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In a typical photon experiment the photon is depicted as moving across the page, say from right to left. Suppose we were actually able to witness such an experiment, from the side (to position of reader to a page). If the photon is actually moving from left to right can I, standing at 90 degrees to the motion, see the photon?

  • Thank you Lubos,(yours is a great site!), Noldorin, Johannes and alexander.desouza I now understand (I think) that a photon consists of something which has no mass, has geodesical movement, and velocity, which can be light speed or even, quote, “..the speed of light is the limit and in certain sense it is infinite..”(Marek). Also, it appear that the light emitted by a photon is not radiative, it is vectorial, in the direction of movement. Thank you gentlemen, for putting me right. Rum stuff this QM. If the photon is traveling at infinite speed, then surely it is everywhere at once, so it is al –  Jan 19 '11 at 00:33
  • Melia: I've converted your response ("answer") to a comment for you, since you don't have enough reputation to post comments yet. Once you get to 50 rep, you can post comments directly. :) – Noldorin Jan 19 '11 at 00:34

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The simple answer is no, the eye can only detect photons by their direct interaction with the retina. In this case, the photon is not "visible", since it is not itself incident on the eye, nor emits "secondary" photons that indicate its position.

Noldorin
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Dear Peter, the pictures are drawn to indicate that the photons are there. They are there even if you don't see them. There are many things that we can't see - or we can't see directly or at a given moment - but they still exist. And of course, you don't see the photons (with a wrong direction) by normal methods - you would have to collide them against something else that you could observe but that would change the propagation of the photons on the page.

However, if you want to be fancy, the answer is that you can actually see photons by other photons. Because of quantum effects (with a virtual box-like loop of an electron, and four attached photons), the electromagnetic waves are slightly non-linear. So light can collide with other light so you could actually "shine" a very powerful beam of light to see another beam.

You can't do it in your kitchen but this phenomenon has actually been tested experimentally at SLAC, California. It works: see this thread:

Scattering of light by light: experimental status

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Luboš Motl
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  • Thank you Lubos,(yours is a great site!), Noldorin, Johannes and alexander.desouza. –  Jan 18 '11 at 23:34
  • This answer is correct of course, but I think it over-complicates things (perhaps it's the wording?). A very simple explanation can be achieved. :) – Noldorin Jan 19 '11 at 00:22
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Being very careful, we'll assume you mean a situation in which there exists only a single photon and two detectors placed at right angles to one another, all of which exists in a vacuum and effectively isolated from the rest of the universe: the answer would be 'no'. In practice of course, very few photons and detectors exist in such perfect isolation.

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Your question is worded such that it presumes the photon to be localized within a size much smaller than the distance to the observer 'at the sideline'. In that case, no, this observer can't see the photon. You can't see a photon that does not hit your eye (in spacetime language the photon needs to travel along the past light cone of your pupil). Now, of course a photon is not a classical particle, it has wavelike characteristics as well. So things are different (the photon can no longer be thought of as localized) when the wavelength of the photon becomes comparable to the distance to the observer. But then you can't say "the photon moves from left to right across the page".

Johannes
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