Did photon obey Uncertainity Principle?

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SUMMARY

The discussion centers on the application of the Uncertainty Principle (HUP) to photons, which are massless particles. The initial assertion that the equation position * velocity * mass must exceed Planck's constant is challenged, as photons do not possess mass. Instead, the correct relationship for photons is defined by the equation E = pc = ħω = ħck, highlighting their momentum despite having no mass. The single-slit diffraction phenomenon serves as a practical example of photons adhering to quantum principles, including the HUP.

PREREQUISITES
  • Understanding of the Uncertainty Principle (HUP)
  • Familiarity with quantum mechanics concepts
  • Knowledge of photon properties and behavior
  • Basic grasp of momentum and energy relations in physics
NEXT STEPS
  • Study the mathematical formulation of the Uncertainty Principle
  • Explore the implications of massless particles in quantum mechanics
  • Investigate the single-slit diffraction experiment and its significance
  • Learn about the relationship between energy, momentum, and wavelength in quantum physics
USEFUL FOR

Physicists, students of quantum mechanics, and anyone interested in the fundamental principles governing particle behavior in quantum physics.

sndtam
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Photon has no mass.So when we apply Uncertainity principle to photon
position*velocity*mass=greater than Plancks constant
So when we use mass as null,the equation implies that 0<h.
That is h is negative.But h is positive.
Please explain me this.
 
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The uncertainty relation follows from the commutation relation of the momentum and position operators. The photon does not carry a mass, but it does carry a momentum.
 
sndtam said:
velocity*mass
This is not momentum. This is an approximation of momentum at low velocity. The correct dispersion relations for the massless particles as the photon is
E=pc=\hbar\omega=\hbar c k
 
Last edited:
sndtam said:
Photon has no mass.So when we apply Uncertainity principle to photon
position*velocity*mass=greater than Plancks constant
So when we use mass as null,the equation implies that 0<h.
That is h is negative.But h is positive.
Please explain me this.

The single-slit diffraction phenomenon is the clearest example of photons "obeying" the HUP.

Zz.
 
Thank you
 
As far as we know, everything obeys quantum principles...including HUP...
 
Naty1 said:
As far as we know, everything obeys quantum principles...including HUP...


I refuse to obey your quantum principles ;)
 
Thanks a lot
 
maverick_starstrider said:
I refuse to obey your quantum principles ;)

A smart dope is in direct violation of my HUP.

- Werner
 

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