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When I first heard about the photons and the double-slit experiment my immediate thought was the following: Alright, energy is not absorbed continuously but in discrete units, photons, but nature somehow needs to keep the desired irradiation levels so it needs to dither, like a printer needs to when it wants to paint gray with black ink. So I began to think what if photons are only an absorption phenomenon of the atoms? And nothing else. So in this way, electromagnetic wave simply becomes a region of elevated energy levels, when this energy reaches an atom there is some chance it got excited, the chance is small, but the stronger the radiation, the bigger the expected mean number of detections at an area.

Now imagine an experiment where you have a single photon source. You push a button and it will emit 1 photon worth of energy. In the model I outlined previously there is a chance it won't be detected at all, there is a chance it's detected once, twice and so on. The distribution is Poission distribution. With a mean value of 1.

We push a button and get a count. But the result can be interpreted two ways:

  • Indeed 1 photon worth of energy emitted per button push, but sometimes more atoms or none of the atoms responded to it. The average number of photons detections was indeed 1.
  • What QM would say: the source was faulty it emitted 0, 1, 2, etc. photons per button push not just 1. But the average number of photons was still 1.

Since all we have is the detection count we cannot really decide what's the truth.

I guess QM is right, but I still wonder is it possible to construct an experiment whose result can only explained by the fact that when 1 photon worth of energy emitted, then there can be only 1 detection and never more than 1?

Calmarius
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4 Answers4

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It is worth contemplating the double slit experiment one photon at a time.

The following sequence of pictures show a typical result observed in a film placed behind the slits in a single-photon two-slit experiment, for increasing exposure times:

singlephotondoublsslit

second slide

second slide

last slide, for economy of space,the original link is broken, but here the time sequence can be still seen .

lastslide

Because we are looking at individual photons the entire experiment has to be done in a dark room to avoid background light. Also the most sensitive part of the experiment (camera and image intensifier) are placed in a black box to avoid scattered light from the laser. The image intensifier is an essential part of this experiment. With the intensifier, every single photon are amplified by a factor of up to one million, so that the signal generated by each photon at the output of the intensifier (a phosphor screen) can be detected with a sensitive film or a CCD camera.

In this case , the probability of the single photon to be absorbed by the intensifier completely is 1, by construction.

but the stronger the radiation, the bigger the expected mean number of detections at an area.

In this case the energy of the single photon will not make a difference as long as it is within the energy levels available on the phosphor screen. One point will appear.

In general the qualifier "stronger" for photons is wrong. Photons can be of high energy or of low energy.

In the experiment described, where photons are released one at a time, there are no buttons, the classical intensity of the source ( proportional to the number of photons) is reduced so that one photon will arrive at a time delta(t). Delta(t) varies because of quantum uncertainties, but a laser will emit a small delta(E) energy photons and there is no problem in deciding that a photon hit the intensifier.

anna v
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It is reasonable to wonder if electromagnetic energy quantization is only quantized for interaction with matter (so also with the photo-electric result), and not intrinsically. But you can think of a photon itself as a quantized packet of electromagnetic energy, and then realize that you need a theory to explain when, where, and how it is involved with matter (and sometimes even scatters off other quantized packets of electromagnetic energy). The theory you end up with, is the standard theory.

But we don't generally have buttons that produce one photon (I know you said it was a thought experiment). What you generally compute is that you know a speed (c) and and the size (L) of the apparatus so how much time (L/c) each photon spends in flight and how many photons per hour and you notice that the total time in flight (actually just the sum of the flight time of each photon) is much less than the total time of the experiment, so you can conclude (too blithely) that likely most if not all of the photons travelled alone. But to know for sure you'd have to pay close attention to the detection. And in fact we can't actually conclude that it is a Poisson process because they aren't actually fully 100% independent, but we can try to make it close because in the one-at-a-time limit they are very close to independent.

So in practice, you can in general divide the time of a normal experiment into regions of time sized exactly so that the expected number of photons is one. But it won't be Poisson. You'd only approach the Poisson for time intervals with much less than one photon expected

The reason for the failure is that when there one photon, the probability of a second one is a bit more than the probability of the first. Electrons do the opposite. So if you want to send electrons through a double slit one at a time, it's easier to conclude they are going one at a time.

Timaeus
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    "generally have buttons that produce one photon" there are single photon sources, those are important in quantum computing and quantum encryptions. So I think we already have a source that can spit a photon when you press a button. – Calmarius Jan 28 '15 at 11:10
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Word absorption means that the photon had to finish its life and convert to excitation energy of the atom in the material. Photon can interact with the matter in a three ways: photo-effect, Compton scattering and pair production. Only in the second process photon survives.

Experimentally, absorption (in terms of photo-effect) is observed in detectors as a single peak of certain energy level (for monochromatic gamma source). Since full energy is absorbed then it must give specific detector response within its resolution.

If photon is not absorbed but scatters inside, it releases a part of its energy which goes into ionization of the atom, this energy is seen as a rather flat spectrum between 0 and photon (peak) energy. It is also possible that photon scatters once or multiple times inside material and then is absorbed, then you will see this also seen as a full energy peak. otherwise it can escape from detector active volume and leave Compton background.

So answering your question, it is not possible to have multiple absorptions, but yes, it is possible to have multiple scattering effects. Then you need to build a array (grid or stack geometry) of relatively thin (thinner that average scattering length) detectors, and then you have chance to observe scattering effects in several layers.

Rafal
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What you ask depends on the detection method. If the detection method is based on the absorption of the photon with electron emission, then 1 photon can be absorbed 1 time. This is the photoelectric effect, and the energy conservation shows that the photon energy cannot be smaller than the ionization energy of 1 electron.

If though we are interested to obtain Compton scattering, a photon with high energy is scattered inelastically by the particles on its way. At high energies, the photon wavelength may be enough small s.t. the photon see the electron in the atom on its way, and not the whole atom. At each scattering event the photon leaves part of its energy to the particle on which it was scattered. The electron gets a recoil due to the collision with the photon, and the photon linear momentum decreases, s.t. its wave-length increases.

Sofia
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