Quantum Chemistry 5.7 - Anharmonicity and Overtones

[@karlknudson9249]

bottom line what three frequencies is nessasary to make h2o disacossiate ? and do these frequencies need to be deflected in a rotational pattern so there would be one exactly were h2o has a harmpnic vibration then a set above then a set below so there would be three frequencies and dwflected off a angluar blade causing a rotational frequencies would that immediately cause molecules to loose there bond ?in h2o?creating hho?

[@KrankenKoff666]

Why is E(n)=1/2*hv(n+1/2) in this episode? In the previous episode there is no factor 1/2 in the beginning of the expression, only E(n)=hv(n+1/2)

[@TMPChem]

Good catch. That's an error. The prefactor of 1/2 shouldn't be there. The energy levels of the harmonic oscillator are E(n) = hv(n+1/2).

[@mctab1]

In 7:10 do you neglect all other terms like C(n+1/2)^3 and so on? I think it can reflect on a curve shape that we get on a potential energy surface

[@nkyu3035]

in video, you said "the more anharmonic it is, the stronger these overtones can be." and in the comment you answered for my question, you said "Greater anharmonicity doesn't necessarily imply a greater presence of overtones." do these mean greater anharmonicity imply the greater presence of overtones(n>=3) but not the greater presence of overtones(n>=2)?

[@TMPChem]

If you start with a perfectly harmonic system and introduce anharmonicity, the strength of all anharmonic transitions (0 -> 2, 0 -> 3, 1 -> 3, etc.) will increase up to a point. Eventually some will continue to increase and others will decrease. This is a somewhat challenging calculation to do in general, but the only hard and fast rules are that in a perfectly harmonic system no overtone transitions are allowed, and when there is non-zero anharmonicity at least some overtone transitions are possible.

[@nkyu3035]

in the case of perfectly H.O., can we say that quantum states of different energy levels have same bond length? that is, the center of bond decides bond length? if not, i'm not sure how we can say that quantum states of different energy levels in the case of anharmonic oscillator have longer bonds compared to the case of perfectly H.O..

[@TMPChem]

The average bond length is a weighted average of all positions weighted by psi_star psi. In the case of a perfectly harmonic oscillator, is the equilibrium bond length for all values of the quantum number n. In anharmonic cases, the short-range wall increases faster than quadratic, and the long-range tail increases slower than quadratic, both of which have the effect of pushing the wavefunction and thus probability density to higher values of x. This effect increases in magnitude as n increases, giving us a higher as n increases. This type of anharmonicity is observed in virtually all molecules, thus this trend occurs in the vast majority of molecules, giving us a larger average bond length as higher vibrational quantum states.

[@nkyu3035]

I guess it means 'yes'? Thank you for the comment!

[@Muck-qy2oo]

You will also get combination tones.

[@TMPChem]

Indeed. Though for some reason, undergraduate textbooks tend to mention overtones in passing, but far less about combination tones.

[@jaimaoasima3698]

why 2nd term is zero? what does it mean by no linear term?? in book it said that l=lo. but which is true for every term. can you please explain it? V(x)= v(r)+De ??

[@mctab1]

the linear term is zero means there is no piece of the graph curve that can be described with linear equation like y=ax+b

[@nkyu3035]

"the transition from Psi_0 to Psi_n" is overtone, right? what i find hard to accept is, why does overtone occur when anharmonicity is weak, that is, n=0? I mean it seems likely if the transition occurs from Psi_5 to Psi_8.

[@TMPChem]

Any transition where delta_n = 1 is a fundamental (0 -> 1, 1 -> 2, 2 -> 3, etc.). Any transition where delta_n > 1 is an overtone (0 -> 2, 0 -> 3, 1 -> 3, 5 -> 8, etc.). There are two factors that control the magnitude of a peak in a spectrum: 1) What is the population of the initial state, and 2) what is the "transition dipole moment" integral between the states (effectively their overlap during excitation by a photon). The population of a state in statistical mechanics is determined by its relative Boltzmann factor. Lower energy states are preferred, with higher energy states less likely. In vibrations, the separation between energy levels (typically 1000's of cm^-1) is often so large that >99% of the molecules are in the vibrational ground state (n = 0). So we could, in principle, see overtones from higher than n_initial = 0, but almost all n_initial = 0. So most overtones we see are 0 -> 2. The transition dipole moment is an expectation value integral of psi_final* transition dipole operator (psi_initial). In a perfectly harmonic system, this integral is zero unless |delta_n| = 1. Greater anharmonicity doesn't necessarily imply a greater presence of overtones, but zero anharmonicity does imply the complete absence of overtones. The math is quite involved, but overtones of |delta_n| = 2 do happen to be the most common in practice, and for reasons mentioned above are almost always 0 -> 2.

[@nkyu3035]

I think i got a rough point though it seems to be an advanced stuff. i'll do a review. Thank you!