I was reading S.H.A.R.C. Manual (Surface Hopping in the Adiabatic Representation Including Arbitrary Couplings) and came across the following paragraph:

"Each state is colored, with one color as contour and another at the core of the line. The contour color represents the total spin expectation value of the state. The core color represents the oscillator strength of the state with the lowest state. See figure 7.3 for the relevant color code. Note that by definition the “oscillator strength” of the lowest state with itself is exactly zero, hence the lowest state is also light grey. This dual coloring allows for a quick recognition of different types of states in the dynamics, e.g. singlets vs. triplets or nπ ∗ vs. ππ ∗ states."

Provided color code

How can one recognize nπ ∗ vs. ππ ∗ from this kind of colour scheme? What does oscillator strength have to do with state character?

  • 1
    $\begingroup$ I've never seen such a plot, but since the oscillator strength of a transition relates to the transition dipole moment, only such states contribute that have the right symmetry for the dipole operator. By projecting out "self-interaction", the picture is decluttered for states that don't have point inversion symmetry. The plot should show clusters. It'd be nice if you had an example of such a plot. I assume it is 2-D. $\endgroup$ – Deathbreath Sep 27 '16 at 14:24

This does not seem to make any sense to me either unless I'm missing something very obvious. Suffice it to say that spin changing transitions are generally strongly forbidden, unless paramagnetic electron spin or heavy atoms make them allowed, and that spin 'forbidden-ness' is stronger than symmetry 'forbidden-ness'. Symmetry forbidden transitions can be made allowed by 'intensity borrowing' (Herzberg-Teller) involving a higher electronic state & a vibration of suitable symmetry, as in benzene.
When both spin & symmetry are forbidden transitions are very weak indeed.
$n\pi ^*$ transitions are often symmetry forbidden (e.g. in carbonyls) and so weak compared to $\pi \pi ^*$. Other than these general rules you just have to look at the individual molecules to work out what is going on, it might be quicker than trying to understand the diagram.

| improve this answer | |

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.