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So when i see white light, is it really just seven different photons? |
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| Dec18-12, 03:54 PM | #1 |
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So when i see white light, is it really just seven different photons?
Im a little confused as to why the eye detects white light as white instead of all the separate colors that compose white light. I mean it isnt a single photon right (multiple single photons of course)? So how do the different photons converge into one color? Also, which may seem like a very ignorant question, how does the eye detect a certain wave length when the photon is in a certain stage of its wave cycle when it instantly hits the eye. Like a blue photon hits the eye, but it is only a quarter of the way through with its cylce, so it hits the eye with wavelength of a red photon.
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| Dec18-12, 04:14 PM | #2 |
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All three overlap to some degree and it is the varying amount of stimulation that each cell receives compared to other cone cells that forms the first stage of color vision. http://en.wikipedia.org/wiki/Color_vision http://en.wikipedia.org/wiki/Trichromat Edit: By the way I just noticed the "Seven different photons" part of your thread title. I just want to make sure you realize that there are more than 7 colors. The visible spectrum is not divided cleanly into colors, it is a gradual shift from one to the other. Each person will see the same wavelength of light a little differently than another person. This is especially true of colorblind people. (Who don't actually see in black and white usually) |
| Dec18-12, 05:22 PM | #3 |
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An addition about the photons: I think you can also generate states where each photon is in a superposition of frequencies (=colors) so even the single photons can be "white".
I don't know however if this is the case in your usual white light. The case of several photons with one wavelength each has already been explained. Another comment about the question about the time-evolution of the electric field. Your eye does not detect the instantaneous electric field, but time-averaged quantities like intensity (=time average of E^2) and colour (which corresponds to the energy of each single photon which corresponds to the frequency). The classical explanation is simply: The oscillaton of the e-field is too fast to detect. Thus any detector including the eye detects the averaged quantities as i said above. Explained quantum mechanically, the electric field is only a derived quantity and should not be considered in the detection process (which is inherently quantum mechanical because in the end single photons get absorbed by single molecules). Instead, what you actually get is a state with many photons whose energy is h*f (where f corresponds to the frequency of the actual macroscopic E-field). The receptors in the eye have some transition between internal states with some energy difference e. The receptor for the color f can absorb a photon if its energy is exactly e=hf. This explains the statement that they eye can detect frequency/photon energy/ color. The intensity simply means that more or less photons where detected. |
| Dec18-12, 05:22 PM | #4 |
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So when i see white light, is it really just seven different photons? |
| Dec18-12, 05:50 PM | #5 |
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If you mix light of different wavelengths, you can produce 'colours' which are not actually part of the spectrum at all. None of the colours we see in real life are ever just one wavelength because most illuminants (sun, lightbulbs, LEDs) produce a range of wavelengths and objects we see have pigments which do not reflect just one wavelength. The only time we see true spectral colours is when we are in a darkened room and we look at the output of a spectrometer or, in rare circumstances, when we see a gas discharge lamp - but even those tend to have emit a spectrum with a number of different lines. |
| Dec18-12, 06:28 PM | #6 |
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in other words: the eye simply isn't accurate enough to separate out the photons if they're too close together (which in an ordinary white light they certainly are) ![]() (btw, i don't think the eye is designed to react to individual photons, i think it needs a "threshhold" number of photons before it takes any notice of them) a photon of a particular wavelength has a particular momentum, and that momentum is what determines how in interacts with anything (including your eye) the momentum is the same throughout the cycle |
| Dec18-12, 09:57 PM | #7 |
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In modern day quantum optics, an eigenstate of the photon number operator is also called (single) photon, at least concerning the n=1 Fock state. This latter entity can be as polychromatic as you want it to be. Of course this coinciding terminology makes things a bit difficult - as if understanding light was not already complicated enough. |
| Dec18-12, 10:02 PM | #8 |
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| Dec18-12, 10:41 PM | #9 |
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Laser light has a very, very narrow linewidth, but it is not infinitely small. You can measure that by measuring the first order correlation function. It is basically the Fourier transform of the power spectral density of the light field. Its decay gives the coherence time of the light which is in turn inversely proportional to the spectral width of the light field. Longer coherence times means spectrally narrower and therefore more monochromatic light.
