Uncertainty principle violation?

1. Jul 14, 2009

espen180

I came up with a scenario in which I think I am violating Heisenbergs uncertainty principle.

Say I build a velocity selector, like the ones found in mass spectrometers, and I fire electrons from an electron gun through it. By measuring the Electric and magnetic fields, I can measure the velocities of the exiting electrons very accurately. The selector is in vacuum, so there is no friction slowdown of the electrons. I now let the electrons pass through a bubble chamber, leaving a trail. Assume the bubble chamber does not slow the electrons down. Now I know that at a certain time, an electron had a certain position (from the trail in the bubble chamber) and I know it had a certain velocity (from the velocity selector), so I know its position and its momentum.

Am I not violating Heisenbergs uncertainty principle now? If not, or if there are errors in my scenario, please explain.

2. Jul 14, 2009

HallsofIvy

HOW do you measure the electric and magnetic fields? The only way I know to measure the field strength at a point is to put some kind of probe there and then the probe itself will alter the field.

That is the whole point of the uncertainty principle. In order to measure anything you must interact with it and that interaction will alter the thing being measured.

3. Jul 14, 2009

ZapperZ

Staff Emeritus
The more obvious problem here is that the HUP isn't about the measurement of a SINGLE position, and a single momentum. You will note that the expressions for $\Delta(x)$ and $\Delta(p)$ have a statistical spread in values. It means that you can either get a spread in a value after a repeated measurement, or your ability to predict the next value of a measurement.

I think I've illustrated this common misconception of the HUP somewhere in here using the single-slit diffraction.

Zz.

4. Jul 14, 2009

Staff Emeritus
And to add yet a third objection, nowhere in this does it actually quantitatively violate the HUP. The HUP is a quantitative statement.

5. Jul 16, 2009

flatmaster

Even if you say you did measure the electron's momentum with the velocity selector and it's position with the bubble chamber, you haven't broken any rules yet.

You would have measured momentum in the x..... Px
And the track tells you the position in the y and z directions.

Technically, you're still aloud to do that.