I am reading about adiabatic and Born-Oppenheimer approximation and I can't understand what steps are exact and what steps are approximations.

Both these approximations are based on the idea that as the nuclear coordinates change the electronic eigenstate remains the same. That is, if, for example, it starts in the ground state it will remain in the ground state as the nuclei move.

First step

The first step of both approximations is to find the solutions of the electronic Hamiltonian parametrized by a number of static configurations of the nuclei. Is that step exact or an approximation?

When we solve the electronic equation we obtain a set of electronic wavefunctions in the form $\{ \phi_i(\mathbf{r};\mathbf{R})\}_{i=1}^\infty$ and energies in the form $\{ E_i(\mathbf{R})\}_{i=1}^\infty$. But is the index really meaningfull? I mean in different values of $\mathbf{R}$ a surface that was indexed as "$1$" in some value, it can be indexed as "$2$" in some other value. Shouldn't we check, after this step, if the surfaces are separated before proceeding to the second step? I mean if they cross then there isn't any adiabaticity involved.

Finally, if this step is approximate, does this mean that nuclei and electron motion can't in principle be separated? In other words, are terms like "electronic states" meaningless?

Second step

In the second step we write the total wavefunction, that is the solution to the full Schrodinger equation as:

$$\Psi(\mathbf{r},\mathbf{R}) = \sum_i\chi_i(\mathbf{R})\phi_i(\mathbf{r};\mathbf{R})$$

Is that step exact?

I understand that the set of wavefunctions that are solutions to the electronic Schrodinger equation form a complete set (for a fixed value of $\mathbf{R}$) where every function $f(\mathbf{r})$ can be represented. Because wavefunctions change with $\mathbf{R}$ we can think that the basis is changing as $\mathbf{R}$ changes so the coefficients of expansions are functions of $\mathbf{R}$. But I can't get why the total wavefunction can be written as the above sum. I must say that I am not familiar with Hilbert spaces so the justification maybe hides there.

Third step

Plugging the above sum into the full Schrodinger equation, multiplying by $\phi_j^*$ and integrating over electronic coordinates leads to a system of coupled differential equations.

Now in this step we approximate the coupling terms and it is said that depending on what terms we equate to zero, we involve either adiabatic or BO approximation. If this is the only step that any approximation is introduced then the first step must be exact.

A final comment about adiabaticity. The motivation behind these approximations is that electronic motion adapts immediately to any change of the nuclear coordinates. But according to adiabatic theorem, this does not guarantees that the electronic eigenstate remains the same. There must be also a gap between the Hamiltonian's spectrum. Why this condition is not explicitly stated (along with the difference in the time scales of motion between electrons and nuclei) when the adiabatic approximation is introduced?


1 Answer 1


In the first step, the nucleus is assumed stationary and point like this is an approximation. Then for a fixed R (position of the nucleus) there would be an infinite number of solutions for eigenfunctions and eigenvalues. The index has nothing to do with R (fixed). The solutions are orthonormal.

In the second step formulating the general solution is setup as a linear combination of the previous eigenfunctions and the spin-functions. This is straight forward core quantum mechanics. No approximation.

The 3'rd step of coupled differential equations is an approximation. This is the mean field approximation, assuming the electrons are subject the mean field of the other electrons.

If you where to move the core (change R) adiabatic (sufficient slowly changing) the fixed R approximation is fine. If the movement is non-adiabatic then the electronic wavefunctions would change.

In many books and texts the standard assumptions of the field are annoyingly often not mentioned.

  • $\begingroup$ So the adiabatic approximation comes into play in the third step where we assume that $R$ changes sufficient slowly and there is enough separation between the potential energy surfaces? $\endgroup$ Commented Jan 12, 2022 at 22:09

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