I had imagined maybe a piece of foil in which the stationary particle is initially positioned. So position would be known and the particle would have no momentum until it is knocked out of the foil by the other particle.
I guess you mean the uncertainty principle applies to the stationary...
Thanks. I guess I am still a little confused. Perhaps a simplified experiment would help:
So keeping to one dimension to avoid trigonometry, the particle to be measured starts at the right of the above diagram with momentum p1 (unknown), hits a stationary particle of known mass and position...
The first detector might be a light gun of some sort perhaps. The reflected photon would be captured after the deflection by a measuring device. I would of thought position can be measured to arbitrary position no matter how acute the angle is between the photon and the particle? Why do you say...
My maths is not great I'm afraid. I guess the amount of velocity is lost is a function of the deflection angle:
1º - a little is lost
0.01º - hardly any is lost
0.000001º - even less velocity is lost
So as the deflection angle approaches zero, the velocity at the second detector approaches the...
You can control the perturbation - it is reduced in size by using a more acute impact angle between the particle and d1. So by using a deflection angle of say 0.01º the perturbation is reduced to almost nothing and the velocity measured at d2 will be almost the velocity the particle had before...
Yes but by making the perturbation angle smaller, the velocity at d2 becomes closer to the velocity at d1. By making the deflection angle arbitrary small, the velocity at d1 can therefore be know to arbitrary precision.
I'm imagining detector 1 as a beam of photons at an acute angle to the particle being measured. So a photon would be deflected and detected to give the position at d1. The deflection angle is tiny so the particle would only be perturbed a little - so we would be able to pick it up in detector d2...
I am not too knowledgeable about QM, so please forgive me if this is a dumb question. I have outlined below an experiment setup for which Heisenberg's uncertainty principle seems not to apply:
Imagine a particle for which we wish to collect the exact position and velocity. We have a detector d1...
A point is defined to have length=0. So the number of points on a line segment length 1 is: (segment length)/(point length) = 1/0 = undefined.
Similarly, a 1x1 and 2x2 square both contain an undefined number of points.
IMO the definition of a point in maths is not too great (defined to have...
It makes intuitive sense to me that a finite region of space should contain a finite amount of information (implying discrete space).
With continuous space, a finite region always contains an infinite amount of information (by my definition)... which seems contradictory.
Then the math from...
OK, let's instead make the particle's position information belong to the region of space that contains the particle. Then it follows that both a small region of space and the whole universe contain the same countable infinity of information (my original argument).
As well as our measurement ability, I think precision (of position) might relate to the nature of the universe:
- Imagine a very simple discrete universe which contained one particle that could be in one of two possible positions. Then position is 1 bit of information.
- A slightly more complex...
- But the particle still has infinite precision (its just we can't measure it accurately) so we could say the particle has infinite (but unmeasurable) information?
- I don't think this effects my argument (it does not matter if we can't measure accurately)
I thought particles had a position when you measure them?
If space is truly continuous each particle must have infinite precision for it position (we could not measure it accurately, but the information would still be there in the system).
- So there is infinite information for one particle...