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How does de broglie-bohm explain obstruction in path?

  1. Jan 15, 2012 #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.
     
    Last edited: Jan 15, 2012
  2. jcsd
  3. Jan 15, 2012 #2

    Ken G

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    It doesn't remain the same, the obstruction changes the interference and that changes the wavefunction, which also changes the probabilities that the particle shows up in various places.
    The wavefunction doesn't know anything, it is a device used by a physicist to make a calculation, so it is the physicist who needs to know about the obstruction to get the answer right. Those who wish to interpret the wave function as something real have a suite of options for deciding how to encode the presence of the obstruction into that real wavefunction, but it doesn't much matter which picture they choose, it is just another calculational device.
    The "pilot wave" handles all that-- there is a conservation principle involved that assures a "full wavefunction" comes through whatever openings there are. In the usual way, without the pilot wave, the conservation principle is upheld simply by how the superposition of amplitudes works, because the amplitudes that you are superimposing are themselves normalized, and superimposing them produces constructive and destructive interference that cancels out in the net, whenever no sources are involved.
     
  4. Jan 15, 2012 #3
    thanks Ken. agreed that wave function is a mathematical device; how it works in reality is anybody guess.


    Is there anything else (for example frequency/energy/spin etc), besides the probabilities/superposition of amplitudes, that gets changed?....in case of obstruction as above

    here... is the "amplitude" ---- the amplitude of probabilities or the amplitude of the wavelength (that gets changed)?
     
    Last edited: Jan 15, 2012
  5. Jan 15, 2012 #4

    Ken G

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    The amplitudes are just probability amplitudes-- what they are probabilities of depends on the "basis" you are using to coordinatize those amplitudes. Things like frequency and wavelength associate with particular choice of "basis" (frequency means you are using an energy eigenfunction basis, wavelength means you are using a momentum eigenfunction basis). The predictions can't depend on the basis you choose, but generally some choices of basis make a lot more physical sense than others, and make the calculations easier.
     
  6. Jan 16, 2012 #5

    Demystifier

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    The answer depends on what do you mean by "photon": particle or wave?

    Here it seems that by "photon" you mean the wave. Well yes, the intensity of the photon wave is halved (because the other half is absorbed by the obstruction, so this wave-function energy evolves into a wave-function energy of the obstruction). However, in Bohmian mechanics the energy of the wave function is NOT a physical energy. Instead, physical energy is the energy of the particle, and that particle energy is not halved; it is either 1 or 0 of the initial energy.

    As I said, the part of the wave function which hits the obstruction is absorbed by the obstruction, so ...

    ... the wave function does get halved.

    It's not recovered. For all practical purposes, the absorption by the obstruction is irreversible.

    Yes, your assumption is correct.

    There is also a classical analogy for all this. The FM radio waves can be thought of as pilot waves for the behavior of your radio-receiver producing the sound. The intensity of radio waves can decrease by various mechanisms, but this will not influence the intensity of the sound you hear from the radio-receiver.
     
    Last edited: Jan 16, 2012
  7. Jan 16, 2012 #6
    thanks Demystifier.

    Now if the "wave" can change the direction of the photon (from 50-50 to each detector to 100% on only detector 1, say in a mach-zhender) it, makes one think that, the wave must have some "force/energy".

    I mean (in a mach-zhender) when there is obstruction the photon will go 50% of the times to detector 1 and 50% of the times to detector 2. However if there is NO obstruction it will go 100% to, say, detector 1.

    i.e. makes one conclude that the wave must have some force that is responsible for the change in the direction of the photon.


    or let's imagine an interesting experiment/setup....where we shoot a photon each (these two photons are separate from the single photon that is travelling the mach-zhender) that is perpendicular each of the arms/path of the mach-zhender. now if the wave were travelling both paths simulations our additional two photons (that are travelling perpendicular to the paths) would be effected
     
    Last edited: Jan 16, 2012
  8. Jan 16, 2012 #7

    Demystifier

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    Yes, it makes sense to say that the wave contains some energy which causes the force on the particle. Indeed, that energy is well described by the Bohmian quantum potential. But the point is that, according to Bohmian mechanics, the object we observe directly is not the wave function but the particle. Consequently, that energy of the wave function is not of a direct physical relevance. It is a helpfull concept, but not an essential one.
     
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