# Is helium liquid at 0 K?

I just saw in the dynamic periodic table that He is liquid at $$-273.15\ ^\circ \rm C$$. Is that true?

How is that even possible? Can someone explain?

• But the question asked I believe is different - even strictly theoretically at 0K helium is still liquid - it does not freeze. You need to apply some pressure to make it freeze (around 25 atmospheres at 1K ). This inability to freeze is mostly due to quantum effects. Zero point energy is just too high. – Jarosław Komar Jun 19 at 6:28
• Ironically SX has tagged this as a "Hot Question" – Jasen Jun 19 at 21:33
• Comments are not for extended discussion; this conversation has been moved to chat. – rob Jun 20 at 19:00

The thermodynamic phase of a material is never a function of temperature only. The correct statement is that helium remains in a liquid state at whatever small temperature achievable in a laboratory at normal pressure. It is well known that $$^4$$He freezes into a crystalline solid at about 25 bar.

Such peculiar behavior (helium is the only element remaining liquid at normal pressure close to $$0$$K) is partly due to its weak interatomic attraction (it is a closed shell noble gas) and partly to its low mass, which makes quantum effects dominant. A signal of the latter is the well-known transition to a superfluid phase at about $$2$$ K.

A theoretical explanation of the avoided freezing at normal pressure could be done at different levels of sophistication. No classical argument can be used, since a classical system would stop moving at zero temperature. A hand-waving argument is related to the zero-point motion of the system, which is large for light particles. A theoretically more robust approach is based on the density functional theory of freezing (DFT).

The qualitative explanation of the freezing process based on DFT is that the difference of free energy between the solid and the liquid is controlled by the competition between the contribution of the change of density (favoring the liquid), and the contribution of the correlations of the liquid phase in reciprocal space, in particular at wavelengths close to the first reciprocal vector of the crystalline structure (favoring the solid). It turns out that the latter contribution is particularly weak in liquid $$^4$$He.

• I've just been assuming the difference between real 0k and lab 0k is big enough that we haven't observed the real freezing point yet. – Joshua Jun 20 at 3:59
• @Joshua all the evidence from experiments and theory is consistent with the stability of the liquid phase down to $0$K. Surprises may always arrive. But it would be a big surprise. – GiorgioP Jun 20 at 5:39
• DFT only considers electromagnetic force interaction. while (it is ok that this method) assumes static nuclei (simulating 0 K), ignoring other forces seems to be 'insufficiently' conclusive imho. +1 for highlighting materials simulation as a way to address this question. /(^_^) – p._phidot_ Jun 21 at 18:11
• @p._phidot_ maybe there is some misunderstanding. DFT of freezing is not a simulation method and does not assume anything about nuclear motion. It focuses on the difference between thermodynamic potentials of the liquid and solid phase, where liquid or solid only refers to the absence/presence of finite reciprocal lattice vecotr components of the one-particle density, – GiorgioP Jun 21 at 18:18
• presence of finite reciprocal lattice vector explains it.. Thanks for the clarification. ( : – p._phidot_ Jun 21 at 18:31

Temperature is interesting in a logarithmic way. There is as much potential for interesting physics between $$1\rm\,K$$ and $$10^{-6}\rm\,K$$ as there is between $$1\rm\,K$$ and $$10^{+6}\rm\,K$$. The difference is that we live in the upper interval, so we have lots of experience and a finely-honed intuition which we call "classical thermodynamics." To explore the sub-kelvin temperature regime, we have to build complicated cryostats and understand quantum mechanics.

Note that $$\rm -273\,^\circ C = 0.15\,K$$ is a positively balmy temperature if you're interested in microkelvin phenomena.

[source]

Here's a phase diagram for helium-3 that shows both the pressure and the temperature on logarithmic scales, going down to $$\rm10\,\mu K$$. (Dear Britannica-from-twenty-years-ago: using a different colored font to represent a unit change is bad design.) For helium-4 the diagram is similar, but the temperatures and pressures of the phase transitions are different: helium-4 and helium-3 have very different masses. The number of distinct phases for helium-4 versus helium-3 is also different, because helium-3 has a different nuclear spin than helium-4 and therefore different statistical properties. There are a number of features on this diagram that we hide from students who are new to thermodynamics, like the critical point on the curve between the normal liquid and the vapor phase.

You can see from this diagram that helium (or at least helium-3) does have a low-temperature solid phase, but not at one atmosphere of pressure. Furthermore, the minimum pressure to support the solid phase appears to be temperature-independent, even at the little triangle of superfluid-A phase. I bet there's an interesting explanation about why the solidification pressure doesn't depend on temperature, but I don't know it.

Helium isn't alone in having a low-density, low-temperature phase which is not solid but is fundamentally quantum-mechanical. It's just that helium's mass is so low that these quantum-mechanical effects happen in a relatively accessible part of the phase space. Superfluidity in helium-4 is closely related to the phenomenon of Bose-Einstein condensation, which happens in rubidium vapor at around $$\rm 10^{-7}\,K$$. Likewise, superfluid helium-3 is a kind of Fermi-Dirac condensate, which has been observed in nanokelvin (!) potassium vapor. Both rubidium and potassium are solids at "normal" pressures, all the way up beyond room temperature, but their low-pressure, low-temperature quantum-mechanical behaviors have a lot in common with superfluid helium.

