I Wave packet experimental detection

Click For Summary
The discussion centers on the concept of wave function "collapse" in quantum mechanics, which remains a contentious topic among physicists. It highlights that the meaning of "collapse" varies depending on the interpretation of quantum mechanics one subscribes to, with no definitive assertion about the physical reality of the process. The conversation emphasizes that "collapse" serves as a mathematical update following a measurement, rather than a description of an actual event. The difficulty of measuring wave-packet dynamics is noted, with examples like the Stern-Gerlach experiment illustrating how measurements can entangle states. Overall, the dissatisfaction with the explanation of "collapse" reflects ongoing debates in the field of quantum theory.
VVS2000
Messages
150
Reaction score
17
TL;DR
Are there any experimental setups that verify the wave packet dynamics we work with in quantum mechanics?
It just came up in my QM class while we were discussing and even my teachers could'nt figure it out
I know the wave function "collapses" when a measurement is made but still not satisfied with it
 
Physics news on Phys.org
Rightfully so. The "collapse" is a very questionable concept and not really needed for the physical interpretation of quantum theory. It's obvious that it depends on the specific measurement made on the measured object, which state this object takes after a measurement has been made. E.g., if you detect a photon in the usual way using the photoelectric effect (e.g., using a CCD cam or a photoplate) this photon gets absorbed and is thus gone for good.

It's of course very difficult to measure "wave-packet dynamics". An example is this:

https://doi.org/10.1103/PhysRevLett.72.3783
https://pure.uva.nl/ws/files/2978244/478_5187y.pdf
 
  • Like
Likes PeroK, VVS2000 and topsquark
VVS2000 said:
I know the wave function "collapses" when a measurement is made
Exactly what "collapse" means depends on which QM intepretation you adopt. Note that discussion of particular interpretations belongs in the interpretations subforum.

In the absence of any particular interpretation, "collapse" is just the mathematical procedure we use to update our model when we know the result of a measurement, and no assertion is made at all about what, if anything, "actually happens".
 
  • Like
Likes PeroK, VVS2000, topsquark and 1 other person
The "collapse" is simply the update of the state after the interaction with a "filter". This idealized "von Neumann filter measurements" are very rarely achieved. An example is the Stern-Gerlach experiment for measuring and preparing spin states of an atom (in the original experiment silver atoms). Here the atom is send through an inhomogeneous magnetic field. According to quantum mechanics the atom moves in different discrete directions depending on the value of the spin component in direction of the magnetic field. Then the position (or momentum) of the atom is entangled with this value of the spin component, i.e., you can just block all atoms which are at positions referring to the spin value you don't want, and thus all atoms going through this filter have a determined spin component in direction of the magnetic field, and you describe them by a corresponding wave function which is a eigenstate of this spin component with the eigenvalue you filtered out.
 
PeterDonis said:
Exactly what "collapse" means depends on which QM intepretation you adopt. Note that discussion of particular interpretations belongs in the interpretations subforum.

In the absence of any particular interpretation, "collapse" is just the mathematical procedure we use to update our model when we know the result of a measurement, and no assertion is made at all about what, if anything, "actually happens".
yeah I know, that's why I told I was not satisfied with that answer
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
Replies
16
Views
769
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
32
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K