-273 degree Celcius

[exchemist]

Well this is a site for kids, but it describes it succinctly:
"So, it's cold. A cold ice cube is still a solid. When you get to a temperature near absolute zero, something special happens. Atoms begin to clump. The whole process happens at temperatures within a few billionths of a degree, so you won't see this at home. When the temperature becomes that low, the atomic parts can't move at all. They lose almost all of their energy.

Since there is no more energy to transfer (as in solids or liquids), all of the atoms have exactly the same levels, like twins. The result of this clumping is the BEC. The group of rubidium atoms sits in the same place, creating a "super atom." There are no longer thousands of separate atoms. They all take on the same qualities and, for our purposes, become one blob. "
http://www.chem4kids.com/files/matter_becondensate.html

"Bose-Einstein condensate (BEC), a state of matter in which separate atoms or subatomic particles, cooled to near absolute zero (0 K, − 273.15 °C, or − 459.67 °F; K = kelvin), coalesce into a single quantum mechanical entity—that is, one that can be described by a wave function—on a near-macroscopic scale."
https://www.britannica.com/science/Bose-Einstein-condensate

Ah OK, I see what you mean. The fact that multiple bosons can occupy the same state means they can all have the same state function - something that is forbidden to fermions. So I suppose that's right: in QM terms they have no separate identity.

But where does the uncertainty principle come in?

 

[Nacho]

Put this atom trap and yourself in a spaceship undergoing acceleration, to a relative speed just under the speed of light, as measured from the lab. Everything on the spaceship gained energy, and the condensate should have also, so would you (traveling along with the atom trap) still see a Bose-Einstein condensate? How about another person that stayed behind in the lab -- would they still observe a condensate that is now in the spaceship?

 

[Q-reeus]

Put this atom trap and yourself in a spaceship undergoing acceleration, to a relative speed just under the speed of light, as measured from the lab. Everything on the spaceship gained energy, and the condensate should have also, so would you (traveling along with the atom trap) still see a Bose-Einstein condensate? How about another person that stayed behind in the lab -- would they still observe a condensate that is now in the spaceship?

Gained uniform translational KE relative to some initial 'rest frame'. And you think that magically translates into gaining thermal energy?! Which is random motions at the atomic/molecular/particle level. We on planet earth are moving at a huge speed relative to say similar inhabitants on a planet in some distant galaxy. Which one has the inhabitants burning up owing to their huge 'energy of motion'?

 

[Nacho]

I see the difference now -- energy of random motions VS linear motion. And I did look up a "temperature" page before asking the question, but didn't quite hit on the difference. The question came about because of what was head scratcher for me. Near absolute zero/motion VS (and) still having relative motion. And I know energy is not needed for that -- that is why I asked the question in terms of acceleration VS relative motion.

 

[DaveC426913]

But where does the uncertainty principle come in?

If they had distinct identities, we would be able to determine a given atom's exact momentum (zero) and its exact position simultaneously. Verboten.

 

[Q-reeus]

If they had distinct identities, we would be able to determine a given atom's exact momentum (zero) and its exact position simultaneously. Verboten.

Incorrect. Uncertainty relation applies just as much to distinguishable particles as to indistinguishable. Say quarks within a nucleon.

 

[DaveC426913]

Incorrect. Uncertainty relation applies just as much to distinguishable particles as to indistinguishable. Say quarks within a nucleon.

I'm not sure exactly what you're objecting to. I mean, how does your statement refute what I stated?

 

[Q-reeus]

I'm not sure exactly what you're objecting to. I mean, how does your statement refute what I stated?

HUP is a fundamental and universally applicable postulate within QM. The onus is on you to provide a reason and/or example of how HUP could ever be violated. References?

 

[DaveC426913]

HUP is a fundamental and universally applicable postulate within QM. The onus is on you to provide a reason and/or example of how HUP could ever be violated. References?

OK, you haven't been following my conversation with Exchemist.

You and I are in agreement; you just didn't realize it.

