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Why doesn't Hydrogen have the same wave functions as Helium

  1. Feb 8, 2010 #1
    We know that Helium can have basically the same solution as the Hydrogen atom if there is only one electron. You get the same equation with a different factor on the potential and the basic solution is the same as the Hydrogen atom. But I'm trying to figure out why it can't work the other way around.

    If you have two hydrogen atoms side by side and one electron, then treating them as two potential wells you get the well-known pair of ground states, symmetric and antisymmetric, with the symmetric state just a little bit lower in energy than the antisymmetric. You can presumably fill this state with one electron, in which case each hydrogen atom should have half an electron.

    Now add a second electron. Shouldn't each of the two hydrogen atoms now be able to incorporate half of the second electron just the way the helium atom does? I can't see how the equations should be any different. So the two hydrogen atoms would end up sharing two half-electrons each.

    It's true that this solution is not the lowest energy state for the two hydrogen atoms. There is a lower energy state where each of them has exactly one whole electron, which is the familiar solution. (There is an even lower state where the two atoms come together and form a molecule, but in this situation we're sort of assuming for sake of argument that the atoms are kept apart.)

    So my quasi-helium states are not the lowest energy state, but they still seem like they should be a valid state that satisfies the relevant equations. It doesn't seem like this happens in practise because the corresponding spectral lines are not present as far as I know. But I can't see why not.
     
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  3. Feb 8, 2010 #2
    Such a "mixed" state for two hydrogen atoms be pretty unlikely (high energy), since the nuclei are separated, I would think. If the two hydrogen atoms are far apart, the electron state needs to have an energy close to a free electron, in order to travel between the atoms with a significant probbility.

    When you put the atoms closer to each other, then the energy of the mixed state becomes lower, and when the atoms are close enough together it will be responsible for the H_2 molecular bond, right?

    Anyway, surely the equations describing 2 hydrogen atoms are very different from one helium atom, since only in one case can the positive charge be considered to be at one point.

    Torquil
     
  4. Feb 8, 2010 #3
    Maybe I should back up a step. Forget about the helium atom for the time being; let's just establish that a single electron can be shared between two protons that are far apart.

    When you write the wave function for a single hydrogen atom, you get a certain solution for the ground state. This is similar to the finite potential well. There is nothing to stop us from writing down two potential wells, separated by an arbitrary distance, and solving the equation. The solution looks exactly the same, only the shape of the wave function is duplicated in each location. You can put one electron into the wave function, and it is shared between the two potential wells. The energy is exactly the same as if you had only one potential well.

    There is one twist to the situation: the wave function in the second well can be the same or it can have the opposite sign of the first well. These are the symmetric and antisymmetric solutions. The symmetric solution has slightly lower energy and hence lower frequency.

    For a solution of the equation, you can put the electron in either the symmetric or the antisymmetric state. Or, you can put it in a superposition of the two. Because the states have slightly different frequency, the wave will slowly go in and out of phase with each other, causing the probability to shift back and forth from one well to the other.

    Whenever the probability is all in one side, it looks like an ordinary hydrogen atom with a solitary proton some distance away. The energy of the proton is zero, and the energy of the hydrogen atom is -13.6 eV. This energy is a constant whether the electron is at one atom or the other, or whether it is shared between both.

    So when the electron is shared, the energy of each atom is the same, and it is -6.8 eV.
     
  5. Feb 8, 2010 #4
    Thanks, I understand a bit better now what you were saying.

    I looked up the H2 molecule in my molecular quantum mechanics book (Atkins). He writes the following about both calculational methods that are discussed (VB=valence bond method & MO=molecular orbital method):
    So I guess this is something like what you were after?

    Torquil
     
  6. Feb 8, 2010 #5

    SpectraCat

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    I am really not clear what you mean by this

    Ok .. I agree that there are two linear combinations of the atomic *orbitals* as you describe, and each linear combination can accommodate *two* electrons. I don't know what half an electron is however ... that is a nonsensical statement. I think you mean to say that the wavefunction, and therefore the probability density, is distributed equivalently between the two wells, giving an equal probability of finding the electron on either nucleus. You will never be able to measure half an electron

    Not sure what you are after here .. this situation is quite different that you describe. First of all, you are ignoring spin, which you cannot do. Are you talking about a triplet or a singlet state in your example? This matters because only eigenfunctions of the spin operator can qualify as valid wavefunctions for the system you describe.

