In a course I am following, we are analyzing the potential wells in molecular absorption spectroscopy.

For molecules with more than one normal mode, we can study the potential wells involved in transitions for each mode and apply the Franck-Condon principle to determine the intensity of each possible electronic and vibrational transition. In these transitions, some of them will not change the geometry of the molecular equilibrium geometry, others will. This leads to think about how the transition change the point group symmetry of that molecule.

Here comes my question: my teacher said that

modes that are non totally symmetric wont maintain the symmetry of the molecule and therefore the potential wells associated with the electronic ground state and excited state will be one just above the other, with the same equilibrium position and just an increase in energy, while totally symmetric modes won't have this property.

Wells for a totally symmetric mode.

Wells for a non totally symmetric mode.

I can't understand why this is true, if there are approximations underlying it and also I can't find any reference for it.

This has been used, later, to distinguish between totally symmetric and non totally symmetric modes in the expression of the transition dipole moment, because for wells not one above the other (totally symmetric) we need to calculate a superposition integral between the two vibrational states, while for totally symmetric modes we know it can be written as a Kronecker delta.


2 Answers 2


This is a bit tricky to explain, the way I imagine it is this: In a totally symmetric mode where, say, each bond extends in phase the potential minimum need not be the same in the excited state and in the ground state but it might be so 'accidentally'. All the forces are balanced. In a non-totally symmetric mode, one set of bonds stretches as the others compress so the potential has to be the same in both states otherwise there would seem to be an internal force being generated that would move the molecule, which is not possible.


Other points are

(a) The spectrum reflects the change in geometry between states and thus is that of totally symmetric vibrations only, but see (d) below.

(b) the excited and ground state have to have the same symmetry,

(c) the asymmetric stretch is a symmetric function of normal coordinates, which is what your lower picture shows.

(d) the non-totally symmetric excited state mode need not have the same frequency as in the ground state, i.e. potential can be wider or narrower. In this case there are selection rules based on the wavefunction's symmetry; both must be the of the same symmetry in the FC factor. The symmetry of vibrational wavefunctions for non-totally sym. modes change with vibrational quantum numbers and it works out that these type of transitions can have $0 \pm2,\pm4..$ transitions only.

  • $\begingroup$ Thank you!!! I'm trying to get into this way of looking at it, but I find it a little difficult. Is there any rigorous demonstration of this? Do you have any reference for it, where I can find a more complete treatment of the subject in general and of this problem in details? $\endgroup$
    – JackI
    Feb 14, 2017 at 16:05
  • $\begingroup$ I don't have a reference at hand unfortunately. Some extra points I now recall (a) the excited and ground state have to have the same symmetry and (b) the asymmetric stretch is symmetric function of normal coordinates , which is what your picture shows. $\endgroup$
    – porphyrin
    Feb 14, 2017 at 17:17
  • $\begingroup$ I've gone through your edit, thank you for your help. I'm continuously trying to understand it, but I can't figure out how can the transition from the electronic ground state to the electronic excited state change the equilibrium position of the nuclei according to the symmetry of the mode I am studying. $\endgroup$
    – JackI
    Feb 14, 2017 at 20:09
  • $\begingroup$ It has been hard, but thinking about your answer and applying it to an easy molecule like water (especially the requirement for the molecule to maintain the same point group symmetry) I've understand what my teacher mean. Thank you! $\endgroup$
    – JackI
    Feb 25, 2017 at 21:24


  • First, ungerade is german for odd and gerade means even.

  • Now, we should discuss energy scales : An electronic transition requires much more energy than a vibrational transition (on the wikipedia figure, the electronic transition is the one that leads to the higher potential well, in which lie the vibrationnal levels)

  • In terms of electronic distance to the nucleus now, we can give some images :

    • From the electronic point of view : It is intuitive that an electron with a higher energy will be on a "higher orbit" (larger orbital) than one with less energy, and thus will be further away from the nucleus. Therefore the offset in the equilibrium position of the electron ($q_{01}$ on the wiki figure)

    • From the vibrationnal point of view : you can imagine that a molecule that vibrates more will have its electrons more prone to go far away from the nucleus.


These were only to be clear about what we are talking. Here comes the answer to you question:

We suppose in the following a transition from the ground state to whatever.

You have to look at the orange curves as the wavefunctions of the electrons. The transition probability from one level to another is proportional to the overlap of their respective wavefunctions : $\langle\Psi_{ground} | \Psi_{excited}\rangle$.

Odd (ungerade) wavefunctions tend to be more confined on the edges of the wells and even (gerade) ones more confined in the center of the well. Because of the offset in the equilibrium position of the electron due to the electronic transition and because the electronic transition is "vertical" in this representation, then you obtain a much larger probability to end up in an odd excited state rather than in an even one. This is easy to visualise thanks to the beautiful wikipedia figure.

  • $\begingroup$ First of all, thank you for your answer! I'm sorry if my question wasn't very clear, I was referring to vibronic transitions, in which both electronic and vibrational states change. Referring to the image you linked, I know that the two big wells are two different electronic states, in which we have different levels corresponding to vibrational states. My issue is with the fact that, in my teacher explanation, the two wells should be one the vertical traslation of the other if they refer to a ungerade mode, while they are sort of "diagonal" (as your picture) for gerade ones. This is what I... $\endgroup$
    – JackI
    Feb 4, 2017 at 13:58
  • $\begingroup$ ... actually can't understand! $\endgroup$
    – JackI
    Feb 4, 2017 at 13:58
  • $\begingroup$ @JackI mmmh, I see your problem and I do not understand that either, I'm no spectroscopy expert moreover. Did your teacher give an example of a simple molecule so that it's easier to think about ? $\endgroup$
    – Mary
    Feb 4, 2017 at 16:22
  • $\begingroup$ @JackI However, I must add that the Frank-Condon principle precisely states what I tried to explain. Consequently, if what you write your teacher said is indeed true, that has little to do with F-C principle I believe. $\endgroup$
    – Mary
    Feb 4, 2017 at 16:28
  • $\begingroup$ I know it is not strictly related to my question, but what I am asking is a way of simplifying the calculation of the Franck - Condon Factor using symmetry. Unfortunately, it has been told to us as really general propriety of every molecule, without a specific example. $\endgroup$
    – JackI
    Feb 4, 2017 at 18:40

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