# Is it theoretically possible to reach 0 kelvin?

I'm having a discussion with someone. I said that it is -even theoretically- impossible to reach 0K, because that would imply that all molecules in the substance would stand perfectly still.

He said that this isn't true, because my theory violates energy-time uncertainty principle. He also told me to look up the Schrodinger equation and solve it for an oscillator approximating a molecule. See that it's lowest energy state is still non-zero.

Is he right in saying this and if so, can you explain me a bit better what he is talking about.

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Claiming an edit to fix the capital letter in Kelvin.. Really? – Edward Stumperd Jul 25 '12 at 12:11
– Christoph Jul 25 '12 at 12:33
@Christoph: Doesn't your reference say that proper names are upper-case? – Nick Kidman Jul 25 '12 at 13:19

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That is so contrary to common usage to be ridiculous. I have never seen anyone write "5 newtons" or "10 joules" or "5 kelvin", though the abbreviations are much more common anyway. – Jerry Schirmer Jul 25 '12 at 17:34

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@JerrySchirmer: No, it is so not contrary to common usage. Just because you don't know the correct usage, doesn't mean you get to hide behind "common usage". The Wikipedia articles on "kelvin" and "newton", among others, clearly know the right usage. The article on "kelvin" even adds "When reference is made to the unit kelvin (either a specific temperature or a temperature interval), kelvin is always spelled with a lowercase k..." Let me ask you this: How do you spell "kilonewton" and "millikelvin"? milliKelvin? Now if anything is rediculous, it's that! – ThePopMachine Jul 26 '12 at 5:35
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## 6 Answers

By the third law of thermodynamics, a quantum system has temperature absolute zero if and only if its entropy is zero, i.e., if it is in a pure state.

Because of the unavoidable interaction with the environment this is impossible to achieve.

But it has nothing to do with all molecules standing still, which is impossible for a quantum system as the mean square velocity in any normalized state is positive.

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Just like Steve B, your giving another reason why it's impossible but not explaining why my reasoning is wrong. In fact you're even admitting that all molecules standing still is impossible. So by the definition of temperature, you're saying my reasoning is right. – Edward Stumperd Jul 25 '12 at 14:12

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@EdwardStumperd: Your definition of temperature is valid only in classical statistical mechanics. But you are not allowed to use it in the quantum realm. – Arnold Neumaier Jul 25 '12 at 14:15
Oh, I was unaware of that. What would be the quantum definition of temperature? – Edward Stumperd Jul 25 '12 at 14:17

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In thermodynamics it is the quantity conjugate to entropy, in other words, the integrating factor for changes of entropy. This is valid both in classical and quantum mechanics. To work out what it means microscopically, one must look at specific models, and then CM and QM differ. In a grand canonical ensemble, it turns out that $k_B$ times the temperature is the inverse of the factor $\beta$ that multiplies the Hamiltonian in the expression $\rho=Z^{-1}e^{-\beta (H+\mu N)}$ for the density matrix. – Arnold Neumaier Jul 25 '12 at 14:23
This works if the temperature nonzero. But the density has a well-defined limit for $\beta\to\infty$ which defines the zero temperature case. The limit is the orthogonal projector to the eigenspace of the ground state energy. in the usual case that the ground state is nondegenerate, the result is a pure state. – Arnold Neumaier Jul 25 '12 at 14:32
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I think you are both wrong.

"The lowest energy state still has non-zero energy" does not mean that the temperature cannot be zero. If the system is in the ground state with 100% probability, then the temperature is zero. It doesn't matter what the ground state energy is.

It's true that all molecules in the substance would stand perfectly still at absolute zero [well, they don't have exact positions by the uncertainty principle, but the probability distribution of position would be perfectly stationary]. But so what? Why would that make absolute zero impossible? [see update below]

Nevertheless, there is no process that can get a system all the way to absolute zero in a finite amount of time or a finite number of steps. There's just no way to get that last little bit of energy out. This is one aspect of the third law of thermodynamics, as discussed in some (but not all) thermodynamics textbooks.

-- UPDATE --

It seems likely that I misunderstood. By "stand perfectly still", I guess you meant "have a fixed and definite position, and a fixed and definite velocity equal to 0". If that's what you meant, then "standing perfectly still" is indeed impossible (because of the Heisenberg Uncertainty Principle). But "standing perfectly still" is not expected or required to happen at absolute zero. Again, a harmonic oscillator which is in the ground state with 100% probability is at absolute zero, but does not have fixed and definite position or velocity.

