New answers tagged photons
3
It is not possible to satisfy both energy and momentum conservation with only one photon.
1
For a two-level system with states $|g\rangle$ and $|e\rangle$, and an electromagnetic mode with Fock states $|n\rangle$, it does make sense to consider the four relevant levels you described. If the electromagnetic mode is resonant with the two-level system energy gap, then $|g, n\rangle$ is degenerate with $|e, n-1\rangle$, and $|e, n\rangle$ is degenerate ...
6
Here is an argument based on spacetime symmetry and the fact that U(1) theory is abelian (therefore photons don't interact with each other) to show that the probability is zero.
Suppose we start from a uniform magnetic field in the $z$ direction. Since there is no preferred direction in the $xy$-plane, the final state photon must be emitted along the $z$ ...
4
To add to @john-rennie 's answer: things get even worse as the environment of the atom makes additional modifications to the energy levels (ranging in the meV). For example, a carbon atom bound to an oxygen will have a slight energy shift compared with a free carbon atom, or a carbon atom bound in a diamond structure. These shifts are very useful in ...
8
The energy levels depend on two things:
the electrostatic attraction between the electrons and the nucleus
the electrostatic repulsion between the electrons
If you take a hydrogen atom, which is what your diagram shows, then there is a single electron and a single proton. The electron is attracted to the proton and there is no electron-electron repulsion ...
5
The diagram you posted shows the electron energy levels for the hydrogen atom only. Other atoms will have different energy levels. Also note that the hydrogen atom itself has an infinite amount of energy levels which the diagram does not really show.
This information is very useful. The energy levels for each atom are unique, and the corresponding ...
0
I'm thinking that light slows down in a medium because photons are being absorbed and then remitted by atoms or molecules in the beam.
That is one possible process and it takes place for example with glass. It happens, what oleg told in his answer: Doing an experiment with polarized light, the glass doesn’t destroy the polarization. The light is coming out ...
0
Is it possible to set up a two photons entanglement so that their global state is $$|\psi\rangle = \frac{1}{\sqrt{3}}(|HH\rangle+|HV\rangle+|VV\rangle),$$ where $H$ indicates horizontal polarization and $V$ indicates vertical polarization?
Yes, this is perfectly possible.
Does the first photon have a 50% or 33% chance to pass through a vertical ...
-2
Question 1:
No, this is not a state which can describe two photons. It is not symmetrized properly. Photons are bosons, so a two-photon wavefunction must be symmetric under exchange of the particle labels. Under exchange of your particle lables, your state becomes
$$
\frac{1}{\sqrt{3}} (|HH\rangle + |VH\rangle + |VV\rangle)
$$
which is not the same as the ...
0
This may be very late but I went through the book you mentioned and I believe a factor of $4\pi\epsilon_{0}$ is missing from equations (6.19) and (6.20) of the book. In particular, to the best of my knowledge, the oscillator strength is a dimensionless quantity and a cross section is in units of area. Multiply the definition in (6.19) by $4\pi\epsilon_{0}$ ...
0
But it's said that after observation that the photon of light that is observed becomes a particle
This is wrong. At the moment, mainstream physics accepts that the underlying framework of all physical theories ( thermodynamics, classical electricity and magnetism, (maxwell's equations) etc) all emerge from the underlying quantum mechanical level ( the ...
2
In physics the word "observe" is usually employed in a technical sense in which consciousness plays no role whatsoever. In particular, in quantum physics, there is no need to bring in the notion of consciousness or conscious reflection on what is seen. It suffices that a mark is made on a photographic film, or a particle detector registers a spark or emits a ...
1
Superconducting nanowire single photon detectors and transition-edge sensors (>95% end-to-end measured at 1550nm) Both operate near a phase transition superconductor - normal metal, where absorption of a photon changes the resistance.
4
Quantum efficiencies of higher than 90 % are no problem in regions of the spectrum.
Photomultiplier tubes (with scintillators)
Solid state detectors
Here is a comparison for the visible: https://www.techbriefs.com/component/content/article/tb/supplements/pit/briefs/29910
Or this new design: https://www.techbriefs.com/component/content/article/tb/...
3
First of, photons do not have identity. It is meaningless to say things like "photons which leave a lens or filter are not the ones which entered". More to the point of your question though: if a photon that is entangled with another system gets absorbed by an atom, that atom is then entangled with that other system. When an atom that is entangled with ...
0
Set up a state $|k_{0},\lambda\rangle$ for a photon of fiducial momentum $k_{0}$ moving along the z axis with helicity $\lambda$.
