Positronium, Hydrogen, Schrodinger Equation

[andrewr]

I have been studying QM and am interested in keeping to the non-relativistic theory for now.

In many experiments of crossed gamma rays in the vicinity of a massive particle (or nucleus), pair generation of an electron and its anti-particle the positron is well described.
In some instances, the electron generated will begin forming an orbital with the positron as a temporary nucleus/atom analogy, but in a short fraction of a second annihilation happens.

There are clearly spin states that cause the lifetime to be extended, as well as the possibility of higher energy orbital transitions delaying the decay. So that normal QM orbitals that one would expect in hydrogen, albeit with different energies than hydrogen, appear to be possible.

In basic QM theory, using the NR Schrodinger equation, usually assumes a massive hydrogen nucleus at the origin -- and the 1S orbitals (ground state) can be worked out easily. In some problems I have observed, the reduced mass is used to aid resolving finer details but having a reduced mass does not in and of itself prevent the diffusion of the electron around the nucleus in accordance with HUP. Sometimes I hear it said that that HUP prevents the electron from "falling" into the nucleus;

Without appealing to the subatomic theories, or relativity, how would one know from just basic QM (NO QCD, QFT, etc). that positronium is not protected from collision by HUP. That is, there clearly will be a reduced mass which will be dead center between the identical mass particles, m/2, but is there a (easy) way to identify where the basic Schrodinger Equation breaks down such that decay happens?

eg: is there a momentum measurement that one can make on the positronium pair that would indicate these two oppositely charged particles have velocities too high to be treated by the non-relativistic Schrodinger equation?

I know the velocities in Gold, cause such a problem, but is that true of positronium?

If not, what is the essential difference based on mass -- which is the only difference that I can identify that would affect basic Schrodinger equation solutions -- that would cause these two particles to be able to get close enough (enough of a probability field) that they could collide?

 

[Bill_K]

The electron does not "fall in" to the positron. What happens is that the wavefunctions of the two particles overlap. This is not anything unique to positronium - you have the same situation in a normal atom too. The electron wavefunction in any S state has a nonzero value at the origin, meaning that the electron spends part of its time inside the nucleus. Depending on the particular nuclide you have, a reaction may occur, in which the electron is captured by the nucleus. This is called K-capture.

One could imagine this happening, for example in a hydrogen atom: the proton might capture the electron, turning into a neutron and a neutrino. However it does not happen in this case, since the neutron's mass is greater than the proton's, and the reaction is not energetically allowed. For positronium, the annihilation into photons is energetically allowed, and that's the difference.

 

[andrewr]

The electron does not "fall in" to the positron. What happens is that the wavefunctions of the two particles overlap. This is not anything unique to positronium - you have the same situation in a normal atom too. The electron wavefunction in any S state has a nonzero value at the origin, meaning that the electron spends part of its time inside the nucleus. Depending on the particular nuclide you have, a reaction may occur, in which the electron is captured by the nucleus. This is called K-capture.

One could imagine this happening, for example in a hydrogen atom: the proton might capture the electron, turning into a neutron and a neutrino. However it does not happen in this case, since the neutron's mass is greater than the proton's, and the reaction is not energetically allowed. For positronium, the annihilation into photons is energetically allowed, and that's the difference.

Hi bill!

I understand that, but what I am interested in is the overlapping wavefunction -- is there a way to tell that the reaction is favorable or not from using Schrodinger's equation, and solving for something?

It is exactly the point that both of them have nonzero wavefunction inside the "nucleus" area that interests me.

Could you show me, in terms of wave-functions, how the mass makes the difference in solutions to Schrodinger's equation ? οr is it impossible to deduce the likelihood from Schrodinger's equation and mass(es) alone ?

Thanks for the reply, by the way.

 

[A. Neumaier]

I have been studying QM and am interested in keeping to the non-relativistic theory for now.

In many experiments of crossed gamma rays in the vicinity of a massive particle (or nucleus), pair generation of an electron and its anti-particle the positron is well described.

But nonrelativistic theories don't allow for pair production or annihilation. These are purely relativistic effects.

 

[andrewr]

But nonrelativistic theories don't allow for pair production or annihilation. These are purely relativistic effects.