    Second if the distance between the two wells if sufficiently large that they don't interact (as I think you are trying to establish here), then the symmetric and anti-symmetric combinations of the 1s orbitals will be effectively degenerate.

    Ok .. now I am lost ... what distinction are you trying to draw between two whole electrons and 4 half electrons distributed among the two wells ... the entire idea makes my head hurt.

    Ok, maybe I see what you are getting at. You are talking about the triplet state of two paired H-atoms at a large internuclear separation. The electrons are unpaired, with one in the symmetric and one in the anti-symmetric combination (if they were both in the same "orbital", their spins would have to be paired in the singlet state). This is actually the ground state configuration .. there is a small energy penalty for pairing the spins in the same orbital to form the singlet state.

    I think you will find that you can recreate your "4 half-electron" combination simply by taking the inear combination of the two possible configurations of the two-electron state I described above ... i.e. the first configuration would have the "left" electron in the symmetric state and the "right" electron in the antisymmetric state, and they would be swapped in the second configuration.
     
  7. Feb 8, 2010 #6
    In the hydrogen molecule ion (2 protons plus 1 electron), the 2 protons are separated by 1.06 Angstroms (2 Bohr radii) and the dissociation (binding) energy is about 2.78 eV, according to Pauling and Wilson, Introduction to Quantum Mechanics, page 330. In the helium atom, the 2 protons are 100,000 times closer together.

    Bob S
     
  8. Feb 8, 2010 #7

    SpectraCat

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    Sure, but he explicitly said he wasn't considering bound molecular states, but rather the states of separated atoms. That said, I am not totally clear on the connection the the helium atom either ... I am waiting for clarification on that.
     
  9. Feb 8, 2010 #8
    I'm trying to solve the simplest possible problem with two electrons. I've chosen the problem of two isolated hydrogen atoms. Yes, we could solve them one at a time, which everyone remembers having done in 3rd year or whenever. But I'd like to see how it looks if you solve it as a two-electron problem. Because you should obviously get the same answer.

    So you solve for the wave functions, and you get the symmetric and the antisymmetric solutions. Now it gets a little funny. You can put both electrons in the lowest state, but only if you make their spins opposite. That seems a little arbitrary if the atoms are far apart. But never mind.

    What really gets me is if you take the six-dimensional wave function of the helium atom, spread it out in space by a factor of two on account of the unit charge of hydrogen vs the double charge of helium, and apply it to this problem, it appears to me that it should be a valid solution of the Schroedinger equation for two isolated hydrogen atoms. Each atom has the same 1/r potential, two half-electrons vying for the same territory with the same mutual interaction energy (scaled back by the same factor)...how is this not a correct solution? That's what I'm trying to figure out.
     
  10. Feb 8, 2010 #9
    The two protons will not share the two electrons, or even one, unless they are so close that they will form a hydrogen molecule ion (with one electron), or a neutral molecule (with two electrons), because these two solutions are a lower energy state than two free hydrogen atoms sharing one or two electrons..

    Bob S
     
  11. Feb 8, 2010 #10
    I don't think there's anything controversial about the two free hydrogen atoms sharing one electron. It seems like very standard potential-well stuff. It only really gets weird when I try to apply similar reasoning to the second electron.
     
  12. Feb 9, 2010 #11
    There are the H2 molecule model using the Bohr orbit.

    See Bohr's 1913 molecular model revisited article (PNAS 2005).
    But I think this molecule model is too simple and incomplete.
    Now isn't 1910's - 1920's.
    So this paper should have used more rigorous methods in calculating the Coulomb force and so on, and found better orbits, I think.

    But as an approximation, this result shows a good tendency.
     
