# Visualisation of H2 molecule wavefuctions needed

1. Feb 8, 2008

### Ulysees

Here are some examples of wavefunction visualisations:

I need the wavefunctions of the H2 molecule at 0 K, ie what the orbitals look like, whether they change with time, etc. Does anyone know the equations?

Does anyone have any pictures of the H2 bond that show orbitals or provide some other visualisation of the wavefunctions?

2. Feb 8, 2008

### Ulysees

And why do the protons stay at a distance and not run repel each other away? In other words, what makes a bond work?

Because E/M forces involved here are not obvious. It's not point masses, you can't approximate them like this. It's wavefunctions having an effect on each other, right?

3. Feb 8, 2008

### Jakell

Potons tend to stick together due to the strong nuclear force. If you get too many together, the protons' electric charge repels others strong enough to make them leave the nucleus (nuclear fission).

As for the H2 molecule at zero kelvin, it doesn't exist. As you get closer to 0K, the wavefunctions fo the atoms woudl spread out, overlap, and combine forming a Bose-Einstein Condensate. For pictures/descriptions of those combined wavefunctions, check out wikipedia and it's links.

4. Feb 8, 2008

### Ulysees

But I'm talking about the 2 protons in an H2 molecule, not protons inside a nucleus. You're not suggesting nuclear forces apply between atoms in a chemical bond?

5. Feb 8, 2008

### peter0302

The strong nuclear interaction has nothing to do with a chemical bond. A chemical bond occurs in order to put the electrons of all the atoms in the lowest energy state (often by filling all the nuclei's valence shells as in the H2 molecule). The reason the protons don't repel each other is simple- the repulsive coloumb force between them is negated by the coloumb force of the electrons.

6. Feb 9, 2008

### Ulysees

> The reason the protons don't repel each other is simple- the repulsive coloumb force between them is negated by the coloumb force of the electrons.

Coulomb forces occur between static charges. With moving charges, and therefore E/M waves being generated, ie the forces come from the E/M waves not the charges directly as in static models, it is not obvious to me what's keeping protons together. Do you know how wavefunctions "attract" or "repel" each other?

I thought visualisation of wavefunction equations is the only way to begin to understand what is really going on at the fundamental level with E/M equations.

7. Feb 9, 2008

### malawi_glenn

The hydrogen atom has a small dipole moment, and hence weak bounds can be made.

That two protons never stick togheter we know from the fact that there is no He-2 bound.

8. Feb 9, 2008

### lightarrow

So you don't believe there is a (coulombian) force between proton and electron in a single hydrogen atom?
If you can believe it, why is it difficult to believe about a force between that electron and both protons in the H2 molecule?

9. Feb 9, 2008

### malawi_glenn

When moving two hydrogen atoms togheter, the spherical symmetry disapears, since the electron repel. This makes each atom get a small electric dipole moment and hence they can bind.

10. Feb 9, 2008

### lightarrow

This can be what happens during the two atoms attraction, but he asked about an already established bond: "And why do the protons stay at a distance and not run repel each other away? In other words, what makes a bond work?"

Last edited: Feb 9, 2008
11. Feb 9, 2008

### malawi_glenn

the electron(s) shield, one proton only "sees" partial charge of the other proton.

12. Feb 9, 2008

### lightarrow

Sorry but I haven't understood how you would describe (in simple terms as we are doing here) the H-H bond.

13. Feb 9, 2008

### malawi_glenn

simple terms? This is as simple I can be without drawing pictures. and working out the state-functions.

14. Feb 9, 2008

### peter0302

In simple terms, there's two positive charges, and two negative charges, which cancel out, and two electrons in the first valence shell of each nucleus, making a stable neutral molecule.

15. Feb 9, 2008

### Ulysees

If they were all tiny spheres in classical physics, and if we simulate this on a computer, it's counter-intuitive to expect the positive and big ones to stick to a fixed distance from each other. It requires that the negative and small ones stay on or near the vertical plane between the two. But then, what are they rotating about? How does the centripetal force match electromagnetic force?

And alright, I didn't know the force caused by the E-field part of an E-M wave is also called coulomb. Surely it must be of comparable stregnth to the B-field force in a molecule, as the electrons move quite fast.

16. Feb 10, 2008

### peter0302

The electrons' behavior is governed by QM. They "want" to stay in that lowest energy state relative to the protons. But the reason the protons do not repel each other away is simply classical physics - their positive charges are negated by the electrons' negative charges.

The question "why don't the protons in the nucleus repel each other" must be answered by QM but the question "why don't the protons in the molecule repel each other" does not.

17. Feb 10, 2008

### Ulysees

> the reason the protons do not repel each other away is simply classical physics - their positive charges are negated by the electrons' negative charges

If the protons are much further apart than the size of electron orbitals, then the total attraction due to E-fields is zero. Therefore the protons have no reason to stay at a fixed distance if the atoms are far apart.

Since the protons stay at a fixed distance, it is necessary that the electric repulsion of the protons is balanced by the total E/M force due to the waves. And for stable equilibrium, this E/M force must grow stronger if the protons "try" to leave.

A force that grows stronger as the protons try to leave? Not explained with anything presented so far.

There's got to be a full E/M wavefunction model of the covalent bond. Anyone know where to look for it please?