Sure, and I am not suggesting using Schrodinger's to analyze the actual production of them, but only it's use with respect to the orbitals that the pair can be in after they are made.

I have not seen an in depth example of such orbitals, but they are referenced in literature talking about positronium, so I am fairly certain they can be solved for. They will, definitely, be different than that for hydrogen as the mass of the positron "nucleus" replacement is much smaller than that of a proton.

What I am wondering about is the wave functions; do they produce any evidence that the velocities attained by the electron would be high enough to need relativistic corrections; or is there any evidence that the overlap of the electron wave function with the positron function is much higher than in Hydrogen so that one could (in principle) find hints as to the instability of these orbitals?

Solutions to Schrodinger's, normalized and conjugate style squared, provide a probability of locating the electron in a certain volume of space. In hydrogen, although the probability is quite high at "origin", the volume of this high probability is vanishingly small. As one integrates over volume, the electron is almost never found in the center of the nucleus -- but reaches reasonable percentages only a short distance outside the expected location of the nucleus. The highest probability being at approximately the Bohr radius for hydrogen.

When looking at the expectation value for the speed of the electron in a volume of space, the velocity would be extremely high in the nucleus volume (a standing wave is a superposition of two traveling waves of opposite directions) -- suggesting that the probability in those volumes is possibly inaccurate using the classical Schrodinger's equation. I was thinking I might try to modify Schrodinger's just a little by replacing the fixed mass used to convert momentum values into energy (h_bar**2/2m multiplier, is equivalent to writing p**2/2m) by a relativistic expression to see how Schrodinger's changed for hydrogen --- eg: if the wave function became smaller/or vanished in the vicinity of the nucleus or not -- But I am not certain such a correction would be valid (I am reasoning by analogy) and so am asking the question non-relativistically to see if anyone has an idea how to determine if an orbital is unstable just using the classical equation.

 

[A. Neumaier]

Sure, and I am not suggesting using Schrodinger's to analyze the actual production of them, but only it's use with respect to the orbitals that the pair can be in after they are made.

In that case, positronium is as stable as the hydrogen atom.

I have not seen an in depth example of such orbitals, but they are referenced in literature talking about positronium, so I am fairly certain they can be solved for. They will, definitely, be different than that for hydrogen as the mass of the positron "nucleus" replacement is much smaller than that of a proton.

The ground state of a charged particle in a Coulomb potential is given by a uniform formula depending only on the reduced mass and the charge. Otherwise everything is identical, also in excited states. Form that you can compute mean velocities and find that they are zero.

 

[andrewr]

In that case, positronium is as stable as the hydrogen atom.

The ground state of a charged particle in a Coulomb potential is given by a uniform formula depending only on the reduced mass and the charge. Otherwise everything is identical, also in excited states. Form that you can compute mean velocities and find that they are zero.

Yes, I see that. Thank you for the thoughtful reply.
S orbitals with no angular momentum are generally standing wave solutions which always return mean velocities of zero; but that doesn't stop the 6S orbital from having "relativistic" corrections in the case of Gold in order to get the correct color to come out.

I'll take your answer as a "no", Schrodinger's itself gives no hint as to the instability.

I considered looking at a relativistic correction by changing the mass of the electron as the KE function of the central potential became sufficiently strong to make it questionable; I haven't worked it through yet, but I suspect it will only worsen the problem by making the electron probability increase in the nucleus area since relativity reduces the effect of electrostatic energy on an electron; so I suppose I'll stop the question at this point, and just work the relativity adjustment through to verify my suspicion..

 

[A. Neumaier]

S orbitals with no angular momentum are generally standing wave solutions which always return mean velocities of zero; but that doesn't stop the 6S orbital from having "relativistic" corrections in the case of Gold in order to get the correct color to come out.

This has nothing to do with relativistic speeds of the electron, but with the fact that the nonrelativistic Coulonb interaction isn't adequate when the interaction energy is more than a tiny fraction of mc^2.

I'll take your answer as a "no", Schrodinger's itself gives no hint as to the instability.

This requires interactions that change particle number. These exist automatically in a fully relativistic treatment.

