Born Oppenheimer Approximation and Product Asnatz

[Derivator]

Hi,

in the derivation of the Born-Oppenheimer-Approximation you have the Hamiltonoperator H=T_n + H_e, where T_n is the kinetic energy of the nuclei and H_e the electronic Hamiltonian.

The Schroedinger equation to solve is:
$H \Psi = E \Psi$

Now, what people do, is the following ansatz:

$\Psi = \sum_k \Psi^n_k \Psi^e_k$

where the $\Psi^e_k$ are soultions of the electronic problem:
$H_e \Psi^e_k = E_e \Psi^e_k$
and $\Psi^n_k is a coefficient that depends on the nucleonic coordinates.$My question:
Using the ansatz:
$\Psi = \sum_k \Psi^n_k \Psi^e_k$
how can you tell, that you don't loose some possible solutions of $H \Psi = E \Psi$?

 

[SpectraCat]

Hi,

in the derivation of the Born-Oppenheimer-Approximation you have the Hamiltonoperator H=T_n + H_e, where T_n is the kinetic energy of the nuclei and H_e the electronic Hamiltonian.

The Schroedinger equation to solve is:
$H \Psi = E \Psi$

Now, what people do, is the following ansatz:

$\Psi = \sum_k \Psi^n_k \Psi^e_k$

where the $\Psi^e_k$ are soultions of the electronic problem:
$H_e \Psi^e_k = E_e \Psi^e_k$
and $\Psi^n_k$ is a coefficient that depends on the nucleonic coordinates.

My question:
Using the ansatz:
$\Psi = \sum_k \Psi^n_k \Psi^e_k$
how can you tell, that you don't loose some possible solutions of $H \Psi = E \Psi$?

Well, you can be sure that you don't lose any solutions within the scope of the Born-Oppenheimer approximation .. i.e. provided that the nuclear and electronic parts of the problem are separable to a good approximation. Solutions that are dependent on coupling between the nuclear and electronic degrees of freedom do exist, and these lead to so-called Born-Oppenheimer breakdown. For example, when two electronic states have a conical intersection, the vibrational motion of the nuclei can become strongly coupled to the electronic state. Similarly, you can have Jahn-Teller couplings between nuclear and electronic motion in molecules with degenerate irreducible representations in their symmetry groups.

 

[Derivator]

thank you

 

[cgk]

I think the trick is that in your last equation, both the set of electronic wave functions and the set of nuclear wave functions form complete basis sets in their respective state spaces. You can thus expand any wave function with coupled nuclear /and/ electronic degrees of freedom into linear combinations of the products of them.

 

[Derivator]

I think the trick is that in your last equation, both the set of electronic wave functions and the set of nuclear wave functions form complete basis sets in their respective state spaces. You can thus expand any wave function with coupled nuclear /and/ electronic degrees of freedom into linear combinations of the products of them.

hmm..., I don't see, why the $\Psi^n_k$ should form a basis of some Hilbertspace. (they are no eigenvectors of some hermitian operator, just coefficients)

Hence $|\Psi^n_k>\otimes |\Psi^n_e>$ does not necessarily form a basis of $\mathcal H_n \otimes \mathcal H_e$ where $\mathcal H_n$ denotes the nucelonic Hilbertspace (that is, in the case of the Born Oppenheimer approximation the space on which the nucelonic kinetic energy operator is acting), and $\mathcal H_e$ the electronic Hilbert space.

 

[DrDu]

The "electronic" Hamiltonian is a hermitian operator in full hilbert space which acts as a differential operator on the electronic degrees of freedom and as a multiplicative operator on the nuclear degrees. So its generalized eigenstates $$\psi_k^e(r;R_0) \delta(R_0-R)$$ form a basis of the full Hilbert space.

 

[Derivator]

So the electronic eigenstates constitue a basis of the full hilbertspace and not only of the electronic subspace? Could you explain, how to see this?

