Position vector r and uncertainty principle

[MHD93]

Hi physicists,

How can two atomic-scale particles interact through gravitation (Gm1m2/r^2) or any other force that is a function of r if r isn't know with complete certainty. Is it that the force itself also comprises uncertainty in its value?

Thanks

 

[Simon Bridge]

In the model where the uncertainty principle is used, we don't use "forces" to describe interactions. The classical equations like the one you used for an example are not used. These are considered to be obeyed only on average.

What happens is that the uncertainty in relative position and momentum leads to an uncertainty in the outcomes of the interactions. Interactions are described by operators on a range of possibilities and the outcome is treated statistically by summing over all possible outcomes.

 

[Many_S_Theory]

Also the gravitational attraction between two sub-atomic particles is absurdly negligible. Although you can demonstrate the effect of say Earth's gravity on quantum mechanics through some neat little interference experiments, I believe Shankar (or is it Sakurai) discusses this.

 

[MHD93]

In the model where the uncertainty principle is used, we don't use "forces" to describe interactions. The classical equations like the one you used for an example are not used. These are considered to be obeyed only on average.

What happens is that the uncertainty in relative position and momentum leads to an uncertainty in the outcomes of the interactions. Interactions are described by operators on a range of possibilities and the outcome is treated statistically by summing over all possible outcomes.

If we use the potential function for such forces (such as the electrical force potential), Doesn't that mean that we use "forces" to describe interactions?

 

[Simon Bridge]

No - it doesn't.
There are lots of different ways to describe situations and interactions - to be true they have to be related to each other. The different descripton lend themselves to different models... eg. compare analyzing a object sliding down a hill via conservation of energy and by Newton's Laws. But, you know, if you want to think of using mv²/2 = mgh is the same thing as using mgsinθ-μN=ma then I'm not going to argue with you ;)

I do want to echo, and add to, @Many_S_Theory's observation - you realize how small the uncertainties in HUP are right? We are not talking about really big variations.

You could argue that "OK, maybe not gravity - but how about electric fields then" in which case I'd want to respond that in the scale where HUP is important, we do not use the electric potential or Coulombs Force Law and so forth ... on that scale the electromagnetic interaction is described in terms of individual photons and Feynman diagrams.

However, I believe:

What happens is that the uncertainty in relative position and momentum leads to an uncertainty in the outcomes of the interactions.

... answers your original question.