    Last edited: Feb 9, 2010
  13. Feb 9, 2010 #12

    SpectraCat

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    It's not arbitrary, it's quantum mechanics. You have stipulated in your example that the two H-atoms are to be considered as a single system, with one wavefunction describing the electrons. That means that the symmetric (S) and anti-symmetric (A) combinations you describe are common orbitals and are governed by the Pauli exclusion principle. So you have to pair the spins if you want both electrons in the same state .. as I said, this costs energy, even at arbitrarily large distances, so the triplet state is lower in energy.

    Ok .. I really have no idea what you are talking about. Your intended reference system of the helium atom is not comparable to the case of two separated H-atoms in any meaningful way. There is a HUGE difference between an electron interacting with ONE nucleus of charge 2, and an electron interacting with TWO nuclei, each of charge one. The first is the +1 helium ion, a simple extension of the one-electron solutions to the H-atom, but with Z=2. The second is the hydrogen molecular ion, a 3-body problem that is formally not possible to solve without an approximation. If you make the usual Born-Oppenheimer approx., then you can solve it using elliptical coordinates as is done in standard QM texts.

    And you keep talking about half-electrons ... as I told you before ... that is nonsensical. There is no such thing as a half electron.
     
  14. Feb 9, 2010 #13
    I can accept if you disagree with where I think the calculation would go, but I don't get it when you say you don't understand what I am trying to do.

    I thought you agreed that for two isolated protons and one electron, the mathematical solution gives you a wave function which is equally shared between the two protons. For the sake of argument, I've been saying that each atom gets half an electron. If this terminology is not the most accurate, it seems to me that it is at least descriptive. If you have a better way to describe what the electron is doing, just apply it to the following paragraph:

    Using my terminology, I would describe what I am trying to solve with the two-electron case as follows: I have two isolated protons and four half-electrons. Each proton gets two half-electrons. The analogy with the helium atom seem pretty obvious; just divide the charge in two everywhere. And I can't see why it wouldn't lead to an exact mathematical solution.
     
  15. Feb 9, 2010 #14

    SpectraCat

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    You are using (fuzzy) logic to rationalize something instead of writing down the math. I understand that you think characterizing equal probability density of a 1-electron wavefunction on two separated nuclei as each atom having "half an electron" is ok ... I am telling you that I think it is not, because it leads to confusion, as you have already noticed.

    However, I think I finally understand what you are saying .. you are not comparing two H-atoms to 1 He atom ... you are comparing two H-atoms to two "half-helium" atoms, which I again think is non-sensical and has led you to your rather odd conclusion.

    In any case, I can perhaps help ease your confusion here. The issue is the 1/|r1-r2| electron-electron interaction term in the potential. As you know, the wavefunctions in this case look like super-positions of ground state 1-electron H-atom wavefunctions (1s orbitals), with essentially no probability density in the space between the atoms. So, the electron-electron interaction term will be *drastically* higher for the case where both electrons are around the same center. In fact, those cases *do* correspond to a helium-atom like configuration, although it is better to call it an H- ion-like configuration, because the nuclear charges is just Z=1.

    In terms of electronic structure calculations, this interaction gives rise to the J (Coulomb) and K (exchange) two-electron integrals in the Hartree-Fock treatment of atomic and molecular energy calculations.

    So anyway, does this answer your question? The states you are talking about do exist in principle, but they correspond to the superposition of the two "separated H- ion plus proton" states, and they are much higher in energy.
     
  16. Feb 9, 2010 #15
    Spectracat, please forgive the apparent sarcasm in what follows. But I'm not quite sure where this ended up.

    So I'm wrong?

    So I'm right?
     
  17. Feb 9, 2010 #16

    SpectraCat

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    EDIT: I didn't say you were wrong, I said you were confused. I think that is accurate.


    I guess you are free to draw whatever conclusions you want ... I put a fair amount of time and effort into analyzing your case an putting it into the *proper* quantum mechanical context. If you already knew the answer, then why did you waste my time?

    EDIT2: FWIW, you are also "wrong" in the sense that your analysis using a helium atom doesn't give the right result for the energies of the superposition states I mentioned in my last post. Applying the Hamiltonian to those states yields half the ground state energy of the H- ion, not half the ground state energy of the helium atom.
     
