Is zero-point energy the real reason behind zero Kelvin being impossible to reach?

Question

I am aware that the third law of thermodynamics entails an infinite number of steps if one wants to bring down the temperature of a system to $0$ K. I understood this to be a classical telling of a more fundamental limit. The more fundamental limit being the zero-point energy itself; that as long as there is zero-point energy in the system, the system can't actually reach $0$ K, and since it's impossible to remove the zero-point energy of a system, it's impossible to reach $0$ K.

However, the Wikipedia page of Zero-point energy says

Therefore, even at absolute zero, atoms and molecules retain some vibrational motion.

Is my understanding incorrect?

Answer 1

Very incorrect, I'm afraid. The ZPE is the energy at absolute zero. The real question is why the infimum on attainable values isn't one of them. The third law is, of course, an inference in 19th century thermodynamic theory; it's a classical obligation on the universe, not a quantum one. Wikipedia's explanation starts from a different statement of the law - that absolute zero has a constant entropy - then shows why this implies it's unattainable. (Laws of thermodynamics typically have multiple formulations, so the issue is understanding what relates them.)

Answer 2

No. There are three types of degrees of freedom in a molecule : translational, rotational and vibrational modes. Thermodynamic definition of temperature deals with translational modes of particles, which is why temperature is defined as a relation to an average kinetic energy of molecule :

$$ \frac {m \overline {v}^{\,2}}{2}\,=\,{\frac {3}{2}}k_{\text{B}}T $$

So as temperature goes down to $0~K$, molecules stops moving across volume of container. However that doesn't mean that they "stands still" stationary at same coordinates. Due to zero-point field, quantum harmonic oscillator will always have at least zero-point energy $E_0 = \frac {1}{2} \hbar \omega_0$ even at 0K. Knowing this we can extract "vibrational ZPE temperature" defined as: $$ \frac {1}{2} \hbar \omega_0 = \,{\frac {3}{2}}k_{\text{B}}T $$

However, this will not be the same thermodynamic temperature which your thermometer shows.