cryogenicsenergytemperaturethermodynamics
I have read that Helium does not freeze at absolute zero under normal pressures.
How could this be possible given that the absolute zero is the lowest attainable temperature and at that temperature, all random movements of the atom stop?
Shouldn't the atoms just stop vibrating and solidify instantly? Why do they possess kinetic energy at absolute zero?
If carefully interpreted and converted to mathematics, the passages are not contradicting one another.
The uncertainty principle guarantees that there is some zero-point energy that can't be eliminated (the second passage) – it is the energy of the ground state of the physical object (atom or a macroscopic piece of a material).
On the other hand, the kinetic energy that increases with $T$ and reaches $E_k=0$ for $T=0$ is the excess energy above the energy of the ground state: if we want $E=0$ for $T=0$, the additive shift $\Delta E$ of the energy has to be chosen appropriately to guarantee this condition. The right additive shift is really the "subtraction of the ground state energy", i.e. of the minimum energy eigenvalue that the physical system may have.
If a physical system is frozen to $T=0$, it means that this physical system is "certain" (100%) to be found in the ground state i.e. the energy eigenstate with the minimum allowed value of $E$. In most contexts, this ground state is pretty much unique. It contains "some motion", by the uncertainty principle, but "the amount of motion above the minimum level allowed by the uncertainty principle" is zero.
In the first passage, Feynman really talks about the overall kinetic energy of the atom as it moves through space. This is indeed $3kT/2$ and strictly goes to zero for $T=0$. No subtraction is needed here. The minimum kinetic energy of the "overall motion of the object through space" is really $E_k=0$ and the corresponding momentum is $\vec p =0$. This doesn't violate the uncertainty principle because, indeed, $\Delta x = \infty$ or $\Delta x \to \infty$. The position of an atom frozen to $T=0$ (which can only be approached in the real world) is absolutely undetermined. The subtle issues of the zero-point energy only arise for the "internal energy" i.e. the relative motion of parts of a bound state (e.g. the motion of electrons around the nuclei). Only for the bound state, we really know that $\Delta x$ cannot be infinite. The constituents' being "bound" means that the distance between them is finite and bounded from above.
For the sake of simplicity, I will answer the question for the bosonic species He(4). Although there are some subtleties for the Fermionic species He(3), due to the presence of total spin-$\frac{1}{2}$, the main message is the same.
The key points are summarized here as follows:
Now we can be quantitative using the harmonic oscillator model. The potential between two atoms is short-ranged repulsive and it becomes attractive for the long-range. Near the potential minimum, the attractive potential can be modeled via the Lennard-Jones potential $-$ $$V(r) = \epsilon_0\left(\frac{d^{12}}{r^{12}}-2\frac{d^6}{r^6}\right),$$ where the parameters $\epsilon_0$ is the trap-depth, i.e., the minimum potential and $d$ is the interatomic separation at the minimum potential.
Since the question involves comparison with other gases, below I put the parameters of He(4) and the closest noble gas neon $$\begin{array}{|l|c|c|} \hline \text{Gas Name} & \text{$\epsilon_0$ [meV]} & \text{$d$ [nm]} \\ \hline \text{He(4)} & 1.03 & 0.265 \\ \hline \text{Neon} & 3.94 & 0.296 \\ \hline \end{array}$$
Now, using the parameters from the above table, we can estimate the zero-point energy in three-dimensions $E_0 = \frac{3}{2}\hbar \omega_0$, assuming an fcc crystal lattice. The oscillation frequency can be estimated as
$$\omega_0 = \sqrt{\frac{4k}{m}},$$ where $$k = \frac{1}{2}\frac{d^2}{dr^2}V(r) = \frac{36\epsilon_0}{d^2}.$$
This expression leads to a $E_0 \approx 7 $ meV for He(4), while the binding energy for atoms is $\approx 1.03$ meV. Therefore the zero-point energy is enough to destroy any crystalline structure of He(4). And this is the reason why He(4) is not found in crystal form, at normal pressure. However, if we compare the binding energy 3.94 meV and the zero-point energy $\approx 4$ meV of neon, we see that the gas can be put into crystal form at relatively small pressure.
To understand the effect of pressure, we look at the following phase diagram of He(4), where we see that the liquid/gas forms continue down to ~0 K, if the pressure remains below 25 atm. The figure distinguishes the two phases He-I and He-II separated by the black line. The superfluid fraction is shown to increase dramatically as the temperature drops.
Best Answer
You have been misled by the idea that temperature is a measure of energy. While this is approximately true at high temperatures, it is not correct at low temperatures. Temperature is actually a measure of entropy; the derivative of entropy with respect to internal energy at constant particle number and volume is inverse temperature. At very low temperatures, quantum mechanical effects become important, and even at absolute zero (0 K), the particles have energy, known as zero point motion. In helium, this zero point motion is large enough to prevent the atoms from sticking together as a solid - it remains a liquid. Above roughly 3.2 MPa Helium-3 becomes solid at high pressure. For Helium-4 it will become solid above ~2.5 MPa. http://ltl.tkk.fi/research/theory/helium.html