# Tag Info

11

It is not a good idea to see a Feynman diagram as some sort of collision process really happening. The diagram is just a term in the perturbative expansion of a quantum mechanical transition amplitude (in other words, a nice "graphical" way to represent a bunch of integrals). The only actual observed objects are two incoming photons with a certain energy, ...

11

No - assuming they don't hit anything they don't decay. The distance dependant "decay" is the drop in the number of photons per volume as the volume gets bigger - it's not a decay of individual photons. It's the same as a crowd dispersing as it leaves a subway exit - nobody is disappearing. Photons can lose energy as they collide with gas or dust in space ...

5

A photon is a unit ("quantum") of excitation of the quantum electromagnetic field. Thinking roughly of the quantum field as a vast collection of quantum harmonic oscillators, each oscillator corresponding to a mode of vibration of the field, we specify the quantum field's state by stating how many quantums above the QHO ground state each mode oscillator is ...

4

So how is it possible to have a quantized outcome from a symmetric continuous event? Easily. So easily that I'll describe the easiest example to me. Which is to describe what happens when a Stern-Gerlach device interacts with a spin 1/2 particle. You could have a particle with any spin whatsoever, but no matter what single particle state you pick it ...

4

The number of photons may indeed be finite because the energy of the photon in ${\rm J}$ is $$E = hf$$ where $h=6.626\times 10^{-34}{\rm J}\cdot {\rm s}$ is Planck's constant and $f$ is the frequency in ${\rm Hz}$. For monochromatic light, the number of photons may be determined from the energy in this simple way because $f$ is a fixed constant: $$N_{\rm ... 4 I should like to capture the excellent comment by user CuriousOne (emphasis mine): There is no such size. A smaller aperture will merely reduce the transmission probability, but there is no known cutoff. Indeed, there is currently a lot of interest in deep sub-wavelength imaging techniques. A photon, by the way, is not an object. It's a quantum, i.e. the ... 3 The reason is that the index of refraction changes with the temperature. Due to the very hot asphalt, there is a strong temperature gradient in the air. This gradient leads to an changing index of refraction with the distance to the road. This is what you see. 3 I know that photons are quantized, they are not continuous. Photons are not quantised, nor are they continuous. They are the charge carriers of the electromagnetic field as arising in quantum field theory. An accelerated charge generates an electromagnetic field whose carriers are, in turn, the photons whose energy might be quantised. So how is it ... 3 Given you've read only QED, this is a highly astute question. Conservation laws in the quantum world work a little differently from classical conservation grounded on Noether's theorem (there is a kind of quantum analogue in the Ward-Takahashi identity). If a quantum entity has state |\psi\rangle, then conserved quantities are measurement means defined ... 2 Why does the wavelength determine a photon's energy? In the 19th century, it was thought that the energy of light was determined only by its intensity. Then, experiments, particularly the photoelectric effect, showed that this was not so: a low-intensity short-wavelength light can cause reactions that intense light of a longer wavelength cannot. Thus, ... 2 Well, actually it doesn't. Knowing the wavelength allows you to calculate the energy, but it does not "determine" it in a causal way. Energy (E), wavelength (\lambda) and frequency (\nu) are related by$$E = h\nu =\frac{hc}{\lambda}$$so if you know the wavelength or the frequency you can determine the energy. I think his use of "determine" confused ... 2 The photon is an elementary point "particle". particle between quotation marks because it is not a classical point particle , it is a quantum mechanical entity. Quantum mechanical entities depending on the boundary conditions display sometime classical point-like elementary particle behavior and sometimes have a probability density for their location ... 2 Kallen-Lhemann representation is just a way to expand in the momentum basis the two point correlation function of a local operator \hat{O}(x), it holds true for massive and massless theories alike except for non abelian gauge theory in which the situation is a bit more complicated. Let's demonstrate the K-L formula with a more general proof: let's start ... 2 The problem for photons is that you cannot observe them time to time. If you do measure their state, photons will collapse to some Eigen state and become classical. In fact, a photon can have a lot of eigenstates in theory. That is special compared to EM for photons. 1 My question is whether individual photons also carry orbital angular momentum? Yes. See https://en.m.wikipedia.org/wiki/Orbital_angular_momentum_of_light If yes, what are the values of orbital angular momentum in one-particle states? To quote the wikipedia page In particular, in a quantum theory, individual photons may have the following ... 1 Yes, single photons can have orbital angular momentum. However, unlike spin, they are not required to have any. Just like in the classical case, the orbital momentum of single photons is determined by the shape of their EM mode- roughly speaking, the wavefront must have a helical aspect to it. In particular, this means that the eigenmodes of light in a 3D ... 1 There are two types of energy involved, and the blurring of this distinction is cause of a huge number of misunderstandings. Light comes in discrete packets called photons. The energy of each photon is proportional to the frequency of the light. On top of that, a light beam can have any number of photons in it, and this gives it its overall power. The power ... 1 As you have correctly stated, the second order correlation function measures the coincidence of two events. The case of g^{(2)}=0.4 means any two events are less likely to happen coincidently than the case of g^{(2)}=0.1, although both of the two cases tending to have antibunching events. 1 The connection of frequency to energy comes when one considers the covariant formulation of the electromagnetic wave propagation. In Panofski and Philips "classical electricity and magnetism" second edition chapter 21. This quote in particular. This associates a zero mass particle with a fourvector, i.e. energy and momentum . Text goes on to explain ... 1 No, because we can think of the classical wave as being made up of a large number of photons. If we have a low-frequency wave with the same energy as a high-frequency wave, it simply means that there are a larger number of low-energy photons. 1 You are looking at two different things assuming they should be the same. In classical electrodynamics the Poynting vector is defined as \mathbf{S} = \mathbf{E}\times \mathbf{H} and enters the variation of energy density as$$ \frac{\partial u}{\partial t} = -\textrm{div}\,\mathbf{S} - \mathbf{j}\cdot\mathbf{E};  according to the functional form of ...

