- #1
San K
- 911
- 1
In de Broglie–Bohm theory (or should we say hypothesis):
the wavefunction travels through both slits, but each particle has a well-defined trajectory and passes through exactly one of the slits.
The wave function interferes with itself and guides the particles in such a way that the particles avoid the regions in which the interference is destructive and are attracted to the regions in which the interference is constructive, resulting in the interference pattern on the detector screen.
Question:
How does De Broglie-Bohm (DBB) explain the fact that when we place an obstruction in one of the paths the intensity/energy of the photon still remains the same even though:
the wave function went through both the paths and one of the paths had an obstruction hence only half of the wave function would have passed? and hence the intensity/energy/something about the photon behavior should be half...
or alternatively how does the wave function know that there is already an obstruction in the path and it hence takes only one path?
or how does DBB explain ...why isn't the wave function getting halved?
or how is the half of the wave function recovered from the blocked path?
we can take the example of a double slit or a mach-zehnder
http://en.wikipedia.org/wiki/Mach%E2%80%93Zehnder_interferometer
note: i am assuming here that the wave-function splits into the two (or multiple) paths.
however i think one has to sort of assume this because the waves do interfere when there is no obstruction.
the wavefunction travels through both slits, but each particle has a well-defined trajectory and passes through exactly one of the slits.
The wave function interferes with itself and guides the particles in such a way that the particles avoid the regions in which the interference is destructive and are attracted to the regions in which the interference is constructive, resulting in the interference pattern on the detector screen.
Question:
How does De Broglie-Bohm (DBB) explain the fact that when we place an obstruction in one of the paths the intensity/energy of the photon still remains the same even though:
the wave function went through both the paths and one of the paths had an obstruction hence only half of the wave function would have passed? and hence the intensity/energy/something about the photon behavior should be half...
or alternatively how does the wave function know that there is already an obstruction in the path and it hence takes only one path?
or how does DBB explain ...why isn't the wave function getting halved?
or how is the half of the wave function recovered from the blocked path?
we can take the example of a double slit or a mach-zehnder
http://en.wikipedia.org/wiki/Mach%E2%80%93Zehnder_interferometer
note: i am assuming here that the wave-function splits into the two (or multiple) paths.
however i think one has to sort of assume this because the waves do interfere when there is no obstruction.
Last edited: