I am studying some quantum mechanics and thermodynamics. I have seen that in QM temperature doesn't appears in the formulas. But it appears in statistical mechanics formulas.

The question is: Temperature appears as a phaenomena of a big number of particles? How big must be that number?


I have seen the link Can a single molecule have a temperature? but there is no agreement on the answers. So, this question might be Can a single molecule have a temperature? revisited

  • $\begingroup$ @JohnRennie Hi! It might be, but I am not sure it is answered there.. $\endgroup$ – user153036 Sep 29 '17 at 4:47
  • $\begingroup$ @JohnRennie look, for example, to the first comment on the first answer.. $\endgroup$ – user153036 Sep 29 '17 at 4:49
  • $\begingroup$ @HernanMiraola That doesn't really make it not a duplicate. It actually makes the situation worse. What if you got an answer you considered "better" on this new question? Then all our answers would be split between multiple questions, and the site loses effectiveness as a reference site. $\endgroup$ – JMac Sep 29 '17 at 10:45
  • $\begingroup$ @JMac yeah you are right..it is my obsession to understand this problem..I am sorry. $\endgroup$ – user153036 Sep 29 '17 at 21:35

For a single particle to have a defined instantaneous temperature is ... unlikely.

Maybe a polymerized blob (vulcanized auto tire?) can have particle character and lots of internal energetic degrees of freedom, and have a temperature that a thermometer could measure. It's a very big molecule, but perhaps still a 'single particle'.

A point particle kinetic energy does NOT define a temperature, in the absence of a stationary reference. Synchrotron beam particles can be ultrarelativistic, but are still cold (or the beam wouldn't be stable). Random motion of those particles is thermal, the velocity of the bunch is not.

A single particle might be at equilibrium with liquid helium, but be inside a fast-moving container. Randomness matters.

  • $\begingroup$ I could understand only a few words of what you say, but thanks for try to help.. $\endgroup$ – user153036 Sep 29 '17 at 21:25

That's to some degree a matter of interpretation and definition.

You yourself have notice so much (on your remark about conflicting answers in the question this one is duplicate of). Even if attributing a temperature to a single molecule is at best a rough approximation, there might be circumstances where it's consistent and useful.

Another example of "lack of agreement" is whether it makes sense to talk about two different temperatures in plasmas: the electron temperature and the ion temperature. The electrons among themselves are in equilibrium, as are the ions among themselves, but the equilibration between both groups happens in a longer time scale, so in practice these temperatures are used in plasma physics, but many deem it inappropriate, as temperature would be defined only in (full?) equilibrium.


Temperature actually corresponds to the root mean square velocity of the collection of particles. If the there is only one particle, then the root mean square velocity is just the velocity of that single particle. So, we can say that even a single particle can possess certain temperature depending on its velocity.

Usually we see the change in temperature of two objects into thermal equilibrium as the statistical probability. The heat exchange between these two object tend to the highest multiplicity (the thermal equilibrium) and the higher the number of molecules, the higher the multiplicity too.

  • $\begingroup$ thanks! but the particle you describe is in a box with lot of particles, isn't it? I mean only one particle $\endgroup$ – user153036 Sep 29 '17 at 4:48
  • 1
    $\begingroup$ Yes, even a single particle can have a temperature. Temperature is the measure of the average of kinetic energy of a particle. So as long as the particle has a kinetic energy, it also has some temperature. $\endgroup$ – Kelvin Onggadinata Sep 29 '17 at 4:51
  • $\begingroup$ I think any equation left you say that..only if we start by the statistical way.. $\endgroup$ – user153036 Sep 29 '17 at 4:53