3
$\begingroup$

We may exactly write the wavefunction of a molecular system as $$ \phi(r, R, t) = \sum_{k = 0}^{\infty} \chi_k(R, t) \psi_k(r, R) $$ where $r$ is the coordinates of electrons, $R$ is the coordinates of nuclei, $\phi_k(r, R)$ is the $k$-th eigenfunction of the electronic Hamiltonian for the electrons and $\chi_k$ is a corresponding nuclear wavefunction.

We then get that the time-dependent Schrodinger equation for the nuclear wavefunctions are $$ i \hbar \frac{\partial \chi_k}{\partial t} = \left[ - \sum_I \frac{\hbar}{2M_I} \Delta_I + E_k(R) \right]\chi_k + \sum_l C_{kl} \chi_l $$ where $I$ denotes a specific nucleus, $E_k$ is the $k$-th eigenvalue of the electronic hamiltonian and $C_{kl}$ is a coupling function defined by $$ C_{kl} = \int dr \psi_k^* \left[- \sum_I \frac{\hbar^2}{2M_I} \Delta_I \right] \psi_l + \sum_I \frac{1}{M_I} \left[\int dr \psi_k^* \left[ -i\hbar \nabla_I \right] \psi_l \right] \left[-i\hbar \nabla_I \right] $$ We may make the adiabatic approximation and assume that the coupling matrix is diagonal such that $$ C_{kk} = - \sum_I \frac{\hbar^2}{2M_I} \int dr \psi_k^* \Delta_I \psi_k $$ I am curious in what real life situations this approximation causes an issue. I know with the full Bohn-Oppenheimer approximation where the coupling is completely ignored there are many cases where this causes issues and I know there are many cases where issues arise when the approximation is made to only consider the ground state $k=0$ but I am not sure what possible issues could arise by ignoring the off-diagonal coupling functions. Are there any chemical reactions in non-extreme, non-relativistic situations that are not adequately described by just a diagonal coupling matrix?

$\endgroup$
1
  • $\begingroup$ You need to give an example. But I would comment that if you are in an diabatic representation you will need off-diagonal terms to describe coupling, but if in the adiabatic case these are already included as these diagonalise the full Hamiltonian and describe the case when nuclei move slowly enough. (The usage is confusing though as sometimes diabatic is also called non-adiabatic although this is also used to describe transitions between adiabats as in Landau-Zener crossings.) $\endgroup$
    – porphyrin
    Aug 2 at 9:07
1
$\begingroup$

With a diagonal coupling matrix, you will not have any transitions between the electronic states of your molecule. The inclusion of a diagonal Born-Oppenheimer correction does not change anything in this regard, though it improves the description of each adiabatic state by itself. So every reaction that explicitly involves an electronically excited state (all of photochemistry) can not be described in the adiabatic approximation. Reason being that after your initial excitation to that excited state, the molecule will never be able to return to the ground state because of the missing non-diagonal couplings. For example, if you want to calculate reaction rates for a photochemical reaction, knowledge of the off-diagonal couplings $\int\psi_k^*\nabla_I\psi_l dx$ is essential.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.