Last edited: Feb 10, 2008
18. Feb 10, 2008

### lightarrow

Infact two hydrogen atoms one meter away from each other, don't feel any attraction
The electrons move in different orbitals (around the two protons and in between them) while the proton's distance is changing, so it's not so difficult to believe that the force could change, at least from a logical viewpoint; the fact this is exactly what happens can't certainly be proved with these simplistic considerations but it can be proved solving the Schrodinger equation.

19. Feb 10, 2008

### Ulysees

You got confused mate. I was replying to this:

> their positive charges are negated by the electrons' negative charges.

That's true at any distance, it is just an incorrect explanation. What's negated is the electric force between the protons, negated by the E/M force coming from accelerating electrons. I'm sure that peter0302 and others can come up with a good explanation in terms of wavefunctions.

Come on guys, you must have studied that. No wavefunction experts?

PS. Schrodinger's equation only says how the wavefunctions evolve, not what the initial conditions are which is what we probably need mainly (I would expect those initial wavefunctions to evolve little or not at all).

Last edited: Feb 10, 2008
20. Feb 11, 2008

### kyuzo

I think you're confusing everyone with your terminology. You keep talking about "E/M", which typically refers to a (classical) electromagnetic field. Non-relativistic quantum mechanics does not treat an electromagnetic field at all, and you don't need relativistic QM to describe the bonding in an H2 molecule. You should probably forget about notions like "E/M forces" and "accelerating elections" in the context of molecular bonding, they're not especially relevant.

I found this image on wikipedia which seems to fit the request in your original post pretty well:

http://en.wikipedia.org/wiki/Image:Dihydrogen-HOMO-phase-3D-balls.png

Of course this doesn't really tell you much about why a covalent bond forms. I suspect what you really want is some sort of movie where the schrodinger equation is used to numerically calculate the time evolution of a system starting with two lone hydrogen atoms in close proximity. I don't know where one would find this sort of thing, but you could definitely theoretically do this, and the result would match the results of molecular orbital theory (or at least they'd better, since MO theory is based on algebraically solving the same equation).

21. Feb 12, 2008

### peter0302

I don't know how what you're saying is different from what I'm saying, except you're saying "E/M force ... accelerating" instead of coloumb force. The two protons feel a repulsive coloumb force between each other. That is negated by the attractive force from the electrons.

Yes, the wavefunction can be calculated. No, I cannot do it. It is pretty complicated when it gets beyond a single hydrogen atom and I do not have a graduate degree in this stuff. That is why, I suspect, you can't find a visualization of it.

My point is that the classical explanation works perfectly well to understand why the protons stay relatively stationary relative to one another. All that really matters in a chemical reaction are the valence electrons. The rest of the atom, i.e. the nucleus and the other electrons, basically mind their business.

Can you get the exact answer with the wave function? Yes. Do you need to? I don't know why you would...

Here. http://www.ptep-online.com/index_files/2007/PP-09-07.PDF [Broken]
Apparently it's even harder than I thought. Figure 1 seems to be what you're looking for but it's obviously a rough sketch.

Last edited by a moderator: May 3, 2017
22. Feb 19, 2008

### per.sundqvist

If you want to plot the $$H_2$$ molecule at T=0K, you could use the approximate 1s solution:

$$\Psi(\vec{r})=c_1e^{-\sqrt{x^2+y^2+(z-a)^2}/a_B}+c_2e^{-\sqrt{x^2+y^2+(z+a)^2}/a_B}$$

where for example the symmetric solution is $$c_1=c_2=1$$. Next let aB=1 and play with the distance between the nucleus =2*a in "units" of aB (Bohr-radii). Note: $$x^2+y^2=\rho^2$$, so you could plot a 2D-plot of $$\Psi(\rho,z)$$.

The 2-electron wave function: $$\Psi(\vec{r}_1,\vec{r}_2)=\Psi_{c_1,c_2}(\vec{r}_1)\Psi_{d_1,d_2}(\vec{r}_2)\pm \Psi_{c_1,c_2}(\vec{r}_2)\Psi_{d_1,d_2}(\vec{r}_1)$$

When you integrate out one electronic coordinate from $$\Psi(\vec{r}_1,\vec{r}_2)^2$$, to get the charge-density, you get more or less the upper expression again.

Per

23. Feb 19, 2008

### Ulysees

That's what I was looking for, thank you sundqvist (is this a central European surname?)

Actually it's even better, it's probability density functions, right? Cause wave functions typically involve spirally rotating complex vectors. But then why do you say that integrating these equations gives the charge density? Did you mean the percentage of charge in a given space?

A spheroidal shape was given earlier, as the orbital. That can't be right from what you've given. What I see in your 1s solution is the distance from each of two points (protons?) being taken as an exponential decay's length, and then you add the two decays together, the result can't be associated with a spheroidal orbital. I wonder if the 2-electron function you mention gives an orbital of that shape.

What are d1 and d2 in the 2-electron function?

Last edited: Feb 19, 2008
24. Feb 19, 2008

### Ulysees

> What are d1 and d2 in the 2-electron function?

I mean why are they assumed to possibly be different from c1, c2, isn't the molecule symmetrical?

25. Feb 19, 2008

### Ulysees

And what does the +/- mean? It can switch between two states? Two types of hydrogen molecule?