# Thermodynamics at absolute temperature [closed]

Assume I have kept a single molecule in vacuum and in absolute zero (0 K) temperature. There is no energy left in the particle, and hence any type of motion is absent. Now, what happens when I increase the temperature of that system? Does the molecule start to move? Does the energy increase? What does temperature even mean in vacuum? I don't know I'm asking a good question, but this troubles me a lot.

## closed as unclear what you're asking by Mithoron, DrMoishe Pippik, Todd Minehardt, tschoppi, TyberiusJan 28 at 15:16

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• Temperature in vacuum is not a thing at all. To have temperature, you must have molecules in the first place. Then yes, as you heat them, they start to move faster. – Ivan Neretin Jan 27 at 15:34

## 1 Answer

The situation you described is not a vacuum because it has a single molecule in it, so what you're asking about is how the concept of temperature applies to an isolated molecule.

First, the idea that you can start at absolute zero even with a single molecule is not consistent because there will always be fluctuations/movement possible in the ground state of a quantum system (the zero point energy or ZPE). The cleanest example of this is the quantum harmonic oscillator: a particle in the ground state has an average position of zero ($$\langle x \rangle = 0$$), but the fluctuations are greater than zero ($$\langle x^2 \rangle > 0$$). Let's think about this in terms of the uncertainty principle. The uncertainty principle is given by $$\Delta p\Delta x \geq \hbar /2$$ where $$\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}$$ and $$\Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$$. Because $$\langle x^2 \rangle >0$$ and $$\langle x \rangle = 0$$, we can see that the uncertainty in the momentum is also non-zero. As we know, the kinetic energy is $$p^2 / 2m$$, so our energy uncertainty is non-zero as well.

Second, and onto the main point of your question, you can certainly define a concept of temperature with only one molecule. Consider that even a single molecule has many different energy levels (i.e. electronic, vibrational, and rotational degrees of freedom are all available). The partition function for our system is still defined as $$Z = \sum_i \exp{\left(-\frac{E_i}{k_bT}\right)}$$ where $$E_i$$ is the energy of the $$i$$th state.(See Note) So, if we start very near absolute zero we will see only contributions from the ground state, but as we increase the temperature of the system we will observe non-zero probabilities of populating excited states. Nothing is broken about the concept of temperature in this situation, the energy scales are just considerably smaller than those in an ideal gas.

Note:

Your initial reaction to applying thermodynamic principles to a single molecule might be to think that it doesn't apply because we only have one molecule and not an ensemble. However, within the density matrix formulation of statistical mechanics, the quantum uncertainty of a system is entirely indistinguishable from the statistical uncertainty of a system.