Vibrational cooling of a molecule

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Malamala
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Hello! I understand that many of the non-optical methods used to cool down degrees of freedom in a molecule (e.g. buffer gas cooling, supersonic expansion) are able to cool down translational and rotational, but not vibrational motion. Is this because the gap between vibrational levels is much higher, so, for example in a buffer gas, one would need to get rid of all that energy with just one collision, which is highly unlikely? However, mathematically, shouldn't we still expect a Boltzman distribution of vibrational levels, too, regardless of their spacing (i.e. they should thermalize with the buffer gas)? Can someone help me understand this issue (if it is indeed an issue) with vibrational cooling? Thank you!
 

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Twigg
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With hot molecules, molecule-molecule collisions turn translational energy into vibrational energy by exciting the molecules into higher vibrational states. This results in a Boltzmann distribution for vibrational state occupancy. If you suddenly cool the translational degrees of freedom, then there will be no source of vibrational excitation. Excited vibrational levels of the ground electronic state decay to lower vibrational states due to electric quadrupole-allowed spontaneous emission on a timescale of 10-100ms lifetime (pretty darn slow). This is why translational cooling =/= vibrational cooling. It takes time for the vibrational levels to settle into a Boltzmann distribution, often longer than you can keep your species trapped.
 
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  • #3
Malamala
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With hot molecules, molecule-molecule collisions turn translational energy into vibrational energy by exciting the molecules into higher vibrational states. This results in a Boltzmann distribution for vibrational state occupancy. If you suddenly cool the translational degrees of freedom, then there will be no source of vibrational excitation. Excited vibrational levels of the ground electronic state decay to lower vibrational states due to electric quadrupole-allowed spontaneous emission on a timescale of 10-100ms lifetime (pretty darn slow). This is why translational cooling =/= vibrational cooling. It takes time for the vibrational levels to settle into a Boltzmann distribution, often longer than you can keep your species trapped.
Thank you! How about rotational levels? If I am to buffer gas cool the molecules, will I have the same problem i.e. the rotational motion will not be Boltzman distributed at the same temperature as the translation motion and it will take a long time (probably even longer than vibrational case, as the lifetimes are longer) for the rotational motion to reach thermal equilibrium?

And why does supersonic expansion lead to rotational cooling, but not vibrational?
 
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Twigg
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Sorry for messy reply. Let me know if anything unclear. Energy splitting is much lower for rotational levels, so its easier for collisions to cause diabatic excitations between rotational states, allowing thermalization. Even at supersonic cool temperatures rotational states thermalize quickly.

Regarding my last post, in hindsight I'm actually not sure what the dominant cooling rate for vibrations is (collisions or spontaneous emission) at supersonic beam temperatures. But spontaneous emission sets the lower limit for how fast the excited states come down.

The advantage of a CBGB here is the lower velocity, meaning for a given beam length you get more time to thermalize (experiment size is finite in real world applications).

Even if you do end up with a high rotational temperature, rotational repumping is easy peasy compared to vibrational repumping, though both take a lot of precious time (which translates into beam length if you don't have the ability to trap your molecules).
 

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