# Doubt on vibronic transitions

When we justify the presence of electronic transitions forbidden by Laporte selection rule, the coupling between electronic and vibrational states help us. Why we say that in this way we are overcoming the Born-Oppenheimer approximation? Possibly I'm searching for an intuitive explanation and/or a theoretical one.

• The BO approximation defines any potential energy surface, and so also the vibrational/rotational levels it has. The selection rules are thus always within this approximation. Commented Aug 26, 2023 at 17:32
• In the BO approximation you decouple electronic and nuclear motion. To obtain potential energy curves, you typically solve the SE as function of nuclear coordinates. Vibronic couplings break down this approximation because the hierarchy of the motions (electronic >> vibration >> Rotation) is no longer valid.
– Paul
Commented Aug 27, 2023 at 5:07
• @porphyrin so, can i say that when i consider vibronic transition i overcome Franck-Condon principle and why? Are Franck-Condon and Born-Oppenheimer related? Thanks for the patience Commented Aug 28, 2023 at 7:23

There are two effects, vibronic transitions and vibronic coupling.

In the BO approximation the wavefunction is a product of electronic $$\psi$$ and vibrational $$\chi$$ wavefunctions, for the ground state and excited state

$$\displaystyle \Psi^g_v=\psi^g(r,R)\chi^g_v(R)\\ \Psi^e_u=\psi^e(r,R)\chi^e_u(R)$$

with vibrational quantum numbers $$v,u$$, electron coordinates $$r$$ and nuclear coordinates $$R$$. The electronic wavefunction depends parametrically on $$R$$, i.e. the electronic energy is calculated separately at each $$R$$, and the vibrational ones $$\chi$$ only on the potential energy in their respective states at position $$R$$.

In vibronic transitions the transition moment for an electric dipole transition is the integral over coordinates $$r$$ and $$R$$

$$\displaystyle \mu^{ge}_{vu}=\int\int \psi^g(r,R)\chi^g_v(R)\; \hat\mu \; \psi^e(r,R)\chi^e_u(R)\,dr dR$$

and where $$\hat\mu$$ is the dipole operator, a vector. In symmetry terms this transforms as $$x,y$$ or $$z$$ and has the form of 'charge times displacement' with displacement vectors $$\hat r, \hat R$$,

$$\displaystyle \hat \mu=-e\sum_i \hat r_i+e\sum_jZ_j\hat R_j=\hat\mu_{elec}+\hat\mu_{nucl}$$

If this operator is put into the transition dipole equation the nuclear term $$\hat\mu_{nucl}$$, which depends on position $$R$$ of stationary nuclei, factors out the integral in electronic coordinates $$r$$ which becomes zero because the ground and excited electronic states are orthogonal, i.e. $$\int \psi^g(r,R)\psi^e(r,R)=0$$ thus we only need to evaluate $$\hat\mu_{elec}$$. The transition dipole therefore becomes

$$\displaystyle \mu^{ge}_{vu}=\int\chi^g_v(R)\,\hat\mu^{ge}\, \chi^e_u(R)\, dR$$

where $$\displaystyle \mu^{ge}=\int \psi_g(r,R)\hat\mu_{elec}\psi_e(r,R)dr$$

and this depends on $$R$$ through the electronic wavefunction. To overcome this difficulty the Condon Approximation is now made, which means evaluating the electronic transition moment at the equilibrium internuclear position, (or its average value) so that it no longer depends on $$R$$ and so is separated from the integral

$$\displaystyle \mu^{ge}_{vu}=\hat\mu^{ge}_0\int\chi^g_v(R) \chi^e_u(R)\, dR$$

and the integral is now the overlap of the vibrational wavefunctions and produces the Franck-Condon factors as the square of the absolute value, $$F=|\langle v|u\rangle|^2$$ where the bra-ket indicates the integral.

The vibronic coupling case is a breakdown of the BO approximation which can often be treated as a perturbation. One particular example is the the case where a non-totally symmetric vibration distorts the electronic state and changes its point group and hence the transition dipole is changed. Benzene is a commonly discusses example where an $$e_{2g}$$ vibration make the lowest transition weakly allowed by borrowing intensity from a nearby allowed transition. This is sometimes called Herzberg-Teller Coupling and /or Intensity Stealing.