cnidocyte
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Basic IR and NMR spectroscopy questions
The idea of quantisation of energy is confusing me. For example in IR, the bond will only absorb energy that is identical to the energy of the bonds
bending or stretching vibrations. Why is this and what does the bond do with that new energy?
Then NMR, the proton will only absorb the exact amount of energy required to jump to the higher energy spin state. Why is this? Why doesn't it absorb
some energy and make a partial jump towards the higher spin state?
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watson.fawkes
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Quote: Originally posted by cnidocyte  | The idea of quantisation of energy is confusing me. For example in IR, the bond will only absorb energy that is identical to the energy of the bonds
bending or stretching vibrations. Why is this and what does the bond do with that new energy?
Then NMR, the proton will only absorb the exact amount of energy required to jump to the higher energy spin state. Why is this? Why doesn't it absorb
some energy and make a partial jump towards the higher spin state? |
Vibrational energy is a form of kinetic
energy, be it from bending or stretching. It's a form of constrained motion, which means that it's a bound state of the molecule (as opposed to a free
state). Bound states in quantum mechanics are quantized in general. The mathematical reasons why bound states are quantized are fairly subtle to get
at in full generality, but have an easy informal explanation: phase matching. Bound states are always geometrically constrained, with zero (or rather
near-zero; that's where the subtlety begins) density outside the molecule. This "distance" must be filled out with an integral number of wavelengths
of the wave function. Since not every wavelength will fit, energy levels are quantized. Closely related to this are cyclic boundary conditions where
the phase must match up with itself (rather than be zero at a boundary); these are hugely different and yield the same quantization principle. This is
the form of boundary condition for the spherical harmonic functions that are the simple solutions for angular momentum in atoms.
The second question is rather more subtle. It's impossible for an atom or molecule to make a partial jump, because there's nowhere to land; that's the
whole principle of quantization. This is not a correct answer to your question, though. It's conceivable that a photon absorption
could excite a molecule into a mixed state, a combination of states with an arbitrary energy. Mixed states are not quantized. So the real question is
this: Why can't photon absorption generate a mixed state?
The first part of the answer is to understand that this is an empirical fact that is not a consequence of the Schrodinger equation. That equation
describes steady-state evolution. Photon interactions are beyond its scope. The full picture says something about transition probabilities between
states before and after an interaction, but it doesn't say what the states after the interaction have to be, nor, certainly, what the prior states
are. But the observational evidence is clear that not only must the "after" state be quantized, but also that the "before" state must be as well. This
fact lies outside the basic Schrodinger picture and requires additional explanation.
For the second part, consider the case where a photon impinges upon a molecule in a mixed state and is absorbed. The empirical evidence is that this
molecule transitions as if it were in some pure state. We can treat this interaction, therefore, as though the interaction forces a measurement of the
initial state. Measurements only measure eigenstates, not mixed states. The final state is also an eigenstate, so we can also treat the
post-interaction state as if it also measured. Therefore, the next part of the explanation is that photon interactions act as if they contain
measurements as an inextricable part of the interaction. The "act as if" clause is important here, because I'm not claiming there are "actual"
measurements going on, whatever an "actual" measurement might mean in full generality.
The third part is to ask the question about why it is that photon interactions act as if they've got embedded measurements. I've got no answer to this
one. So ultimately, I haven't provided a complete reason, but merely reduced one "why" question to another.
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