Therefore, you need infinitely long coherence time to get perfectly monochromatic light, which is of course unphysical. But laser light is pretty much the closest to ideal monochromatic light you can get. |
| Dec19-12, 02:58 PM | #10 |
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| Dec19-12, 04:48 PM | #11 |
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Is this just muddying the waters here, my friend? Are you havin a virtual larf?
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| Dec19-12, 07:10 PM | #12 |
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As a rough sketch as to why single photons can be polychromatic consider the way single photons are typically created. The easiest way lies in using some system which offers some blockade effect for an optical transition. That may include single quantum dots, nitrogen-vacancy centers in diamond or single atoms. In all of these cases, you have some states which can only be occupied by one carrier, maybe some excited state of the atom or the exciton ground state of the quantum dot. Now as the carrier goes from the excited state to the ground state, it will take some time until the carrier can be brought back to the excited state again. For this duration you can be sure that only one photon has been emitted and no other photons will follow for a certain duration, so you have a single photon. For some systems, like the mentioned NV-centers in diamond, it is possible that the excited state does not only couple to one certain ground state, but to a continuum of states, maybe via emission into the spectrally broad phonon sideband. The emitted photon will then be in a superposition of many possible emission channels at different energies and therefore be polychromatic (in the sense that it contains probability amplitudes corresponding to all of these different energies). |
| Dec19-12, 07:57 PM | #13 |
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Mentor
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We're getting way off topic here. So, getting back on topic,
Yes: Our rods are sensitive to individual photons. No (part 1): Nonetheless, we will see nothing if just one lone photon activates a rod and then nothing else happens for a while. There's a lot of signal processing that happens in the eye. One of those things is noise rejection. A single photon would be indistinguishable from noise. So even with the rods, we essentially can't detect individual photons. No (part 2): Our rods don't see color. They're monochromatic. Go out far from the city on a moonless night. Wait half an hour or so and you will be able see (barely) because there's still some ambient light from the stars. You won't see any colors. It's our cones that see colors; see post #2. Our cones need a significant photon flux before they become active. Look at a big sheet of white paper in sunlight and you'll have tens of millions of photons activating your cones every second. Reading in low lighting conditions results in thousands to tens of thousands of photons activating your cones every second. As far as our perception of white is concerned, all that matters is the extent to which those three different kinds of cones are activated. A white light source, one whose spectrum truly is flat, will activate those cones in a certain way. The exact same effect can result from the right combination of red, green, and blue lights. Obviously this does not have a flat spectrum. Because we only have three kinds of sensors, this anything but white light still looks "white". |
| Dec20-12, 04:09 AM | #14 |
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This is beginning to make some sense to me but there are some loose ends. For such a photon to be absorbed (detected) it would need to deliver all its energy into the detecting system. It would need to match the 'transmitter' characteristics and have some sort of 'spectrum' in its own right. What could its energy be? Would it be a version of hf; a kind of h∫(F(f)df expression, giving a 'massively energetic' photon? In your description, are you really saying that this sort of photon is in the minority (I guess you are)? These photons would not be like your regular photons which are totally anonymous - able produced by one process, possibly frequency shifted on their journey and then absorbed by a totally different process - as with light produced thermally in a star and then absorbed in a particular gas atom transition. What sort of fractional bandwidth are we talking about? Is this any different from the familiar photoelectric emission, in which a range of photo-electron energies will result from a broad range of incident frequencies? How much of this is, in fact, just a way of re-stating the description of a process in terms of what we could call photons and then having to bend the model to fit? I guess this often happens in Science though, on the way to a better model. |
| Dec20-12, 06:59 AM | #15 |
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what does it mean that a photon has a certain color.does one want to describe color as something like color of quarks?
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| Dec20-12, 07:43 AM | #16 |
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"Colour" in quarks is just a random name, chosen by quirky, hippy researchers (no offense to any who happen to read this! ) in the same way that they chose "Strangeness".
It hasn't anything to do with the colours we see. |
| Dec20-12, 02:17 PM | #17 |
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http://www.physicsforums.com/showthread.php?t=660085 |
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