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Born Oppenheimer Approximation and Product Asnatz

  1. Jul 11, 2011 #1
    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:
    [itex]H \Psi = E \Psi[/itex]

    Now, what people do, is the following ansatz:

    [itex]\Psi = \sum_k \Psi^n_k \Psi^e_k[/itex]

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


    My question:
    Using the ansatz:
    [itex]\Psi = \sum_k \Psi^n_k \Psi^e_k[/itex]
    how can you tell, that you don't loose some possible solutions of [itex]H \Psi = E \Psi[/itex]?
     
  2. jcsd
  3. Jul 11, 2011 #2

    SpectraCat

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    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.
     
  4. Jul 11, 2011 #3
    thank you
     
  5. Jul 11, 2011 #4

    cgk

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    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.
     
  6. Jul 12, 2011 #5
    hmm..., I don't see, why the [itex]\Psi^n_k[/itex] should form a basis of some Hilbertspace. (they are no eigenvectors of some hermitian operator, just coefficients)

    Hence [itex]|\Psi^n_k>\otimes |\Psi^n_e>[/itex] does not necessarily form a basis of [itex]\mathcal H_n \otimes \mathcal H_e[/itex] where [itex]\mathcal H_n[/itex] 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 [itex]\mathcal H_e[/itex] the electronic Hilbert space.
     
    Last edited: Jul 12, 2011
  7. Jul 12, 2011 #6

    DrDu

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    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 [tex] \psi_k^e(r;R_0) \delta(R_0-R)[/tex] form a basis of the full Hilbert space.
     
  8. Jul 13, 2011 #7
    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 [itex]-\frac{1}{2}\nabla^2_1-\frac{1}{2} \nabla^2_2[/itex]. Say that the [itex]|1_n>[/itex] are eigenstates of [itex]-\frac{1}{2} \nabla^2_1[/itex] and [itex]|2_n>[/itex] are eigenstates of [itex]-\frac{1}{2} \nabla^2_2[/itex].
    In this case, your (DrDu) argumentation works as well:

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

    However, the basis of the hilbert space on which [itex]-\frac{1}{2}\nabla^2_1-\frac{1}{2} \nabla^2_2[/itex] is operating should be given by all the vectors [itex]|1_n> \otimes |2_m>[/itex].
     
  9. Jul 13, 2011 #8

    SpectraCat

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    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 [itex]\Psi^n_k[/itex] and [itex]\Psi^e_k[/itex]?

    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, [itex]g_k(\vec{R_n})[/itex], 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.
     
  10. Jul 13, 2011 #9

    DrDu

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

     
  11. Jul 13, 2011 #10

    DrDu

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    Or, to clarify notation: let [itex]h_1(x_1,x_2)[/itex] be an operator acting in the hilbert space and depends parametrically on [itex]x_2[/itex], then [itex]\mathcal{H}_1[/itex] and [itex]H_1 [/itex]being the corresponding operator "lifted" to the full hilbert space [itex]\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2[/itex]. [itex] x_2 [/itex] is no longer a parameter but an operator acting in [itex]\mathcal{H}_2[/itex] hence [itex] H_1 \neq h_1 \otimes \mathbf{1}_2[/itex] !
    Then [itex]|\psi_n^e(x_2)\rangle[/itex] is an eigenstate of [itex]h_1(x_1,x_2)[/itex] and [itex] |\psi_n^e(x_2)\rangle \otimes |x_2> [/itex] is the eigenstate of [itex]{H}_1[/itex] .
     
    Last edited: Jul 13, 2011
  12. Jul 13, 2011 #11
    Yes, k is just the index of the states (it is used as the index k in the following: if [itex]\Psi_1 , ... \Psi_N[/itex] constitute a basis, then an arbitrary [itex]\Psi[/itex] which is element of the linear span of this basis can be expanded as [itex]\Psi = \sum_k c_k \Psi_k[/itex])

    [itex]\Psi^e_k[/itex] are the eigenvectors of the electronic hamiltonian.

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


    hm, this or to be more precise [itex]H_1 = h_1 \otimes \mathbf{1}_2[/itex] was my conecpt of how to lift the operator [itex]h_1[/itex] from [itex]\mathcal{H}_1[/itex] to [itex]\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2[/itex]
     
    Last edited: Jul 13, 2011
  13. Jul 13, 2011 #12

    DrDu

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    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 [itex]\mathcal{H}_2[/itex]. Mathematicians would speak of a "fiberisation".
     
  14. Jul 15, 2011 #13

    DrDu

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    I just wanted to give the expression for the electronic Hamiltonian [itex] H_1[/itex] in full Hilbert space:
    [tex] H_1 =\int dx_2 (h_1(x_2) \otimes |x_2\rangle \langle x_2|) [/tex] . When [itex]h_1 [/itex] is independent of [itex]x_2[/itex], this expression can be seen to reduce to the direct product.
     
  15. Jul 17, 2011 #14
    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 [itex]\Psi = \sum_k \Psi^n_k \Psi^e_k[/itex], you can derive by using the adiabatic approximation a differential equation, which will give you the [itex]\Psi^n_k[/itex]. (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 [itex]\Psi = \sum_k \Psi^n_k \Psi^e_k[/itex] as the total wave function, but rather [itex]\Psi = \Psi^n_l \Psi^e_l[/itex], that is, they neglect all terms in [itex]\Psi = \sum_k \Psi^n_k \Psi^e_k[/itex] except the l-th term. Is there any justification, that this further approximation introduces only small errors?

    Edit: See for example
    from http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Chemical_Bonding/Nuclear_Potential_Energy_Curves#The_Born-Oppenheimer_Approximation_and_Adiabatic_Approximation [Broken]
     
    Last edited by a moderator: May 5, 2017
  16. Jul 18, 2011 #15

    DrDu

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    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.
     
  17. Jul 18, 2011 #16
    yeah, i've looked at Born and Oppenheimers article (http://www.ulb.ac.be/cpm/people/scientists/bsutclif/bornop.pdf [Broken]), 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.
     
    Last edited by a moderator: May 5, 2017
  18. Jul 18, 2011 #17

    DrDu

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    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.
     
  19. Jul 18, 2011 #18
    I only understand, that this explains why we can neglect terms like [itex]<\Psi^e_i | \nabla |\Psi^e_j>[/itex]or [itex]<\Psi^e_i | \nabla^2 |\Psi^e_j>[/itex]. 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 [itex]\Psi = \Psi^n_0 \Psi^e_0[/itex] instead of an expression like [itex]\Psi = \sum_k \Psi^n_{k,0} \Psi^e_k[/itex].
     
    Last edited: Jul 18, 2011
  20. Jul 19, 2011 #19

    DrDu

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    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?
     
  21. Jul 19, 2011 #20
    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...)
     
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