# If translational energy is a continuum, why molecules don't absorb any energy quantity?

I am just starting with quantum mechanics. Here an image where I show possible transitions but only should excite translational movement.

Can a molecule absob ANY quantity of energy if it goes to translation? I think I am misunderstanding something.

In the image are electronic, vibrational and rotational states. Translation should be the white space, because it is almost a continuum. The arrows are pointing out different translational states transitions, but somehow I am proposing that rotations and vibrations are not excited, only translation (although it doesn't seems possible)

• Your diagram is extremely unclear. What do the lines mean? Is any part of the diagram meant to signify a continuum? – Emilio Pisanty Oct 4 '17 at 14:57
• I am sorry. Numbers in bold at right are electronic states, left vibrational states, and the small ones at right are rotational. Translation is the whole space between @EmilioPisanty. I will try to be clearer – santimirandarp Oct 4 '17 at 15:04

## 1 Answer

For one, it's important to keep in mind that no transition is infinitely sharp, because that would mean that the frequency is defined to infinite precision and that is unphysical. Instead, all observed transitions have a finite linewidth, which comes from two different sources:

• There are external factors (known as inhomogeneous broadening) such as Doppler shifts over a distribution of possible velocities, magnetic field variations, or collisions with other molecules in the gas, all of which will make the line broader. If you're careful enough, these can generally be removed.

• There is also an intrinsic component (also known as homogeneous broadening), which comes from the fact that transitions never take forever, as they involve a transfer of population from excited states to the ground state or vice versa, and if they take a finite time then they cannot be infinitely sharp in frequency.

Either way, all of those lines need to be thought of as having a finite thickness. This is fine if the lines are far apart, but once you start having a forest of thinly-spaced lines of different descriptions, then inevitably you'll end up in a situation where you have an effective continuum, because the underlying states are discrete but you're unable to resolve them.

That said, even when that is not an issue, you still have motional broadening that comes from the fact that the centre-of-mass translational degree of freedom can take up some of the energy of the transition.

For simplicity, it's best to think of a discrete system with well-separated eigenstates (it doesn't matter whether it's an atom with electronic degrees of freedom, a molecule where you've included vibrational and rotational states, or something different, all you need is that their separation be bigger than their intrinsic linewidth) where we add in the centre of mass's translation as a relevant degree of freedom.

In such a case, it is possible for some of the energy to go to the translation (and I wrote a starter guide on how you describe that at How does one account for the momentum of an absorbed photon?), but the process is not very efficient and it only has a nonzero cross-section over a very small bandwidth. This is in large part because, if the molecule is in free space, the amount of energy you can deliver to it is constrained by the momentum change involved in the process (equal to the photon momentum); this is a fixed quantity, and once you divide by the molecular mass, you tend to get tiny energies deliverable to the translational degree of freedom.

Or, at least, that's the case for atoms and molecules undergoing electronic and vibronic transitions. For nuclei the story is a good bit different, because gamma radiation really packs a good momentum punch, and if you're not careful then the energy transfer associated with that punch will result in a large motional broadening of what you might hope to be sharp spectroscopic lines. Luckily, though, we have the Mössbauer effect to transfer that momentum to a surrounding lattice when we do things in the solid phase, which eliminates that motional broadening.

• So the whole energy of the molecule is not really cuantized? I mean, always system absorb part of the incoming frequency for translational movement although it may has low probability? Is it possible to go to a higher translational level withouth changing vibrational/rotational energy? – santimirandarp Oct 4 '17 at 16:45
• Energy is quantized, as a general rule, whenever the relevant degree of freedom is confined. When the COM translation is not confined, then total energy is not strictly quantized, but since the coupling to COM motion is weak then energy is quantized to a good approximation (where the good in 'good' is proportional to the weak in 'weak'). And yeah, it's perfectly possible (in principle) to deliver energy directly to the COM motion, though for neutral systems in practice you'll often involve, at least at the level of virtual transitions, additional DOFs. – Emilio Pisanty Oct 4 '17 at 16:55
• Any short conceptual video or text to see about these things? – santimirandarp Oct 4 '17 at 17:04
• "These things" is an exceptionally broad field, particularly without an idea of where you are technically. (And, for stuff that's this technical, you can mostly forget about videos.) I don't have anything relevant off the top of my head. – Emilio Pisanty Oct 4 '17 at 17:47
• I have studied those things but have a terrible mess in my head. And I don't know if I understood your general idea about my question. I have read Szabo Ostlund in electronic structure and part of Terrell Hill in statistical th..Well..thanks anyway, help is always welcome. – santimirandarp Oct 4 '17 at 18:10