Heisenbergs Uncertainty for light

In summary, the uncertainty principle states that it is impossible to know both the position and momentum of a particle with absolute certainty. It applies only to observables that don't commute.
  • #1
duffbeerforme
6
0
Hi, I was just wondering if Heisenbergs uncertainty principal is about position and velocity (or energy and time),.. then how come we can know that a photon travels at light speed (its velocity) exactly and also know that it hit some detector (its position).
If we know the velocity exactly then the light must be everywhere at the same time right?
 
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  • #2
duffbeerforme said:
Hi, I was just wondering if Heisenbergs uncertainty principal is about position and velocity (or energy and time),.. then how come we can know that a photon travels at light speed (its velocity) exactly and also know that it hit some detector (its position).
If we know the velocity exactly then the light must be everywhere at the same time right?
To be precise the uncertainty principle which you're referrring to relates the uncertainty between position and momentum. It seems to me (although I'm not 100% sure) that since its quite possible to measure the momentum p of the photon and its energy E to arbitrary precision (i.e. the energy operator commutes with the position operator) then one can deduce the precise value of the momentum. The relation is p = (E/c^2)c = E/c and therefore c = E/p.

Pete
 
  • #3
I'm not an expert, but I think I can answer this one.

As the person above me said, the uncertainty relation doesn't apply to any two observables you can think of. It only applies to observables that don't "commute". So it would apply to position and momentum, but not to position and velocity.

If you consider the position and momentum of a photon, you will see that neither variable is known precisely. The momentum of any given photon is actually a cloud of momenta. So, let's say you think you have photons of a given wavelength (and therefore, a given momentum). It turns out that what you actually have is a bunch of different wavelengths that differ slightly from each other. Think of the photon as a bunch of waves with different wavelengths grouped together in a "wave packet" like here:
http://www.st-andrews.ac.uk/~bds2/ltsn/Edinburgh/wave/index.html

Likewise, the point where the photon contacts the detector is not a precise point but is a small cloud of points.

The smaller the position cloud is, the bigger the momentum cloud will be, and vice-versa.
 
  • #4
Usaf Moji said:
I'm not an expert, but I think I can answer this one.

As the person above me said, the uncertainty relation doesn't apply to any two observables you can think of. It only applies to observables that don't "commute". So it would apply to position and momentum, but not to position and velocity.
There is no such thing as a velocity operator and as such there is no observable corresponding to such an operator. I guess you could define a velocity operator by using the expression p = mv and rewrite it as v = p/m. Since momentum has an operator and m is a constant then there doesn't seem be be a problem with it. However it is obvious from this definition that the velocity operator commutes with the momentum operator.

As far as measuring position as a function of time and deducing the speed from those measurements then I don't see how that is possible since detecting the position of a photon involves destroying the photon. Two successive measurements of position would then seem to be impossible.

Pete
 

1. What is Heisenberg's Uncertainty Principle for light?

Heisenberg's Uncertainty Principle for light, also known as the uncertainty principle of optics, is a fundamental concept in quantum mechanics that explains the limitations of our ability to measure the position and momentum of a photon (a particle of light) simultaneously. It states that the more precisely we know the position of a photon, the less precisely we can know its momentum, and vice versa.

2. How does Heisenberg's Uncertainty Principle apply to light?

Heisenberg's Uncertainty Principle applies to light in the same way it applies to all quantum particles. It states that it is impossible to know both the position and momentum of a photon at the same time with absolute certainty. This is due to the wave-particle duality of light, which means that light can behave as both a particle and a wave, and thus cannot be accurately measured by classical physics laws.

3. What are the implications of Heisenberg's Uncertainty Principle for light on scientific research?

The uncertainty principle for light has significant implications for scientific research, particularly in the field of optics and quantum mechanics. It means that there will always be a certain level of uncertainty and unpredictability in the behavior of light, making it necessary for scientists to use statistical methods and probabilities to study and understand light's properties and interactions.

4. Can Heisenberg's Uncertainty Principle be violated?

No, Heisenberg's Uncertainty Principle cannot be violated. It is a fundamental law of nature that has been proven through various experiments and observations. While it may seem counterintuitive or contradictory to our classical understanding of physics, it is an essential principle that governs the behavior of all quantum particles, including light.

5. How does the uncertainty principle for light relate to the observer effect?

The uncertainty principle for light is closely related to the observer effect, which states that the act of observing or measuring a quantum particle can affect its behavior. In the case of light, the act of measuring its position or momentum will inevitably disturb its state, making it impossible to accurately measure both at the same time. This is a result of the inherent uncertainty and probabilistic nature of quantum particles.

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