- #1
texta
- 6
- 0
Hi,
I have a problem with the uncertainty principle. The way I understand it, Heisenberg used ideas from classical mechanics and the concept of wave/particle duality to show a contradiction in classical mechanics, i.e. that it is impossible to know with exact precision both the momentum and position of a particle.
I've read about the Heisenberg microscope thought experiment which leads to:
[tex]\Delta[/tex]x [tex]\Delta[/tex]p [tex]\geq[/tex] h [tex]/[/tex] 4pi
Let's assume that instead of a gamma ray we use an extremely low energy photon to determine very precisely the momentum of the electron under the microscope. This means that the position of the electron would be very uncertain.
But, we know the time the photon was emitted. We also know the time that we observe the electron in the microscope. So the total distance covered by our photon could not be more than c * (observation time - photon emission time).
Example:
An electron with mass K, is moving at velocity V, with an uncertainty in the velocity of M.
[tex]\Delta[/tex]p = KVM
[tex]\Delta[/tex]x = h / 4 pi * KVM
If the photon's energy was low enough, the value of M would be small enough such that [tex]\Delta[/tex]x is greater than c * (observation time - photon emission time). But if the photon hit the electron, it means that the following must be true:
0 < electron x position < c * (observation time - photon emission time).
Can somebody please explain where I have gone wrong, this is really bugging me!
I have a problem with the uncertainty principle. The way I understand it, Heisenberg used ideas from classical mechanics and the concept of wave/particle duality to show a contradiction in classical mechanics, i.e. that it is impossible to know with exact precision both the momentum and position of a particle.
I've read about the Heisenberg microscope thought experiment which leads to:
[tex]\Delta[/tex]x [tex]\Delta[/tex]p [tex]\geq[/tex] h [tex]/[/tex] 4pi
Let's assume that instead of a gamma ray we use an extremely low energy photon to determine very precisely the momentum of the electron under the microscope. This means that the position of the electron would be very uncertain.
But, we know the time the photon was emitted. We also know the time that we observe the electron in the microscope. So the total distance covered by our photon could not be more than c * (observation time - photon emission time).
Example:
An electron with mass K, is moving at velocity V, with an uncertainty in the velocity of M.
[tex]\Delta[/tex]p = KVM
[tex]\Delta[/tex]x = h / 4 pi * KVM
If the photon's energy was low enough, the value of M would be small enough such that [tex]\Delta[/tex]x is greater than c * (observation time - photon emission time). But if the photon hit the electron, it means that the following must be true:
0 < electron x position < c * (observation time - photon emission time).
Can somebody please explain where I have gone wrong, this is really bugging me!