# Schrodinger equation molecules

1. Aug 12, 2013

### Big-Daddy

How do I write a full Schrodinger equation, pre-approximation, for a mixture? Let's say 75% H2 and 25% He by number of particles.

I already know the form and very basic applications of the Schrodinger equation and the Hamiltonian. What I want to know is, the specifics, such as how to specify the potentials. I'd like to be able to write the full equation. Does this PDF specify the potentials properly: http://www.phys.ubbcluj.ro/~vchis/cursuri/cspm/course2.pdf If so, then what do we have to do to this Hamiltonian to set up an equation that is ready for us to solve in principle?

I'm aware that exactly solving such a thing isn't possible, I just want to see the form.

Last edited: Aug 12, 2013
2. Aug 12, 2013

### Simon Bridge

The pdf shows you the general setup for a system of nuclei and electrons.
That would be the general approach - basically each particle moves in the potential of all the other particles together.

If you had 4H2 and 1He, then you have a mixture of 9 nuclei and 9 electrons ... so you'd need 18 position vectors.

The continuation basically makes assumptions that are fairly valid for molecules, and if you are interested in the electrons.

3. Aug 12, 2013

### Jorriss

I don't think you write the Schrodinger equation for a mixture of 75% H2 and 25% He. In principle, you add up how many electrons and nuclei there are, write the Hamiltonian with a coloumb potential between each pair, then solve. Your particular mixture will correspond to a particular solution I suspect.

4. Aug 16, 2013

### Big-Daddy

Why 18 position vectors? Shouldn't it be 19 (9 nuclei and 10 electrons)?

Jorriss, my issue with that is that it suggests there would be a discrete number of possible mixtures, in other words that a mixture of 80% H2 and 25% He might show exactly identical properties to a mixture of 82% H2 and 28% He. I cannot think this is true.

Going from the full form of the Hamiltonian, how do we then write the full form of the Schrodinger equation such that we can directly solve for the set of possible wave-functions and energy of each? The Hamiltonian in the PDF does not appear to contain any wave-functions directly.

5. Aug 16, 2013

### Jorriss

Assuming the 2nd one is 82/18, they would, I believe, correspond to do different solutions. Those two solutions are not energetically equivalent so why would they show identical properties?

What is wrong with there being a discrete number of mixtures? If one actually solved for the spectrum this way it would give *more* solutions for the mixture than we usually think of because we think of many, many configurations of a mixture as being degenerate.

This is very crude though, this is not how one should properly think of quantum mechanics of many particle systems. It might be more elucidating for you to go study Helium from a QM book and then extend the idea.

If you have H, then you solve the Schrodinger. The PDF, though I did not look at it, likely contains no wave-functions because this probably is completely unsolvable so we dont have any wave-functions.

Last edited: Aug 16, 2013
6. Aug 16, 2013

### Simon Bridge

You can't. That's why we have perturbation theory.
That's why you won't see formal "solve the Schrödinger equation" treatments, for more than a single particle, without some sort of trick to it.

7. Aug 17, 2013

### Big-Daddy

Ok but I'd like to at least see the equation in its full form, so that I can then understand the approximations that come from it on my own time. The inability to actually get the solutions out of the equation is not what I'm worried about for now, though I will shortly consider that as well; I just want a full picture of what we're starting with.

My question is basically pertaining to writing the equation in a form which could, in principle, be solved if a solution existed. The document presents a convincing Hamiltonian as a function of the coordinates of each particle, but contains no wave-function, nor does it express energy, so I'm unsure how to go from the full Hamiltonian shown on page 3 to the Schrodinger equation form of that Hamiltonian.

8. Aug 17, 2013

### Simon Bridge

The way I remember it, a solution to the full equation does exist - we see it in our experiments.
However, we cannot arrive at it from the full equation, even in principle, because of the form the equation takes.
We do not arrive at predictions by writing down the complete hamiltonian (say) and then making approximations to simplify it, what we do is write down the math describing simpler situations and hope that the differences are are small.

Consider: it is possible to write down Newtons laws for three freely gravitating bodies - but no analytic solutions can be found even in principle. We say "the three body problem is insoluble".

We know some solutions must exist, however, because the solar system manages to function. If we want to do astrophysics we take advantage of the vast distances involved to isolate pairs of bodies - knowing that the effects of distant bodies will be small and hoping they will be small enough. The effects of other bodies are included as "perturbations". So we think of planets orbiting stars and moons orbiting planets.

We do something very similar in quantum mechanics - only, in this case, it is the two-body problem that is insoluble ... which is why we use stuff like Hartree-Fock (the single-particle energy levels are worked out, and populated by rules.).

In QFT we used to say that the one-body problem is insoluble, because of the self-interaction. Which is what renormalization is for.

If you write down the full Shrödinger equations for each of N interacting particles, you will end up with N simultaneous 2nd order differential equations to solve. Your reference has an example for you to follow. Enjoy.

