# Uncertainty and the speed of light?

By Uncertainty Principle speed of a particle cannot be constant. then speed of a single photon not?

Dale
Mentor
2020 Award
By Uncertainty Principle speed of a particle cannot be constant
No. I am not sure how you came up with this, so I don’t know how to address it.

willem2
By Uncertainty Principle speed of a particle cannot be constant. then speed of a single photon not?
According to uncertainty principle it's impossible to know the position and the momentum of a particle exactly simultaneously. The product of the uncertainties is >
ħ/2. If you would exactly know one of them, the uncertainty in the other must be infinite, wich isn't possible. (If you detect a particle at all, you must have some idea where it is).
For particles not moving at the speed of light, the speed depends on the momentum, and if you knew the exact speed, you would know the exact momentum, and you would know nothing about the position at all, wich is impossible.
However photons can have different momenta, even if they all move at the same speed, so there is no problem.

Dale
Staff Emeritus
According to uncertainty principle it's impossible to know the position and the momentum of a particle exactly simultaneously.

That is untrue. It is a statement about ensembles of systems, not about individual particles.

That is untrue. It is a statement about ensembles of systems, not about individual particles.

If there was a way to determine the position and momentum of a single particle with arbitrary precision, it would violate it for an ensemble as well. Just repeat.
In a statement like σx σp >= ħ/2 we're talking about a probability distribution for the position and momentum of a single particle.

Nugatory
Mentor
If there was a way to determine the position and momentum of a single particle with arbitrary precision, it would violate it for an ensemble as well. Just repeat.
I can measure the position of a particle with arbitrary precision and I can measure the momentum of a particle with arbitrary position, but this does not allow me to do the same for an ensemble. When I perform my arbitrarily precise measurements on each member of the ensemble, I will find that the results are distributed according to the uncertainty principle. As an extreme example, I can first measure the momentum of every particle in the ensemble and then discard those whose momentum does not fall in a specified and arbitrarily narrow range; this is a preparation procedure for an ensemble of particles with known momentum. Now nothing stops me from measuring their position; but I will find a spread of position values consistent with the uncertainty principle.
In a statement like σx σp >= ħ/2 we're talking about a probability distribution for the position and momentum of a single particle.
Talking about the probability distribution of a single measurement is like talking about the probability distribution of a single sample. Analogously, when I state that a tossed coin is biased to come up heads 60% of the time, I'm not saying anything about any one toss, I'm making a claim about the ratio of heads to tails in a hypothetical ensemble of identically prepared coins.

Dale
vela
Staff Emeritus
Homework Helper
I can measure the position of a particle with arbitrary precision and I can measure the momentum of a particle with arbitrary position, but this does not allow me to do the same for an ensemble. When I perform my arbitrarily precise measurements on each member of the ensemble, I will find that the results are distributed according to the uncertainty principle. As an extreme example, I can first measure the momentum of every particle in the ensemble and then discard those whose momentum does not fall in a specified and arbitrarily narrow range; this is a preparation procedure for an ensemble of particles with known momentum. Now nothing stops me from measuring their position; but I will find a spread of position values consistent with the uncertainty principle.
But you're assuming in your example that the uncertainty principle applies to each particle. @willem2, I believe, is saying if the uncertainty principle doesn't apply to a single particle, as Vanadium implied, then you can determine the momentum and position of the particle to arbitrary precision at the same time. So to repeat your example, do this for a bunch of particles, discarding the ones that don't fall in the arbitrarily narrow ranges of position and momentum at the same time. You now know the momentum and position of the ensemble precisely.

Talking about the probability distribution of a single measurement is like talking about the probability distribution of a single sample. Analogously, when I state that a tossed coin is biased to come up heads 60% of the time, I'm not saying anything about any one toss, I'm making a claim about the ratio of heads to tails in a hypothetical ensemble of identically prepared coins.
You could also repeat the measurement multiple times with the same coin, and the underlying probability distribution will emerge.

From my understanding, the uncertainty principle does apply to a single particle. If the particle is in some state, you can calculate ##\Delta x## and ##\Delta p## based on that state, and the product of those two quantities will be greater than ##\hbar/2##. It has nothing to do with making measurements. It has to do with the waviness of nature at the quantum level.

Doc Al
Mentor
If the particle is in some state, you can calculate ##\Delta x## and ##\Delta p## based on that state, and the product of those two quantities will be greater than ##\hbar/2##. It has nothing to do with making measurements. It has to do with the waviness of nature at the quantum level.
My understanding is that ##\Delta x## is a measure of the spread of position measurements on an ensemble of particles all in that state. Similarly, ##\Delta p## is the spread of momentum measurements made on a similarly prepared ensemble. Their product is a property of the state (and thus the ensemble).

(I'm essentially agreeing with the post by @Nugatory above.)

Dale
Mentor
2020 Award
But you're assuming in your example that the uncertainty principle applies to each particle.
That is not how I read his comment. To me it seems that he is talking about arbitrarily high precision for the position and momentum of a single particle.

vela
Staff Emeritus
Homework Helper
That is not how I read his comment. To me it seems that he is talking about arbitrarily high precision for the position and momentum of a single particle.
Simultaneously?

Dale
Mentor
2020 Award
Simultaneously?
Yes

A photon can have a fairly precise momentum (to the extent that you can accurately determine its wavelength), but it can't be "localized" in a way that's compatible with the usual formulation of the Heisenberg Uncertainty Principle. The principle does apply, but the mathematics are different for massless particles. Nice summary here: https://arxiv.org/abs/1205.0516