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Heisenberg’s uncertainty principle is very commonly used to justify why we can’t get to$\ 0$ Kelvin by saying it is impossible to make particles stand still because they will either have an undetermined position or an undetermined momentum.

However, in solid state physics, it is common refering to a system where all the particles are in the lowest possible energy levels (in the ground state) as being at$\ 0$ Kelvin.

If we have a system composed of a set of harmonic oscilators, all of them in the ground state, than the “system at ground state” picture would tell us the system is at$\ 0$ Kelvin. On the other hand, if we considered the “uncertainty implies no$\ 0$ K" picture, the zero point motion of the oscilators would be used to argue that the system is not at$\ 0$ Kelvin.

How would this tension be solved? In this situation, how would we undoubtedly state whether the set of harmonic oscilators in the ground state is at$\ 0$ K or not?

Qmechanic
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Lucas Francisco
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1 Answers1

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The picture that the Heisenberg uncertainty principle has to do with temperature is simply misleading. Temperature has to do with thermodynamics. So a correct formulation of what you are asking is given by the third law of thermodynamics, a consequence of which is that you cannot achieve zero Kelvin in finte number of steps, or alternatively, with finite cost (something experimentalists know well).

Heisenberg uncertainty principle can sometime be used to provide an estimate for the energy of the ground state. This is is only a little more accurate than dimensional analysis. In any case this shows that Heisenberg uncertainty principle is perfectly compatible with ground states which in turn means that the temperature is zero Kelvin.

lcv
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