field with the energy = ionization energy in order to stimulate the jump in ground state.

I would be grateful to read your comments if any of this ways is possible and/or to read new suggestions.

I am trying to refer to your proposals of recombination, one by one.

"The velocity of the electron and its trajectory may be well known so the velocity and the trajectory of the ion may be chosen so that they come at almost parallel paths and the ion has the same velocity. So they will be almost at rest to each other."

FORGET ABOUT !  The electron in an atom is a QUANTUM OBJECT, the concept of trajectory DOESN'T EXIST for it, i.e. the velocity and position at each time are not well defined.

Also, a bound state in an atom has NEGATIVE energy. When you bring an electron in the vicinity of an ion, even if its velocity is zero, and therefore its energy is zero, you have to get rid of an amount of energy in order to make its total energy negative.

Let's see your proposals of how to do so:

1.In the cavity that you suggest, if the release of the ionization energy is not possible, the recombination WON'T TAKE PLACE.

2.Which trajectory? An electron at zero velocity has a huge wavelength, i.e. you don't know where it is in the space.

3. Linear momentum conservation won't allow what you say. We agreed that the electron we bring close to the ion has zero velocity, i.e. zero energy, otherwise an even bigger energy than the ionization energy has to be rid of. So, for the beginning the system ion + electron has total linear momentum equal to zero .The ionization energy cannot be taken by the nucleus, because that would give the nucleus a recoil. The formula p^2 = 2ME says that for a given energy E, the bigger is the mass of the ion, the bigger will be the recoil linear momentum. Then for linear momentum conservation a photon has to be emitted.

4. To attach the ion to some net of atoms, the ion may not remain an ion anymore. Electrons from other atoms may begin to rotate around this ion too. The picture becomes more complex. Of course, an electron will be missing from the complex, but the electron of which atom in the complex?

5. The two ions will repel one another and move away from one another, But, I repeat, the electron is a wave. In the middle between the ions the attraction field is zero. Close to one or another of the ions you return to the problem of how to get rid of the ionization energy.

6. Do you want to bombard your ion and electron with photons? Will the two particles remain at rest?

This is some GENERAL remark that can help in case of similar questions:

The ground state of an atom is a STABLE STATE. Without intervention from outside, i.e. supplying the ionization energy or more than it, the ground state remains as such FOREVER.

Now do you know what a UNITARY transformation is? If a system, e.g. the ion and the electron, can be described by a Hamiltonian with REAL eigenvalue, i.e. in simple words, the system has a well-defined energy, it evolves UNITARILY. What is relevant to us is that a unitary transformation is REVERSIBLE.  The system of the free electron and ion can be described by such a Hamiltonian and has a well-defined real energy (the energy of the electron is zero, and if the ion is at rest it only has its internal energy).,

Now, if this system, left alone, could transform into a ground state of the atom without sending a photon to the environment, the REVERSED transformation could happen too: without investing the ionization energy from outside, the ground state could transform into a free electron and an ion.

The concept of UNITARY transformations gives you a simplifying TOOL, so, you won't have to work hard on all sort of versions.

Good luck !

Sofia, Thank you for considering my proposals.

I think there is some misunderstanding. You wrote:

“The electron in an atom is a QUANTUM OBJECT, the concept of trajectory DOESN'T EXIST for it, i.e. the velocity and position at each time are not well defined.”

I dont agree that one can not bring an electron and an ion on parallel paths in the physical sence. The Heisenberg relation dpdx >h means that if the velocity is known to say 0.001 % than the uncertainy in x is about 10^(-30) at least. So in mathematical sense there is no trajectory but physically there is some. Also in a very early paper about alpha decay Mott showed that after first observation of the paricle (first bubble in the ionization chamber) the particle follows a line trajectory (according to QM!) as is experimentally obvious from considering the tracks.

In fact I dont want to bring electron+ion in a certain point but only to cover the probability clouds.

I dont see what is the difference between To achieve parallel paths for the ion and electron from making two protons collide in proton- proton coliders.

Also You wrote:

“Let's see your proposals of how to do so:

1. In the cavity that you suggest, if the release of the ionization energy is not possible, the recombination WON'T TAKE PLACE.”

No, I mean the cavity permits only release of photons with energy greater than ionization energy or equal to the ionization energy. The idea is that the cavity prohibits jumps to the other possible levels but enhences only the jump into the basic level (a photon is released of course).

