Can scattering experiment be used to determine location and momentum of an electron?

  • Context: Graduate 
  • Thread starter Thread starter sumitgarg1
  • Start date Start date
  • Tags Tags
    Compton scattering
Click For Summary
SUMMARY

The Compton scattering experiment can be utilized to determine the location of an electron by measuring the wavelength change (λ' - λ) and the scattering angle (θ) of a photon interacting with the electron. The relationship is defined by the equation λ' - λ = h(1 - cosθ) / (m * c). Additionally, the uncertainty principle is applicable, as the momentum (Δp) and position (Δx) of the electron are complementary observables, indicating that precise measurements of one will inherently affect the other.

PREREQUISITES
  • Understanding of Compton scattering principles
  • Familiarity with the Heisenberg uncertainty principle
  • Knowledge of photon behavior and properties
  • Basic grasp of quantum mechanics concepts
NEXT STEPS
  • Study the mathematical derivation of the Compton scattering formula
  • Explore the implications of the Heisenberg uncertainty principle in quantum mechanics
  • Investigate experimental setups for measuring scattering angles and wavelengths
  • Learn about complementary observables in quantum physics
USEFUL FOR

Physicists, quantum mechanics students, and researchers interested in the behavior of particles at the quantum level, particularly those studying scattering experiments and the implications of the uncertainty principle.

sumitgarg1
Messages
1
Reaction score
0
It seems Compton scattering experiment can be used to determine location of electron by hitting it with photon.
As per Compton scattering. if we measure λ' , λ and θ accurately; the location of electron can be determined.(in theory).

λ' - λ = h(1-cosθ)/m*c

also when θ=π then :

Δp = h*/λ -h*/λ'

Δx = c * Δ t / 2 where Δ t = time interval of photon discharge from source and receiving it on detector.

In this scenario ; can someone please explain how uncertainty principle applies to Compton scattering??
 
Physics news on Phys.org
I haven't read the Stack Exchange thread, but at a heuristic level, looking at your expressions for ##\Delta p## and ##\Delta x##, the former involves wavelength and the latter involves time. Those are complementary observables just as momentum and position are complementary observables. So you haven't found a way to avoid the uncertainty principle; you've just exchanged one pair of complementary observables for another.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K