This makes intuitive sense to me but I have never heard anyone confirm it. Is it one photon per wave length or does it approximate one wavelength or something else?
19930
At the end of this nice video (), she says that  is a chain reaction of electric and magnetic fields creating each other so the chain of wave moves forward.  I wonder where the  is in this explanation. What is the relation between electromagnetic wave and photon?
120728
I was wondering if a photon is divisible.   If you look at a photon as a particle, then you may be able to split it (in theory).  Is it possible and how do you split it?
54716
Is a photon of 400nm in wavelength "smaller" than a photon of 1km in wavelegth?
94947
A photon is emitted from a source and reflected off an object (or objects) until it hits the human eye. The color of the object we see depends on the photon wavelength. If photon travels with constant speed how does the human eye know how far the object from which the photon was reflected?
32911
Reading about photons I hear different explanations like "elementary particle", "probability cloud", "energy quanta" and so forth. Since probably no one has ever seen a photon (if "seen" it supposedly - and rather conveniently - ceases to exist) but many experiments seem to verify its properties (or are they maybe adjusted to fit the experiment). I thus can't help wondering if the "photon" is then just a physical/mathematical tool with inexplicable properties (like zero mass - but affected by gravity fields - and constant speed c in space) invented to explain some otherwise unexplainable phenomem and to supplement the elementary particles and their interactions. In short: Are they real or imaginary!  Does anybody know? Or maybe the answer is "blowing in the wind" because to most physicist it probably just does'nt matter as long as it works (as the alternative healers say). Sorry if I seem a little sarcastic here and there.
40920
The HUDF used to be the deepest image of the universe ever taken by the Hubble telescope, the furthest star in this image is 59000 light years away.  The star in question:   Now Imagine a light emitting object like a star and an observer like a telescope, 59000 light years apart. I represented everything in 2D, but in 3D I suppose it would be the same but with spheres:    Now picture the photons that the object is emitting, in particle form. They will all have one direction and they will go towards this direction at C, given that there are no massive objects in between their paths:    Notice how the photons have separate paths, because they have different directions, now obviously there should be a lot more photons but for simplicity I drew it like that. My point is, given enough distance shouldn't the probability of hitting a photon be so small that seeing an object becomes nearly impossible? Like so:    Here the observer can detect the photons, but if it was slightly off it wouldn't:    So how can we detect objects so far away? Is there something about light's wave form that I'm not considering, is this how light is emitted at all?  Update:  Looking at other duplicates questions I've seen two answers:  1- There are just too many photons being emitted that the "gaps" are just too small to make any difference.  2- This isn't really how photons are emitted, instead a quantum field is spread out radially and we only detect the photon when we interact with the field.  Does this mean that there are objects that we can't see, even if given enough time for the light to reach us (presumably outside of the current observable universe)?  Or do the gaps not exist at all because of answer number 2?
43352
Since electromagnetic energy is carried by photons and moves in forms of waves, does it mean that a single photon when propagating through space doesn't follow the straight path but instead always moves up and down, up and down like a wave. If so another question arises the speed of propagation of light in vacuum is fixed meaning that it will always take the same amount of time for it to travel from point A to point B, but if a photon always moves up and down it will also mean that it travels longer distance than the distance between A and B and so it ill travel faster than light propagates, is it even possible, could you please clarify these concepts for me?
16766
How is the following classical optics phenomenon explained in quantum electrodynamics?   Reflection and Refraction   Are they simply due to photons being absorbed and re-emitted? How do we get to Snell's law, for example, in that case?  Split by request: See the other part of this question .
9068
I know that if we solve the Maxwell equation, we will end up with the phase velocity of light being related to the permeability and the permittivity of the material. But this is not what I'm interested in - I want to go deeper than that. We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with $c$ in the spacing. The rare atoms will disturb the light in some way. So I am interested in how the atoms affect the light. Photon absorption-emission theory Some textbooks that I read explain it in a way kind of like this: In a material the photons are absorbed by atom and then re-emitted a short time later, then they travel a short distance to the next atom and get absorbed&amp;emitted again and so on. How quickly the atoms in a material can absorb and re-emit the photon and how dense the atoms decides the apparent speed of light in that material. So the light appears slower because it has a smaller “drift speed”. Interference theory But recently I realize an alternative explanation: Atoms respond to the light by radiating electromagnetic wave. This “new light” interferes with the “old light” in some way that results in delayed light (advanced in phase), this can easily be shown by using simple phasor diagram. Consequently effectively the light covers a smaller phase each second, which gives the impression of a lower phase velocity. However the group velocity is changing in a complicated way. I think that the first explanation does not explain the change in phase velocity of light. if we consider light travelling into a slab of negative refractive index non-dispersive material, let’s say the light is directed perpendicular to the slab. The phase velocity’s direction will be flipped, but group velocity’s direction in the material will not change. Only the second explanation can explain the flipped phase velocity direction. I guess that the velocity that we get in the first explanation is actually belongs to the group velocity. It makes sense to me that the front most of the photon stream determines the first information that the light delivers. So the question is What really cause the phase velocity of light to be decreased?  &quot;drift velocity&quot; of photons (they aren't the same photons, they are re-emitted all the time) phase difference between absorbed and emitted light something else  And also, I still don't really understand detailed explanation of the absorption-emission process for small light's wavelength (for large lambda compare to the atoms spacing, the photons will be absorbed by the phonons). The dispersion relation that we know is continuous and also some material is non-dispersive, therefore the absorption process must occur in all frequency for a certain range. So definitely it doesn't involve the atomic transition, otherwise it will be quantized. My guess is that the relevant absorption process gets smooth out by the dipole moment. What makes the spectrum continuous? EDIT: link for dispersion relation:
65189
discusses the recent invention of a "quantum radar" that works by bouncing one photon off of something, storing a photon that's entangled with it, and watching for interference on the stored photon that indicates the radar photon has bounced off an object.  Most of that makes sense, (about as much as anything in quantum physics makes sense, at least,) but... how exactly do you "store" a photon?  To my mind at least, the concept of storage involves putting something in a location and keeping it there, unmoving, but photons are not capable of not moving at the speed of light.  So what's going on in this scenario?