Now, instead of working with states of sharp momentum and helicity, we can consider states of sharp total angular momentum by using the projection operator onto the states of a group irrep.
Suppose we have a group $G$. The ...
1
You are correct in thinking that both the electron and ion share the photon's energy. Due to conservation of momentum and the fact that the ion Is much more massive than the electron, the ion receives only a tiny share of the energy.
1
A photonic quantum state can consist of multiple photons, as you correctly implied. If there is a fixed number of photons in the state, then we call it a "number state" or a "Fock state." Each photon in the Fock state can carry its own set of spatiotemporal and polarization (spin) degrees of freedom. Some people say that all the photons in a Fock state must ...
2
The intensity is not related to the wavelength if that is what you mean when you say length. The intensity is proportional to the number of photons.
0
A molecule in the gas may absorb the photon with energy $E=\hbar \omega$. Now, the electrons typically want to fall down to lower energy states when they are excited. When they do this they either release a photon with the same energy as before, OR the electron may give its energy to rotational and/or vibrational stages. This process is referred to as non-...
0
According to foundational principles in Quantum Mechanics, a photon cannot directly heat a gas molecule.
A photon adds orbital energy to an electron, which rapidly re-releases an identical photon in a random direction.
A molecule's temperature is almost completely due to the kinetic energy of the atomic nuclei, as that is where almost all of the mass ...
0
The diagrams of diffraction you have shared don't show light as a longitudinal wave, they are a completely different representation of waves. The lines represent wave fronts NOT compression & rarefaction. Wave fronts can be thought of as corresponding to wave crests, but really mean points on waves that are in phase with one another.
0
I tend to think particle-wave duality this way:
The actually physical state of the system is some abstract mathematical object. In old fashion quantum mechanics, it's a wave function (or ray or whatever). In quantum field theory, it's the quantum field (super position of many classical field).
What do we mean by a free particle? One way to informally "...
0
It's up to you how you think about electromagnetic radiation. The more EM phenomena you observe, the more you will relate those phenomena to light as particles or aa waves.
In order to explain photoelectric effect, Einstein suggested that light is made up of photons
The photoelectric effect is one of these phenomena which where not explainable with light ...
0
Yes, they're absorption edges. From this documentation:
4.10. The Output Table
The atomic numbers and fractions by weight of the atomic constituents are given above the table. The main body of the table is supplied with enough headings to be self-explanatory. The left-most column gives the designations of the absorption edges (K, L1, L2, L3, M1, M2, ...)...
0
We can observe a particle only by its observables, such as charge, energy-momentum, angular momentum or spin. Probability density is not an observable. The Born rule for electrons states the probability distribution for finding an electron is $|\psi|^2$, but in fact this is the nonrelativistic Noether charge density divided by $e$. The continuity equation ...
1
With the invention of lasers we no longer need the photoelectric effect to have an easy experimental demonstration that light is composed of particles:
Have a look at this single photon at a time double slit experiment:
Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, ...
0
The photoelectric effect can be explained without photons. First, let us define “photon.” It is not a thing like a wave packet. Please, it is not a detection click either. It is a dualistic phenomenon. A photon will go one way or another at a beam-splitter, but if you re-converge the beam you will develop an interference pattern. That is a rough quote ...
0
I don't see the contradiction. The wavelength of light stretches with time in the way that you wrote. The wavenumber $k=2\pi/\lambda$ decreases accordingly.
What we mean when we say that the CMB has a redshift of 1089 is that it was emitted (i.e. released, when electrons and protons bound into atoms) at a time when the universe was 1090 times smaller than ...
1
You need to remember Hubble's discovery - the farther away an object is, the more the light it emitted has been redshifted between it and you. Because light moves at finite speed, this also means that the time between when the light was emitted and observed also correlates with redshift. So, when someone says something "has a redshift" or is "at a redshift", ...
0
As stated by others, the fact that photons have spin 1 cannot be proven. I believe that is not what you are asking. You are asking about the mathematical formalism to describe photon spin.
The reason that this formalism does not exist is that conservation of photon spin is incompatible with gauge invariance. There exists no gauge invariant expression of ...
3
I don't have a good reference for this, so I tried working it out myself. My analysis has some loose ends, but it at least suggests a plausible answer to the question.
Clarification about Born's rule
The general version of Born's rule refers to observables and states. It applies to everything from relativistic QFT to non-relativistic single-particle QM. ...
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