Visualisation of H2 molecule wavefuctions needed

[Ulysees]

Here are some examples of wavefunction visualisations:

http://upload.wikimedia.org/wikipedia/commons/c/cf/HAtomOrbitals.png

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?

[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?

[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.

[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?

[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.

[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.

[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.

[lightarrow]

> 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.

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?

[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.

[lightarrow]

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.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?"

[malawi_glenn]

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

[lightarrow]

the electron(s) shield, one proton only "sees" partial charge of the other proton.Sorry but I haven't understood how you would describe (in simple terms as we are doing here) the H-H bond.

[malawi_glenn]

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

[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.

[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.

[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.

[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?

[lightarrow]

> 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.Infact two hydrogen atoms one meter away from each other, don't feel any attraction :smile:

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. 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.

[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).

[kyuzo]

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).

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).

[peter0302]

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 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.


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?

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...

[Edit]
Here. http://www.ptep-online.com/index_files/2007/PP-09-07.PDF
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.

[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

[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?

[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?

[Ulysees]

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)

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

[per.sundqvist]

Hi, for the lowest total spin S=0, \uparrow\downarrow, the total charge density n is symmetric. Let's scale x,y,z with aB, then the charge density is given by:

n=\Big[\exp(-\sqrt{\rho^2+(z-\frac{a}{2})})+\exp(-\sqrt{\rho^2+(z+\frac{a}{2})})\Big]^{2}

where a is the distance between the nucleus. If you apply a strong magnetic field, the electrons will be spin-polarized, so that the total spin will be S=(1/2+1/2)=1, \uparrow\uparrow. In that case you would have a overlap and mixing term (overlap=integral of the two exponentials) in the charge density. The distance "a" has to be optimized in order to minimize the total energy, but you could play with it when you plot it. Also note that n is unnormalized, but it does not matter when you plot it.

By the way I'm from northern Europe.

Best,
Per

[monish]

I think what you've done here is to bring two ordinary hydrogen atoms together
and just written the equation of the overlapping wave functions. In fact, the wave
functions have to become distorted by the mutual proximity of the two atoms. I don't
see this in your solution. So I don't think it shows how the interaction leads to
a bonding force.

[Ulysees]

Hi, for the lowest total spin S=0, \uparrow\downarrow, the total charge density n is symmetric.

Can you explain the implications of total spin S=0 please? Does spin mean rotation around some axis, and no spin mean no rotation? Without rotation one would expect the electrons to fall to the nuclei.

[per.sundqvist]

monish: Yes its a very simple model. The aim is to plot it. But if you do for example an "exact" Hartree-Fock calculation of the problem and then plot it, I doubt that you could see much difference (in a plot) from the solution I have given!

Ulysees: Spin is an intrinsic magnetic moment (\vec{m}=I\cdot A \hat{n} current x Area) originated from an electron. The nature of it is not so simple as "rotation", but is related in nature to charge in rotation. The magnetic moment of the electron, the spin, are matrix operators (Pauli-matrices in 1-electron case), it comes out from the Dirac equation and so on. You could find a lot on it on wikipedia.

Best
Per

[per.sundqvist]

A simple explanation is that the 2 electrons have opposite spins s_1=\uparrow, s_2=\downarrow. Then the Pauli principle says that you could fill 1 orbital (the symmetric one in this example) with 2 electrons. If they have the same spin you have to put them in different orbitals (symmetric and anti-symmetric in H2-case), otherwise the total wave function =0, which is non-physical.

[Ulysees]

Per, are you also missing the force between the protons, it should play a role in determining the optimal distance a.

Also, how do we work out energy from charge distribution/probability density function/wave function? Is it just like in classical physics?

[per.sundqvist]

Well if you give a trial wave function, like the one I have given in previous post, (with c1=c2=1) you could formally get the total energy as:

E=\frac{e^2}{4\pi \epsilon a}+\frac{<\Psi\mid\hat{H}\mid\Psi>}{<\Psi\mid\Psi >}

where the Hamiltonian (a differential operator) is:

\hat{H}=-\frac{\hbar^2}{2m}\left(\nabla_1^2+\nabla_2^2\righ t)-\frac{e^2}{4\pi \epsilon}\sum_{j=1}^{2}\left(\frac{1}{\mid\vec{r}_ j-a\hat{z}/2\mid}+\frac{1}{\mid\vec{r}_j+a\hat{z}/2\mid}\right)
+\frac{e^2}{4\pi \epsilon\mid\vec{r_1}-\vec{r_2}\mid}

[Ulysees]

Well if you give a trial wave function, like the one I have given in previous post, (with c1=c2=1) you could formally get the total energy as:

E=\frac{e^2}{4\pi \epsilon a}+\frac{<\Psi\mid\hat{H}\mid\Psi>}{<\Psi\mid\Psi >}

where the Hamiltonian (a differential operator) is:

\hat{H}=-\frac{\hbar^2}{2m}\left(\nabla_1^2+\nabla_2^2\righ t)-\frac{e^2}{4\pi \epsilon}\sum_{j=1}^{2}\left(\frac{1}{\mid\vec{r}_ j-a\hat{z}/2\mid}+\frac{1}{\mid\vec{r}_j+a\hat{z}/2\mid}\right)
+\frac{e^2}{4\pi \epsilon\mid\vec{r_1}-\vec{r_2}\mid}

How do you do the first "|" here? < psi | H | psi >

Also, previously ri was the variable of a function psi(ri). But now you seem to be giving a formula for energy like this:

energy = E(r1,r2)

Aren't we integrating over r?