9. Aug 17, 2013

### Jorriss

Ok, I think I may see some of the issue now. Once you have the hamiltonian, you solve the Schrodinger equation with that Hamiltonian and in solving the SE you solve for the wavefunctions and energies. This is why you never see those - actually solving it ie getting the wave function and energies, is completely intractable for all but the simplest cases.

10. Aug 19, 2013

### Big-Daddy

To SimonBridge:
I would rather understand exactly how the system is laid out, even in a complicated and intractable case, in general before going on to the approximations.

To Jorriss:
But the Hamiltonian as written in the PDF has no explicit mention of the wave-function, only of coordinates for each particle. That's why I would like to understand how one goes from the Hamiltonian in this coordinate form to the equation with wave-function and energy terms.

I appreciate that this is not the stage at which approximations begin - they come earlier - but I would like to understand this level in principle before looking at approximations that are in practice necessary for most systems.

11. Aug 19, 2013

### Jorriss

Have you taken a course on differential equations? Is an expression such as Lf = g, where L is a linear operator and f, g are functions familiar?

12. Aug 20, 2013

### Simon Bridge

@Big Daddy:
You appear to be laboring under some misunderstandings about the Schrodinger equation - at a fundamental level.

The wavefunction and the hamiltonian are different things.
Specifically, the hamiltonian does not contain in any way the wavefunction.

If the hamiltonian is H, then the Schrodinger equations is $\text{H}\Psi = i\hbar\frac{\partial}{\partial t}\Psi$
The wavefunctions $\Psi$ are the functions that make that expression true.

Last edited: Aug 20, 2013
13. Aug 20, 2013

### kith

Your Hamiltonian depends on N electrons and M nuclei. So the (time-independent) Schrödinger equation looks like this:
$$\text{H} \psi(\textbf{r}_1,...,\textbf{r}_N,\textbf{R}_1,...,\textbf{R}_M) = E \psi(\textbf{r}_1,...,\textbf{r}_N,\textbf{R}_1,...,\textbf{R}_M)$$

Last edited: Aug 20, 2013
14. Aug 20, 2013

### Big-Daddy

No. I have studied some differential equations but I do not know what is meant by a linear operator. Nor is it obvious to me how to go from this Hamiltonian - which I can understand - to the Schrodinger equation from which one will potentially calculate the wave-functions and corresponding energies.

15. Aug 20, 2013

### Simon Bridge

There's your problem, you are missing a chunk of math.

Have you seen how to use the Schrodinger equation for a single particle?

16. Aug 20, 2013

### Jorriss

You understand then that the Schrodinger equation is a differential equation? And the wavefunctions are the solutions? How then could the Hamiltonian depend on the wavefunction?

17. Aug 21, 2013

### kith

I have posted the correct equation directly above this comment. What part of it is unclear?

18. Aug 21, 2013

### Big-Daddy

Yes. I have seen the form of the time-independent equation for the 1-electron, 1-nucleus case.

I think I understand your two questions. But I don't know what you mean by the Hamiltonian depending on the wave-function. http://users.aber.ac.uk/ruw/teach/237/hatom.php shows the full form of the equation for a hydrogen atom in polar coordinates (I'm fine with Cartesian) directly under "Separating the radial from the angular part".

So then, do we merely need to multiply every term in the Laplacian by ψ (e.g. replace every ∂2/dx2 with ∂2ψ/dx2, and same for y and z, for Cartesian coordinates) and then simply multiply the other terms in the Hamiltonian by ψ as well. And then equate that to energy * ψ and (in base principle) solve for all possible ψ that satisfy the equation and the corresponding energies?

How come the Hamiltonian in the PDF is missing the reduced Planck constant / 2μ, which is present in the hydrogen atom's Hamiltonian according to the website I just linked?

The part which to me looks like the Hamiltonian is multiplied by the wave-function, which is still confusing me and I'm not sure about (check my above post where I try to explain what I'm thinking). But in the case of energy, E*ψ is a correct interpretation of the RHS?

19. Aug 21, 2013

### kith

Yes, that's it. (Although you mistakenly wrote "Laplacian" in your first sentence)

It is written in atomic units.

What do you mean by "in the case of energy"? The RHS of the (time-independent) Schrödinger equation always contains the energy.

20. Aug 21, 2013

### Big-Daddy

So I have to rewrite the ∂2/∂n2 (where n can be x, y, z, or something else depending on the coordinate system - and will appear, individually, for the 3 coordinates of each particle, across the summation) as ∂2(ψ(coordinates of all particles)))/∂n2 and then multiply the latter 3 terms by ψ as well, then solve - would I need to convert it back from atomic units in order to solve, say, for the 1-electron system if I wanted to? Or, keeping it in atomic units, would I simply end up with an analogous solution in terms of atomic units?

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