I will also comment your considerations of the other suggestions based on REVERSIBILITY PRINCIPLE soon.

So you are looking for a way to resonantly recombine an electron with an ion to the ground state of the atom. There might be some studies of recombonation at a storage-ring facility on this. There has been some studies on recombination at CRYRING, University of Stockholm, maybe you could try and see if they have published anything.

Hi Ilian,

I don't understand you. You say,

"if the velocity is known to, say, 0.001% then the uncertainty in x is about 10^(-30) at least."

Your saying that the velocity is known up to 0.001% means that Δv/v = 0.001. I can't use this information in the uncertainty principle, because this principle doesn't work with relative uncertainties as  Δv/v and Δx/x . It works with Δp and Δx .

About the alpha track in a bubble chamber, this is a BAD example. A VERY energetic particle, as the alpha emitted from radio-active nuclei, do INDEED behave as particles and not as waves. The wavelength of a particle is equal to λ = ħ/p = ħ/(2mE)1/2, In the case of an alpha particle emitted with a typical energy of 4MeV, λ ~ 10-13cm, i.e. VERY small,

As to proton-proton colliders, do you know a particle consisting in two protons?

Good luck,

Sofia

Ilian, you asked about conditions whereby a DIRECT  transition from ionized continuum to a ground state is feasible.  Sure, here is one:  find an atom/ion with a potential being a DELTA-function -- and viola! -- you are right on money! (Explanation: DELTA-function potential has always only ONE confined state, which is by default also its ground state.) Of course, I am semi-kidding, such atoms don't exist in nature.

But, to a degree, you can come close to that: noble gas atoms (helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and  radon (Rn)) posses an interesting property of having relatively narrow gap between  their first excited level and the photo-ionization limit. The best in this respect, as one may anticipate, is helium (He) : its first excited quantum level is at about 20 eV, whereas the  photo-ionization limit is about 24 eV (which is by the way, the highest in the world of neutral atoms). There is sort of quantum desert between ground state and 20 eV.  You can get pretty fast from ionized continuum down to the first excited level, after which your electron will have to go into a "free fall". This is one of the reasons why  helium (as well as other "nobles") is so popular in the field of very high harmonics generation, sub-femtosecond and atto-second pulses, etc.

I disagree with Kaplan's answer. Especially noble elements are stable. I return on a fact that I mentioned before. If such a process were possible, i.e. to place an electron on an ion without some photon to take the ionization energy away, the INVERSE process would occur spontaneously. In particular, an atom of a noble element would pass into an ion with the electron eliminated.

Sorry !

Sofia, it is good thing to be so self-assured, but the way I see it you are greatly unfamiliar with the physics of ionized gases, or at least laser physics or contemporary nonlinear optics, and perhaps physics in general -- and greatly unaware of that. Once helium atom is ionized, it becomes a very active object, and is of great use in laser technology (He-Ne laser, as an example), where it is ionized by electron discharge and its excited quantum level provides a pumping channel  for a laser scheme. More recently, it became a work-horse in extreme nonlinear optics for very high harmonics generation and atto-second pulse generation. You are just starting your way in science (I hope, it will continue), so you may want to be aware  that most often arrogance comes hand-in-hand with ignorance. If you think that something is strange, it is always a good idea to ask questions first...

Alexander, how far do you want to go? How far? I simply protest! 

Please reconsider your saying that I am not familiar with physics and just starting, and words as arrogance, etc. If you are willing to talk to me, keep the discussion purely scientific.

Simple conservation laws:

I remember that your problem was to obtain recombination without discarding the excess of energy by photon emission. One suggestion of yours was that this excess be taken by the recoil of the finally obtained atom. Unfortunately, the energy and linear momentum conservation won't allow.

But, assume the contrary and let's write these laws. On the left sides I write the quantities before recombination, and on the right sides, after recombination.

The symbols of the velocities mean here ABSOLUTE values, and the sign minus in the equation 2 signifies that I took the particular case when the two particles move on the same axis in opposite directions. Now, multiply the equation 1 on both sides with (m+M)/2. One gets

            m(m+M)ve2 + M(m+M)Vion2 = (mve - MVion)2

 Reducing similar terms and moving everything to the left side, one gets

           mM(ve2 + Vion2 + 2veVion) = 0 .