32375
I haven't yet gotten a good answer to this: If you have two rays of light of the same wavelength and polarization (just to make it simple for now, but it easily generalizes to any range and all polarizations) meet at a point such that they're 180 degrees out of phase (due to path length difference, or whatever), we all know they interfere destructively, and a detector at exactly that point wouldn't read anything.  So my question is, since such an insanely huge number of photons are coming out of the sun constantly, why isn't any photon hitting a detector matched up with another photon that happens to be exactly out of phase with it? If you have an enormous number of randomly produced photons traveling random distances (with respect to their wavelength, anyway), that seems like it would happen, similar to the way that the sum of a huge number of randomly selected 1's and -1's would never stray far from 0. Mathematically, it would be:  $$\int_0 ^{2\pi} e^{i \phi} d\phi = 0$$  Of course, the same would happen for a given polarization, and any given wavelength.  I'm pretty sure I see the sun though, so I suspect something with my assumption that there are effectively an infinite number of photons hitting a given spot is flawed... are they locally in phase or something?
61910
I understand the equation explaining angles of refraction related to wavelengths. But I'm looking for a more intuitive approach/visual explanation? Perhaps related to QM?
73896
If you emit a single photon, wait an interval, emit a second photon, wait the same interval, and keep repeating, is the resultant stream of photons describable by the frequency at which you emitted each one? Or does each photon have its own "frequency", independent of the rate of emission? I thought the former description explained the Doppler effect.
31799
This was discussed in an answer to a  but I think that it deserves a separate and, hopefully, more clear answer.  Consider a single photon ($\lambda$=532 nm) traveling through a plate of perfect glass with a refractive index $n=1.5$. We know that it does not change its direction or other characteristics in any particular way and propagating 1 cm through such glass is equivalent to 1.5 cm of vacuum. Apparently, the photon interacts with glass, but what is the physical nature of this interaction?  Let's not consider high-order effects like Rayleigh/Raman scattering.
73509
I've tried to search for this answer but found no proper explanation.  My previous ideas were...  1) It is not a point source, so the different points on a filament lamp may cancel out the light from other points on a filament lamp  2) It emits light rays at random (even though I've read this, I don't exactly understand what it means. Does it mean a single point on the filament may emit a photon one second and then the next second not emit anything?)  Are these reasons along the right lines? Could someone give a proper explanation as to why a filament lamp is an incoherent light source?
45792
I read that it makes a difference whether you calculate $\frac{dE(\lambda) }{d \lambda}=0$ or $\frac{dE(\omega)}{d \omega}=0$ in the sense that the maximum energy density with respect to the wavelength does not coincide with the frequency maximum that one would assume to be at $\omega_{max} = \frac{c}{\lambda_{max}}$. Actually, I do not understand why this is so ( Now, I am only interested in a pure physical explanation, mathematically this is clear and due to the chain rule). Does anybody know how to explain this odd thing?  Somehow I feel that the core of the question has not been completely answered. Although I can easily look up wikipedia, where the need for the chain rule is explained, I am rather interested in understanding where the following argument breaks down, which seems to be why this question has caused some confusion in the past(as you can see by googling this question):  So we have $E(\omega)$ the energy radiated at a given frequency. Now this function has a maximum somewhere, so there is a frequency where a maximal amount of energy is emitted. In other words: If you add the sum of the photon's energies at each frequency that are emitted, you will notice that the maximum is reached at this frquency. Now $E(\lambda)$ tells you basically the same for the wavelength, but again: We know where at which frequency the maximal amount of energy is radiated, so we know the corresponding wavelength.   I think there is a need to explain this.   If anything is unclear, please let me know.
70045
If I am not wrong when light hits for example white wall most of the photons are absorbed and transformed into heat and few of the photons at certain wavelength are reflected from the object. So white wall reflects photons at wavelength corresponding to white color so we can see it.  1.Can someone clarify if it is correct?  2.What happens if that photons reflected from white wall hit another white wall (are they reflected once? If that so can they be reflected forever?)  3.What happens when I "shoot" photons of certain wavelength into object which color corresponds to that photons wavelength e.g. Blue light into blue wall