• Re the flat curve in the p-T plane, this is because there is no latent heat of melting so the Clausius-Clapeyron equation predicts the boundary to be flat. This is because as zero temperature the entropy of both the solid and liquid phases must go to zero (by the third law). (Source: Nature volume 165, pages829–831 (1950).) – jacob1729 Jun 19 at 19:44
• Ok, He is liquid at $10^{-9} K$. What is at $10^{-18} K$? Is it still liquid? – peterh Jun 20 at 16:58
• @peterh If I understand the comment by jacob1729, the solid- liquid phase boundary should remain independent of temperature once the superfluid state has been reached. – rob Jun 20 at 17:35
• Ok, so in other words, we aren't actually talking about absolute zero, we are talking about a range of temperature that is > 0K but just can't be resolved in a diagram that only uses centiKelvin as resolution. Literal 0 K isn't accessible experimentally, is it? Do we actually expect Helium to remain liquid at normal pressure, no matter how close to 0K we get? – kutschkem Jun 21 at 8:35
• @kutschkem The inaccessibility of absolute zero is one statement of the third law of thermodynamics. Our models of superfluidity and Bose-Einstein condensation are about what happens when a macroscopic system reaches its quantum-mechanical ground state, so we are predicting there aren’t any lower-temperature phases to change to. – rob Jun 21 at 11:42

From the Wikipedia article on helium:

Unlike any other element, helium will remain liquid down to absolute zero at normal pressures. This is a direct effect of quantum mechanics: specifically, the zero point energy of the system is too high to allow freezing. Solid helium requires a temperature of 1–1.5 K (about −272 °C or −457 °F) at about 25 bar (2.5 MPa) of pressure.

So yes, helium is liquid down to absolute zero (unless the pressure is high enough to cause it to solidify), and the reason is because of quantum effects - in particular, the "zero point energy", or the lowest possible energy a system can have while still obeying the Heisenberg uncertainty principle.

Yes, it is true.

If you cool off a gas, you can make it condense into a liquid. For some gases, this is easy because the temperature at which they turn into liquids is fairly high. For example, for water vapor (steam) this happens at +100 C. But for helium gas, it does not turn into a liquid until it has been chilled down to -269 degrees C. Wikipedia has a pretty good article about this.

• no my doubt is like at 0K the motion of all the particles should stop including helium right. if that happens then it will be a solid. but the fact that they say Helium will be liquid means that it can still flow. How? – Christina Melita Jun 19 at 7:00
• @ChristinaMelita, all motion does not stop at superlow temperatures. Quantum mechanics steps into the picture and messes things up- see Rob's answer below. – niels nielsen Jun 19 at 17:04

Other answers have highlighted the remarkable properties of super-cold helium. The apparent conflict between liquid (implying motion) and absolute zero (implying no motion) might still be troubling, though, so I'll add some basic points that apply to all substances, not just to helium.

Other answers have already pointed out that quantum physics is essential at such low temperatures. Instead of saying that all motion stops, we can say it in a more quantumy way like this: absolute zero ($$T=0$$ Kelvin) has been attained if the system is in the state of lowest total energy. Lowest total energy implies zero momentum... but if the momentum is zero, then how can the substance flow? The apparent conflict dissolves when we remember these basic points:

• Even if all of the atoms in a sample of helium could have zero momentum in one reference frame, they wouldn't have zero momentum in other reference frames. Clearly, no matter how cold the sample might be, its coldness cannot prevent it from being in relative motion compared to something else in the universe (like, say, the laboratory), because motion is relative. The conceptual significance of this is highlighted in another answer by knzhou.

• What if we try to ignore that issue by working in the sample's "rest frame"? The quotation marks are a warning! In quantum physics, the (badly named) uncertainty principle implies that zero momentum and localized in a finite region of space are mutually contradictory conditions. Absolute zero is not compatible with being contained in a laboratory of finite size. The same is true for any specific nonzero momentum, so the reference-frame issue highlighted in the previous paragraph doesn't make this go away — not for liquids or solids or anything-else-ids. Absolute zero is absolutely unattainable.

• The ground state of helium doesn't have all particles with 0 momentum. If it were a BEC this would be true, but of course a BEC wouldn't be a liquid. Interactions prevent the state you are thinking of from being an eigenstate, and thus it is not the ground state. – jacob1729 Jun 19 at 19:34
• @jacob1729 You're right, of course, but that doesn't change the message. (Note the words "Even if..." in the third paragraph.) A state of zero total momentum would be translation-invariant, so nothing would "flow" in that case, but overall relative motion would still be possible by Lorentz symmetry because the substance has nonzero mass, and a perfectly sharp value of the momentum (zero or otherwise) is inconsistent with being localized anyway. Those messages still apply in the presence of interactions. – Chiral Anomaly Jun 19 at 21:04