Since all the quantities in the left side are positive, the equality is IMPOSSIBLE. Thus, one cannot escape the photon emission.

Good luck 

Sofia, I am not exactly looking for this.

I am looking for better conditions to force electron to make transition to the ground level with  greater possibility than to the band of higher levels.

But nevertheless looking back to reversability approach you refer to

consider the following: The atom can be split to ion and electron in

strong static electric field. Of course the field will loose this amount

of ionizatiion energy but no (at least real one) photon will be involved

in the process. So reversability tells then that it would be possible to

recombine ion+ and e- in an electric field so as no photon would be

emitted but the energy of the field should increase.

Nevertheless I understand that recombination in this way would be very

difficult and I am not very interested in it.

So in fact I think that recombination in a cavity is most appropriete

to diminish the rate of recombination to exited states and increase the

rate to ground state. (As the gap in He between ground level and exitation

band is 20 eV maybe it would be possible to tune the cavity

approprietely).

As it concerns Nevis Francis Mott article on alpha tracks the prove has

nothing to do with the energy of the particles but only with wavefunction

collapse. I propose you read it though I dont know where to find it.

I also dont understand your problem with % for velocity and Heisenberg principle, . Just take 10 m/s. I dont mind if I was not correct to diminish Dx by ten orders.

Ilian, my reply here concerns only reversibility. I am not sure that you understand it properly, so, I want to clarify the issue.

Reversibility means that a process that goes from left to right, in our case  e- + ion+ -> atom0 , can go in inverse direction, i.e. atom0 -> e- + ion+.

If there is some additional ingredient that you HAVE TO introduce in order to help the process to go in the opposite direction, then the process is IRREVERSIBLE.

Nothing can be swept under the carpet. If the process  e- + ion+ -> atom0  were possible, then the energy invested by the strong electric field for splitting the atom into electron and ion, were such an ingredient.

The rest of the issue I leave to you.

(About the issue with the Heisenberg principle, I leave it for tomorrow. Now it is midnight in my country, and "tomorrow is another day").

Best wishes,

Sofia 

Alexander, thank you very much for pointing out to the Helium (by the way

it was my favorite priory but for much other reason).

Unfortunately I am not sure what you are suggesting about the problem for recombination to ground state without any transition to the exitation band. Helium looks very promissing with the gap of 20 eV and the 4 eV exitation band. But I dont know what is the probability to make recombination to the ground state

directly. Many members here suggest that the e- will first fall to an

exited state and than fall into ground state,

I would be thankful if you can point some research on the probabilities

of recombination to the different levels of Helium.

I think they might well depend on velocity of e-. In my opinion if an e=

is present for a short interval in the vicinity of the ion dT than it is

more probably to emit more energy dE and hence make a jump to the ground

level instead to the 4 eV exitation band (dE.dT>h). If time of presence

is h/4eV than it seems that such transitions would be prohibited for sure

but a transition to 24 eV is free to happen.

One more remark: I understand that you want very much to avoid the higher levels band. Of course, I don't want to discourage you, but just to call your attention on some fact.

Assume that the electron approaches the ion tangentially, so that when it is the closest to the ion it passes through the ground level region. Now, if the electron velocity is high, it will cross that region and get out from the ion without recombination. To the contrary, if the electron velocity is low, prior to cross the ground level region it will meet the free electron region. As its kinetic energy is low, the attraction of the ion may slow down the electron movement and from now on, the electron will fall, first on the higher levels, then on lower levels, etc., with the appropriate emissions of photons.

So, just take this into consideration.

About the uncertainty principle I remember what it was: the question was if we can get at once a good precision in a particle position and velocity. The owner of the question indicated a relative precision of the velocity, and I called his attention on that the principle involves absolute precisions, i.e. not Δv/v and Δx/x but Δp and Δx.

Since small particles have low mass, and since Δp = mΔv, even if one gets a low Δv, one can get a much lower Δp, s.t. the uncertainty in position ħ/2Δp, will be big.

I wish you very good luck,

Sofia

Dear Ilian,

I see you work very hard. Won't it be a good idea what is the basic problem that you have? Maybe there is a way around, without looking for direct landing to ground level.

Have a good day,

Sofia

As there few people following this question and I think I got a resolution thanks to Alexander Kaplan and Sofia Wechler I decided to publish it as a new question in order to get more opinions. I must make some calculations though.