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I'm having a discussion with someone. I said that it is -even theoretically- impossible to reach $0$ K, 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 Schrödinger 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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    $\begingroup$ Claiming an edit to fix the capital letter in Kelvin.. Really? $\endgroup$
    – Edward Stumperd
    Commented Jul 25, 2012 at 12:11
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    $\begingroup$ @Christoph: Doesn't your reference say that proper names are upper-case? $\endgroup$
    – Nikolaj-K
    Commented Jul 25, 2012 at 13:19
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    $\begingroup$ 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. $\endgroup$
    – Zo the Relativist
    Commented Jul 25, 2012 at 17:34
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    $\begingroup$ @JerrySchirmer It was a surprise to me too, because I've been capitalizing it for a long time. On the other hand I have not been capitalizing ohm, so I can hardly claim consistency in the matter. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 26, 2012 at 0:53
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    $\begingroup$ @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! $\endgroup$
    – ThePopMachine
    Commented Jul 26, 2012 at 5:35

8 Answers 8

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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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    $\begingroup$ @EdwardStumperd: Your definition of temperature is valid only in classical statistical mechanics. But you are not allowed to use it in the quantum realm. $\endgroup$
    – Arnold Neumaier
    Commented Jul 25, 2012 at 14:15
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    $\begingroup$ 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. $\endgroup$
    – Arnold Neumaier
    Commented Jul 25, 2012 at 14:23
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    $\begingroup$ 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. $\endgroup$
    – Arnold Neumaier
    Commented Jul 25, 2012 at 14:32
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    $\begingroup$ @BenCrowell: glasses are no true equilibrium states, hence retain a nonzero entropy if it were possible to cool them down to zero temperature. $\endgroup$
    – Arnold Neumaier
    Commented Sep 12, 2013 at 10:06
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    $\begingroup$ @MarcelKöpke: Temperature is mathematically well-defined defined only in the thermodynamic limit, i.e., for an infinite system. It becomes an approximate concept for finite systems, but being approximate, the question of reaching exactly T=0 becomes meaningless. Also note that the slightest interaction of a subsystem with its envirenment turns its state into a mixed state, whereas in the absence of degeneracy, T=0 means pure. We cannot shield a subsystem of any significant size from its environment. $\endgroup$
    – Arnold Neumaier
    Commented Mar 30, 2015 at 12:25
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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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  • $\begingroup$ 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. $\endgroup$
    – Edward Stumperd
    Commented Jul 25, 2012 at 13:03
  • $\begingroup$ 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. $\endgroup$
    – Steve Byrnes
    Commented Jul 25, 2012 at 15:17
  • $\begingroup$ 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. $\endgroup$
    – Edward Stumperd
    Commented Jul 25, 2012 at 15:57
  • $\begingroup$ 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. $\endgroup$
    – Steve Byrnes
    Commented Jul 25, 2012 at 19:17
  • $\begingroup$ 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. $\endgroup$
    – Ron Maimon
    Commented Jul 26, 2012 at 1:43
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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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  • $\begingroup$ whatttt. hotter than infinite hot = colder than absolute zero? $\endgroup$
    – Polydynamical
    Commented Aug 24, 2020 at 16:47
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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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  • $\begingroup$ So I was right. What was he talking about then? $\endgroup$
    – Edward Stumperd
    Commented Jul 25, 2012 at 12:21
  • $\begingroup$ 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) $\endgroup$
    – anna v
    Commented Jul 25, 2012 at 13:08
  • $\begingroup$ Ah ok, I know about degrees of freedom and what not, but I didn't realize that that was what he was talking about. $\endgroup$
    – Edward Stumperd
    Commented Jul 25, 2012 at 13:21
  • $\begingroup$ 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". $\endgroup$
    – Edward Stumperd
    Commented Jul 25, 2012 at 13:26
  • $\begingroup$ yes, it seems so to me too. $\endgroup$
    – anna v
    Commented Jul 25, 2012 at 13:30
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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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  • $\begingroup$ You need a system to be undisturbed for a long time to have a well defined energy. $\endgroup$
    – Ron Maimon
    Commented Jul 26, 2012 at 1:44
  • $\begingroup$ 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? $\endgroup$
    – Marty Green
    Commented Jul 26, 2012 at 1:59
  • $\begingroup$ 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. $\endgroup$
    – Ron Maimon
    Commented Jul 26, 2012 at 2:08
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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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    $\begingroup$ 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. $\endgroup$
    – Emilio Pisanty
    Commented Nov 8, 2012 at 15:32
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What people don't understand is that the laws of thermodynamics are not exact in the same way as, for example, energy conservation is. They are only quite probable, meaning that for a finite system there always exists a non-zero propability to break them.

So, even though it's quite improbable to reach $T= 0 \ \textrm{K}$, in principle it is possible. With quite some huge effort, e.g. immense heat baths and heat pumps working for thousands of years, the propability $p$ for the system to be at zero temperature within a finite time $\Delta t$ could reach $p \approx 1$. Then it's only a matter of waiting for chance.

The argument about the harmonic oscillator not reaching $E=0$ in the ground state is no argument against zero temperature, since temperature is more or less the mean excitation energy per degree of freedom. This also corresponds to the fact that potential energy is always only know up to a constant. If $H$ ist the Hamiltonian for the oscillator then $H + \textrm{const.}$ is just as good.

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Literally saying, I like common sense more than physics, because that is more applicable in daily life.

Theoretically, I can say anything but in real life, it might not be so. You see, to make anything colder I must introduce an object more colder than that to transfer the energy it already has. Now since that is not possible, (even theoretically it can't be so) nothing can be colder than 0K.

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    $\begingroup$ There are other ways of cooling things than by heat conduction. Refrigerators cool their working fluid by letting it expand. $\endgroup$
    – user4552
    Commented Jun 1, 2013 at 4:16
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    $\begingroup$ I dont think u should open with that line..however, even if that is not to be considered, I would say that your answer is wrong as you are using circular reasoning. Think more about what you just typed, and try to edit the answer. $\endgroup$
    – Saurabh Raje
    Commented Jun 1, 2013 at 9:51

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