Uncertainty Principle at T=0

[andrewm]

If \Delta x \Delta p > \frac{\hbar}{2}, what happens at T=0? Since "all motion stops" must we have \Delta x diverge?

Or is the zero-point motion allowed to occur at T=0, and only classical kinetic energy is zero?

[christianjb]

Good question.

A particle in a well has a momentum distribution even at T=0 (obtained from the magnitude squared of the FT of the wf).

The puzzle then, is how can a particle have a non-zero momentum at zero temperature? A partial answer is that the uncertainty principle puts a constraint on how small the momentum distribution can be: a narrower momentum distribution would force the wavefunction to become more delocalized in real space- which eventually increases the energy as the wf is pushed into sampling higher energy regions of the potential. So, reducing the kinetic energy comes at a cost of increasing the potential energy.

This is also why particles have 'zero-point energy' at T=0.

[Gerenuk]

Note that T=0 means the system is at its absolutely lowest energy state and no other. From there you can draw conclusions.

But in a infinite free space I suppose that the particles will be indeed completely delocalized. They will stand still, but you don't know where.

[WarPhalange]

It is my understanding that we can kind of see this happening with getting helium to record low temperatures. I forget how it works exactly, but they take a picture of it or something and you can see some kind of I guess reflection of the light, smaller temp -> bigger area of light reflected.

Or something like that.

[LogicalTime]

what happens at T=0? Since "all motion stops"


Error! you have assumed that at T=0 all motion stops, this is a common misconception. Remove assumption and proceed robot!

[andrewm]

A partial answer is that the uncertainty principle puts a constraint on how small the momentum distribution can be: a narrower momentum distribution would force the wavefunction to become more delocalized in real space- which eventually increases the energy as the wf is pushed into sampling higher energy regions of the potential. So, reducing the kinetic energy comes at a cost of increasing the potential energy.

This is also why particles have 'zero-point energy' at T=0.



Thanks, that's an easy way to remember where zero point energy comes from!


Error! you have assumed that at T=0 all motion stops, this is a common misconception.


I guess I need to start taking my grade-school physics with a grain of salt, don't I? :smile:

[HallsofIvy]

If \Delta x \Delta p > \frac{\hbar}{2}, what happens at T=0? Since "all motion stops" must we have \Delta x diverge?

Or is the zero-point motion allowed to occur at T=0, and only classical kinetic energy is zero?
That is precisely why you can't have a temperature of absolute 0.

(That's really just another wayt of saying what Logical Time said.)

[Count Iblis]

That is precisely why you can't have a temperature of absolute 0.

(That's really just another wayt of saying what Logical Time said.)


I don't agree with this. At T = 0, the system will be in the ground state. It is true that you cannot get to the ground state if the system isn't already in it using thermodynamic means only (if you want to cool the system down, you need something cooler, and if you let the system perform work to exact energy, it's entropy will stay the same at best, while at absolute zero the entropy must be equal to zero).

But the laws of physics do not forbid a system to be in its ground state. When we do thermal physics and talk about temperature, we are describing a system of many degrees of freedom statistically, and we assume that we only have access to a few external parameters (e.g. the volume) to control the system.

[ThomasT]

Interesting topic. Here's a couple of links dealing with

entropy:
http://en.wikipedia.org/wiki/Entropy_statistical_thermodynamics)

and Nernst's theorem:
http://en.wikipedia.org/wiki/Nernst%27s_theorem

Along the lines of what christianjb and others wrote, I'm thinking in terms of the steps required in the experimental production of lower and lower temperatures as being constrained by the uncertainty relations. So, insofar as there exists a real fundamental quantum of action, then the production of 0K temperature materials is impossible.

[Count Iblis]

An example of a system at T = 0 would be a computer that has all its bits initialized to zero, the |0,0,0,0,0,0,....0> state can be said to have a temperature of zero.

[christianjb]

Let's not forget Nernst's beautiful proof that there is no 4th law of thermodynamics.

Three people developed the 1st law, it took two people to develop the 2nd law, and Nernst was the sole originator of the 3rd law. Thus, there can be no 4th law, because it would have to be developed by zero people (which we know from induction).

[atyy]

Let's not forget Nernst's beautiful proof that there is no 4th law of thermodynamics.

Three people developed the 1st law, it took two people to develop the 2nd law, and Nernst was the sole originator of the 3rd law. Thus, there can be no 4th law, because it would have to be developed by zero people (which we know from induction).

Hence there is a 0th law. I suppose that means a -1st law is also possible?

[ThomasT]

I recently posted a reply which I thought was just as silly as the three preceding it (#'s 10, 11 and 12).

It seems to have been deleted. I'm wondering why? Perhaps it was deemed the most silly? :smile:

[LogicalTime]

I recently posted a reply which I thought was just as silly as the three preceding it (#'s 10, 11 and 12).

It seems to have been deleted. I'm wondering why? Perhaps it was deemed the most silly? :smile:

Maybe you are living life backwards. Is your entropy decreasing?