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Is temperature of a single molecule defined?

This question just cropped up in my mind as I have often heard of laws being violated when it comes to the scale of a single molecule. Does this happen in case of temperature too?


marked as duplicate by akhmeteli, Kyle Kanos, Brandon Enright, Waffle's Crazy Peanut, John Rennie Jan 1 '14 at 9:22

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There are ways to define "temperatures" for single molecules or atoms, but they are subtly different than the way in which temperature is defined for bulk material. Temperature is essentially a statistical quantity as averaged over the various internal states of the material, and for large composite systems at equilibrium, the internal energy levels are Boltzmann distributed, meaning that there is a probability distribution for the possible states the system can occupy. Higher energy states are less likely than lower energy states, with the exact relation determined by the "temperature", which tells you the total average energy content of the material.

For single molecules, you can define temperature in the same way, namely by the occupancy of the various energy levels, but this can lead to bizarre ramifications, such as negative temperatures in lasing materials, and multiple definitions of temperature for molecules whose quantum states are not at equilibrium (often you'll see authors make a distinction between "translational temperature" and "vibrational temperature", for example).

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    $\begingroup$ I am just a high school student who does not know much about thermodynamics. So from the point of view of a layman, is temperature of single molecule defined? $\endgroup$ – user2369284 Dec 26 '13 at 17:33
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    $\begingroup$ @user2369284: It depends on how you define "temperature". However, if you're using the usual definition of temperature, then the answer is no, since temperature is a statistical quantity, and doing statistics on a sample size of 1 molecule is somewhat meaningless. $\endgroup$ – DumpsterDoofus Dec 26 '13 at 17:50
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    $\begingroup$ And what is the definition. $\endgroup$ – user2369284 Dec 26 '13 at 17:51

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