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You explained why you think he is wrong, but -while giving a vague direction of what may be another reason why it's theoretically impossible to reach 0K- you didn't explain why you think my reasoning is wrong. – Edward Stumperd Jul 25 '12 at 13:03
You didn't tell us your reasoning. Why do you think it is impossible for all molecules in the substance to be perfectly still? In your question you gave no reason whatsoever for this belief. I can't explain why your reasoning is wrong if I don't know your reasoning. – Steve B Jul 25 '12 at 15:17
I know it's not a full explanation as I don't give a reason why it would be impossible for all molecules in the substance to be perfectly still (1), but that wouldn't matter for the question as long as (1) is true and we use the classical definition of temperature. However, all of this is irrelevant now after Arnold's answer. – Edward Stumperd Jul 25 '12 at 15:57
the problem is that the words "stand perfectly still" is ambiguous. If you had said WHY it is impossible to "stand perfectly still" then it would have helped us understand what you meant by "stand perfectly still". It seems my first guess at what you meant was wrong. I updated my answer. – Steve B Jul 25 '12 at 19:17
This answer is fine, but there is a small issue--- how do you determine an (isolated) system is in it's ground state with certainty? The more degrees of freedom you have, the harder it is. If you have an atom, you can determine it is in its ground state, perhaps with something approaching certainty, but there will be radiation surrounding the atom. If you make a cavity to cool the radiation to nothing, you will have to cool the cavity, and so on, so that the third law tells you that there will never quite be certain its in the ground state. – Ron Maimon Jul 26 '12 at 1:43

For a temperature to be definable and measurable the distribution of the kinetic energies of the molecules in the medium under discussion should be known.

The process of cooling involves removing thermal energy from a system. When no more energy can be removed, the system is at absolute zero, which cannot be achieved experimentally. Absolute zero is the null point of the thermodynamic temperature scale, also called absolute temperature. If it were possible to cool a system to absolute zero, all motion of the particles comprising matter would cease and they would be at complete rest in this classical sense. Microscopically in the description of quantum mechanics, however, matter still has zero-point energy even at absolute zero, because of the uncertainty principle.

The uncertainty principle assures that molecules cannot stay perfectly still and continue being in a certain position , i.e. in the material under study. Certainly not all molecules of the material, this would be necessary to define a 0K temperature.

The solution with the vibrational degrees of freedom that molecules may have is not conclusive , though sufficient as proof for that the specific material that displays these vibrational modes cannot go to 0K. It is the HUP that is general for all materials.

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So I was right. What was he talking about then? – Edward Stumperd Jul 25 '12 at 12:21
he was talking about molecules where the temperature is also determined by the rotational and vibrational degrees of freedom, missing in single atom gases for example. So although it is correct that a molecule will always have some vibrational energy and thus contribute to temperature that way, and thus not reach 0k, it is not a universal argument, since there exist single molecule materials, with no vibrational rotational degrees of freedom. ( I gave a link for degrees of freedom) – anna v Jul 25 '12 at 13:08
Ah ok, I know about degrees of freedom and what not, but I didn't realize that that was what he was talking about. – Edward Stumperd Jul 25 '12 at 13:21
Also: if that is true isn't he just trying to prove the exact thing that I am saying: "it is impossible for all molecules to stand perfectly still, which would be necessary to achieve 0K, so reaching 0K is impossible". – Edward Stumperd Jul 25 '12 at 13:26
yes, it seems so to me too. – anna v Jul 25 '12 at 13:30

I wonder why the measurement postulate has not been mentioned so far. Consider a cubical microcrystal of sodium chloride containing 64 atoms (4 on each side). If we cool it off so it is as close to absolute zero as possible, then we can represent its state as a superposition of pure states. One of those states is the ground state. If we then measure its energy, is there not some finite probability that it will be found in its ground state?

The atoms will not be stationary. They still have their zero-point energy. But in the ground state the temperature of the crystal is absolute zero.

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You need a system to be undisturbed for a long time to have a well defined energy. – Ron Maimon Jul 26 '12 at 1:44
Ron, you haven't addressed my point about the measurement postulate. Isn't the act of measurement supposed to force the system into a definite eigenstate? – Marty Green Jul 26 '12 at 1:59
How long to prepare the system and do the measurement? If you want to be sure it's in the ground state, you have to leave the system alone forever. The third law is asymptotic, the longer you are willing to wait, the closer you can come. – Ron Maimon Jul 26 '12 at 2:08

from WP-negative temperature

In physics, certain systems can achieve negative temperature; that is, their thermodynamic temperature can be expressed as a negative quantity on the kelvin scale.

A substance with a negative temperature is not colder than absolute zero, but rather it is hotter than infinite temperature. As Kittel and Kroemer (p. 462) put it, "The temperature scale from cold to hot runs: +0 K, . . . , +300 K, . . . , +∞ K, −∞ K, . . . , −300 K, . . . , −0 K."

. The inverse temperature β = 1/kT (where k is Boltzmann's constant) scale runs continuously from low energy to high as +∞, . . . , −∞.

from Positive and negative picokelvin temperatures :
... of the procedure for cooling an assembly of silver or rhodium nuclei to negative nanokelvin temperatures.

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This is what my Science teacher said on the matter. Nothing can reach absolute zero because Energy is linked to Mass, in the sense that if there is no energy, there is no mass. It would disappear. That can't happen due to other laws, so 0K can't be reached.

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This is not quite true. 0K is not reached when there is absolutely no energy, but rather when the system in question is in its lowest energy state as allowed by the physical laws it has to obey. Thus for a gas in a box, if all the particles are stationary, then their motional degrees of freedom are at 0K even though they have energy in the form of mass - they just don't have kinetic energy. – Emilio Pisanty Nov 8 '12 at 15:32
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