# Heisenburgs Uncertainty Principle in a narrow tube?

1. Apr 17, 2013

### cmcraes

The principle states: δxδp≥h/4π
I understand what it means and ive seen in proven both theoretically and expiamentally, so im not questioning the inequality here.

My question is:
If there was a very narrow tube that had a vacuum inside it and a laser attached at one end that shoots single photons. The tube is black on the outside so no light can get inside, there are 'perfect mirrors' all around the inside so no light could be lost, why cant we know a photons position and momentum at the same time now? I mean assuming we know its energy when it leaves the laser, its velocity, and the dimentions of the of the vacuum tube, why cant we always know the photons displacement and momentum at any instantanious moment in time? Thank you!

2. Apr 17, 2013

### ZapperZ

Staff Emeritus
What you have constructed is a waveguide.

The thing about waveguides here is that, if the geometry is "wrong", your laser light will not propagate, and it will be reflected. In fact, this is one aspect of constructing waveguides that is extremely important - can your waveguide sustain the EM mode that you are trying to propagate.

Now, with that in mind, let's look at what you are trying to do. For your setup to make a suitable test of the HUP, the diameter of the tube must be comparable to the wavelength of the light source. If not, if it is "too big", then this really doesn't test anything and your $\Delta x$ is huge. My guess here is that if you make the diameter of your tube that small, you end up with no propagation down that tube, i.e. you end up with a structure that includes wavelength of that light you are trying to propagate within its cutoff.

Zz.

3. Apr 17, 2013

### cmcraes

Thanks! Makes much more sense now

4. Apr 17, 2013

### cmcraes

What if this was done with an electron or neutrino?

5. Apr 17, 2013

### fluidistic

You cannot know anyway the momentum/energy of the photon with an arbitrary accuracy even though it came from a laser.