    Last edited: Feb 9, 2010
  18. Feb 12, 2010 #17
    No, those aren't the states I'm talking about. When you solve the H- ion you get something essentially different from the helium atom because the forces don't scale equally. You have the same interaction force between the electrons but only half the central attraction of the nucleus. You definitely don't "recover" an equivalent to the helium atom or anything else by taking the superposition of this system with a bare proton. That's not how superposition works.

    What I've done is identify a system where all the forces scale exactly the same way, so the solution should be a scaled replica of the helium atom. I don't know why I'm having difficulty making myself clear on this. It is possible that I am incorrect in the calculation, but as far as the physical system which I am analyzing, I don't know how I could describe it any better than I already have.
     
  19. Feb 13, 2010 #18
    In both the helium atom and the hydrogen molecule, the two electrons' orbitals have the symmetrical structure.
    But there are some differences between them, I think.

    The helium has only one nucleus, so both the two electrons of the helium are moving around all over the helium atom.
    If we pay our attention to the "Coulomb force" in the hydrogen molecule, in the space which is closer to the nucleus 1, the Coulomb force from the nucleus 1 is always stronger than the Coulomb force from the nucleus 2. (The electron-electron interaction is smaller than the electron-nucleus interaction, because the two electrons are mainly at the opposite positions of the nucleus.)

    So one electron of the hydrogen molecule mainly belongs to one of the two nuclei rather than always moving around all over the hydrogen molecule (H2).

    Actually, the bonding energy of H2 is only 4.7467 eV, which value is much smaller than the absolute value of the helium ground state energy (-79.005 eV).
    And the internuclear distance of H2 is 0.7414 A(angstrom), which is longer than the Bohr radius. (In the hydrogen molecule ion, it is much longer, as Bob S says in #6.)

    There is a difference in the nuclear movement,too. This motion of H2 is probably more complex than the helium.
    But if we try to consider this effect correctly and compare these energies, even in the helium atom, it is difficult to calculate the energy including the nuclear movement.(See this thread.)

    Because the Schroedinger equation doesn't have the clear orbitals, so it can't judge whether the nucleus is stopping or not. (For example, when the nucleus is just at the center of the two electrons, the nucleus stops, and this effect vanish only at this point.)
     
    Last edited: Feb 13, 2010
  20. Feb 15, 2010 #19
    But my question wasn't about the helium atom or the hydrogen molecule. It was about two separated hydrogen atoms.
     
  21. Feb 15, 2010 #20
    Was my quote from the Atkins book in post #4 relevant for your problem? You didn't comment on it, so I don't know if you missed it or not :smile:

    Torquil
     
  22. Feb 15, 2010 #21

    SpectraCat

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    Ok, you kind of missed my point, so perhaps I didn't explain it very well. I'll try again.

    In the case of two non-interacting H-atoms, as we are considering here, there are a total of six ways of placing the electrons into the ground state atomic orbitals (i.e. no electrons in 2s or higher). There are four degenerate cases corresponding to the cases where each nucleus has 1 electron: up-up, up-down, down-up and down-down, where up and down refer to the orientation of the electron spins on some space fixed axis. The energy for each of these states is -27.2 eV (2 x -13.6 eV, neglecting fine structure). The other two cases are the ion-pair states I described previously, where both electrons are around a single nucleus. The energy for those cases is just the energy of the free hydride ion, or about -14.36 eV (estimating from electron affinity of H-atom of 73 kJ/mol).

    So you see, the ion pair states are *way* higher in energy. In the current model, those are the only states that have an electron-electron repulsion term in them. So, since helium has electron-electron repulsion, you will have to build those states into whatever linear combination you might take to get your model.

