# Uncertainty relation and absolute zero.

1. Sep 20, 2004

### misogynisticfeminist

Is there a certain fixed degree of uncertainty? Like for example, we can't reach absolute zero because we would know both the momentum and position of the particle. But if it is say, 3-4 K above absolute zero, there would be a really small degree of uncertainty involved, because its quite localized and its momentum is quite small as well.

So, is there a point of "minimum" uncertainty? and would it be useful in experiments? Is my understanding of absolute zero and the uncertainty relation flawed around here?

Last edited: Sep 20, 2004
2. Sep 20, 2004

### humanino

The Heisenberg undeterminacy principle is not related to temperature. You can find the way the constant is derived : $$(\Delta a)(\Delta b) \geq \frac{1}{2}\langle \psi |[\hat{A},\hat{B}]|\psi\rangle$$ [thread=39172]here[/thread]

3. Sep 21, 2004

### misogynisticfeminist

I'm really sorry, but I can't really absorb the mathematics inside there. i can't really understand stuff if its using too much maths to explain it. Is there a non-mathematical way to describe it?

Thanks...

4. Sep 21, 2004

### marlon

$$(\Delta p)(\Delta x) = \frac{h}{2\pi}$$

This is the uncertainty relation between momentum p and position x. You have a analogue relation between energy E and time t.

Now, as humanino pointed out these equations do NOT depend on temperature, yet p depends on temperature when looked at kinetic degrees of freedom of a gas of particles for example.

But if you would lower the energy (lower p for that matter) then the uncertainty on position will become bigger. A very good localization only occurs at high-energy and this is why in elementary QM-textbooks you will find a paragraph on the gamma-ray-microscope and not on the IR-ray-microscope, as an explanation for what happens in the uncertainty relation of Heisenberg.

regards
marlon

5. Sep 21, 2004

### misogynisticfeminist

I've heard of Heisenberg's Gamma Ray microsope. But why is it that when the energy is higher, the particle is more localized? Besides it being due to the uncertainty relation, won't a high-energy photon mess up the momentum and position of a particle much more than a low energy one?

6. Sep 21, 2004

### marlon

No, because the uncertainty in the target-particle's position is equal to the wavelength of the incident radiation divided by the sinus of the half-angle subtended at the parget-particle by the lens.
Basically if you wanna keep the uncertainty as little as possible then the wavelength of the incident radiation (the photons) needs to be as small as possible thus the more energy for the photon the better.

Indeed deBroglie states that momentum p is equal to: p=h/wavelength

regards
marlon

7. Sep 22, 2004

### misogynisticfeminist

Hmmm, ok, actually, the gamma ray microscope has something to do with visibility if I'm not wrong. In order to see a particle, we can't use a photon with associated wavelength of visible light, we can, but its gonna be blured. And if we use a photon with an associated gamma ray wavelength, its gonna be clearer but it momentum would still be affected.

But won't the incoming photon mess up the momentum, but also the position of the particle as well? Its like billard, I've hit the white ball on the 8 ball, the 8 ball now has a diff. momentum but also a diff. position, and the higher the energy of the white ball, the more the momentum and position of the 8 ball is messed up. Or is the relation between billard balls and particles a little too deterministic?

I've just realized I have not really understood the gamma ray microscope.

8. Sep 22, 2004

### marlon

Don't compare the collision of a photon with some particle with the collision between two balls. The properties of photons (i am referring to the particle-wave interpretation and the fact that a photon does not have mass) are totally different of those of the billiard balls...

What is it that you do not understand about the gamma ray microscope ???

There are two factors here :1) keep the wavelength as small as possible so that the uncertainty on the particle's position is as small as possible

2) The wavelength cannot be too small though because of the accuracy of the image. I mean, if the wavelength of an incident photon get's too small you won't be able to distinguish two separated objects from each other...Thus you get a blurry image...

regards
marlon