Timeline for Photon-electron absorption probability
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2019 at 23:50 | comment | added | Victor Novak | Yeah, I guess deeper understanding of mathematics (which I'm lacking) could help me express myself much better. It is still a major obstacle for me. | |
| Dec 9, 2019 at 15:27 | comment | added | user137289 | The reason why I wrote my answer after this one is that I did not like the phrase: "If a photon impinges on the atom." It echoes the language in the question, but it is rather imprecise, and it does not convey the right image. For example, cross sections can be very large. In the Rutherford experiment, the Coulomb scattering cross section is not even finite (in the theory of a bare nucleus). | |
| Dec 9, 2019 at 12:42 | comment | added | Emilio Pisanty | The probability of interaction is independent of what happens after the interaction (i.e. of whether the state decays afterwards or not). It is perfectly possible to have a 100% probability that the photon will be absorbed (i.e. half of one Rabi oscillation, on a Jaynes-Cummings picture) which subsequently decays. The two are independent aspects of the process. | |
| Dec 9, 2019 at 11:35 | comment | added | anna v | @EmilioPisanty corected my age. I am completely unfamiliear with quantum optics , and am not going ( at 80 years of age) to start studying to see where the misunderstanding is . I am fairly sure that what I say is correct for the hydrogen atom and the interactions the question asks. about. . In my QM books, if quantum optics comes with a state that is 100%, it should be a stable solution of the QM equations, not an interaction, as the question asks. | |
| Dec 9, 2019 at 11:14 | comment | added | Emilio Pisanty | @annav That is a perfectly legitimate position ─ if one is willing to accept it when the field has moved past one's ability to follow it in detail. In this case, this answer is wrong and inconsistent with our current understanding (which, again, is 40+ years in the making). I'm sorry to be blunt, but there's no way to sugar-coat this that isn't disingenuous. | |
| Dec 9, 2019 at 10:54 | comment | added | Emilio Pisanty | To emphasize: this is standard material in quantum optics, and it has been a key component of the literature for upwards of 40 years. It can, if needed be expressed in the old-school language of cross-sections (for the full 3D case, anyways ─ the cases of restricted dimensionality in waveguides and cavities obviously require modification to that language, again as in ages-old standard material), but even that is absent from this answer. | |
| Dec 9, 2019 at 10:51 | comment | added | Emilio Pisanty | @annav Apologies, but that is yet again hogwash; the only sense in which it's true is the parts that are platitudes about probabilities in the real world with no connections to QM. The HUP has nothing to do with it. | |
| Dec 9, 2019 at 10:37 | comment | added | anna v | @EmilioPisanty Sorry but I have to remind you of Hesenberg's uncertainty principle. In quantum mechanics . It could be 99.999999%, never 100. Only stable levels have a probability of finding the electron at that level 100%, because of the nature of the wavefunctions. | |
| Dec 9, 2019 at 9:45 | comment | added | Emilio Pisanty | "It cannot be 100% because of the nature of probabilities" - this is hogwash, I'm afraid. It's perfectly possible for single-photon states to have 100% interaction with atomic systems. This turns out not to happen in 3D, but it's an everyday tool in cavities. "because of the nature of quantum mechanical scattering amplitudes" is a non-explanation. | |
| Dec 8, 2019 at 19:29 | comment | added | anna v | Note that the probability of finding the electron at a certain radius is not measurable for radii larger than the ones shown. . This is for the stable atom, that is shown in the solution. One would have to solve the scattering of photon+hydrogen amplitude, which will be a different function. | |
| Dec 8, 2019 at 18:47 | comment | added | Victor Novak | Thank you for a great source! I assume that by "calculable probability" you mean this distribution: hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html - ? So, if I got you correctly: even if we somehow will take "prepared" hydrogen atom (with its electron on, say, n = 1 level), and if we will be going to impinge this hydrogen atom with the "right" photon - we will never succeed to set "right colliding conditions" for absorption to be 100% certainly happen, is that right? | |
| Dec 8, 2019 at 17:32 | history | answered | anna v | CC BY-SA 4.0 |