Visualisation of H2 molecule wavefuctions needed

[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:

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

 

[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:

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

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?

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


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?