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Photodetectors at the slits in Double Slit Experiment

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  1. Mar 11, 2015 #1
    Hey everyone, I was just wondering how the detectors on the slits operate. The ones which supposedly observe the particle without altering it's trajectory too greatly. I have no idea how this would work with a photon. I'm assuming it's easier with an electron or something.
     
  2. jcsd
  3. Mar 11, 2015 #2
    As far as I'm aware, photon detectors absorb the photon so any detected at the slits wouldn't make it to the screen.
     
  4. Mar 11, 2015 #3
    Ok. Well then of course there's not going to be an interference pattern. I thought something more interesting was going on, like somehow a weak measurement was being made and then the interference pattern disappears and leaves two clumps of particles.
     
  5. Mar 11, 2015 #4
    There is a correlation between the degree of interference and the amount of information available about the 'which-way' info of a quantum system (see for example Zajonc and Greenstein 'The Quantum Challenge' (2nd edition), pgs 108-113).

    I would think 'weak measurements' gather only partial information about the path the system took, but I'm not too informed about the nature of weak measurements and what information is actually gather -- I'm sure someone with more knowledge will chime in.
     
  6. Mar 11, 2015 #5
    That's pretty cool. I am very curious as to how this 'which-way' info is obtained. I am leaning towards an interpretation in which the particle nature of waves is really just a result of the high energies that we use to measure the waves with, since we need higher energy to produce smaller wavelengths.
     
  7. Mar 11, 2015 #6

    bhobba

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    Its simple - you put a detector in the path. How they work - I will leave that up to experimental types. But at the screen where the pattern forms in the double slit lets say it's a photographic plate and we have photons. The photon interacts with the plate, becomes entangled with it, and you get a mixed state giving the probability atoms at a certain position on the plate being affected. When you develop the plate later those positions show up.

    You are caught in a very common misconception - the wave-particle duality - its a myth - although a very very common one:
    http://arxiv.org/pdf/quant-ph/0609163v2.pdf

    Note what it calls a wave, as it actually states, is in fact a wave-function, which is a lot different to a wave in the usual sense. Please put wave-function where it says wave. If the difference isn't clear start a new thread and me and/or others will explain the gory detail.

    In fact what particles are is explained bt Quantum Field Theory (QFT) - they are excitations of an underlying quantum field eg all electrons are excitations of an electron quantum field that pervades all space. What excitation means it a bit subtle - but it's related to creation and annihilation operators of a quantum harmonic oscillator:
    http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

    It turns out a quantum field mathematically when you do a Fourier analysis on it, the 'parts' of that analysis, are mathematically the same as the quantum harmonic oscillator and the creation and annihilation operators create and destroy objects that in every way behave like particles.

    Thanks
    Bill
     
  8. Mar 11, 2015 #7

    Nugatory

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  9. Mar 11, 2015 #8
    I think it's to do with the energy carried by the photons themselves - that's the ones which you send out from the detector to the slits to detect whether there are any passing particles.

    In order to detect which slit the particle has gone through you would need to use a photon with a wavelength comparable to the distance between the 2 slits (or else your resolution would not be high enough to tell which slit the particle has gone through). But a photon with this wavelength would carry an energy Ephoton=hc/λphoton, which would be absorbed by the passing particle as you try to detect it.

    With the photon wavelength you are using, Ephoton would be just great enough that it increases the particle's kinetic energy (thus its momentum pparticle), such that it de Broglie wavelength λparticle=h/pparticle would decrease to such an extent that your slit width is no longer small enough for the particle's wave behaviour to be observed.
     
  10. Mar 11, 2015 #9

    DrChinese

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    You can determine which path information without absorbing all of the photons at a slit. You place polarizers across each slit. If they are oriented parallel, there IS interference. If they are oriented orthogonal (crossed), there is NO interference. Any setting in between will give you a mix of the 2, so you can dial it as you like.

    And if you polarize the source light to an angle midway between the 2 slit settings when they are 90 degrees apart (so 45 degrees relative to the source), you can see that there is no good way to determine why it goes through one and only one slit. Except that there is no self-interference between "paths" in one case and there is in the other.
     
  11. Mar 11, 2015 #10
    Wow it's really amazing how these two effects converge to this strange immeasurability. In the first case it seems to be that if the wave can cohere between the two slits, then it is immeasurable between them. The second one seems to say that as soon as the wave begins to cohere rotationally, the same effect occurs. Is it right to say that wave coherence (and thus interference) among paths or states leads to immeasurability (to certain degrees) between those paths?
     
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