    Anyway, this is the concept I was trying to get across before. If we are going to think about your example in terms of concepts we recognize from simpler systems, the we are pretty much forced to break it down in these terms. I suppose that you could build an ansatz around the helium atom instead of hydrogen, but it is easy to see that it will have problems describing some common systems. Take for example an isolated H-atom ... how could your model cope with that? Even if we assume it makes sense to talk about half-electrons, it is not consistent with any experimental measurements to think of hydrogen consisting of two half-electrons orbiting a proton. For one thing, it implies that there is a non-zero contribution from electron-electron repulsion in a one-electron system! So, even if you were able to make the math work out somehow, I don't see your model as being at all intuitive, which means it would have little predictive or even pedagogical value.

    But perhaps I am wrong .. I would be interested to see if there is some way you can cook this all together to get the right energy for your two H-atom test case.
     
  23. Feb 15, 2010 #22
    Torquil, I'm terribly sorry I didn't respond to your original post but honestly it deals with a different problem than the one I was asking about. SpectraCat was closest to dealing with my actual question so I got involved in a discussion with him. He's just come back with some interesting points which I'm going to try to respond to now.

    For starters, I have to commend you for coming up with the correct value for the energy of the H- ion, apparently by reasoning from chemical thermodynamics. I found the same value last week in this paper by A. R. P. Rau:
    www.ias.ac.in/jarch/jaa/17/113-145.pdf

    The other thing I'd like to point out is that when you solve the system of two isolated hydrogen atoms with two protons, you also get novel states that don't occur when you simply solve the hydrogen atoms one at a time. So in that sense we are on the same page.
    How can you be sure that there are not more unexpected solutions?

    Here is where you seem to lay down an unjustified condition: namely, that if I am proposing helium-like states then I must create them as superpositions of the six states you have allowed. This seems pretty arbitrary. In particular, the linear combination which you suggest of

    1/sqrt(2)|H- over here and H+ over there> + 1/sqrt(2)|H+over here and H- over there>

    does not give the same solution as I am proposing: for one thing, the energy of -14.3eV is different from the -19.5 eV which I get for a combination of mini-heliums (based on the helium ground state energy of -78eV).

    You can get some insight as to why these species (hypothetical or not) are different if you look at Rau's paper. In H-, the two electrons really don't occupy the same orbital: the second one is farther from the nucleus. Of course you then symmetrize the function so the electrons are indistinguishable, but there really are in a sense two different wave functions. If I'm undertanding the implications correctly, it means you can symmetrize them or you can choose to antisymmetrize them and presumably this means for the antisymmetric combination, you are allowed to put the electrons in the triplet state. This is very different from helium: one electron does exactly what the other does and you can't antisymmetrize the spatial wave functions because then everything would go to zero. So the total wave function is symmetric and the electrons have to go in the singlet state.


    Well, wasn't that my question to begin with: why do I seem to get new solutions for the four-body system that I didn't get when I analyze it as a pair of isolated two-bodies?

    You make a great deal of my phrase "half-electrons". True, quantum mechanics does not recognize half-electrons, but it places no restrictions on the wave function of an electron being distributed in two widely separated locations. This is the only sense in which I have ever used the phrase "half-electrons".

    But does the math work or doesn't it? If it works, it seems to me there should be consequences.
     
  24. Feb 15, 2010 #23
    Let's talk about Spectracat's superposition of states, quoting my description from the previous post:

    1/sqrt(2)|H- over here and H+ over there> + 1/sqrt(2)|H+ over here and H- over there>

    I hope this is clear enough. Let's recap: I suggested solving two isolated protons as a four-body system. The expected hydrogen atom solutions come up, and also apparently this variation on the hydrogen negative ion. SpectraCat is doubtful that my "mini-heliums" are a valid solution.

    Looking at the wave function above, we can see that you can just as well put a minus sign and get an antisymmetric version. Then, using the familiar procedure, you can combine the symmetric and antisymmetric to return the traditional, location based representation: a negative ion here and a bare proton there, and vice versa.

    What I'm going to suggest is that the mixed function above is awfully close to the first excited state of Helium. In other words, an exact replica of the 1s2p state spread out in space. The energy of the hydrogen negative ion comes out quite close to the expected one quarter the value for helium (-14.5eV based on helium, -14.36 eV for the actual H- ion).

