B Why do we know all photons have same speed as the speed of light?

[fxdung]

With Fock state, how can we define probability the divice find out photon at a position?

 

[vanhees71]

That's not so easy to describe. One way to detect photons is to use the photoelectric effect. In this case you can use the usual dipole approximation for the interaction of a photon with an atom (within the detector), assuming that the wavelength of the photon is much larger than the typical size of the atom. E.g., to typical atomic size is of the order of an Angstrom, i.e., $10^{-10} \; \text{m}$. If you have light or photons in the visible range of the spectrum the wavelength is around $500 \; \text{nm}=5 \cdot 10^{-7} \; \text{m}$, such that for this application you can use the dipole approximation.

The mathematical analysis then gives for the detection probability that it is proportional to the energy density of the em. field at the position of the detector, i.e., it's proportional to the expectation value $\langle \vec{E}^2(t,\vec{x}) \rangle$.

It's very often a good approximation to think of the photons simply as an electromagnetic wave and the probability to detect a photon as given by the corresponding intensity of the electromagnetic wave.

In this rough picture the difference between a single-photon state and a classical em. wave (which in terms of quantum electrodynamics is represented by a socalled coherent state of high intensity) is that you don't get a continuous distribution of light but just one single point on a detector screen (a photoplate or in our times a CCD cam). You never know, where an individual photon will be detected but only the probability to detect it at a given position on the screen. A single photon is either completely absorbed and detected or it's not absorbed and stays undetected. There's no way that a single photon can be detected at two or more spots at once, and that's all that's left from what was considered "particle like" in the (short) era of "old quantum mechanics". Only if you repeat an experiment (like shooting single photons through a double slit) you build up a distribution of photon-detection points which then follows the probability distribution predicted by QED. What's "wavelike" in this pattern is the fact that you get an interference pattern as with classical electromagnetic waves when looking at this pattern build up from many photons.

What was hand-wavingly and pretty desperately known as "wave-particle dualism", and Einstein himself, who promoted the photon picture as "a heuristic point of view" only, was very puzzled by it, is today resolved by modern quantum mechanics and the assumption that all that can be known about some object (here a photon or the electromagnetic field) is in what quantum state they are "prepared", and this implies only statistical knowledge about the outcome of measurements (here the position of the detection of em. fields/photons on a screen).

 

[snorkack]

Can you name just one that measures c in a vacuum and gets a spread of values?

No, and that´ s my point. We know what it looks like when photons have different speeds, and it always appears related to the effects of medium on photon, not intrinsic properties of photon. Whereas rest mass of photon would show up as low frequency radio waves having index of refraction that diverges at low frequencies independent of medium.

 

[lomidrevo]

We know what it looks like when photons have different speeds

Do we know? Are you sure? Because I have no idea.

Whereas rest mass of photon would show up as low frequency radio waves having index of refraction that diverges at low frequencies independent of medium.

That doesn't make any sense to me. How would that suggest a rest mass of photons?

 

[vanhees71]

It's correct that in-medium photons have a different spectral function than free photons in the vacuum, but that doesn't make the massive.

The only place where photons get really massive is within superconductors, where the electromagnetic gauge symmetry gets "Higgsed" through the formation of a "Cooper-pair condensate".

You can see this from the point of view of the BCS model (Bardeen, Cooper, Shriever): Start from a model of a superconductor first neglecting the electromagnetic field. Formally you have a gas of electrons with an effective attractive interaction (through the electron-phonon interaction). Whenever you have fermions with an effective attractive interaction there's a phase transition leading to spontaneous symmetry breaking and the formation of Cooper pairs and a non-vanishing ground-state expectation value for the corresponding (quasibosonic) field, leading to condensation.

Now you gauge this model in the usual way to get the electromagnetic interaction in. The so far global spontaneously broken symmetry is now a local gauge symmetry, which cannot be spontaneously broken. What happens instead is that the gauge field turns out to be massive by "eating up" what was the Nambu-Goldstone mode of the spontaneously broken global symmetry, which means that the photon gets massive and the Nambu-Goldstone mode is gone.

 

[snorkack]

Do we know? Are you sure? Because I have no idea.

Diffraction of light is due to photons of different frequency having different speed. But it always looks due to interaction with medium.

That doesn't make any sense to me. How would that suggest a rest mass of photons?

Because of the relation
E²=p²c²+m²c⁴
Assuming zero m, it simplifies as
E=pc
It approaches E=pc when E>>mc²
But it would change and the speed of particle would fall as E approaches mc². And the speed would have no lower bound.

Compare photon with neutrino. Since neutrinos are hard to observe and low energy neutrinos especially so, it is hard to measure neutrino speed. Indeed, even though some neutrinos have rest mass as shown by oscillations, we do not have evidence that they all do. But in contrast to low energy neutrinos, low energy photons are easy to observe - longwave radio waves! And they do not show evidence of medium independent, vacuum dispersion which photon rest mass would cause.

 

[lomidrevo]

Diffraction of light is due to photons of different frequency having different speed.

No, it isn't. @Lord Jestocost alredy mentioned in above post that "different speed" is only apparent. I also suggest to read the Feynman's lecture linked by him, it explains why.
Btw. to describe refraction or diffraction of light you can safely stay within domain of classical EM field theory, no need to complicate life by jumping to QED and talking about photons.

Because of the relation
E2=p2c2+m2c4
Assuming zero m, it simplifies as
E=pc
It approaches E=pc when E>>mc2
But it would change and the speed of particle would fall as E approaches mc2. And the speed would have no lower bound.

I am not sure what you are trying to do here. I hope you know that "speed of light is the same constant for all observers" is a postulate of special relativity. You just cannot use the results of SR (like $E^2 = p^2c^2 + m^2c^4$ ) to try to "simulate" how it would be if constancy of speed of light is violated. The theory is build on this fact. The equations would not be valid if $c$ could vary. According to SR, for light you get exactly $E = pc$, point.

 

[snorkack]

I hope you know that "speed of light is the same constant for all observers" is a postulate of special relativity.

You could just as well call it a "postulate of special relativity" that speed of neutrinoes is the same constant for all observers. Except, oops, neutrinoes oscillate... which doesn´ t actually affect special relativity. Special relativity merely requires a special maximum speed to exist, and does not require any particles to actually travel at that speed - whether it is all neutrinoes, one of three neutrinoes, no neutrinoes but all photons or not photons either (but only gravitons) - SR is unaffected.
However, if any photons inherently, in vacuum, traveled slower than light like most neutrinoes do, the problem is that it would mess up electromagnetic fields, especially at the low frequency end. Therefore we can verify that light travels exactly at speed of light, not only approximately at the speed of light like most neutrinoes.

 

[vanhees71]

Neutrinos are not massless. They are quite complicated critters too ;-)).

 

[Dale]

At this point the question in the OP has been answered. So we will close this thread rather than go off on more tangents. @fxdung in the future please try to not hijack your own thread. Keep the discussion to the question in the OP and try to actually respond to the other participants instead of hammering us with a barrage of new questions