# Just how quantized are energy levels?

One thing which has bothered me for a while about the idea of quantized energy levels is that it's clear just how quantized they are.

Let me give an example. If I was looking the energy some electromagnetic wave would need to an $\ce{O-H}$ in water from the 0th vibrational energy level to the 1st vibrational energy level, it would be around $\mathrm{3750~cm^{-1}}$. Now, for some electromagnetic wave to cruise by and give enough energy to make that transition, how exact must the energy of that wave be?

Meaning, does it need to have exactly $\mathrm{3750~cm^{-1}}$ of energy? Well, what if the transition was actually $\mathrm{3750.3~cm^{-1}}$. Then that wave doesn't have enough energy anymore, and the likelihood of some wave having exactly $\mathrm{3750.3~cm^{-1}}$ seems a lot less likely than just being very close to that value.

So what happens in actual transitions that the molecule only takes the exact amount of energy it needs? And what happens to whatever leftover energy there might be from a wave that had enough energy to excite this water to it's first vibrational energy level? Is there ever leftover energy like I just mentioned? Also, do two waves of say $\mathrm{1875~cm^{-1}}$ each ever meet up and excite this same transition? This to me also seems so unlikely because that would involve constructive interference exactly of the two waves.

First off, let's think of the light as photons, instead of waves. That makes it clear that yes, there are two photon and multi-photon processes. As you surmise, the selection rules for these are much more complicated than a single-photon process.

Your main question is about "how exact" an absorption needs to be. In reality, it depends on the system.

Atomic Absorption Spectra

For atoms, there are no vibrational or rotational transitions, only electronic (or perhaps nuclear). So atomic spectra are indeed very tight lines, e.g., the Rydberg formula.

Diatomic Molecules

Now you have one vibration and some rotational levels. The latter are important, since you can go up or down some rotational levels with a vibrational transition. (And if you have an electronic transition, you can usually go up vibrational and/or rotational levels.)

So the spectra is a bit more complex than single lines. For example you can do rotational-vibrational spectroscopy.

In this case, the individual transitions are still pretty sharp - not completely lines, but $0.3\:\mathrm{cm^{-1}}$ probably won't match any particular transition.

Larger Molecules

Now you have very small spacing in the rotational lines, so if you take a vibrational absorption spectra (e.g., IR spectra) you'll see that most molecules have some width to the absorption peaks, so $\pm 0.3\:\mathrm{cm^{-1}}$ will still fit within the peak, just not as strongly.

• So what exactly is happening that accounts for the widening of the peaks? I imagine it could be a larger number of things, but how does that still fit under the definition of quantized energy levels? – jheindel Aug 30 '15 at 4:30
• @jheindel If you look at the spectrum above it's just a bunch of discrete lines. There's the quantisation. The widening of the peaks is just as Geoff explained: in a large molecule, each vibrational energy level will have a number of rotational levels (the spacings between rotational energy levels are smaller). So you get a bunch of peaks with different intensities – orthocresol Aug 30 '15 at 9:27
• There are various other sources of line broadening, most notably Doppler broadening (molecules or atoms moving parallel or antiparallel to the light source (or observer)). Removing this and other broadening sources, ultimately one is left with the uncertainty principle. Just like momentum and position are mutually exclusive (complementary) observables ($\Delta x \times \Delta p \geq \hbar \div 2$), in the energy-time domain, $\Delta E \Delta t \geq \hbar \div 2$). – Fedor Aug 30 '15 at 20:40