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Performing the Double Slit

  1. Jun 23, 2005 #1
    I have seen this experiment done with a laser, but I want to know how hard it is to do the version of the experiment where only a single particle is fired toward the slits at a time. Is it possible for me to carry out such an experiment, or does such a thing require lots of cash?

    Also, from what I've read, the experiment has been performed with "detectors" of some sort on each slit to see which one the particle goes through, and with the detectors there the particles only land in the region defined by the sum of squares of the probabilities from each slit, but without the detectors the particles are described by the square of sums.. :confused: What kind of detectors are used that still allow the particles to continue on their journey otherwise untouched?
     
  2. jcsd
  3. Jun 23, 2005 #2
    I doubt that the experiment is cheap or easy to perform. There was a video of some research team (at Hitachi or some other japanese(?) company) doing a double slit experiment with single electrons posted somewhere on PF. Maybe on this forum.

    You can't measure which slit is used without somehow interacting with the particles. Simplest way would be to keep only one slit at a time open. That way you'll know which slit the particle has passed through once you detect it.
     
  4. Jun 23, 2005 #3
    Oh. I thought 'performing the double-slit' meant something else. Sorry for the intrusion.
     
  5. Jun 23, 2005 #4
  6. Jun 28, 2005 #5
    It's not actually that hard to do a single-photon version of the experiment. First of all, you need some filters that will absorb most of the photons in the laser beam. A few pieces of darkly coloured glass will do the trick, or you can buy specially made filters from an optics supplier. You will also need one of those devices that measures light intensity (I forget what they are called). Needless to say, you have to do the experiment in complete darkness, and keep the entire setup covered for the duration of the experiment.

    The first thing to do is check the intensity of the laser and figure out how many filters you need to ensure that there will only be one-photon in the apparatus most of the time. From laser physics, we know that the number of photons emitted follows a poisson distribution and you know the laser frequency and dimensions of the apparatus, so this is fairly easy to work out. Then you set up your apparatus with a photographic plate where the screen would normally be, wait a long time and develop the plate. You might want to expose different plates for different amounts of time to see how the pattern builds up from individual detection events.

    Now, the tricky bit is to convince yourself that the pattern was really created by single photon events and not, for example, by the small number of events where two or more photons are in the apparatus at the same time. Unfortunately, you cannot eliminate these entirely because of the Poisson statistics. Analysis of the signal to noise ratio of your interference pattern can be used to do this.

    I actually performed this experiment as an undergrad. Of course, we used a slightly more sophisticated setup involving very stable lasers and very sensitive photomultiplier tubes for detection. However, the setup described above is sophisticated enough to get decent results.

    You cannot really place detectors at the slits in this setup, but as an alternative you can repeat the experiment with one of the slits covered and then the other. You should be able to see that the resulting patterns do not add up to the two-slit pattern.
     
  7. Nov 22, 2009 #6
    Let's say a single photon goes through both slits and interferes with itself so as to completely cancel out (showing up as a darker line in the interference pattern)- Then where has the energy of the original photon gone?

    Then imagine a few milliseconds of time later (an aeon in the quantum world) another photon goes through both slits and reinforces itself (showing up as a brighter line in the interference pattern)- Then where does the extra energy come from?
     
  8. Nov 22, 2009 #7

    DrChinese

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    A photon cannot cancel itself out in total, just in certain areas. There is conservation of energy.

    The brighter and darker areas correspond to the probability amplitudes, but there is always a whole photon or not.
     
  9. Nov 22, 2009 #8
    When you launch a photon that constructively interferes with itself, it will land on the screen as a single photon; not two photons. When you launch a photon that destructively interferes with itself, it will not land on the screen.

    1. How can you know if a single photon 'cancels itself out' at the slit if you never get to observe this single photon in the first place (after all, we presumed the photon destructively interfered with itself at the slit, meaning it will never be detected)?

