New answers tagged

2

Engtangled photons: first you must understand that the photon is the particle obtained when the modes of the electromagnetic field are quantized, and that they are created and destroyed as discrete quanta of energy, in agreement with Planck's relation, $E=hf$, where $f$ is the frequency of the electromagnetic field corresponding to the quantized mode; that ...


0

"What entangled photons really are?" Photons are measurements of the state (changes) of a quantum field. Let's take your knowledge about the hydrogen atom as a starting point. Let's say your atom is in a p-state. The atom then changes into an s-state. Its angular momentum and energy change. Angular momentum and energy conservation demand that both ...


1

Here is probably the simplest argument that I have heard of. Detection of a photon from a thermal source gives a rise to a probability to detect several more in a short interval of time due to stimulated emission. Assume that you have some atoms in a medium that emits light, and they are in an excited state. If you know that one atom emitted a photon, this ...


2

Okay, so just to be clear I am going to consider processes in which a photon and an atom at some energy level go in, and the photon and atom exchange energy (and momentum) such that a photon with a shifted (either higher or lower) energy comes out, while the atom ends up in a different internal electronic state than it started in. A general diagram looks ...


2

Photon experience infinite time dilation and hence, time is stationary for it. Does photon experience time  photon travels at c through the three spatial dimensions. All of its velocity is directed through the three spatial dimensions. Thus Brian and Einstein are stating that a photon must be stationary in the fourth dimension. For if the photon had any ...


1

This answer is not rigorous, just an afterthought: Remember that a photon has to be fully absorbed first, it will not absorb and emit simultaneously. This would leave us with an electron that is momentarily in a forbidden energy state. From that state it could emit a photon to jump to what should be its correct (i.e., allowed) energy level. Uncertainty will ...


-1

I believe if an atom receives too much energy it can be ionized. There are allowed energy levels but above these levels is the region for free electrons. The energy levels there are not quantized and can receive any energy. Photons with higher energy can put the total energy above the allowed energy levels therefore ionizing the atom. Another way to look at ...


0

From page 319, the $n$th coefficient of the expanded Taylor Series has a divergence degree $D = D_0 - n$, where $D_0$ is the degree of divergence of the amplitude considered (photon-photon, in this case). So we have for the first non-vanishing term $n = 4$ and so $D = 0 - 4$ .


1

Wavelength is used as a convenience. It's much easier to imagine a photon with a 500 nm wavelength than to comprehend a photon oscillating 600 trillion times per second. But in reality that's all it is is a photon moving at the speed of light and oscillating 600 trillion times per second as it goes along. The photon completes one cycle every 500 nm. Many on ...


3

Just as a supplement to ACuriousMind's answer, it is worth noting that buried in the bottom of their paper they actually show what the "spin 1/2" eigenstates are in terms of the regular basis: $|j=1/2\rangle=\frac{1}{\sqrt{2}}(|1, -1 \rangle + |0,1\rangle$) $|j=-1/2\rangle=\frac{1}{\sqrt{2}}(|-1, 1 \rangle + |0,-1\rangle$) where $|l, \sigma\rangle$ is the ...


6

It probably does not mean anything. That paper concerns the quantization of electromagnetic waves in less than three spatial dimensions. In fact, there are a number of decades-old results showing that it is often possible to evade the spin-statistics relationship in lower-dimensional systems. While these kinds of results (including this new one) may be ...


13

Nothing is happening. At least, nothing except that a new generalized quantity suggestively called "angular momentum" was defined and subsequently measured. But nothing we know about the usual angular momentum of photons is changed by this in any way. Standard total angular momentum is $J = L + S$, where $L$ is the orbital and $S$ the spin angular momentum. ...


0

It is clearly acausal that for a photon to be emitted it must be absorbed. That's action at a distance. That is not what time symmetry means. That view is known to not be correct, and there are no problems with advanced and retarded potentials, it is known what they mean and how to deal with them, and it is not this. Wheeler and Feynmans paper is clearly ...


3

This sounds like the "retrocausation" in the Wheeler–Feynman absorber theory. Since the only invariant quantity in relativity is the relativistic interval, which is zero along light like curves, all "place-instants" of photon's existence are technically not separated from each other in the (pseudo) metric, and hence causal, sense. This means that photon ...


0

Firstly, as @AnubhavGoel has mentioned, I'm assuming you have mixed up the terms "elastic" and "inelastic" here as what you have said is nonphysical. Now by definition in an elastic collision total kinetic energy is conserved. In a photon, $E_k = pc$, so the total momentum of the photon is also conserved. Secondly, I'm assuming the symbol $I$ you have used ...


