In quantum mechanics we learn the the classical picture of a vibrational excitation is wrong. The atoms will 'vibrate' only while the molecule is in a coherent superposition of two vibrational states, and will find itself in a stationary state once this coherence is broken.

Let's say I have a carbon monoxide in vacuum, and I excite it from v = 0 to v = 1. In order to describe the electron distribution of the molecule, I would expect the electron distribution to be given by the sum of the electron distributions for the CO at each bond distance, weighted by the nuclear wave function.

So... An excited CO in vacuum would look similar to a stationary molecule of C-o-o, where the little o's indicate that the electron density is smaller than that of a regular oxygen. Is this interpretation correct?

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    $\begingroup$ Vibrational excitation changes the bond geometry in a molecule. Definitely, always. I'm not sure where you want to go with your question however. $\endgroup$
    – Karl
    Jul 31 '18 at 21:40
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    $\begingroup$ And why exactly must atom positions always be stationary while not in a transition? Born-Oppenheimer approximation? Not saying you're wrong, but I've never looked at it that way. $\endgroup$
    – Karl
    Jul 31 '18 at 21:46
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    $\begingroup$ Well of course " an electron in a 2p orbital is considered to be 'stationary'" because the energy of the electron is stable, not because the electron isn't "moving." You seem to be conflating wave models of quantum mechanics and particle models of classical physics and wanting consistency between the two which just doesn't exist. $\endgroup$
    – MaxW
    Jul 31 '18 at 22:06
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    $\begingroup$ @Max That doesn't make much sense. The expectation value is constant even for a classical oscillator if there is no excitation (the energy of the system is constant). It doesn't mean it is not moving. $\endgroup$
    – Greg
    Aug 1 '18 at 1:21
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    $\begingroup$ Rather than essentially changing your question and in course invalidating existing answers, please ask a new question linking back to the other establishing context. $\endgroup$ Aug 2 '18 at 12:58

Yes they change 'shape'. A diatomic only in the sense that the bond extends and contracts, but for a triatomic or polyatomic molecule there are bond stretches and bends. The number of different ways these vibrations change the 'static' shape of a molecule, such as may be measured by x-ray diffraction, depends on the symmetry and this is described by the point group. Only certain combinations of bond stretches and bends are allowed and these are called 'normal modes'.

Some other points.

A molecule will vibrate even when it has zero point energy, i.e. when $v=0$; by the Heisenberg principle this must happen. If it were stationary then it would have infinite energy because its position is determined exactly. The electron density, if it were to be measured would thus be an average of that at the zero point energy. Calculations of this made at zero potential energy would thus be slightly different. Slightly because the bond extension is only a few percent of the bond length at this energy, and also most nuclear wavefunction density is located at the minimum energy.

We assume that the Born-Oppenheimer condition applies, briefly that electrons being far lighter than nuclei instantaneously adjust to the slow nuclear motion as a molecule vibrates. This allows us to draw the potential energy profile, for example a Harmonic or Morse potential. When we draw the familiar wavefunctions for vibrational motion these are the nuclear wavefunctions at each energy and are the stationary solutions to the Schroedinger equation and we interpret these as meaning that the molecule has a probability of being at each position given by the square of the wavefunction.

In a long time spectroscopic experiment (e.g. using a spectrophotometer) only the spectrum is measured, i.e. energy gaps between levels.

In experiments using a femtosecond laser pulse$^*$ it is possible to excite several vibrational wavefunctions almost instantaneously, i.e. in a time shorter than a vibrational period, in this case a 'wavepacket'$^{**}$ is produced and this 'particle' moves left and right inside the potential. This motion has been observed by using a second femtosecond laser pulse as a probe. As the energies of the vibrational levels are not all the same the wavepacket, which starts off in phase, will become out of phase then 'in phase' again and so goes on repeatedly recurring until it is disturbed, say by a collision with another molecule. (In these time resolved experiments Fourier transforming the data gives the spectrum.)

$^*$ As a femtosecond pulse is short it has a wide spectral width and can therefore excite several energy levels.

$^{**}$ A wavepacket is the direct summation of the vibrational wavefunctions multiplied by a time dependent term related to the vibrational frequencies. Thus the wavepacket evolves in time. The wavepacket recurs at regular time intervals due to the time dependent terms coming into phase just as they were at time zero. This is a property of a collection of oscillators; children on swings would show the same effect.

  • $\begingroup$ Thank you! But this is not precisely what I had in mind. I have edited my question to make what I mean a bit clearer :-) $\endgroup$
    – Max
    Aug 2 '18 at 0:28

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