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In QED Feynman describes the photomultiplier experiment as a proof that a photon behaves as a particle. The logic is as follows: with monochromatic single photon light source a photomultiplier coupled with a speaker is either silent or responds with sounds of the same loudness. Hence light is a particle which either hits the multipliers plate or not.

However,the multiplier's action is based on the electrons, which are particles. Why there cannot be an alternative explanation, that light is a wave, whose energy either suffices to knock out a single electron,that initiates the cascade, or does not and in that case the photomultiplier remains silent.

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    $\begingroup$ One electron cannot initiate a cascade, light has to hit many electrons $\endgroup$
    – anna v
    Sep 24, 2021 at 10:46
  • $\begingroup$ indeed, "three or four" writes Feynman. How do we know that photon never knocks out one or two electrons, if they cannot initiate a cascade? $\endgroup$
    – Daniel Kislyuk
    Sep 24, 2021 at 10:51
  • $\begingroup$ Great question, and you're totally correct, this doesn't prove that light is indeed comes in discrete packets of energy. However, there are other examples to show this. $\endgroup$
    – Ofek Gillon
    Sep 24, 2021 at 10:56
  • $\begingroup$ actually, no, Feynman writes that in the point of initial contact it is a single electron. $\endgroup$
    – Daniel Kislyuk
    Sep 24, 2021 at 10:56
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    $\begingroup$ For example, send very weak light into a beam splitter, put detectors on both outputs and measure the correlation between the detectors with no time delay. You'll see it's 0 $\endgroup$
    – Ofek Gillon
    Sep 24, 2021 at 10:58

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Term photomultiplier refers more to the principle of detection than to what we count, which is the central point of the question here. In this respect it is better to differentiate between a single-photon detector and a photon counter, which are doing somewhat different job (although the former is usually a part of the latter).

Single photon detection is not the same as photon counting. The counting error typically increases as $$\sigma_n\propto\frac{1}{\sqrt{n}}$$ where $n$ is the number of photons detected. Thus, indeed, detecting with certainty presence of a single photon is nearly impossible. However, detecting the arrival of the 1001th photon after we counted a thousand of them can be done with high precision.

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  • $\begingroup$ here the term photomultiplier refers to a particular type of device used to detect single photons and it is the term used by Feynman in his book. Could you please elaborate how exactly the discussion of counting error contributes to the question of whether we can conclude that a photon is a particle on the basis of the electrostatic photomultiplier experiment. $\endgroup$
    – Daniel Kislyuk
    Sep 24, 2021 at 11:50
  • $\begingroup$ @DanielKislyuk Do you agree that being countable is a particle-like property? Waves, on the other hand, have continuous amplitude and intensity. $\endgroup$
    – Roger Vadim
    Sep 24, 2021 at 11:52
  • $\begingroup$ I do agree. My argument is that since electrostatic photomultiplier functioning is based on particles it can quantize the wave input and it is impossible to conclude on the basis of this particular experiment, whether photon is a particle or a wave. $\endgroup$
    – Daniel Kislyuk
    Sep 24, 2021 at 11:55
  • $\begingroup$ @DanielKislyuk I think this is actually the point: that one photon produces exactly one electron, i.e., that the output is quantized. How would you get the same outcome with a continuous wave? $\endgroup$
    – Roger Vadim
    Sep 24, 2021 at 12:02
  • $\begingroup$ If those photons as waves had energies at the level sufficient to knock out one electron, slightly below and slightly above, the result would have been exactly the same, wouldn't it? $\endgroup$
    – Daniel Kislyuk
    Sep 24, 2021 at 12:17
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This is a long comment on the discussion in comments:

I would not expect Mr. Feynman to make silly mistakes in his books, so I'm really looking for counter-arguments to my question. –

It is not a mistake, because if one puts down the photoelectric effect etc, this is what happens, and so it could just be that he had in mind an argument that is not explicitly stated.

Talking of mistakes and Feynman :

I have been lucky to be at summer schools and workshops where Feynman took part, So I want to share a story I have written elsewhere before that shows that even giants in physics can be wrong at times.

This story is in the 1980s, after Feynman had his first operations and when he was feeling well enough to go to conferences, in Crete, at a theoretical workshop. It is the second day he is in Greece, the first being spent going around the “must sees” of Athens with yours truly, so definitely jet lagged.

It is after sunset, and many eminent physicists are having a drink on a veranda overlooking the Aegean, and several Greek ones from other disciplines attracted by the pole of Feynman (as yours truly).

There is small talk and a lull in conversation and suddenly the full moon draws the attention on the horizon in all its glory, reflected in the waters. Feynman looks and observes: ” that must be the West then”.

The interest in the story is not the slip of Feynman, it is the reaction of the physicists. Feynman said it after all, so that must be the west. I do not remember if t’Hooft was there at the time, he sure was at the workshop. It remained for yours truly, youngest and most irrelevant in the company to sputter: “but,but, but….”.

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