Look back at my post 45. The last word is Verboten. i.e. violation of HUP is forbidden.

To add some clarification to my post:

If they were to have distinct identities, then we would be in a position to determine a given atom's exact momentum (zero) and its exact position simultaneously. And that is verboten - forbidden - due to HUP. So, the atoms smear out, losing their identities, thus not violating HUP.

 

[Q-reeus]

OK, you haven't been following my conversation with Exchemist.

You and I are in agreement; you just didn't realize it.

Look back at my post 45. The last word is Verboten. i.e. violation of HUP is forbidden.

To add some clarification to my post:

Yes fine to a point but in #45 you tied it down to whether particles have distinct identities or not. Which is why I gave a counterexample of quarks within a nucleon. Within the nucleon at least two distinct hence distinguishable quark species exist yet HUP still applies. Similarly for a hydrogen atom - electron + proton.
HUP is independent of particle distinguishability. We evidently agree that at the opposite end, in Bosonic condensed systems like superconductors or superfluids or BEC's, complete indistinguishability applies. Momentum certainty means complete loss of knowledge of particle position, all in accordance with HUP.

 

[exchemist]

Yes fine to a point but in #45 you tied it down to whether particles have distinct identities or not. Which is why I gave a counterexample of quarks within a nucleon. Within the nucleon at least two distinct hence distinguishable quark species exist yet HUP still applies. Similarly for a hydrogen atom - electron + proton.
HUP is independent of particle distinguishability. We evidently agree that at the opposite end, in Bosonic condensed systems like superconductors or superfluids or BEC's, complete indistinguishability applies. Momentum certainty means complete loss of knowledge of particle position, all in accordance with HUP.

Yes, this sort of addresses the obverse of my concern.

I follow the concept that if many particles all occupy the same quantum state then there is one state function for them all, so they have in effect merged into a single quantum entity. I had never really thought about it that way (we chemists did not spend any time on Bose-Einstein condensates I regret to say) but it seems to make sense.

But this merging is due to the common wavefunction concept, which does not invoke the uncertainty relation. I should have thought the uncertainty relation could only be applied to this single state function, because we have just argued that that is the only state function present in the system. So we would have limits on the simultaneous determinacy of the momentum and position, or the energy and lifetime, of this single merged state.

I should have thought that we can't on the one hand say there is only a single merged state, with loss of identity of the particles, and on the other claim we can apply non-commuting observable operators to the state functions of the individual particles, when their individual state functions no longer exist.

 

[Q-reeus]

Yes, this sort of addresses the obverse of my concern.

I follow the concept that if many particles all occupy the same quantum state then there is one state function for them all, so they have in effect merged into a single quantum entity. I had never really thought about it that way (we chemists did not spend any time on Bose-Einstein condensates I regret to say) but it seems to make sense.

But this merging is due to the common wavefunction concept, which does not invoke the uncertainty relation. I should have thought the uncertainty relation could only be applied to this single state function, because we have just argued that that is the only state function present in the system. So we would have limits on the simultaneous determinacy of the momentum and position, or the energy and lifetime, of this single merged state.

I should have thought that we can't on the one hand say there is only a single merged state, with loss of identity of the particles, and on the other claim we can apply non-commuting observable operators to the state functions of the individual particles, when their individual state functions no longer exist.

Yes thanks - you are quite right. The proper conceptual handle is that of a single wavefunction. But one can still infer the general momentum-position behaviour of constituent atoms within such a condensate. As discussed in fourth para onward this article: https://www.andor.com/learning-acad...an-introduction-to-bose-einstein-condensation

 

[exchemist]

Yes thanks - you are quite right. The proper conceptual handle is that of a single wavefunction. But one can still infer the general momentum-position behaviour of constituent atoms within such a condensate. As discussed in fourth para onward this article: https://www.andor.com/learning-acad...an-introduction-to-bose-einstein-condensation

I notice the way they phrase this is not very clear-cut. They speak of a "macroscopic quantum state" in which the individual particles "have phase coherence in their wavefunctions". This seems some way short of saying they lose their identity.

