Talk:Absolute zero

real world conditions

What do you mean by real world conditions? I believe that on Earth (in the outdoors) the lowest temperatures are around −50 to −60 °C. However in the laboratory people routinely measure at nanokelvin temperatures (10⁻⁹ K), which is very close to the absolute zero.--Paul Wormer 09:51, 14 December 2009 (UTC)

Doesn't the 3rd Law of Thermodynamics say it's unattainable? Peter Jackson 09:55, 14 December 2009 (UTC)

I see that Ro changed the wording somewhat, but I still don't know what "real world" is supposed to mean. Do you mean "outside the laboratory" or "in nature"? Indeed the 3rd law states the unattainability and 10⁻⁹ K is very close to zero, but not equal to zero. Some experiments are already done at a few hundred picokelvin, so IMHO the non-attainability is pretty academic. --Paul Wormer 14:36, 14 December 2009 (UTC)

Translational motion?

I'm asking this not because I suspect it is wrong, but because I did not understand it upon reading the article.

At absolute zero, the particles should be still, but cannot be because of Heisenberg, at least from the quantum-mechanical viewpoint. Surely

$\Delta x\times \Delta p_{x}\geq {\frac {\hbar }{2}}$

should mean that particles cannot be perfectly still? Currently, it is not clear if the definition includes the particles being completely still. Could someone who understands this in greater detail write a more easily understandable introduction, please? Johan A. Förberg 22:11, 28 July 2010 (UTC)

Johan, it all depends on what you call still (at rest, not varying in time). The sentence the particles are still has a completely different meaning in quantum mechanics than in classical mechanics. The quantities in Heisenberg's relation do not depend on time (are stationary) because they are derived from stationary wave functions and time-independent operators. Whereas x and p_x do depend on time in classical mechanics, they do not in quantum mechanics (in Schrödinger's picture they are time-independent linear operators). Quantum mechanical states (wave functions) of physical systems that do not experience external time-dependent fields are stationary (that is, are essentially time-independent).

As is stated in the article, in quantum statistical mechanics zero kelvin means that all particles are in their stationary ground (lowest energy) state, in which they have "smeared-out"—but time-independent—distributions in x and p_x. Expectation values of x and p_x computed from these "smeared-out" distributions do indeed satisfy Heisenberg's uncertainty relation. In classical statistical mechanics, on the other hand, the particles are at rest (still) at zero kelvin, i.e., all their physical properties (position, velocity, etc.) are time-independent and "sharp" (the opposite of "smeared-out"). --Paul Wormer 09:33, 29 July 2010 (UTC)

Thanks for this one, Paul! --Daniel Mietchen 16:00, 29 July 2010 (UTC)

I think I understand better now. I put a wikilink to quantum mechanics on the page so that others will as well. Thank you! Johan A. Förberg 07:45, 30 July 2010 (UTC)

Good to see you back, Paul ... even if for just a short while!! Best regards, Milton Beychok 15:58, 31 July 2010 (UTC)