    2. The 'brighter line' of the interference pattern is just a collection of individual photons. Think about it: One photon constructively interferes with itself, causing it to land on the screen in a certain place (what place will that be? Well, the place where there will eventually be the 'brighter line,' of course!).
    Next photon is launched, and nothing happens; we presume it destructively interfered with itself.
    Third photon is launched and it constructively interferes with itself, landing in what will soon be one of your 'bright lines' of the interference pattern.
    The fourth photon is launched and nothing is seen on the screen; it is presumed it destructively interfered with itself.
    The fifth photon is launched, and it constructively interferes with itself, landing on what is beginning to look like an interference pattern.

    As you can see, if you launch enough photons, you will end up with the characteristic interference pattern, with no photon exhibiting 'extra energy.'
    When you launch a photon that constructively interferes with itself, it will land on the screen as a single photon; not two photons. When you launch a photon that destructively interferes with itself, it will not land on the screen.

    You ask, "But what about the photons that destructively interfered with themselves? What happened to their energy?" The answer is this: Observe the total energy that is on the screen in the form of photons that constructively interfered with themselves and compare this to the total energy that was emitted by the laser. Assuming lasing efficincy of 100% (just pretend), the total output spent by the laser will equal the total photon energy that landed on the screen, thus conservation of energy is maintained.

    Dr. Chinese, do you see a problem with this last paragraph?
     
    Last edited: Nov 22, 2009
  10. Nov 22, 2009 #9

    DrChinese

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    When you send a single photon through the double slits, you get exactly one hit on the screen. The photon does not destructively interfere with itself in such a way that it disappears.

    The self interference comes from the evolution of the photon's wave state, as there are components going through each slit. These components then interfere constructively AND destructively with each other. They reinforce in some places or tend to cancel in others. As this happens, there is an increased or decreased probability respectively that the photon will follow that path. Of course, we don't know which is actually taken.
     
  11. Nov 22, 2009 #10
    Even with a perfectly uniform laser beam and a stack of filters, you have no way of knowing whether you are putting through one photon at a time or simply a continuous e-m wave of low intensity. That's because you can't experimentally distinguish whether the Poisson statistics occur at the point of emission or the point of detection.
     
  12. Nov 22, 2009 #11
    Another reason in favor of me enrolling in QM classes at this University: The evolution of the pilot wave (correct?) will cause an increase or decrease in probability of the photon taking a path that effects a hit on the screen, but I don't see it. I can't see how a 'cancelled' photon (one that didn't follow its evolved path) can still make it to that screen.
     
  13. Nov 22, 2009 #12

    DrChinese

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    The photon isn't canceled, just that path is. It is "steered" away from areas of low probability (where cancellation occurs) and towards areas of high probability.
     
  14. Nov 22, 2009 #13
    Can't a single photon be represented as a wave function? And can't a hit on the screen be viewed as a collapse of that wave function? If this is so, then indeed we know we're sending EM waves through, one photon at a time.
     
  15. Nov 22, 2009 #14
    Understood. Thanx! :smile:
     
  16. Nov 23, 2009 #15
    No, a hit on the screen doesn't necessarily mean the wave function collapsed. Let's take the screen to be a photographic plate. Your theory is that when a silver halide crystal changes color, it must have absorbed a single photon. An alternate theory would be that the probability of the crystal changing color is proportional to the intensity of an electric field, which may be uniformly spread out over the whole plate. No photons. Statistically, these two interpretations give the same result. So the "hit" on the screen doesn't prove you have detected a single photon.
     
  17. Nov 23, 2009 #16

    f95toli

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    No, photons can -in general- not be represented by ordinary wavefunctions.
    Moreover, a single photon state (as with all number states) is very different from e.g. a coherent state generated by a laser or a thermal state generated by for example blackbody radiation. These states have different properties and obey different statistics. They are if you want different "types" of light.
     
  18. Nov 23, 2009 #17

    Cthugha

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    If you really want to know whether you have prepared a single photon state you need to insert a 50/50 beamsplitter into the beam and put two avalanche photodiodes capable of single photon counting in the beams leaving the beamsplitter. Now you can measure the cross correlation between the exit ports of the bs. If the normalized cross correlation coefficient is near 0 (you never have simultaneous clicks in both diodes) you have a single photon state. If the normalized cross correlation coefficient is near 1 (clicks in both detectors are statistically independent) you have a weak laser beam.
     
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