# A contradiction of Heisenberg uncertainty principle?

1. Jun 25, 2006

### AlbertEinstein

A contradiction of Heisenberg uncertainty principle??

Suppose an atom is cooled to 0 K (Practically impossible). However theoretically, all its motion will cease ; and therefore it will be possible to determine the exact position of the atom and exact velocity (which is zero).Doesn't this contradict the Heisenberg uncertainty principle which states that it is impossible to determine the exact position and velocity of an object.

2. Jun 25, 2006

### GDogg

That may be so in the classical sense.

But at 0K quantum-mechanical systems are really in their ground states with a kinetic energy greater than 0.

3. Jun 25, 2006

### Hootenanny

Staff Emeritus
For further information Einstein, google for "zero point motion"

4. Jun 25, 2006

### ZapperZ

Staff Emeritus
In addition to what Hoot said, there is also a measureable effect called the deBoer effect seen especially in the noble gasses where parameters such as specific heat has to be corrected due to an INCREASE in the kinetic energy of the atoms or molecules as temperature DECREASES. So the notion that all motion ceases as T approaches zero is a fallacy.

Zz.

5. Jun 25, 2006

### Farsight

6. Jun 25, 2006

### DaveC426913

Atoms near 0K don't behave like atoms, they blend or 'smear' together into a Bose-Einstein Condensate. It becomes meaningless to talk about locations of specific particles.

7. Jun 25, 2006

### masudr

nb. only if those atoms are bosons; some atoms have spin half (eg. silver)

8. Jun 26, 2006

### spin_spin

At 0 K the mass of the particle is zero. When there is no particle how will you measure its temperature

9. Jun 26, 2006

### Quaoar

Um. Particles do not lose mass as they lose thermal energy.

10. Jun 26, 2006

No, not "practically" impossible, just plain impossible--due to this law of physics [http://www.infoplease.com/ce6/sci/A0861526.html]:

The Third Law of Thermodynamics
"A postulate related to but independent of the second law is that it is impossible to cool a body to absolute zero by any finite process. Although one can approach absolute zero as closely as one desires, one cannot actually reach this limit. The third law of thermodynamics, formulated by Walter Nernst and also known as the Nernst heat theorem, states that if one could reach absolute zero, all bodies would have the same entropy. In other words, a body at absolute zero could exist in only one possible state, which would possess a definite energy, called the zero-point energy. This state is defined as having zero entropy".

I see no point to the discussion of this thread, e.g. ..." suppose an atom is cooled to absolute zero..." there is just not anything to "suppose". A state of zero-point energy is just plain impossible because it requires a body to reach absolute zero, which is impossible due to the third law of thermodynamics.

11. Jul 3, 2006

### MrPootys

isnt the question a little ironic, seeing as the heisenberg uncertainty principle was the first to infer the idea of zero-point energy being the lowest state of a quantum particle since the motion and position cannot be known.

12. Jul 4, 2006

### DaveC426913

Could someone please verify, explain and/or provide a reference for this?

13. Jul 4, 2006

### masudr

I've done the Stern Gerlach experiment. And I used silver atoms in the ground state, and it split into 2 beams (not 3 or 4 etc) thus showing that the values of spin angular momentum it could take were $\pm\mbox{\frac{1}{2}}\hbar.$ But you can't really take my word for it.

This is an independent source (but admittedly also the internet, but not very uncredible): http://galileo.phys.virginia.edu/classes/252/Angular_Momentum/Angular_Momentum.html

It mentions that someone did the S-G experiment using Hydrogen atoms in the ground state, and discovered it must also have angular momentum $\pm\mbox{\frac{1}{2}}\hbar,$ from which they eventually concluded that it was due to the electron.

14. Jul 5, 2006

### Schrodinger's Dog

I remember talking to someone about this and he said pretty much what you guys have said about what happens as you approach absolute zero, the problem was he ascertained what would happen at absolute zero, and that motion would still exist, I had always heard that it could not be reached at least in theory so I pointed out the logical fallacy of stating what would happen at absolute zero as if it was true. It then took me four posts to explain why he was being illogical.

So guys quick question what would happen if we reached absolute zero would there still be motion

I think that's the problem some times with learning about stuff at degree level as this guy had, people who study further are much more leary of making statements of fact about situations which have never been experimentally validated, but degree students appear more cock sure, maybe it's just the few I've spoken to and I freely admit my understanding of the subject is far from complete, but I do know what constitutes a logical fallacy. by the way he was backed up two other students from other parts of the globe, so I'm assuming this type of jumping the gun learning, at least in this small case, is widespread. Or that someone knows something that I don't which is as ever always a possibility.

EDIT: I think what is more likely though is that people take what might happen at absolute zero as what will happen, either because the lecturer isn't clear enough or they've jumped to a conclusion.

