Timeline for What actually is white light?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2021 at 17:37 | comment | added | The Vee | Really, it's nothing that wouldn't be said in any QM textbook, if you accept the view that $E = ħω$ is a relation between observables rather than classical numbers. | |
| Jan 10, 2021 at 17:36 | comment | added | The Vee | @garyp E is an observable. It will come with a probability distribution. So does any other observable in quantum mechanics, but as opposed to many other examples, E does not have a discrete spectrum so in reality it can never have an exact value (eigenvectors). In interaction with matter, e.g., absorption, a wavefunction collapse may happen, so seemingly an "exact" energy may be observed, but nothing prevents it from having an arbitrarily broad statistics, as nicely summed by S. McGrew's comment below mine. | |
| Jan 9, 2021 at 22:23 | comment | added | S. McGrew | Keep in mind that the wavefunction represents a probability distribution. If a detector (e.g., an atom) is in the path of a broad-spectrum wavefunction, and the detector is only sensitive to a very narrow frequency range, the likelihood that it will detect the photon depends on the fraction of the wavefunction's spectrum that falls within that range. | |
| Jan 9, 2021 at 21:01 | comment | added | garyp | @TheVee Then $E = h\bar{\nu}$? How can $a+$ deliver the fraction of energy needed to make up the energy of the photon? When the white light interacts with an isolated atom, what happens? One particular $a-$ operates, the one at the transition energy? What happens to all the other modes. If you are considering an ensemble I might understand better. I have it in my head that $a+$ raises the excitation of a mode by $h\nu$, and a mode is monochrormatic by definition. We shouldn't discuss it here, but if you can point me to a resource I'd be grateful. | |
| Jan 9, 2021 at 17:46 | comment | added | The Vee | @garyp A preferred basis then, huh? To me a single excitation means a+ applied on vacuum, and that a+ can be any nontrivial linear combination of an arbitrary number of other creation operators. In fact, it's the monochromatic excitations that can't exist, because they aren't normalizable! | |
| Jan 9, 2021 at 1:18 | comment | added | S. McGrew | It's pretty easy to set up an experiment that shows photons to have a finite spectral width, such as doing single-photon interferometry in which the relative path lengths are varied. E.g., if the wavefunction of a single photon is monochromatic, its spatial extent along its path must be infinite, but in fact the experiment will simply produce a result that echoes the temporal coherence (which is essentially the inverse of the bandwidth) of the source. | |
| Jan 9, 2021 at 0:32 | comment | added | garyp | The most popular way of expressing the concept of photon is that it is the excitation of an EM mode. This implies single frequency. I'm aware that there are expressions of photon that allow a distribution of frequencies, but they are not commonly used, and are not as useful as the single-frequency expression. | |
| Jan 8, 2021 at 14:56 | comment | added | S. McGrew | @leftaroundabout, My statement was that *the wavefunction of any photon has a finite spectral width", not that the photon itself does (unless "photon" refers to the photon wavefunction). There is a whole fascinating and enlightening conversation to be had around this concept. If you'd like, we can continue in chat. | |
| Jan 8, 2021 at 11:49 | comment | added | leftaroundabout | I daresay “any photon has a finite spectral width” is, if not just wrong, then at least more confusing than helpful. Photons don't contain the information of how they were created in any meaningful (measurable) sense. They're just excitations of the EM field. Spectral uncertainty comes from certain properties of the whole system under consideration, not from anything about the individual photons. | |
| Jan 8, 2021 at 1:44 | comment | added | Brian Bi | @Prof.Legolasov I think there are people on this site who know the properties of blackbody radiation. I don't know anything other than what I said in my previous comment. Your question will get more attention if you post it as a new question. | |
| Jan 8, 2021 at 1:43 | comment | added | Prof. Legolasov | @BrianBi still, there must be a way to characterize the widths of individual photons, or at least the distribution of widths | |
| Jan 8, 2021 at 1:31 | comment | added | Brian Bi | @Prof.Legolasov The light we receive from the Sun is mostly blackbody radiation, and blackbody radiation is not a bunch of "white" photons. Such a distribution would have low entropy, since all the photons would be in the same quantum state, whereas blackbody radiation has high entropy. | |
| Jan 8, 2021 at 0:55 | comment | added | Prof. Legolasov | @S.McGrew I find it hard to believe that a measurement is impossible in principle for sunlight. Even if it is impossible in practice, we should be able to predict this using astrophysics | |
| Jan 7, 2021 at 16:55 | comment | added | S. McGrew | @Prof.Legolasov Unfortunately it is not possible to measure the spectral width of a single photon from the Sun, or from any other source. The spectral widths of the wavefunctions of a large number of identically produced photons from a source can be determined statistically. | |
| Jan 7, 2021 at 16:37 | comment | added | anna v | @Prof.Legolasov may this answershttps://en.wikipedia.org/wiki/Sunlight#Composition_and_power | |
| Jan 7, 2021 at 16:00 | comment | added | Prof. Legolasov | What are the spectral widths of the photons emitted by the Sun? Are those mostly narrowband photons of mixed frequencies or broadband homogenous “white” photons? | |
| Jan 7, 2021 at 15:31 | history | answered | S. McGrew | CC BY-SA 4.0 |