To see where my problem is, i'll give an example:

Take 2 non interacting free particles. The hamiltonian is given by $-\frac{1}{2}\nabla^2_1-\frac{1}{2} \nabla^2_2$. Say that the $|1_n>$ are eigenstates of $-\frac{1}{2} \nabla^2_1$ and $|2_n>$ are eigenstates of $-\frac{1}{2} \nabla^2_2$.
In this case, your (DrDu) argumentation works as well:

"The $-\frac{1}{2}\nabla^2_1$ Hamiltonian is a hermitian operator in full hilbert space which acts as a differential operator on the $|1_n>$ kets and as a multiplicative operator on the $|2_n>$ kets."

However, the basis of the hilbert space on which $-\frac{1}{2}\nabla^2_1-\frac{1}{2} \nabla^2_2$ is operating should be given by all the vectors $|1_n> \otimes |2_m>$.

 

[SpectraCat]

Upon further reflection, perhaps I don't understand what you are trying to express with your "product ansatz" ... I thought I could tell from context, but the mathematical notation seems a bit strange. In other words, what is k indexing (it looks like it is just the states), and what is represented by $\Psi^n_k$ and $\Psi^e_k$?

Based on what I *think* you mean from context now, I think that Dr. Du is correct, because the electronic wavefunctions depend parametrically on the nuclear coordinates, and thus span that space as well. For example, what happens to your manifold of k electronic states when you make a small change to the nuclear coordinates? You get a completely different set of electronic eigenstates for each set of nuclear coordinates. Therefore, since your nuclear "coefficient function" (which would probably be more clearly denoted as something like, $g_k(\vec{R_n})$, to distinguish it from the nuclear eigenstates) is defined at every nuclear configuration, you will also have a complete set of electronic eigenstates at every nuclear configuration, thus you have a complete set of states.

 

[DrDu]

If you use the $-\frac{1}{2} \nabla^2_1-\frac{1}{2}\nabla^2_2$ hamiltonian, the eigenstates will be
$|1_n> \otimes |2_m>$ with $|1_n>$ and $|2_m>$ being momentum eigenfunctions. If you use $-\frac{1}{2} \nabla^2_1$, your eigenstates will also be of the form $|1_n> \otimes |2_m>$, but with $|1_n>$ being a position eigenstate. $|2_m>$ will still be a momentum eigenfunction. However, if you add a potential, $|2_m>$ will become position dependent on 1. This position dependence can be non-trivial (Berry phase).

So the electronic eigenstates constitue a basis of the full hilbertspace and not only of the electronic subspace? Could you explain, how to see this?

To see where my problem is, i'll give an example:

Take 2 non interacting free particles. The hamiltonian is given by $-\frac{1}{2}\nabla^2_1-\frac{1}{2} \nabla^2_2$. Say that the $|1_n>$ are eigenstates of $-\frac{1}{2} \nabla^2_1$ and $|2_n>$ are eigenstates of $-\frac{1}{2} \nabla^2_2$.
In this case, your (DrDu) argumentation works as well:

"The $-\frac{1}{2}\nabla^2_1$ Hamiltonian is a hermitian operator in full hilbert space which acts as a differential operator on the $|1_n>$ kets and as a multiplicative operator on the $|2_n>$ kets."

However, the basis of the hilbert space on which $-\frac{1}{2}\nabla^2_1-\frac{1}{2} \nabla^2_2$ is operating should be given by all the vectors $|1_n> \otimes |2_m>$.

 

[DrDu]

Or, to clarify notation: let $h_1(x_1,x_2)$ be an operator acting in the hilbert space and depends parametrically on $x_2$, then $\mathcal{H}_1$ and $H_1$being the corresponding operator "lifted" to the full hilbert space $\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2$. $x_2$ is no longer a parameter but an operator acting in $\mathcal{H}_2$ hence $H_1 \neq h_1 \otimes \mathbf{1}_2$ !
Then $|\psi_n^e(x_2)\rangle$ is an eigenstate of $h_1(x_1,x_2)$ and $|\psi_n^e(x_2)\rangle \otimes |x_2>$ is the eigenstate of ${H}_1$ .

