The idea of an adiabatic potential energy surface or adiabatic state is closely related to the Born-Oppenheimer approximation. This has been discussed on this site before in some detail. People frequently talk about diabatic states as well. It seems based on the reading I have done that a diabatic state is not really counter to an adiabatic state as one might expect from the names, but is sort of a different representation when looking at processes involving multiple potential energy surfaces.

So, my question is, how are adiabatic and diabatic states/surfaces/processes (what word should I be using here??) similar and how are they different? Can both pictures be used to describe the same processes all of the time or only some of the time (or perhaps I'm mistaken and they are never the same)?

Also, I think this is just terminology, but where does a nonadiabatic process fit into all this? I assume this just means there is a crossing of surfaces and the Born-Oppenheimer approximation breaks down, but that isn't quite the same as what I'm asking above I think.

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    $\begingroup$ If I understand correctly it is about at what point you add the interaction between the two states. In the diabatic case one forms two PES then the off -diagonals are the couplings. In the adiabatic the coupling is added when forming the PES. Non-adiabatic is about how to treat the crossing point, as in Landau-Zener. Does not arise in the adiabatic case as the surfaces are already separated and so do not cross. I stand to be corrected on this, however. $\endgroup$ – porphyrin May 21 '18 at 10:32
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    $\begingroup$ I'd be interested in seeing a clear explanation to this as well. If your bounty doesn't work out, I might add a post to the Canonical Q&A on meta. $\endgroup$ – Tyberius Jun 19 '18 at 18:37

A potential energy surface (PES) is generally the electronic energy written as a function of the nuclear coordinates. This is obtained by solving the Hamiltonian, considering explicit dependence on the electronic coordinates $r$, and parametric dependence on the nuclear coordinates $R$.


Solving this, gives us the electronic energy as the function of the nuclear coordinates, $E(R)$. A plot of this function is the PES. This PES is obtained within the Born-Oppenheimer (BO) approximation, which allows us to assume that the kinetic energy of the nucleus does not matter. The physical interpretation is that the nucleus is very heavy compared to the electrons and, therefore, so much more slower that it can be considered almost stationary. This PES is generally written in the adiabatic basis, where the diagonal terms represent the electronic energy and the off-diagonal terms are the nuclear kinetic energy, $T_n$.

The off-diagonal terms can be neglected, as we already saw that $T_n$ is nearly zero. But in cases where two PESs come very close together (conical intersections and avoided crossings), there is coupling of the nuclear and electronic coordinates. This means that the nuclear kinetic energy can no longer be neglected, and there is a breakdown of the BO approximation. Further, the numbers become very high when the two states (PESs) cross as in a conical intersection and become too hard to handle. In such cases, a unitary transformation is done to change the basis from an adiabatic to a diabatic basis. By doing this, the diagonal term, $T_n$, gets converted to a potential energy term and is easy to handle.

Therefore, adiabatic and diabatic PESs differ by the basis they are written in. They can both be used to describe the same process, but there is no real need to use the diabatic representation when the BO approximations holds.

They can be referred to as adiabatic/diabatic PESs, or states, or surfaces, or representations. The term 'process' here is incorrect as it refers to thermodynamics terminology.

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    $\begingroup$ Since posting this question I have done some reading and what you have written is my understanding as well. The difference is whether you choose a basis that diagonalizes the kinetic energy operator or potential energy operator. Thanks for the clear answer. $\endgroup$ – jheindel Aug 10 '18 at 17:30

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