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I was reading G.I.Taylor's Single - Photon Double Slit experiment.

Now, at a time only a single photon gets emitted. What is the probability of it at a certain point of the screen to hit?

The answer is written rather abruptly:

Although the photon can hit anywhere, the probability is more, where the brightness(bright fringes) i.e. Intensity is more.

Now, this means intensity existed before photon actually hit the wall? How can it be possible?

When photon hits the screen, the energy gets transferred to the screen. So, after photon collides, intensity does come in discussion.

But if we accept the answer, it is like saying that energy came somehow on the screen before the photon; after that photons would strike there where the intensity of the energy is more. So, who is actually transferring the energy: photon or the waves? If it is wave that transfers energy as being evident from the above quoted argument, then what photons are doing actually ?? What are they meant for if they are not responsible for transferring energy??

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  • $\begingroup$ The photons (non-trivially) constitute the classical wave - these are not seperate entities, so I don't understand the question. $\endgroup$ – ACuriousMind Feb 1 '15 at 13:13
  • $\begingroup$ The author of that quote is simply saying that the most probable regions for a single photon to hit turns out to be proportional to the intensity of the fringe patterns that you would expect classically. (And therefore the classical result can be explained in terms of individual photons interfering with themselves) $\endgroup$ – lemon Feb 1 '15 at 13:19
  • $\begingroup$ @lemon: Then sir, photons actually interact with the screen,right? $\endgroup$ – user36790 Feb 1 '15 at 13:27
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    $\begingroup$ They must interact with the screen to produce an image, yes... $\endgroup$ – lemon Feb 1 '15 at 13:48
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This is a probability statement based on observation. If I look at traffic accidents on three different junctions, and count for a year, I may say "accidents are 4.3 more likely on this junction than on that junction". All things being equal, the following year you will indeed see that ratio of accidents, and so insurance companies can start to compute the risk (probability) of individual cars having accidents depending on the route they take.

Now to your experiment. The appearance of fringes tells us that more photons have arrived at one location than another. We can therefore say "a photon traveling through these slits has a higher probability of arriving here than there" - not because the intensity somehow causes them to be redirected, but rather because we now have a reliable measurement of what actually happens - and we can derive the probability of an event from this observation.

The photons carry the energy. Their behavior is like that of a wave. Energy is not transferred before the photon hits the screen. But after 1 million photons have hit the screen, I have a pretty good probability distribution with which I can compute likely places where the 1000001th photon will hit.

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  • $\begingroup$ So, when the 1st photon was about to hit, there were no fringes on the wall; so every point had equal probability. . . $\endgroup$ – user36790 Feb 2 '15 at 4:20
  • $\begingroup$ . . . So, after many photons had hit, we can say the next photon would tend to hit at those region where the previous photons had clustured ie. The Brigh Fringes. $\endgroup$ – user36790 Feb 2 '15 at 4:22
  • $\begingroup$ Not quite. If no photons have landed you may not know what the probability is - so you could bet on any position equally. However the distribution is fixed by the setup even before the first photon passes through the slits. $\endgroup$ – Floris Feb 2 '15 at 4:49
  • $\begingroup$ Where the first one would hit is ascertained by what,sir? Then as said by Lemon above, the photon would tend to hit where we get bright fringe classically. $\endgroup$ – user36790 Feb 2 '15 at 5:09
  • $\begingroup$ The photon will hit in accordance with the wave function. You don't know the wave function without observing many photons but the photons know it from the slit geometry. Yes the probability is as given by the fringes - the fact that you don't know the probability doesn't change what that probability is. If I throw a non standard (biased) die you don't know the probability that I roll a 6. But if you observe 1000 rolls you get a good idea. However that same probability existed before the first throw - you just didn't know it. $\endgroup$ – Floris Feb 2 '15 at 12:41

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