Under normal atmospheric pressures, liquid helium does not freeze even when cooled very close to absolute zero. This is attributed to the uncertainty principle or due to zero-point energy. But the quantum uncertainty or zero-point energy is not an exclusive feature of liquid helium only. Then, why should it stop the freezing of helium but not that of other liquids? If it is strong in helium, then why?


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:

  • The energy contribution from the zero-point motion is seven times larger than the depth of the attractive potential between two He(4) atoms. Therefore, the zero-point motion is enough to destroy any crystalline structure of He(4).
  • Helium is special because of the combination of its small mass and the value of binding energy.
  • The zero-point energy for the other gases is either comparable or far smaller than the depth of the attractive potential that holds the atoms.

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.

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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.

enter image description here

  • 1
    $\begingroup$ For completeness it would better useful to discuss the pressure at which Helium does solidify. $\endgroup$
    – KF Gauss
    May 26 '20 at 19:18
  • $\begingroup$ Your current expression for Lennard-Jones potential grows unboundedly at infinity and vanishes at zero. $r$ and $d$ should be swapped. $\endgroup$
    – Ruslan
    May 26 '20 at 20:45
  • $\begingroup$ Also, your expression for $\omega_0$, when using your expression for $k$, yields a quantity with units $\frac{\mathrm{m}\sqrt{\mathrm{kg}}}{\mathrm{s}^2}$, which is not a unit of (angular) frequency. $\endgroup$
    – Ruslan
    May 26 '20 at 21:02
  • $\begingroup$ @KFGauss, thank you for the suggestion. I added some details and the phase diagram. Cheers! $\endgroup$ May 27 '20 at 0:16
  • 1
    $\begingroup$ Could you add a citation to your source of LJ parameters and functional form of $V$? These seem to give a bit different results from those in Helium Cryogenics by Steven W. Van Sciver, $\S3.2.1$ Intermolecular Interactions. In particular, with the data from this book I get zero-point energies of $6.05\,\mathrm{eV}$ for helium and $4.61\,\mathrm{eV}$ for neon, and binding energies resp. $0.881\,\mathrm{meV}$ and $3.08\,\mathrm{meV}$. $\endgroup$
    – Ruslan
    May 27 '20 at 5:48

Helium, at pressures below 25 atm and absolute zero, does not freeze because it's zero point energy is high enough to stop it from going into a solid phase, and hence is stable as a liquid. Other gases in general do not have such high zero point energies, and thus transition from liquid to metal as the temperature plunges.

As to why it Helium has a high zero-point energy, the analysis is very complicated, but in 1935, F. London did a (sophisticated back of the envelope) calculation that explained the phenomenon and in 1950, C.L. Pekeris increased the accuracy of prediction by one order of magnitude.

London had essentially summed this up as :

One can roughly take account of the decisive contribution of the zero point energy which is due to the quantization of the mean free path. Closest packed structure has been found to be stable only under pressure and this seems to explain why solid helium can only exist, even at the absolute zero, under pressure. If no external pressure is applied, a configuration with the coordination number four has proved to have considerably lower energy. In seems that this configuration gives a rough model of the liquid modification of helium which is stable at the absolute zero.


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