Jump to content

How Is Uncertainty Principle Uncertain?

Featured Replies

If, say we had a particle at rest at point A, wouldn't that violate the Uncertainty Principle because we know it's location, Point A, and it's speed in the time dimension which is C. Or am I just ignorant here? :confused:

[math]\Delta{x}\Delta{p}\geq{\frac{\hbar}{2}}\Rightarrow{\Delta}x{\geq}\frac{\lambda}{2}[/math] that may be slightly wrong as i did it from memory. what are your speed and position relative to? saying it is a a certain speed is (as swansont said) like asking "what is the difference between a duck?".

Yes, that statement would violate the Uncertainty Principle of QM. Luckily, there is no uncertainty principle in Relativity so it´s ok to say something like that as long as you stay within that scope.

 

For QM: The axioms of QM state that a particle is completely described by it´s wavefunction f(x) and it´s probability density in space is |f²|. The important point here is that the particle is completely described by f(x) alone. Unlike classical theory you don´t need the momentum to define the particle´s state.

Now, any state can be made up on a combination of plane waves g(p)=exp(i kx) which are states with definite momentum (and thus definite velocity). If you try to make up a function f(x) which is nonzero only at a certain point A out of a linear combination of g(p) you will notice that all p contribute to this sum. So a particle at a certain point has all momentums and thus not one definite one.

How can one be certain that the particle is still there at point A ?

If, say we had a particle at rest at point A, wouldn't that violate the Uncertainty Principle because we know it's location, Point A, and it's speed in the time dimension which is C. Or am I just ignorant here? :confused:

 

suppose that you are in a frame in which the particle is at rest, and that the particle isn't being subjected to any forces. Then it will remain at rest in the frame, until such time as something external acts upon it, at which time it will accelerate in that frame.

 

The way physicists use the uncertainty relation, connects to their measurements of the position of it. But if they aren't trying to determine its position, then it will just sit there at rest.

 

It doesn't care whether or not you "know" it is at rest.

 

But if you were to fire something at it with speed c, and then have that thing reflect, and come back to you with speed c, you would be able to calculate how far away it was when it was hit, provided you measure the time it takes for the thing you fired to return to you. Then since you know you fired it at speed c, and you measured t, ct would give the distance it was away from you. But when the thing hits it, you have to have conservation of momentum, and conservation of energy. Some "energy" will be imparted to the particle upon its being 'hit'.

 

So therefore, that which returns to you will have less "energy" then it started with.

 

The point is, it will no longer be at rest. So that by the time you know where it was, you will no longer know where it is.

 

This is the kind of logic inherent in interpreting the uncertainty principle.

 

So to answer your question... yes, you are right... in the customary sense of the interpretation.

 

Regards

 

PS: There is also the question about what is meant by the term particle. If you have a body, then the uncertainty relation would apply to the center of mass of the body, which is a point, not a particle.

 

There are other issues which need to be addressed, but thankfully they are not infinite in number.

suppose that you are in a frame in which the particle is at rest' date=' and that the particle isn't being subjected to any forces. Then it will remain at rest in the frame, until such time as something external acts upon it, at which time it will accelerate in that frame.

 

The way physicists use the uncertainty relation, connects to their measurements of the position of it. But if they aren't trying to determine its position, then it will just sit there at rest.

 

It doesn't care whether or not you "know" it is at rest.

 

But if you were to fire something at it with speed c, and then have that thing reflect, and come back to you with speed c, you would be able to calculate how far away it was when it was hit, provided you measure the time it takes for the thing you fired to return to you. Then since you know you fired it at speed c, and you measured t, ct would give the distance it was away from you. But when the thing hits it, you have to have conservation of momentum, and conservation of energy. Some "energy" will be imparted to the particle upon its being 'hit'.

 

So therefore, that which returns to you will have less "energy" then it started with.

 

The point is, it will no longer be at rest. So that by the time you know where it was, you will no longer know where it is.

 

This is the kind of logic inherent in interpreting the uncertainty principle.

 

So to answer your question... yes, you are right... in the customary sense of the interpretation.

 

Regards

 

PS: There is also the question about what is meant by the term particle. If you have a body, then the uncertainty relation would apply to the center of mass of the body, which is a point, not a particle.

 

There are other issues which need to be addressed, but thankfully they are not infinite in number.[/quote']

I understabnd that all measured electrons, for example, are measured as point particles with no minimum measured radius, or size. Even electrons that have just been subjected to two hole diffraction are seen on the interference screen pattern as point particles only. Is it not true that the uncertainty spreading in the particle location has absolutely nothing to do wth the spatial limits of the electron size? The uncertainty is a theory imposed attribute, not on the particle, but on the observer's ability to measure certain local attributes of that particle.

 

All observation indicate the size of the electron is very small, immeasurably small. the wave length limits do not perturb the physical size of the electron as much as the accelerated electron may oscillate at a higher rate and in this sense appear to decrease in size as the frequency of oscillation increases. This certainly seems like a reasonable description of the electron dynamics, but the "real" size of the particle electron has not been maeasured. Within the limits of current experimental results the electron may be considered a point particle. Notwithstanding the periodic mathematical contrivances that need be imposed, ad hoc, to expalin away two hole diffraction phenomenon, for instance, electrons are physically measured as point massive objects.

 

Geistkiesel

Archived

This topic is now archived and is closed to further replies.

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Configure browser push notifications

Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.