    So the question is: if this procedure works, why can't we get an even lower value (more stable) for the H- ion by using my ground-state helium model? I think the answer may be that parity screw you up. Notice that for the excited state of helium, phi(r1,r2) is a different state from phi(r2,r1). So you can take sums and differences and make a new basis. But for the ground state, swapping r2 and r1 returns exactly the same function. So you can't do the same trick with sums and differences.
     
  25. Feb 15, 2010 #24

    SpectraCat

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    Well, it's just a cheap parlor trick (definition of electron affinity), but thanks. :smile: It's hard to find tabulations of gas phase ion energies, but electron affinities are easy to find.

    What "novel" states are you talking about? Unless I misunderstand the question, you only get the six cases I mentioned above (more on that below), if we neglect electronically excited states.

    It's really not arbitrary at all. The stipulation here is that the H-atoms are non-interacting .. therefore their energies are degenerate, and their individual wavefunctions are just the ground state H-atom wavefunctions (1s-orbitals). The added wrinkle of electron spin produces the four possibilities I listed above. The only other choices for putting 2 electrons around the nuclei are the ion pair states (again, neglecting electronically excited states). What else could there be? The H-atom orbitals are a complete set of solutions, so once you have exhausted the ways of populating the 1s orbital, there are no more possibilities without involving the n=2 (or higher) shell.

    Now, once you start to bring the H-atoms closer, and they start interacting, those higher-lying orbitals do start to get mixed in to the solutions (a la configuration interaction), along with the ion-pair states, and things get very messy. You can approximate this variationally, or with perturbation theory, but it isn't a lot of fun ... better to let a computer do it. :wink:

    Yeah, this is just one of those manifestations of the problems I have been indicating you will get with this model. Basically, a He-atom is a crappy model for an H-atom, even given the stipulations you have introduced. It has built in electron-electron repulsion, which may not be appropriate in all situations (such as the one we are considering). Also, even though you are correct that the forces scale properly, what about other considerations, such as the average radius, and the nuclear screening? Those are going to be wildly different from the He-atom case. So, I don't think it is at all clear that you can simply take 25% of the He-atom energy as the correct energy for this system. You would need to solve the Coulomb and exchange integrals for this specific case, and come up with some way to deal with the electron correlation. It's gonna be messy, and worse, that is what one ends up doing for molecules *anyway*, even when we use computational techniques based on expansion in the (simpler) basis of 1-electron atomic orbitals that is typically chosen. So it is hard to see what advantages derive from your treatment ...

    Do you think the situation is any different for the He atom? If you designate a "primary" electron, then the "secondary" electron in He will experience a reduced nuclear charge due to screening effects, so you end up in the same place. The effect is more dramatic in the H- case to be sure, but it is just a question of degree ... the phenomenology is the same in both cases.

    No, this is not correct. The electrons are paired in a singlet state in both cases. The triplet configurations correspond to excited states ... again, they are closer lying in the H- case, but the logic is the same.


    I think it is because you have introduced a bunch of extra terms (e.g. electron-electron interaction) that are not present in the physical system under consideration.

    Fair enough .. it seemed like you were using it in a different context. Just remember thought that it is just as nonsensical to talk about the energy (or any other observable) for "half the wavefunction". Those quantities are defined in terms of integrals over the *entire* wavefunction. That is why I think the "half-electron" idea is a bad one, even in the context you describe above.

    I think that unfortunately it doesn't, for the reasons I have outlined above.
     
  26. Feb 15, 2010 #25

    SpectraCat

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    Yup. That is the second ion-pair configuration I mentioned.

    No, I really don't think so. First of all, the first excited state of helium is 1s12s1. Second, the H- ion states are singlets, and the first excited state of helium is a triplet, so the symmetry is wrong. Also, not sure where you got that 14.5 eV number .. and it is simply wrong to take 25% of a helium result as the correct result for this system, for any state, for the reasons I laid out in the previous post.

    No, this isn't correct. If the wavefunctions are properly antisymmetrized (which they must be in the end), you can never get a different wavefunction back from swapping two elecrons ... you get the same wf back, multiplied by -1.
     
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