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7

I would tell them to re-read and understand that paper, and know that few spectroscopists would disagree with it. The point is that far too many people use the word "photon" without knowing what a photon really is or under what context the word can be used. For the vast majority of applications a semi-classial conception of the radiation field is adequate. ...


4

The issue here, I believe, is not existence of photons, but the fact that people may choose terminology and concepts they find appealing. The word photon has been coined long ago for an idea that is quite far from the current views on light and the meaning of the word has been evolving many decades. Its current use in textbooks and papers is quite broad ...


4

The work function is the difference between the energy of the most excited electrons in the conduction band and the vacuum. It's the energy required to take an electron from the conduction band and remove it to outside the metal. To a first approximation the work function is independant of the intensity of the light, because it's an intrinsic property of ...


4

The spin of a photon can be $+1$ or $-1$ but not zero. The spin is directly related to the circular polarisation of the light beam. I think spin $+1$ is clockwise polarisation and $-1$ is anticlockwise, though my success rate at remembering which way round it is rarely exceeds 50%. Linear polarisation is a sum of left and right circular polarisation i.e. it ...


3

The photon as an elementary particle is special: the quantum mechanical wave equation whose solutions squared will describe the probability of finding the photon at a given phase space point is the Maxwell equation that describes the classical electromagnetic wave, except it is the potential form of it and it acts as an operator on the wave function of ...


3

No, it doesn't mean that the photons don't go any further. It means that when the temperature gradient inside the star reaches a threshold, the gas becomes convectively unstable. Heat is transferred more efficiently by moving parcels of gas than by transferring photons from hotter regions to cooler regions. So, the photons continue to diffuse outwards, but ...


3

Wire grid polarisers allow radiation to pass that has it's electric field polarised perpendicularly to the direction of the wires. The explanation is that the component of the light polarised parallel to the wires sees the grid as if it were a solid conductor and therefore most of it is reflected and the rest absorbed in the first couple of skin depths. In ...


2

This link says why photon electric field interactions are "trivial" : In a vacuum, the classical Maxwell's equations are perfectly linear differential equations. This implies – by the superposition principle – that the sum of any two solutions to Maxwell's equations is yet another solution to Maxwell's equations. For example, two beams of light pointed ...


2

The interaction between two charged partices occurs through a change of momentum. All very well, but the next question is how we calculate the momentum exchange, and this is where quantum field theory comes in. The interaction between two electrons is described as a disturbance in the quantum fields involved. Quantum field theory gives us an expression for ...


1

1) There exists the classical electromagnetic wave as described by Maxwell's equations. 2) The photoelectric effect showed that these electromagnetic waves are composed of photons, with energy E=h*nu , where nu is the frequency of the classical wave. Single photon experiments have been performed by limiting the intensity of the beam to one photon at the ...


1

Of course linear polarization is an observable, we measure it. Linear polarization along an axis x, and perpendicular to x, are the eigenstates of the observable "polarization along x". In fact, by measuring polarization along x, we measure the projection of the electric field along the axis x.


1

The polarization states can be effectively defined in terms of the Stokes operators $\hat{S}_i$ for $i=[0,3]$ $\hat{S}_0 = \hat{a}^{\dagger}_{x}\hat{a}_{x}+\hat{a}^{\dagger}_{y}\hat{a}_{y}$ $\hat{S}_1 = \hat{a}^{\dagger}_{x}\hat{a}_{x}-\hat{a}^{\dagger}_{y}\hat{a}_{y}$ $\hat{S}_2 = \hat{a}^{\dagger}_{x}\hat{a}_{y}+\hat{a}^{\dagger}_{y}\hat{a}_{x}$ ...


1

Don't think about individual single photons bouncing around. The energy (heat) diffuses slowly, exchanging energy with neighboring particles by various means (smashing into each other and radiation that only makes it as far as the next particle before being absorbed). Different photons are emitted and absorbed over and over, as well as physical collisions ...


1

Red shifts happen because of various causes. there also exist blue shifts: Conversely, a decrease in wavelength is called blueshift and is generally seen when a light-emitting object moves toward an observer or when electromagnetic radiation moves into a gravitational field. Now on redshifts: Some redshifts are an example of the Doppler effect, ...


1

The photon is not deflected by the charge. It's deflected by the spacetime curvature. The spacetime curvature is related to the stress-energy tensor (by Einstein's equation), so if you change the stress-energy tensor you change the curvature. A static black hole, described by a Schwarzschild metric, has a very simple stress-energy tensor since it's ...


1

The formal analogy between a mode of the radiation field and a particle in a harmonic potential stems from the fact that both systems have the Hamiltonian (in appropriate units) $$ H = \frac{1}{2}P^2 + \frac{1}{2}\omega^2 X^2,$$ where the variables $X$ and $P$ obey canonical commutation relations $[X,P] = \mathrm{i}\hbar$. For the radiation field, these ...


1

As amplification: The emission of photo-electrons is a quantum effect: one electron receives the energy from a single photon. If the energy of that photon, proportional to its frequency, is sufficient to overcome the work function, then photo-electrons will be produced. If the frequency is too low, then no photo-electrons will be produced. Increasing ...


1

Spontaneous Parametric Down Conversion is a process that produces pairs of photons. The process goes like this: a strong beam of ultra-violet (UV) photons, is sent upon a down-conversion crystal (placed inside your black-box). Inside the crystal, the UV photon is SPLIT into two photons, named (I don't know why) SIGNAL and IDLER. In general the two photons ...


1

Here is a suggestion of how to produce the state in your second formula. Choose to make holes in a screen, NOT at the two points A and B of which I spoke in my first answer, but at two other points, e.g. on top of the upper cone, and on the bottom of the lower cone. Then, pass the upper beam through a beam splitter $BS_1$. Block the transmitted wave, let ...


1

Given two like charges -- two electrons for example -- does moving them farther apart release a photon? Photons are released when charged particles are accelerated. . Moving them apart may generate photons if there is acceleration. If the velocity is constant, no radiation. Electrons in conduction bands of metals are in a quantum mechanical state and ...


1

No. Light is a disturbance of electromagnetic field and from particle aspect, photons aren't charged particles. Assuming you're asking about light in vacuum (that's when you can see sole effects caused by electromagnetic field), the external electromagnetic field can't affect light because it's a linear field (you can Google to know what linearity means). ...


1

Photon counting statistics cannot always be explained by classical fields. In these experiments, the state of the field is monitored continuously by a photodetector. I believe these represent one of the clearest experimental demonstrations of the quantum nature of the radiation field. For example, in observing the emission of photons from a single atom, ...


1

Now per QED, electrical charges interactions are effected by photons. Suppose you are one of the two charges. How do you know to attract or repel the other charge? You want something that does not exist - intuitive picture of physical process within a theory which is a demonstration of how far can one go with mathematisation of experience and ignoring ...


1

Each photon has an oscillating electric and a magnetic component. In vacuum the components are always perpendicular to each other and also perpendicular to the motion of the photon (see this sketch). When photons are unpolarized, these oscillations are equally distributed in 360 °. A polarization grid can rotate about 50% of the photons so that they can pass ...


1

Strictly speaking a photon cannot be localized and the single particle "wavefunction" (as well as it's corresponding position operator $\hat{r}$) only exists in an approximate sense. The reason for this is quantum electrodynamics (QED), which is the theory that contains photons, is a quantum field theory (QFT) rather than the (non-relativistic) quantum ...



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