0

I will use most of @Ted answer to describe 'hot' but I will ask a more basic question: I think the best way to think about it is that the sentence "the photons have cooled" is simply describing a fact, not explaining that fact. At early times, the photons at any given location had a thermal (blackbody) distribution corresponding to a high ...


0

Photons are created and destroyed (they're usually thought of as packs of energy which are emitted and absorbed by other particles, as Anders pointed out) independently of what happens to them between those events. As they move very fast, any proper time interval for them is close to null. Notice that this doesn't mean they're immortal, this means that any ...


-3

Well photons is basically made of electromagnetic radiation basicly it is pure energy so it is not made of quarks that bounce around inside a atom producing gravity and time dilation so that's why it still experiences no time dilation also this particle is a point particle it doesn't feel the effect of the 3rd dimension that we live in it is a 0 dimensional ...


2

"Photons experience no time" is a conclusion drawn from taking the limit of time dilation as $v$ approaches $c$. This transformation never gets you to light speed, i.e. never represents a massless particle, only gets asymptotically closer. Taking that same limit, length contraction causes distances to reduce to nil. A particle moving at the speed of light ...


-4

you can't physically destroy a photon but when a photon is moving through free space it looses energy over time atoms also can assorb a photon and take some energy away but as a photon looses its energy it would start breaking apart until there is nothing left


1

Borrowing the concept of a "black box" from engineering, we have a photon "going in" into the box and a photon coming out of the box. We take a "picture" of the photon going in (its amplitude and wavelength), and we take a "picture" of the photon coming out (its amplitude and wavelength), and we compare the "pictures." If the "pictures" are equal, then we ...


4

You misunderstand special relativity. For objects that are moving at large speeds, the time runs more slowly for the object compared to the observer who measured the speed. To observe the motion of the object, you don't have to go to its coordinate frame and observe from there. You observe from the outside, that's how you measured the speed in the first ...


2

Let me explain light. Classically, light is electromagnetic radiation. There exists a field permeating all of spacetime called the electromagnetic field. Charges create curvature in this field. When charges accelerate, waves are created in this field. These waves are what we perceive as light. A little more specifically, let us examine Maxwell's equations ...


3

Yes, both the internal potential energy and the internal kinetic energy of a bound system (in the rest frame of its center of mass) contribute to the bound system's inertial mass according to $E=mc^2$. For a paper discussing the evidence that this is true for internal kinetic energy in particular, see Kinetic Energy and the Equivalence Principle.


2

Theoretically, the answer is yes. However, looking at the practicality of the situation, the answer is no. The photons can not be contained inside the box unless they are either 1. Created inside the box itself, or 2. They are trapped beforehand, and then brought inside the box. In 1., they do not add to inertia because, they are created using energy ...


6

Yes, mass and energy are equivalent. A more competent relativist might be able to give you the complete description, but to first order you can say that the mass of an object is simply the total energy in its volume divided by c^2. That mass is equivalent to the inertial mass by the weak equivalence principle, which is a cornerstone of GR. That is to say, ...


23

Yes! In fact, this kind of phenomenon is very common. For example, the mass of a proton is much greater than the sum of the masses of the constituent quarks. Much of the extra mass comes from the gluons that bind the quarks together; like photons, gluons are massless, but they contribute to the inertia.


2

So the entire electromagnetic force can be described as having these objects which interact by exchanging virtual photons. Photons -- light -- in some sense are the electromagnetic force. It is therefore unsurprising that light can be absorbed by a particle -- an electron, say -- as a sudden "push" which launches the electron in some new direction with ...


-1

Quantum mechanics is a theory that can only predict probability distributions. It cannot predict trajectories. It is ruled by differential equations which have as solutions the wavefunctions, and the complex conjugate square of the wavefunction gives the probability of a specific, photon, electron, to be at (x,y,z,t) given the boundary conditions of the ...


1

According to quantum physics, when certain different polarizers are placed over the slits in the double-slit experiment (for instance, one vertical and one horizontal polarizer, or one circular clockwise and one circular counter-clockwise), thus "marking" each photon with which-way information, the photon indeed passes through only one slit, resulting in no ...


0

The limit c comes from Maxwell's equations for electromagnetic waves in vacuum. This does not know about photons, is pure mathematics given E and B fields and permittivity and permeability of free space. $$c= \frac1{\sqrt{\mu_0\varepsilon_0}}= 2.99792458\times 10^8~\mathrm{ms^{-1}}\;.$$ If photons do have a mass within the experimental limit, Quantum ...