 

[andrewr]

This has nothing to do with relativistic speeds of the electron, but with the fact that the nonrelativistic Coulonb interaction isn't adequate when the interaction energy is more than a tiny fraction of mc^2.

That's a very interesting point, that I don't understand completely.

Based on wave-number (k), which has an inverse of wavelength, the uncertainty in the location of the electron below that length (roughly 1 wavelength) is indeterminate. The same is true of its velocity. So, I wouldn't think we could say anything about whether the electron was moving or not -- rather, we are stuck with saying such motion is undetectable by measurement through some kind of uncertainty. QM only detects motions where the phase of the wave function goes imaginary -- eg: traveling wave solutions, and these require non-point like volumes of the wave to "move".

In water and other "media", there is nothing stopping the carrier molecules from "diffusing" at high velocities even if the overall wave is made to be a standing wave, where the net motion is oscillating but not traveling in any net direction.

An electron has a rest mass, m₀~=9.10938 x 10^-31Kg.
This mass is understood to be some kind of constant that does not smear out in space uniformly -- because the wavefunction "collapses" to a location (or a local vicinity) whenever the electron is "detected". But the electron can gain mass in relativistic motion.

When you say "speed" of the electron, are you referring only to measurable translational motion -- or do you include oscillation?

An interaction energy greater than a fraction of mc**2, is going to be one where the electron has gained mass -- which would normally be associated with *some* kind of motion in classical physics.
In the gold atom, the S orbitals become smaller in radius -- and the p,d,f, orbitals are non-linearly expanded and contracted depending on how close each part of the orbital is to the nucleus. This tendency causes a relative shift in the total energy needed for each orbital -- and changes the calculation for the color of Gold as a physical effect.

Overall, the trend is that these corrections are on the order of the KE (1/2mv**2) in NR physics (Schrodinger's basis) vs. 1/2mv**2 + 3/8mv**4 + 5/16mv**6 ... which is a relativistic expansion.

The velocity in these two formulas originally came from classical translational motion; Is there another way to calculate this "interaction energy" without referring it back to a velocity calculation historically?

I can understand not wanting to bring the word "speed" of the electron into the discussion, as it is confusing when no one can agree on any kind of model for this "speed". So I would prefer to speak about it in terms of KE and total Energy, just as the Hammiltonian formulation does in the Schrodinger Equation -- and stick to probabilities.

Would I be stating things correctly to say that at higher KE of the Hammiltonian, that the mass of the electron as used in the Hammiltonian needs to be corrected for interaction energy? That such interaction, in accordance with the Lorentz transformation, is going to tend to decrease the phase rate requred to represent that energy, and therefore in S orbitals, is going to tend to increase the spatially instantaneous wave number of the wavefunction nearer to the nucleus?
(shorter wavelength near the nucleus/lorentz contraction).

Or would even this kind of statement be inadequate to qualitatively describe the relativistic effects, causing confusion to readers?

This requires interactions that change particle number. These exist automatically in a fully relativistic treatment.

mmm... Klein Gordan and all are overly complicated for the kind of things I want to do. Basic chemistry works mostly with the lighter elements in any event. I am mostly just wanting to see if there is an easy way to detect when the classical approximation that Schrodinger's uses is going to break down -- so I can at least know when my simple calculations will need to be re-evaluated, and the results significantly inaccurate.

I suppose, I need to calculate "where" in a wavefunction the interaction energy is sufficiently large to invalidate the results. eg: compute some kind of electron volt threshold where the electron's total energy vs. the potential energy of the central nuclear charge causes significant distortions (>1%) in the interaction calculated classically.

I really appreciate your response, and I think if you could just clear up my vocabulary / give me an idea of how close/far I am from making sense, that would be enough for me to move on.

Thanks again.
--Andrew.

 

[The_Duck]

The electron and positron in positronium are basically nonrelativistic. You use the Schrodinger equation to find the ground and excited states, which look the same as for the hydrogen atom, with the only difference being the different reduced mass. Completely ignoring relativity in this way gives you good accuracy in predicting, say, the energy levels of positronium.

However you can't see annihilation in this way; pair annihilation is something that exists only in quantum field theory. Once you develop that, you use the Schrodinger calculation to tell you the probability of the electron and positron being in the same place as an input to a QFT calculation that gives you the decay rate of positronium.