 

[Derivator]

Upon further reflection, perhaps I don't understand what you are trying to express with your "product ansatz" ... I thought I could tell from context, but the mathematical notation seems a bit strange. In other words, what is k indexing (it looks like it is just the states), and what is represented by $\Psi^n_k$ and $\Psi^e_k$?

Yes, k is just the index of the states (it is used as the index k in the following: if $\Psi_1 , ... \Psi_N$ constitute a basis, then an arbitrary $\Psi$ which is element of the linear span of this basis can be expanded as $\Psi = \sum_k c_k \Psi_k$)

$\Psi^e_k$ are the eigenvectors of the electronic hamiltonian.

$\Psi^n_k$ are just expansion coefficients initially (which finally will turn out to be the eigenvectors of the born - oppenheimer equation)

$H_1 \neq h_1 \otimes \mathbf{1}_2$

hm, this or to be more precise $H_1 = h_1 \otimes \mathbf{1}_2$ was my conecpt of how to lift the operator $h_1$ from $\mathcal{H}_1$ to $\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2$

 

[DrDu]

hm, this or to be more precise $H_1 = h_1 \otimes \mathbf{1}_2$ was my conecpt of how to lift the operator $h_1$ from $\mathcal{H}_1$ to $\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2$

Yes, I know, but this only works in the trivial example without kinetic energy. As soon as you add a x_2 dependent potential to h_1, it doesn't act non-trivially on $\mathcal{H}_2$. Mathematicians would speak of a "fiberisation".

 

[DrDu]

I just wanted to give the expression for the electronic Hamiltonian $H_1$ in full Hilbert space:
$$H_1 =\int dx_2 (h_1(x_2) \otimes |x_2\rangle \langle x_2|)$$ . When $h_1$ is independent of $x_2$, this expression can be seen to reduce to the direct product.

 

[Derivator]

Hi folks,

thanks for your help so far.

I came across another difficulty, related to the above ansatz:

When using the ansatz for the total (electron + nuclear) wave function $\Psi = \sum_k \Psi^n_k \Psi^e_k$, you can derive by using the adiabatic approximation a differential equation, which will give you the $\Psi^n_k$. (This is the Born-Oppenheimer equation, which describes the 'movement' of the nuclear wavefunction in a pseudo potential given by the eigen values of the electronic wave function.)

Now, in the end, people don't use $\Psi = \sum_k \Psi^n_k \Psi^e_k$ as the total wave function, but rather $\Psi = \Psi^n_l \Psi^e_l$, that is, they neglect all terms in $\Psi = \sum_k \Psi^n_k \Psi^e_k$ except the l-th term. Is there any justification, that this further approximation introduces only small errors?

Edit: See for example

A common approximation is to assume that this expansion can be represented by one term in the summation, with only one ak coefficient being non-zero, and hence nuclear motion occurs on only one electronic state.

from http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Chemical_Bonding/Nuclear_Potential_Energy_Curves#The_Born-Oppenheimer_Approximation_and_Adiabatic_Approximation

 

[DrDu]

Yes, the reason lies in the difference of electronic and nuclear masses. Why don't you read the original article by Born and Oppenheimer? Wikipedia will direct you also to several translations of the article.

 

[Derivator]

yeah, I've looked at Born and Oppenheimers article (http://www.ulb.ac.be/cpm/people/scientists/bsutclif/bornop.pdf ), but unfortunately, they've used a completely different method. they didn't use the above ansatz, they did a kind of perturbation theory similar to Rayleigh's one.

