Double Slit Experiment - Measurement Problem

AI Thread Summary
The discussion centers on the impact of measurement on the double slit experiment, specifically regarding molecular fluorescence and direct illumination. Both methods involve interactions that alter the momentum of electrons, thereby destroying the interference pattern. In the fluorescence scenario, when an electron excites a molecule, the subsequent photon emission disrupts the electron's trajectory, similar to the effects seen in direct illumination. The key factor in both cases is the transfer of energy and momentum during interactions, which leads to the loss of the interference pattern. Ultimately, any measurement that involves photon interaction with electrons results in a collapse of the wave function, eliminating the characteristic interference.
Spherical Cow
Messages
9
Reaction score
0
Question
Suppose molecules are placed next to the slits that fluoresce when electron passes, emitting a photon that can be imaged. Explain why this is no better than direct illumination, and explain how the two-slit interference pattern is destroyed.

My Answer
In direct illumination, a narrow beam of light is passed through one slit. Each time an electron passes through the slit, it will scatter a few photons, which can be detected. Collision with the photons will change the electron's momentum, thus destroying the interference pattern.

My Question
How is the interference pattern destroyed with the molecular fluorescence experiment? Is it due to the emitted photons interacting with the electrons? I.e. same reasoning as the direct illumination experiment?
 
Physics news on Phys.org
Well, what happens when matter fluoresce?
 
The electron excites the molecule. When the molecule relaxes to its ground state, it releases photons.

Is it the transfer of energy from the electron to the molecule that disrupts the interference pattern? Or is it the interaction between the electron and photons (i.e. electron losing momentum)?
 
When the electron transfer energy to the molecule, it transfer its kinetic energy. Specifically, it transfers kinetic energy by colliding with the electrons of the molecule, which makes it lose momentum.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top