2

You should remember one thing : electromagnetic field is just a spatial representation of how electric charges interact with each other, and by "interact" I actually mean "exchange some energy". Electrostatic and magnetostatic energies Lets imagine that we want to build "from scratch" a given charge distribution $\rho(\textbf{x})$. That means that we ...


2

Yes, and it is unavoidable. Let's consider an intrinsic semiconductor for simplicity. If the semiconductor is at the absolute zero of temperature, then all electrons will be in the valence band. At any non-zero temperature there is a chance that some electrons will have been promoted by thermal agitation into the conduction band. These electrons will ...


19

It is a matter of definition of "same". Classically one can define "same" condition of particles by labels stuck on them. Light classically is a wave, and same needs a new definition. We apply the everyday definition by identifying the light beam with the source. The light leaving the sun is the same light arriving on earth. The light reflected from the ...


11

Photons are boson, so it follows the Bose-Einstein statistics which is only true if the particles are truly indistinguishable. If you can distinguish between two photons, then it will follow the classical Boltzmann statistics which is not what happen in experiments. That means photons with same properties are the same. Even in your situation with photon ...


12

Your question is based on the assumption that a photon is a fundamental object i.e. that photons are something we can point to and say here is photon 1, here is photon 2, and so on. The trouble is that quantum field theory particles are somewhat elusive objects. This is particularly so for particles like photons that are their own antiparticles because such ...


2

What are photons? Photons get emitted every time when a body has a temperature higher 0 Kelvin (the absolute zero temperature). All bodies, surrounding us (except black holes) at any time radiate. They emit radiation into the surrounding as well as the receive radiation from the surrounding. Max Planck was the physicist who found out that this radiation has ...


-3

EM Waves are basically spinnings of photons. To create a EM Wave, you basically move some electrons in a directional way(think of an antenna). Electron is a charged particle as well as protons. Charged particles emit photons, and if you emit photons in an ordered way such as this: You are seeing a dipol antenna. When you apply negative voltage(intense ...


-2

The energy is carried in individual photons. A photon with twice the frequency has twice the energy. x-rays are made of photons with higher frequencies. The energy is transferred as kinetic energy as with the photoelectric effect. The energy of a photon is calculated as E=hf (Energy=Plank's constant x the frequency).


1

To be clear, Maxwell's equations are known as "Lorentz-invariant" equations, which means that they take the same form in every Lorentz-transformed frame of reference. Special relativity actually came about from studying Maxwell's (classical) equations without charges or currents. Then we get: $$\nabla \cdot \mathbf{E}=0$$ $$\nabla \cdot \mathbf{B}=0$$ ...


4

Classical electromagnetism is perfectly compatible with special relativity. In classical E&M, light is an electromagnetic wave and there is generally no useful formulation in terms of particles. The most widely used technique to combine quantum mechanics with special relativity is relativistic quantum field theory. The relativistic QFT that ...


0

The very word, photon, belongs to the quantum mechanical regime. It is one of the elementary particles in the standard model of particle physics. Elementary particles are described with quantum mechanical wave functions, which are complex function. The complex conjugate square gives the probability of finding the particle at (x,y,z,t). In the case of the ...


0

It has to do with the total energy or power of the EM wave you're interested in, as well as the frequency of the wave. As a simple example, a 3mW laser at 500nm wavelength will produce roughly 7.55*10^15 photons per second. From how large this number is, it's not difficult to see how light will usually be made up of an extremely large number of photons. For ...


3

In quantum mechanical domain these type of question does not have meaning. Every single photon is associated with a wave and vice versa. But to talk whether an electromagnetic wave contains a single photon or not is an ambiguous statement. When people say an electromagnetic wave necessarily contains many photons it only means that a incident beam of ...


0

I finally figured it out, at least for the simple case where the two atomic states have the same angular momentum: In this case the photons are always opposite in angular momentum (meaning they are either both left or both right handed). Regarding linear polarization, it depends on the parity of the system. In one case the linear polarizations are 100% ...


2

The color of the photon is related to its frequency $f$, which can be related to the energy of the photon by the expression $E = hf$, where $h$ is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ...


0

No, but a mercury lamp does. Each mercury and calcium atom emits two indistinguishable photons out of a single atom, instead of one, like the rest of the atoms. The photons emitted by a single atom at the same time are entangled. Do not expect entangled photons out of atoms that emit a single photon at the time. Also the mercury lamp must not have a ...



Top 50 recent answers are included