Uncertainty principle regarding a narrow optical fibre?

[bcrelling]

If we take a narrow aperture(such as the one used in the double slit experiments) and attach a length of optical fibre to it, surely now we know the position (confined by the width of the fibre) and also the momentum(defined by the direction of the fibre).

How does the uncertainty principle apply here?

 

[jambaugh]

You are comparing position in one direction with momentum in the orthogonal direction. [Px , Y ] = 0... so you can simultaneously measure to arbitrary precision the position in one direction and the momentum perpendicular to that direction.

 

[bcrelling]

You are comparing position in one direction with momentum in the orthogonal direction. [Px , Y ] = 0... so you can simultaneously measure to arbitrary precision the position in one direction and the momentum perpendicular to that direction.

I'm not sure I understand.

The position is known, as the photon traveling down the optical fibre is contrained on two axes by the optical fibre(x, y). As for the remaining axis in the forward direction(z), the position is calculable by knowing when the photon was emitted from source at that it will certainly be traveling at speed C and can only go a certain distance at such a speed.

As for momentum, we know the direction of the wire and the speed of the photon.

Sorry if I'm missing something obvious.

 

[jambaugh]

I'm not sure I understand.

The position is known, as the photon traveling down the optical fibre is contrained on two axes by the optical fibre(x, y). As for the remaining axis in the forward direction(z), the position is calculable by knowing when the photon was emitted from source at that it will certainly be traveling at speed C and can only go a certain distance at such a speed.

As for momentum, we know the direction of the wire and the speed of the photon.

Sorry if I'm missing something obvious.

Yes, velocity and momentum are not the same thing, especially for photons which all travel at speed c. Rather in this case it is more a matter of frequency.
There are also issues of modality in the waveguide behavior of the optical fiber. It doesn't quite "just run down the line" but "bounces off the walls" a bit.

There are corresponding energy time uncertainty relations so when you get down to timing the photon you are talking about an interval of time when it was emitted into the wave guide which is inversely proportional to its energy uncertainty. As you use timing to know position you use energy to know momentum, both being linked by the constancy of c.

Remember that photons are not point particles. They are quantized emissions absorptions of EM field. You can "emit a single photon" in principle over the course of a year. It isn't necessarily emitted (even a single one) instantaneously. You must take a year to emit a single photon if you insist on emitting that photon in a very precise momentum=frequency band (order of energy= hbar/year).

Contrawise if you are emitting a single localized (in space and in time of emission) photon it must have a broad uncertainty in its energy & momentum.

[edit] The books have it right. Trust the books. That is in the Regan sense, "trust but verify!".

 

[DimReg]

I'm not sure I understand.

The position is known, as the photon traveling down the optical fibre is contrained on two axes by the optical fibre(x, y). As for the remaining axis in the forward direction(z), the position is calculable by knowing when the photon was emitted from source at that it will certainly be traveling at speed C and can only go a certain distance at such a speed.

As for momentum, we know the direction of the wire and the speed of the photon.

Sorry if I'm missing something obvious.

A common misunderstanding when it comes to the uncertainty principle is that people imagine making both a position and momentum measurement, and therefore think they can measure both. The truth is, you CAN measure both, and it's exactly when you do measure both that the uncertainty principle is most interesting. It tells you that the product of errors in your measurements will always be non-zero, even in ideal situations. So you can actually measure the position and momentum simultaneously to rather good precision (the minimum uncertainty is very small).

From this, without closely looking at your proposed setup, I conclude that you will always have a small amount of error, no matter how well you do your experiment.

 

[Harry Wilson]

From this, without closely looking at your proposed setup, I conclude that you will always have a small amount of error, no matter how well you do your experiment.

That's not how uncertainty works. It does not describe experimental errors, but rather errors in prediction.