It is not clear to me whether they mean there is a single "group" wavefunction, as you and I have been discussing, based on Dave's description, or whether they are implying the individual entities are still distinct but in just in phase and sharing the same space.

 

[Q-reeus]

I notice the way they phrase this is not very clear-cut. They speak of a "macroscopic quantum state" in which the individual particles "have phase coherence in their wavefunctions". This seems some way short of saying they lose their identity.

It is not clear to me whether they mean there is a single "group" wavefunction, as you and I have been discussing, based on Dave's description, or whether they are implying the individual entities are still distinct but in just in phase and sharing the same space.

Having only a very rudimentary grasp of the basic concepts, best if I simply refer to a technical article like: https://people.smp.uq.edu.au/MatthewDavis/thesis_MJ_Davis.pdf
See 1.2 'BEC in an ideal gas' pp 2-4 (12-14 logical). The approach there at least is in terms of particle occupation numbers - ground state vs excited states. So the concept of constituent particle persists at least in a statistical sense. It remains true that individual particles in the ground state cannot be isolated as such and only the coherent behaviour of the whole manifests.

 

[Dinosaur]

From Oystein Post #35

Seems like it would be much easier to determine position. Explain what you mean.

The above relates to my remark

Since they are almost motionless, their positions becomes indeterminate.

The Heisenberg Uncertainty Principle states

. . . that we cannot measure the position (x) and the momentum (p) of a particle with absolute precision. The more accurately we know one of these values, the less accurately we know the other.

BTW: Einstein, Heisenberg, & others consider momentum a more fundamental quantity than velocity. Hence the above description of the UC Principle. We less brilliant folks tend to be more comfortable expressing the Principle in terms of velocity & position.​

If the particles are almost motionless, their velocity is very accurately determined (it is close to zero). This requires the position to be less accurately known.

Id Est: Each particle seems to occupy a larger than normal volume. There is no reason to believe that the particles have gotten larger. Hence the situation is viewed as the position of each particle being less determinate than usual.​

Also from Oystein Post #35

Explain that also

The second remark relates to my remark

The particles become almost indistinguishable, losing their individual identity.

The uncertainty of position makes the volumes of space occupied by some of the particles seem to overlap, making it difficult (?impossible?) to distinguish among them.

 

[iceaura]

Is it theoretically possible to arrange things so that the position and velocity of a particle are both known to arbitrary precision, and the mass thereby become uncertain?

 

[origin]

Is it theoretically possible to arrange things so that the position and velocity of a particle are both known to arbitrary precision, and the mass thereby become uncertain?

My (limited) understanding is that it is only complementary properties are involved in the uncertainty principle.
Complemenatrity

 

[exchemist]

Is it theoretically possible to arrange things so that the position and velocity of a particle are both known to arbitrary precision, and the mass thereby become uncertain?

Origin is right, although the principle of complementarity does not seem to be quite synonymous with the uncertainty principle.

Position and momentum, and energy and lifetime, are examples of pairs of "conjugate variables". There is a fundamental uncertainty relation between any pair of conjugate variables, so that one cannot measure both at once with arbitrary precision.

As I understand it the original description of the principle of complementarity was tied up with the inevitable interaction between the measured system and the measurement apparatus. The uncertainty relation does not rely on that, but on something fundamental to QM, namely that conjugate variables are Fourier transforms of each other (a result of the wavelike nature of matter).

 

[Q-reeus]

The question in #56 was can momentum p = γmv be split between γm and v in such a way as to make v certain and γm totally uncertain. But they are not conjugate variables so it makes no QM sense. In the case of a photon it might be said |v| = c is certain, but Feynman's path integral approach to QED cast's doubt on even that, and anyway p =k E/c = khf/c leaves a single relevant variable for p, namely frequency f (there being no causal connection between f and k). So nothing 'conjugate' to work with.