Schrodinger equation molecules

[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.

 

[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.

 

[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.

 

[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.

 

[Jorriss]

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.

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.

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.

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 don't have any wave-functions.

 

[Simon Bridge]

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?

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.

 

[Big-Daddy]

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.

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.

 

[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.

 

[Jorriss]

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.

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.

 

[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.

 

[Jorriss]

To SimonBridge:
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.

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?

 

[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.

 

[kith]

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.

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

 

[Big-Daddy]

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?

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.

 

[Simon Bridge]

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.

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

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

 

[Jorriss]

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.

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?

 

[kith]

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.

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

 

[Big-Daddy]

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

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

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?

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?

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

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?

 

[kith]

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?

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

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?

It is written in atomic units.

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?

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

 

[Big-Daddy]

It is written in atomic units.

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?

 

[kith]

It doesn't matter which unit system you use. Usually, the calculations are simpler if you use a unit system which takes into account typical magnitudes of physical quantities for your problem. In atomic (and molecular) physics, the atomic units accomplish this.

In the lab, many people use the SI system. So often, someone has to convert in order to compare calculations with experiments.

 

[Big-Daddy]

Ok, the PDF shows on page 1 a couple of changes made to the Hamiltonian to go from the SI unit form to the atomic unit form. Could you outline a few others (for that specific Hamiltonian) just to give me an idea?

 

[kith]

I don't see where your problem is. The pdf is quite self-explaining. For background information on the units, there's the wikipedia article. Maybe you don't know how to construct a Hamiltonian in the first place?

 

[Big-Daddy]

No, I have a basic idea of what a Hamiltonian is. I'm asking about how to convert the units for the Hamiltonian case, if I want to write that general-body Hamiltonian in SI units. I found this PDF also representing a multi-body Hamiltonian: http://www.phys.ubbcluj.ro/~vchis/cursuri/cspm/course2.pdf (equation 7 on page 2), but there are two discrepancies. Firstly there seems to be no operator in this second Hamiltonian for the kinetic energy of the nuclei (the other 4 operators are there); secondly, this second Hamiltonian appears to have randomly placed several (1/2) signs in front of the operators for repulsion between electrons, and between nuclei respectively, which are not present in the first Hamiltonian (the first Hamiltonian being the one in the PDF linked by my OP). Can you explain these differences?

 

[kith]

No, I have a basic idea of what a Hamiltonian is. I'm asking about how to convert the units for the Hamiltonian case, if I want to write that general-body Hamiltonian in SI units.

All terms have the dimension of energy. You have to combine as many of the base quantaties of the atomic units (m_e, hbar, q_e and ε₀) as needed to get this dimension. For example, a kinetic term like ∂²/∂x² has the dimension of 1/[l]². In order to get [E]=[p]²/[m] you need to multiply by hbar²/m_e because [hbar]=[l][p].

But this is more time-consuming than simply constructing the Hamiltonian in SI units which is why I asked if you don't know how to do this.

You included the same pdf as in the OP btw.

 

[Jorriss]

I didn't look at the new link, Big-Daddy, but if there is no kinetic energy term, they are probably just examining the potential or the nuclear terms. If the potentials differ by factors of 2, it's probably a counting thing. Compare the summation bounds in the cases that differ.

 

[Big-Daddy]

All terms have the dimension of energy. You have to combine as many of the base quantaties of the atomic units (m_e, hbar, q_e and ε₀) as needed to get this dimension. For example, a kinetic term like ∂²/∂x² has the dimension of 1/[l]². In order to get [E]=[p]²/[m] you need to multiply by hbar²/m_e because [hbar]=[l][p].

But this is more time-consuming than simply constructing the Hamiltonian in SI units which is why I asked if you don't know how to do this.

You included the same pdf as in the OP btw.

Many apologies. http://www.physics.metu.edu.tr/~hande/teaching/741-lectures/lecture-03.pdf

No, I could not construct the Hamiltonian in SI units that is shown in atomic units on the first PDF. But this (the second PDF) should be a good starting point.

I didn't look at the new link, Big-Daddy, but if there is no kinetic energy term, they are probably just examining the potential or the nuclear terms. If the potentials differ by factors of 2, it's probably a counting thing. Compare the summation bounds in the cases that differ.

They claim it is a molecular Hamiltonian, though? How can they simply leave out kinetic energy of the nuclei after that?

 

[kith]

No, I could not construct the Hamiltonian in SI units that is shown in atomic units on the first PDF.

Why not construct it from scratch? If you know how to construct a Hamiltonian in principle, the molecular Hamiltonian is no rocket science. Just try it and ask specific questions if you get stuck.

They claim it is a molecular Hamiltonian, though? How can they simply leave out kinetic energy of the nuclei after that?

This is explained on the first page of your pdf. Please read the sources you quote from before you ask questions.

 

[Big-Daddy]

Thanks. Then the remaining questions would be, why the scaling factor of (1/2) in the 2nd PDF, but not the 1st? The explanation given in the 2nd doesn't quite explain to me why the factor is missing in the 1st.

Other than that, it looks to me like I'd only need to add the term for the KE of nuclei back in and transform it into SI units, and then the Hamiltonian in the 2nd PDF is the SI unit version of the Hamiltonian in the 1st. Correct?

 

[kith]

Thanks. Then the remaining questions would be, why the scaling factor of (1/2) in the 2nd PDF, but not the 1st? The explanation given in the 2nd doesn't quite explain to me why the factor is missing in the 1st.

The summation is different. In some terms in the second Hamiltonian, the contributions of the indicidual particles are counted twice. He takes that into account by multiplying by 1/2.

Other than that, it looks to me like I'd only need to add the term for the KE of nuclei back in and transform it into SI units, and then the Hamiltonian in the 2nd PDF is the SI unit version of the Hamiltonian in the 1st. Correct?

Yes

 

[Big-Daddy]

So in SI units all I have to do with the kinetic energy term, bracketed T_n, is replace it with this:

$$- \sum^M_{A=1} ({\frac{h^2 \cdot m_e}{2 \cdot m_A} \cdot \nabla_A^2})$$

Where m_e is the mass of an electron, m_A is the mass of nucleus A and h (sorry I couldn't find h bar but that's what I meant) is the reduced Planck constant.

And put that into the second Hamiltonian and I would have created a complete SI unit version of the first?

 

[Jorriss]

Big-Daddy you seem to have some very big misunderstandings. What book have you been studying quantum from?

 

[Big-Daddy]

I don't study quantum mechanics on its own yet. I have just read the chapters in Atkins' Physical Chemistry.

Is my above expression correct?

 

[Jorriss]

I don't study quantum mechanics on its own yet. I have just read the chapters in Atkins' Physical Chemistry.

Is my above expression correct?

If it is supposed to be kinetic energy, what are the units on the expression you put forth?

In any event, physical chemistry texts are really horrid on the whole at introducing quantum mechanics. If you want a book that is better, but still chemistry oriented, I'd recommend getting McQuarrie's Quantum Chemistry.

 

[Big-Daddy]

Sorry, I noticed my answer isn't dimensionally sound. How about this one:

$$- \sum^M_{A=1} ({\frac{h^2}{2 \cdot m_A} \cdot \nabla_A^2})$$

I will look into certain books which show some development of more quantum ideas.