1

A photon is a quantum mechanical elementary particle and follows quantum mechanical formulae, not classical ones. In quantum mechanics the only way an elementary particle can change direction is through an interaction with another elementary particle or field. The interaction is shown with feynman diagrams which give the integrals that have to be calculated ...

1

How can photons/particles/objects/things be massless? Photons aren't massless the way people think. A photon has a non-zero "inertial mass" and a non-zero "active gravitational mass". But it doesn't have a "rest mass" because it's never at rest. You can't slow down a photon like you can slow down an electron. Or speed it up by pushing it. Rest mass does ...

1

The edge provides a boundary condition that the EM field must satisfy. The total EM field is "aware" of the boundary. "Photons", being quantized excitations of the EM field, are created (emitted) and destroyed (hit a screen) only where the EM field exists. If you are trying to think of photons as particles, forget it. You'll end up with all sorts of ...

1

One thing that you can be sure of is, for a large enough LED, you will get poisson statistics to a very good approximation. Neither bunching nor anti-bunching. The reason is simple: One photon comes from a certain part of the LED, the next photon is likely to come from a totally different part of the LED and head in a totally different direction. There's no ...

1

You've mixed up which time dilates for which observer, and written yourself into a paradoxical corner: if going closer to the speed of light slows time for the object going that speed, and if time slowing down means going slower, then the conclusion is that "the faster you go, the slower you go." Which obviously doesn't make sense as you've pointed out. ...

1

Is there any size of photon if so what is it? The photon is an elementary particle among the others which form a basis for the standard model of particle physics. The Standard Model of elementary particles (more schematic depiction), with the three generations of matter, gauge bosons in the fourth column, and the Higgs boson in the fifth. The ...

1

The transformation equations you specify are not correct since they do not respect unitarity. The condition of unitarity (or energy conservation) for the action of the beam-splitter gives the following transformations: $\hat{c}=\sqrt{\tau}\hat{a}+\sqrt{1-\tau}\hat{b}$ $\hat{d}=\sqrt{1-\tau}\hat{a}-\sqrt{\tau}\hat{b}$ The minus sign in the second equation ...

1

Here's a simple explanation of where it can happen: Photon Sphere and a more complex one: Wikipedia. A photon-sphere can exist around a Neutron Star too. But a pure circular path is impossible, just as a pure circular orbit is impossible. There's always going to be some eccentricity to any orbit and in the case of light, it will either quickly ...

1

Light is made up of photons which are elementary particles. The standard model of particle physics fits the data very well with the hypothesis that all elementary particles are point particles. In an experiment where the size of a slit is varied and the cross-section for a photon to pass through the slit is measured two consideration should enter when ...

Only top voted, non community-wiki answers of a minimum length are eligible