# Hamiltonian for mixtures

1. Jul 14, 2013

How do I write the non-approximated Schrodinger equation Hamiltonian for a mixture containing 25% by partial pressure of H2 gas and 75% by partial pressure of He gas, at 100 KPa pressure and 298 K?

2. Jul 14, 2013

### Jano L.

If you want exact Schroedinger equation, you have to take all charged point-like particles - nuclei and electrons - as basic constituents. The number of molecules is irrelevant.

In order to write a meaningful Schroedinger equation, you have to have the particles enclosed in some finite volume. You can model this situation by a potential well described by some function $V(x,y,z)$. You can model the molecules as a system of charged particles with total potential energy $U$ given by the Coulomb electrostatic energy. The Hamiltonian is

$$\hat H = \sum_a \frac{p_a^2}{2m_a} + U + \sum_a q_a V(\mathbf r_a).$$

The index $a$ runs over all nuclei and electrons. The masses and charges are arbitrary and you can choose them such that they correspond to your problem.

Then the Schroedinger equation for the function $\psi(\mathbf r_1, \mathbf r_2, ...)$ is

$$\partial_t \psi = \frac{1}{i\hbar} \hat{H} \psi.$$

3. Jul 14, 2013

What does $p_a$ represent, and don't we have to specify the function $V(\mathbf r_a)$?

Shouldn't there be temperature and pressure dependence? Shouldn't there be dependence on how much of each gas is present in the mixture?

4. Jul 14, 2013

### Jorriss

$p_a$ is the momentum of particle a.

Temperature and pressure are statistical mechanical quantities. I suspect, though one may correct me on this, that pressure can still be reasonably interpreted as the negative derivative of the energy with respect to volume?

"How much" of each gas is present will correspond to different solutions. He is more stable than hydrogen and will correspond to lower energy solutions. I don't really know if that is the right way to think about it though - saying the system is 3/4ths He and 1/4th H2 feels like a classical way of thinking about it.

Last edited: Jul 14, 2013
5. Jul 14, 2013

### Jano L.

These are the quantities that restrict the set of applicable wave functions for your situation. But these restrictions can be applied only after the possible wave functions, solving the Schroedinger equation, are found. The equation itself remains unchanged.

6. Jul 15, 2013

I see. So once all of the solutions are found for the equation (each solution corresponding to one wave-function), how do I then apply a temperature/pressure condition or a mixture composition condition (in this case, 25% H2 and 75% He) to find which wave-function will be representative of the sample under those conditions?

I'm guessing all wave-functions which remain when the set of all solutions is restricted firstly by the temperature/pressure condition and secondly by the mixture composition condition will be the possible isomers under those conditions. And of these, whichever isomers correspond to a lower energy wave-functions will be more prevalent than others under those conditions.

7. Jul 16, 2013

And can one go directly from a molecular wave-function to the molecular orbital set for that molecule?

8. Jul 16, 2013

### Jorriss

There's no reason to get molecular orbitals if you have the actual wavefunction. MO's are an approximation on the order of hartree fock theory.

9. Jul 16, 2013

Ok then. And let's say I want to know if the H2 in this mixture (i.e. mixed in with some He) will react with O2 under certain statistical mechanical conditions (i.e. temperature and pressure). How in principle do we write the equations to figure this out? Solving them as you've said is an entirely different matter.

Can we specify a product and get the rate of the reaction we've written? Or, can we get the expected and likeliest product out of the two reactants, if they will indeed react under the given conditions?

10. Jul 16, 2013

### Jorriss

So you're moving out from quantum mechanics then. If you want to know IF they'll react, regardless of the rate, you'd want to compare the free energies. The free energy difference can give the relative proportion of reactants and products.

Rates are a lot more complicated. If you're comfortable starting to discuss approximations I can tell you a few to look into.

11. Jul 17, 2013