[per.sundqvist]

Ok here is the definition:

<\Psi\mid\hat{H}\mid\Psi>=\int d^3r_1\int d^3r_2 \Psi(r_1,r_2)\hat{H}\Psi(r_1,r_2)

The integration is simplified by the fact Psi is a product of independent wave function phi(r1)*phi(r2) etc.

Hope it helps,
Per

[Ulysees]

So it looks like potential energy for each possible pair of charges, infinitesimal or not.

No kinetic energy?

No other form of E/M energy?

[per.sundqvist]

the kinetic energy is T=-\int\Psi\nabla^2\Psi d^3r

Here is a picture of H2 molecule i played with now (just 1 electron though,
but with two protons). I solved it by the package comsol/femlab in 3D:

Per

[per.sundqvist]

Hmm, I missed the hbar^2/2m factor in the kinetic term and it is for 1 particle only, i.e., Psi=Psi(r_1)

\nabla^2\Psi=d^2\Psi/dx^2+d^2\Psi/dy^2+d^2\Psi/dz^2

here is a better picture which is symmeric and without noicy "cut-surface":

/Per

[dr_d_is_cool]

nuclear action happens bewteen all atoms, just not at the scale big enough to cause any effects... and i wasnt aware that a coulomb had anything to do with the H2 molecule

[CPL.Luke]

sorry if this has been addressed before, but part of the reason a covalent bond forms is that the two electrons want to be closer to each other due to a quantum mechanical effect, bosons like to be closer to each other and electrons in certain orbitals become bosons (when their in certain orbitals their spin adds to an integer) and so those electrons can bind in molecules, of course this is only part of how a molecule forms with E&M forces also playing a role.

it begins to get sketchy whenever you're talking about forces in QM as QM doesn't understand the concept of a force, except in the classical limit

[lightarrow]

Ok here is the definition:

<\Psi\mid\hat{H}\mid\Psi>=\int d^3r_1\int d^3r_2 \Psi(r_1,r_2)\hat{H}\Psi(r_1,r_2)


You forgot a "*" in the first Psi.

[XVX]

The quantum mechanical exchange force from the overlap of wavefunctions establishes the covalent bond.

The expectation value of the square of the separation distance between two particles for identical particles is different than distinguishable particles.

Identical particles can have an expectation value greater or less than that of distinguishable particles depending on their symmetry.

Symmetric spatial wavefunction:

p --> 2e <-- p

Antisymmetric spatial wavefunction:

e <--p p-->e

When considering the full wavefunction we have to include spin. Thus, to get covalent bonding, the electrons have to be in the antisymmetric spin singlet state to make the full wavefunction antisymmetric for fermions with the symmetric spatial wavefunction causing the charge density to be greater between the protons.

[per.sundqvist]

"You forgot a "*" in the first Psi."

Yes you are right, but the anzats I gave in a previous post was all real wave functions, so the complex conjugate is not necessary.

About electronic bosons it is formally correct that 2 pair of electrons could be seen as bosons, but is not clear if this is something you in practice really use?

\Psi(\vec{x}_1,\vec{x}_2,\vec{x}_3,\vec{x}_4)=+\Ps i(\vec{x}_3,\vec{x}_4,\vec{x}_1,\vec{x}_2),

Best,
Per

[Ulysees]

> "You forgot a "*" in the first Psi."

Yes you are right, but the anzats I gave in a previous post was all real wave functions, so the complex conjugate is not necessary.
Are you sure that formula is for real wave functions?

I thought we avoided the actual complex wave function (lowercase psi) by using the probability density function instead (uppercase psi = magnitude(psi)^2).

[XVX]

Uppercase psi, which is the full wavefunction, including the spinor, is symmetric when your dealing with bosons.

Cooper pairs are coupled electrons that form a boson state. Covalent bonding is not Cooper pairing but the exchange force.

[Ulysees]

Uppercase psi, which is the full wavefunction, including the spinor, is symmetric when your dealing with bosons.

Cooper pairs are coupled electrons that form a boson state. Covalent bonding is not Cooper pairing but the exchange force.

Any examples of the maths that go with these would be appreciated.

The function that Per as given as uppercase psi looks like probability density function to me (or charge density).

[Ulysees]

Here's Per's function again:

\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}

He said it's the approximate 1s solution. Looks like the sum of 2 independent atoms. Anyone got the exact solution for the molecule?