 

[A. Neumaier]

So, I wouldn't think we could say anything about whether the electron was moving or not -- rather, we are stuck with saying such motion is undetectable by measurement through some kind of uncertainty.

It is much better to view the electron as a portion of the quantum electron field - then intuition starts working again properly.

When you say "speed" of the electron, are you referring only to measurable translational motion -- or do you include oscillation?

I only used it since you asked for it. I'd never use it except for electrons in a beam, where the speed is something reasonably well-defined.
Formally, the speed of the electron is p/m, where p is the momentum operator of the electron. This is precise enough to calculate its expectation.

In the gold atom, the S orbitals become smaller in radius -- and the p,d,f, orbitals are non-linearly expanded and contracted depending on how close each part of the orbital is to the nucleus. This tendency causes a relative shift in the total energy needed for each orbital -- and changes the calculation for the color of Gold as a physical effect.

Not in a nonrelativistic calculation. The correct color appears only when one used in place of the Coulomb interaction an interaction with the right relativistic corrections.

Would I be stating things correctly to say that at higher KE of the Hamiltonian, that the mass of the electron as used in the Hamiltonian needs to be corrected for interaction energy?

The mass used is always the rest mass. To see what matters, look at the relativistic correction terms: spin-orbit coupling, spin-spin coupling, and the like. If their presence has a significant effect on the spectrum, a relativistic treatment is needed.

if there is an easy way to detect when the classical approximation that Schrodinger's uses is going to break down -- so I can at least know when my simple calculations will need to be re-evaluated, and the results significantly inaccurate.

For ordinary atoms, iron is usually taken to be the limit. But it depends on how much accuracy you want...

 

[epenguin]

:shy: Can an outsider join this with crude question that may however interest non-physicists?

OK first thought, positronium whilst it exists is analogous to a hydrogen atom and that was all, I never needed to think about it.

Now on second thoughts trying to picture the charge distribution (probability bla bla). In usual H atom pictures, the proton charge is very concentrated and the electron charge is all around it. The usual pics of atoms just show us the electron density only (OK?). But in positronium the two particles are equivalent so they can't very well be both surrounding each other!? So an [STRIKE]un[/STRIKE]semieducated guess is they would look something like a hydrogen molecule with fairly small net charge densities mostly around axis between e^- and e⁺ maxima? Or maybe not because wherever there is most positron there would be most electron attracted to it?? So like one diffuse sphere with no net charge anywhere in the unexcited state, a bit more feature in different excited states? I wonder if or where there is a pic or simulation showing distribution of net charge and the clouds of probability of each particle? I googled quickly and only found these which are not that illuminating and I don't know how authoritative. http://www.blacklightpower.com/FLASH/Positronium.swf http://www.blacklightpower.com/theory/animations.shtml#fundamental (Offhand I don't even know why it depicts the independent particles as disks.)

 

[A. Neumaier]

Now on second thoughts trying to picture the charge distribution (probability bla bla). In usual H atom pictures, the proton charge is very concentrated and the electron charge is all around it. The usual pics of atoms just show us the electron density only (OK?). But in positronium the two particles are equivalent so they can't very well be both surrounding each other!?

In the ground state, both the electron field and the positron field surround their common center of mass in a spherical cloud with fuzzy boundaries. One has a smeared version of planet around sun (hydrogen) versus double star (positronium). In both cases, the real center is the center of mass, but in the first case, this almost agrees with the center of the heavy particle.

 

[alxm]

For ordinary atoms, iron is usually taken to be the limit. But it depends on how much accuracy you want...

People go heavier than iron all the time, with effective core potentials.
I'm doing some on molybdenum at the moment. I'd say 6th row and up has large enough spin-orbit effects to require a full relativistic treatment.

That said, andrewr here isn't actually interested in using existing methods. he's already declared those to be 'old', 'flawed', and with some unspecified 'well-known problems'.

 

[A. Neumaier]

People go heavier than iron all the time, with effective core potentials.
I'm doing some on molybdenum at the moment. I'd say 6th row and up has large enough spin-orbit effects to require a full relativistic treatment.

