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A. Neumaier
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How do they respond to dim coherent laser light?f95toli said:There are CQED methods that allow you to "sense" the number of photons in a Fock state dispersivly.
How do they respond to dim coherent laser light?f95toli said:There are CQED methods that allow you to "sense" the number of photons in a Fock state dispersivly.
Exactly. Now we completely agree :-)).A. Neumaier said:And in the case of prepared 1-photon states one knows the presence of the single photon only because of the preparation, not because of the detection! The detection only informs one about the presence of a local burst of the energy density (sufficient to fill the cup, in the classical analogy). Together with the knowledge about the preparation one infers that the photon arrived.
I was talking about photons in a beam extending over a laboratory desk, and their detection. Your setting is quite different.f95toli said:Blais, Alexandre, et al. "Cavity quantum electrodynamics for superconducting electrical circuits: An architecture for quantum computation." Physical Review A 69.6 (2004): 062320.
These systems can be used to create and manipulate just about any photon state you can imagine, including obviously simple Fock states.
This can be done for the state of any source - one doesn't need any assumptions on whether or not it contains ghostlike particles.f95toli said:Note also that is is possible to perform tomography of photon states using "classical" detectors by using methods borrowed from radio engineering (IQ mixing) . It is is of course an indirect method since you can't "see" individual events, but as long as you are able to average over many events you can re-construct the state of the photons generated by your source.
A. Neumaier said:The density matrix of an ##n##-level system has ##n^2-1## independent degrees of freedom, and it is not difficult to find this many independent observables such that, from the measurement of their mean (by repeated observation of individual systems emitted by the source), the density matrix can be reconstructed by solving a linear system of equations.
consider the following. One performs a "box quantization" of the electromagnetic field by treating each cavity mode of the electromagnetic field as a harmonic oscillator. Here though you are actually starting with cavity modes of the electromagnetic field rather than modes of the Schrödinger wave function but it goes the same way. Singling out a particular mode, from Maxwell's equations it is found one can treat the electric field as the generalized coordinate and the magnetic field as the conjugate momentum. (Of course other approaches are possible including treating the magnetic field as the coordinate or more commonly quantizing the vector potential, but this approach illustrates the relation of wave function to EM field.)sciencejournalist00 said:Or is the electromagnetic wave considered to be the wavefunction of the photon?
What about the double-slit experiment, and the interference fringe pattern?A. Neumaier said:This is not correct. it is well-known that photon location cannot be described by a probability distribution
The point where a beam hits a screen determines only the transversal position, not the longitudinal one. Transversal position is reasonably well-determined in the paraxial approximation.tade said:What about the double-slit experiment, and the interference fringe pattern?
I'm sorry, I don't understand this terminology.A. Neumaier said:The point where a beam hits a scree determines only the transversal position, not the longitudinal one. Transversal position is reasonably well-determined in the paraxial approximation.
If a beam goes into the z-direction, the transversal position are the x,y coordinates only, while z is the longitudinal position. The paraxial approximation defines what a beam is in term of a field.tade said:I'm sorry, I don't understand this terminology.
There are commuting operators for transversal position but not for longitudinal position.tade said:How does this relate to probability waves, or that photons cannot be described by a probability distribution?
Ok, is that considered a transversal probability distribution?A. Neumaier said:If a beam goes into the z-direction, the transversal position are the x,y coordinates only, while z is the longitudinal position. The paraxial approximation defines what a beam is in term of a field.There are commuting operators for transversal position but not for longitudinal position.
Sending particles through a double-slit apparatus one at a time results in single particles appearing on the screen, as expected. Remarkably, however, an interference pattern emerges when these particles are allowed to build up one by one (see the image to the right).
This demonstrates the wave-particle duality, which states that all matter exhibits both wave and particle properties: the particle is measured as a single pulse at a single position, while the wave describes the probability of absorbing the particle at a specific place of the detector. This phenomenon has been shown to occur with photons, electrons, atoms and even some molecules, including buckyballs.
No, and the problem is in the words "being there" - wording it that way is already assuming that the photon has a position and hence a probability distribution for that position. There's no substitute for actually learning and using quantum electrodynamics, but if you want a non-rigorous and intuitive sort of understanding, you could say that the bright fringes are bright because there is a higher probability that the electromagnetic field will deliver energy to those areas, or you could say "appearing there" instead of "being there".tade said:Aren't the bright fringes bright because photons have a higher probability of being there, and the dark fringes are dark because the photons have a low probability of being there?
Nugatory said:No, and the problem is in the words "being there" - wording it that way is already assuming that the photon has a position and hence a probability distribution for that position. There's no substitute for actually learning and using quantum electrodynamics, but if you want a non-rigorous and intuitive sort of understanding, you could say that the bright fringes are bright because there is a higher probability that the electromagnetic field will deliver energy to those areas, or you could say "appearing there" instead of "being there".
The wave function is given by the Silberstein vector field.tade said:I see.
Am I right to say that a photon cannot be described a scalar wavefunction?
A. Neumaier said:The wave function is given by the Silberstein vector field.
See http://arnold-neumaier.at/ms/optslides.pdftade said:Ahhh, thank you.
Thanks.A. Neumaier said: