Graduate Bohmian Trajectories: Intersections & Young Experiments

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SUMMARY

Bohmian trajectories, as discussed, cannot intersect themselves due to the dependence of speed on position, a principle that applies universally to well-behaved initial-value problems in differential equations. The discussion clarifies that trajectories are analyzed in space-time rather than three-dimensional space, meaning crossing the same spatial point at different times does not constitute an intersection. The conversation also touches on the implications of boundary conditions in Bohm's equations of motion and the nature of momentum in close proximity, emphasizing that while trajectories cannot intersect, they can approach each other closely without violating the principles of Bohmian mechanics.

PREREQUISITES
  • Bohmian mechanics fundamentals
  • Understanding of differential equations and initial-value problems
  • Concept of space-time versus three-dimensional space
  • Fermi exclusion principle in quantum mechanics
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  • Study the implications of boundary conditions in Bohmian mechanics
  • Explore dynamical systems and phase diagrams
  • Investigate the relationship between momentum and trajectory in quantum mechanics
  • Read David Bohm's original papers on quantum theory and trajectories
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Physicists, quantum mechanics students, and researchers interested in the foundations of quantum theory and the implications of Bohmian trajectories.

PaleMoon
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i read that a bohmian trajectory (in this interpretation) cannot intersect itself because the speed depends on the position. there is no visualization problem in a Young experiment with trajectories from the slits to the screen.
it becomes harder when a particle is trapped in a small region by a potential which make it remain there a long time before a possible tunnelling. how does it work then?
 
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PaleMoon said:
i read that a bohmian trajectory (in this interpretation) cannot intersect itself because the speed depends on the position. there is no visualization problem in a Young experiment with trajectories from the slits to the screen.
it becomes harder when a particle is trapped in a small region by a potential which make it remain there a long time before a possible tunnelling. how does it work then?
The trajectory is considered in space-time, not in 3-space. Crossing the same space point at different times is not an intersection in space-time.
 
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A. Neumaier said:
The trajectory is considered in space-time, not in 3-space. Crossing the same space point at different times is not an intersection in space-time.
Exactly!
 
I am sorry. my question was very incorrectly asked.
i wonder why in bohmian model two different trajectories cannot intersect. i know that the speed
at one point in space time only depends on the configuration space point and that intersection of two curves would be a problem. but can two such trajectories be very very close without problem?
is there some force repelling them. (i have the same problem with fermi exclusion principle when the positions only differ by an infinitesimally small distance)
 
PaleMoon said:
why in bohmian model two different trajectories cannot intersect.
This is not special to Bohmian mechanics; it holds for any reasonably well-behaved initial-value problem for differential equations. Read some introductory literature about dynamical systems and the phase diagrams representing their flow!
 
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Geodesics can meet. Bohmian trajectories cannot.

Bohm himself writes this

It is in connection with the boundary conditions appearing in the equations of motion that
we find the only fundamental difference between the psi-field and other fields such as electromagnetism
 
please read the bottom of page 5 in Bohm's paper
i repeat my question.
Is there something that prevent two points in a small neighborhood to have orthogonal momentums (2 different trajectoiries might be tangent if they meet and exist!)
 
PaleMoon said:
Is there something that prevent two points in a small neighborhood to have orthogonal momentums
If the neighborhood has not a zero size, then nothing prevents it.
 
is it possible a t = 0 that all the momentums are equal to p = 1_x> (unitary ane parallel to x) except in the plane y = o where the momentum would be 1_z> ?
 
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PaleMoon said:
is it possible a t = 0 that all the momentums are equal to p = 1_x> (unitary ane parallel to x) except in the plane y = o where the momentum would be 1_z> ?
No.
 
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  • #11
Oh yes i see now but can you elaborate?
 
  • #12
PaleMoon said:
Oh yes i see now but can you elaborate?
If you see it, then why do you need need an elaboration?
 

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