As I had said, it depends on the accuracy wanted, and hence on the application. Astronomers need accurate spectral data and employ for Fe spectrum calculations the mass correction, Darwin term and spin-orbit correction resulting from the Dirac equation.

 

[epenguin]

In the ground state, both the electron field and the positron field surround their common center of mass in a spherical cloud with fuzzy boundaries. One has a smeared version of planet around sun (hydrogen) versus double star (positronium). In both cases, the real center is the center of mass, but in the first case, this almost agrees with the center of the heavy particle.

Almost? I have a problem seeing why not exactly. Aren't all the electron clouds in the H atom symmetrical around the nucleus (proton)? And why not too in positronium, from the argument I gave?

 

[A. Neumaier]

Almost? I have a problem seeing why not exactly. Aren't all the electron clouds in the H atom symmetrical around the nucleus (proton)? And why not too in positronium, from the argument I gave?

I was referring here to the classical version (where things are immediate) and its quantum operator analogue (where things are less immediate). The center of mass operator that must be separated away to get the reduced 1-particle picture with a radial potential is q = (m_c q_c + m_e q_e)/(m_c+m_e), where c refers to the ''central'' proton or positron. This is never equal to q_c.

 

[andrewr]

It is much better to view the electron as a portion of the quantum electron field - then intuition starts working again properly.

:) intuition would work properly if I understood what you meant by quantum electron field.

I only used it since you asked for it. I'd never use it except for electrons in a beam, where the speed is something reasonably well-defined.
Formally, the speed of the electron is p/m, where p is the momentum operator of the electron. This is precise enough to calculate its expectation.

Yes, I was thinking that relativistic expansion terms are all momentum or speed based. The Hamiltonian indicates that there is kinetic energy of some kind and potential energy. So, any relativistic expansion is going to act like either the mass increases (wavelengths decrease) or like the potential has less effect.

Not in a nonrelativistic calculation. The correct color appears only when one used in place of the Coulomb interaction an interaction with the right relativistic corrections.

I apologize for being unclear. I was talking about a relativistic correction when I made the comment.

The mass used is always the rest mass. To see what matters, look at the relativistic correction terms: spin-orbit coupling, spin-spin coupling, and the like. If their presence has a significant effect on the spectrum, a relativistic treatment is needed.

Yes, but the expansion terms are computing "as if" the mass increases in volumes where the KE of the Hamiltonian is large. Naturally, the expansion uses the rest mass. Without jumping to the Dirac equation, I can do a first order correction for Hydrogen by including the first term of the relativistic expansion like this:

$$m_{0}c^{2}\left[\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}-1\right]\sim=m_{0}\left(0.5v^{2}+\frac{0.375}{c^{2}}v^{4}+...\right)$$

If I pull out only two terms of the expansion, I am able to get the fine structure of Hydrogen reasonably well -- and all I am doing is augmenting the momentum calculation in Schrodinger's (mass * velocity) with a relativistic term.

Unfortunately, the numerical methods I am using for this first try become prohibitive with the third term of the expansion. From what I can tell, Gold requires that third term -- because the KE is high enough that the second term isn't enough. I want to be able to work with tungsten and cobalt, so I am pretty sure that I am going to have to find a different way to calculate that third relativistic term..

For ordinary atoms, iron is usually taken to be the limit. But it depends on how much accuracy you want...

Well, I think I had best just get the basics working first -- and worry about increasing the accuracy later...

I looked at the article you posted. It has some nice results that I might be able to use to compare with when I get to Iron.

Thanks again for the help. I appreciate it.
--Andrew

 

[A. Neumaier]

:) intuition would work properly if I understood what you meant by quantum electron field.

It is the quantum version of a classical field, just as a quantum particle is the quantum version of a classical particle. Thus the intuition about a quantum field is: classical field plus quantum corrections of order O(hbar).

Unfortunately, the numerical methods I am using for this first try become prohibitive with the third term of the expansion. From what I can tell, Gold requires that third term -- because the KE is high enough that the second term isn't enough. I want to be able to work with tungsten and cobalt, so I am pretty sure that I am going to have to find a different way to calculate that third relativistic term..

The Dirac-Fock method might be a good alternative. See, e.g., Phys. Rev. A 49, 1724–1729 (1994)