Last edited: Jul 5, 2006
15. Jul 11, 2006

### roundedge

I think to assert that there would still be motion at absolute zero is a contradiction in terms. The concept of absolute zero is that the atom has zero thermal energy. Thermal energy is essentially equivalent to kinetic energy. No kinetic energy means no movement (rotational, vibrational, directional). Don't think of absolute zero as a place to be at, but rather a state the atom is in.

16. Jul 11, 2006

### vanesch

Staff Emeritus
It is not true that temperature = 0 means kinetic energy = 0. This is only true in *classical* physics when the hamiltonian depends only in a *quadratic way on the canonical momentum* (the equipartition theorem).
http://www.scielo.br/pdf/bjp/v30n1/v30n1a19.pdf

Have also a look at:
http://theory.ph.man.ac.uk/~judith/stat_therm/node54.html [Broken]

Temperature is defined in all generality as (the inverse of) the derivative of the logarithm of the number of allowed states to energy. The logarithm of the number of allowed states is usually called the entropy.

Last edited by a moderator: May 2, 2017
17. Jul 11, 2006

### vanesch

Staff Emeritus
Well, you can, in theory, talk about the situation of zero ENTROPY: it is in fact the situation you encounter when studying small systems when you (think you) know the state perfectly. A single simple harmonic oscillator (on the blackboard) in the ground state is a situation of zero entropy, for instance.
Nothing stops you from considering then, the theoretical situation of 10^20 simple harmonic oscillators in the common ground state. That's a zero entropy state too.

The third law of thermodynamics simply states that at zero entropy, temperature is zero too, and a consequence is that you cannot reach, in a finite number of interactions with non-zero entropy systems, the ground state of a system perfectly without some amplitude for the non-ground state. Nobody tells you that you cannot CONSIDER that state, but what is told by the third law of thermodynamics is that a system in a zero entropy state (= ground state) cannot be in interaction with anything else, unless it is ALSO in the ground state. So you cannot interact with a system of zero entropy, without destroying that state somewhat. In how much you destroy it is depending on the system, and it might very well be that for what you want to study, it doesn't make any difference.

However, an *isolated system* can BE in a zero-entropy (ground state) state. But you cannot get a system of non-zero entropy, by a finite number of interactions, into a zero-entropy state, simply because somewhere along the chain, you'll put it in interaction with a non-zero entropy system. But this might still be neglegible for what you want to do.

Depends on what you call "motion". If you mean: "changing expectation values of position with time", then, no of course, because in the ground state, all expectation values are independent of time. If you mean: "zero expectation value for kinetic energy", then the answer is yes, as is the case for a harmonic oscillator already.

The zero-entropy state is simply the quantum-mechanical ground state of the system. So, it is not that it is a non-existent or a forbidden state, it is simply that there is no way to reach is perfectly starting from a non-zero entropy state in a non-zero entropy environment in a finite number of steps.

That said, there is no problem reaching "effective zero entropy" for a certain set of degrees of freedom, if the spectrum is discrete near the ground state: it is sufficient to lower the entropy enough for the probability for a non-ground state (of these degrees of freedom) to be present to be neglegibly small. Said degrees of freedom are then "frozen out" (like molecular vibrational degrees of freedom, for instance).

So when studying a certain aspect of a physical system, related to a certain set of degrees of freedom, one can always approach as much as one wants, the state that is the ground state for those degrees of freedom - in which case, it wouldn't make any difference if we were really AT 0 K or not.

An example at room temperature: at room temperature, electrons and ions form neutral atoms and molecules: the ionisation degrees of freedom which would turn gases into plasmas are essentially frozen out. So when studying gases at room temperature conditions, one doesn't have to take into account ionisation of the gas (although there IS a very small probability for it ionizing). This won't change anymore significantly if you cool the gas further.

18. Jul 11, 2006

### Careful

***
The third law of thermodynamics simply states that at zero entropy, temperature is zero too, and a consequence is that you cannot reach, in a finite number of interactions with non-zero entropy systems, the ground state of a system perfectly without some amplitude for the non-ground state. ***

Hmmm, I thought that the third law of thermodynamics stated that for reversible processes the Limit_{T -> 0} dQ/T = 0, implying that at zero temperature isothermal and adiabatic processes are the same. Nobody says the entropy at absolute zero needs to equal 0, S = constant though. Neither does this imply that at absolute zero no classical motion exists since we all know that the classical equipartition principle cannot be consistently upheld. Thanks for the reference, was not aware of that one :-)

Careful

Last edited: Jul 11, 2006
19. Jul 11, 2006

### vanesch

Staff Emeritus
But you wouldn't object, I guess, to:
"consider an atom in its ground state, blah blah blah".

Nevertheless, the two phrases are equivalent.
If it is cold enough that the relevant degrees of freedom are essentially in their ground state, this is equivalent to the "zero temperature" case concerning these degrees of freedom.

20. Jul 11, 2006

### vanesch

Staff Emeritus
You are entirely correct (nitpicker !). The constant entropy is given by the logarithm of the number of degenerate ground states that exist.
If the ground state is unique, then S = 0. I wanted to limit the explanation to this case but should have said so.