Gavroy
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hi...does anybody know here...how you can calculate the probability density of photons?
Gavroy said:hi...does anybody know here...how you can calculate the probability density of photons?
Gavroy said:hi...does anybody know here...how you can calculate the probability density of photons?
Gavroy said:hi...does anybody know here...how you can calculate the probability density of photons?
A. Neumaier said:There is no such thing, since photons do not have a position operator.
See the entries ''Particle positions and the position operator'' and ''Localization and position operators'' in Chapter B1 of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#B1
Rap said:Thanks for the link - but what would you say is being calculated by dividing the spectral energy density at some frequency by Planck's constant times that frequency?
See Secs. 3.4 and 4.1 ofGavroy said:hi...does anybody know here...how you can calculate the probability density of photons?
A. Neumaier said:It is frequency dependent, hence cannot be ''the'' number density. The only meaningful measure of the ''amount of photon stuff'' in some region S is the integral over S of the energy density.
Rap said:Yes, but for a small frequency interval df, would it not be the number density of photons with frequency f to f+df?
One shouldn't do physics like Procrustes http://en.wikipedia.org/wiki/Procrustes to match one's own preferences.Demystifier said:See Secs. 3.4 and 4.1 of
http://xxx.lanl.gov/abs/0904.2287 [Int. J. Mod. Phys. A25:1477-1505, 2010]
This is a very general probabilistic interpretation of relativistic many-particle states,
The paper doesn't address this (and the associated problem of gauge invariance) at all.Demystifier said:which includes photons.
A. Neumaier said:For essentially monochromatic light, you can pretend that it is by defining it in this way, but what's the use of it?
A. Neumaier said:The energy density (intensity) is very useful in optics, and sufficient without having an additional dubious concept of number density that is already meaningless for white light.
Rap said:White light has a certain energy density (U) as a function of frequency (f). Integrate U(f)/hf over all frequencies and you have the photon density.
This technical problem is easy to overcome, similarly to plane waves (momentum eigenstates) in standard QM. You can normalize it in a large but finite spacetime box, or you can use a more rigorous rigged-Hilbert-space techniques.A. Neumaier said:... interpretation as a density in space-time is incompatible with the standard Schroedinger view, since the wave function of a Schroedinger particle cannot be normalized such that the space-time integral is 1.
The paper presents a general discussion of arbitrary spin. The gauge invariance is not discussed explicitly, but it is relatively easy to do: Instead of summing over spin indices, you sum over the physical (2 for photons) polarizations.A. Neumaier said:The paper doesn't address this (and the associated problem of gauge invariance) at all.
For a useful probability interpretation one needs a representation of the wave functions as psi(x) where x is space position and momentum and angular momentum are represented in the standard way, so that |psi(x)|^2 can be viewed as a particle density, and the result is covariant under Euclidean motions.Rap said:Well, I wasn't considering "use" but I'm trying to understand the implications of there being no position operator for photons. When you say "pretend that it is", where does the pretense fail? I mean, what logical inconsistency results from assuming it is a photon density?