 

[DrDu]

Yes, if you use the general ansatz for the wavefunction, you get some terms where the nuclear kinetic energy operator acts, once or twice on the electronic wavefunction. These terms lead to transitions between the electronic levels and are called non-adiabatic terms. They are weighed with m/M, hence they will generically be small and can be neglected (there are always exceptional situations, e.g. conical intersections where this isn´t true). Then considering a single electronic wavefunction is sufficient.

 

[Derivator]

I only understand, that this explains why we can neglect terms like $<\Psi^e_i | \nabla |\Psi^e_j>$or $<\Psi^e_i | \nabla^2 |\Psi^e_j>$. But I don't see, why from this we can conclude, that parts of the wavefunction itself can be neglectet. That is, I can' t see why from this adiabatic approximation (neglecting these terms in the first sentence of this post) it follows, that for example the total ground state is given by $\Psi = \Psi^n_0 \Psi^e_0$ instead of an expression like $\Psi = \sum_k \Psi^n_{k,0} \Psi^e_k$.

 

[DrDu]

Think of the Hamiltonian as a matrix in electronic state space. When you neglect the non-diagonal matrix elements (the non-adiabatic couplings) the matrix becomes diagonal. Of what form are the eigenvalues of a diagonal matrix?

 

[Derivator]

well, the eigenvalues are the entries on the diagonal.

(sorry, i can't focus on this stuff today, i just was told, that my grandfather died... nevertheless, i have to prepare for the final exams next week...)

 

[DrDu]

I am very sorry for your grandfather.

The point is that not only will the eigenvalues become equal to the diagonal elements but also will the corresponding eigenfunctions become vectors with only one non-zero entry. In the case of the Born-Oppenheimer approximation, this means that a time independent solution of the full Schroedinger equation is built up from only a single electronic wavefunction as long as the energetic spacing between neighbouring electronic states is much larger than the norm of the non-adiabatic couplings between these states.

 

[Derivator]

The point is that not only will the eigenvalues become equal to the diagonal elements but also will the corresponding eigenfunctions become vectors with only one non-zero entry.

This is true, if one expands the eigenfunctions in the eigenbasis. But, actually you never do this. They are expanded with respect to the state space vectors.

It seem to me, that I do not understand the essential point, again... arg...

 

[DrDu]

What you do is to set up
$H_{jk}=<\Psi^{el}_k|H|\Psi^{el}_j>_{el}$
where the "el" suffix on the brakets means integration over electronic coordinates, only.
This is a matrix in the electronic state space and a differential operator for the nuclear coefficients.
Hence, the wavefunction will be represented as $( \Psi^{n}_{1}(R), \Psi^{n}_{2}(R) \ldots)^T$

As long as you can neglect the non-adiabatic terms, the matrix will be diagonal and the vector will have only one non-zero entry.

 

[Derivator]

hmm, i still don't see it. What is you Hamiltonoperator? Is it the total Hamiltonian? If this is the case, it's matrix elements look like the left hand side of the following equation:

http://img717.imageshack.us/img717/1989/capturetfm.png

now you neglect the non-adiabatic coupling terms, and you get the born oppenheimer equation.
Solving this equation, you get the nuclear wave functions $\Psi^n_i$ (not only one $\Psi^n_j$) and thus, you can write the total wavefunction as $\sum_i \Psi^n_i \Psi_i$, where $\Psi_i$ is the solution of the electronic problem.

 

[SpectraCat]

hmm, i still don't see it. What is you Hamiltonoperator? Is it the total Hamiltonian? If this is the case, it's matrix elements look like the left hand side of the following equation:
http://img717.imageshack.us/img717/1989/capturetfm.png

now you neglect the non-adiabatic coupling terms, and you get the born oppenheimer equation.
Solving this equation, you get the nuclear wave functions $\Psi^n_i$ (not only one $\Psi^n_j$) and thus, you can write the total wavefunction as $\sum_i \Psi^n_i \Psi_i$, where $\Psi_i$ is the solution of the electronic problem.

Think about what you are saying. You can make the same statement for any eigenfunction-eigenvalue problem .. it is equivalent to saying that the nuclear eigenstates form a complete basis for describing the nuclear motion. However, if we neglect non-adiabatic coupling between the electronic surfaces due to nuclear motion, then there is no NEED for the more general solution. All nuclear motion happens on a single electronic surface, and so the matrix will be diagonal, as DrDu already pointed out.

 

[Derivator]

I'm too stupid for this...

Is it correct, that in the adiabatic approximation all $\Psi^n_i$ except one (say $\Psi^n_j$) is equal to 0? Thus $\sum_i \Psi^n_i \Psi_i = \Psi^n_j \Psi_j$ ?

(The problem is, that I still don't see, that all nuclear wavefunctions (except one) vanish, because of the fact, that the hamiltonian is diagonal.)

I mean, I understand, that the eigenvectors of a diagonal matrix only have 1 entry which is different from 0, but this one entry should be equal to 1. The eigenvectors of a diagonal matrix looks like (0,0,...,1,0,...,0)

 

[SpectraCat]

You're not stupid .. just maybe a little confused.

First, let's clarify some notation here. What do you mean when you write $\Psi^n_j$? Based on your initial descriptions in this thread, I took it to mean the entire space of possible nuclear configurations for the electronic state j. Now it seems that you are using it to denote a single nuclear eigenstate on the effective potential energy surface represented by electronic state j. Of course there in general many nuclear bound states (i.e. ro-vibrational wavefunctions) supported for a given electronic surface.

Anyway, my comments in this thread have always assumed the first definition. If you have switched to the second, then I believe that you need to change your notation, so that your nuclear wavefunctions have a separate index for the different bound states on a given electronic surface: Then the following condition would hold true, assuming the Born-Oppenheimer approximation. The total wavefunction (electronic + nuclear) for a molecule could always be written as: $\Psi = \phi_j\sum_k\chi_{j,k}$, where $\phi_j$ denotes the electronic wavefunction and $\chi_{j,k}$ is an eigenstate of the nuclear Hamiltonian, including the vibrational and rotational motions, solved on the electronic potential energy surface described by $\phi_j$.

 

[Derivator]

hi spectracat,

$\Psi^e_k$ respectively $\Psi_k$ are the eigenvectors of the electronic hamiltonian. (Sometimes, I omitted the label 'e')

$\Psi^n_k$ are just expansion coefficients initially (which finally will turn out to be the eigenvectors of the born - oppenheimer equation)

 

[SpectraCat]

hi spectracat,

$\Psi^e_k$ respectively $\Psi_k$ are the eigenvectors of the electronic hamiltonian. (Sometimes, I omitted the label 'e')

$\Psi^n_k$ are just expansion coefficients initially (which finally will turn out to be the eigenvectors of the born - oppenheimer equation)

Right, and that notation implies that for each choice of k, there is only one $\Psi^n_k$. That is not correct if those will eventually be eigenstates of the nuclear Hamiltonian, which is why I suggested the notation change in my last post.

 

[Derivator]

ok, let's use your notation.

The ansatz is $\Psi = \sum_k\sum_i\phi_i\chi_{i,k}$. Now, using the adiabatic approximation, why is the final solution given by $\Psi = \sum_k\phi_j\chi_{j,k}$?

Using the adiabatic approximation leads to the Born-Oppenheimer equation (better: many Born Oppenheimer equations, one for each electronic surface). The solution of this equation is given by the $\chi_{i,k}$. I don't see, why all $\chi_{i,k}$ (except one) (with respect to the index i) should vanish.

Well, you could say, that one uses only one electronic surface. But in this case my question is, why $\Psi = \sum_k\phi_j\chi_{j,k}$ is a good approximation, that is, why are all $\chi_{i,k}$ (except one) small, so that we can neglect them.