Timeline for Standing Spherical Wave normalization after decomposition in Plane Waves
Current License: CC BY-SA 3.0
15 events
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| Dec 14, 2016 at 20:48 | comment | added | Andrew | If you want a less wishy washy answer, in the absence of other information often the best choice (for time-independent situations) is to work with eigenfunctions of the Hamiltonian which are normalized in either the Kronecker-delta or Dirac-delta sense (indeed this is one step in defining what you mean by a photon, identifying these modes). Once you decide to go with use these basis functions, the normalization is fixed. | |
| Dec 14, 2016 at 20:43 | comment | added | Andrew | @AntonioFazzolari Thanks! Glad I can help. Indeed, the normalization is the key step! (I sometimes think half of theoretical physics is normalizing modes properly). Your question about observations doesn't have a very satisfying answer, I'm afraid. You can choose to normalize your mode functions however you want. However, if you make a bad choice then you will have to work a lot harder to compute observable quantities later on. The crucial factor of $k$ in $\psi_{k\ell m}(\vec{x})$ is ultimately nothing more than a particularly convenient choice. | |
| Dec 14, 2016 at 18:19 | comment | added | Antonio Fazzolari | Yeah, now it's really clear, and also quite elegantly shown. I think that the key step is to choose and appropriate normalization, and algebra works flawlessly. So thank you for helping me with your skills. I would like to add one more question. Is there any experiment, or any other physical evidence that the SWs, for example, associated with photons in a spherical cavity, udergo this particular normalization? Surely not energy considerations, since energy depends only on frequency... | |
| Dec 14, 2016 at 17:55 | vote | accept | Antonio Fazzolari | ||
| Dec 14, 2016 at 7:01 | history | edited | Andrew | CC BY-SA 3.0 |
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| Dec 14, 2016 at 5:59 | comment | added | Andrew | @AntonioFazzolari I added some comments above... let me know if that helps! | |
| Dec 14, 2016 at 5:58 | history | edited | Andrew | CC BY-SA 3.0 |
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| Dec 14, 2016 at 4:14 | history | edited | Andrew | CC BY-SA 3.0 |
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| Dec 13, 2016 at 18:36 | comment | added | Antonio Fazzolari | I followed your reasoning and it seems correct, showing in fact that my assumption can be wrong. Btw I was not questioning that SW are delta normalized, but just asking if there is some multiplicative factor w.r.t. the "cartesian" delta obtained from PW normalization. Can you plz indicate to me the formula showing that the norm of $j_{lm}(kr) $ gives you a pure delta in radius? The only I found was a $\pi\delta(k-k')/2/k^2$ , moreover, do we really need a delta in radius? or in momentum? I think the 2nd one, being basis functions "indexed" by k,l,m. | |
| Dec 13, 2016 at 5:15 | comment | added | Andrew | Reading the note you added to the question, I think that comparing the norms in the two bases you should get $\delta^3(\vec{x})=\delta^3(\vec{x})$, which indeed looks like $\infty=\infty$ when $\vec{x}=0$. Perhaps what you're getting at is that in Cartesian coordinates natural for the PW basis, $\delta^3(\vec{x})=\delta(x)\delta(y)\delta(z)$, whereas in spherical coordinates natural for the SW basis $\delta^3(\vec{x})=(\delta(r)/r^2) (\delta(\theta)/\sin\theta) (\delta(\phi))$. I can add a bit more about this to the answer too if you want. | |
| Dec 13, 2016 at 5:02 | comment | added | Andrew | ...If that's what you wanted to know, I can add a bit to the answer showing how you get the 3D delta function in the SW basis. | |
| Dec 13, 2016 at 5:02 | comment | added | Andrew | @Antonio Fazzolari: No worries, I figured boson was a typo and I know what you meant by 'pulverize' (I just thought it might be useful to point out). Apologies I only told you stuff you already know. I'm not quite sure I understand what you mean by 'norm'. It's true that $\psi_{\ell m}(\vec{x})$ is delta function normalized. You can see that in either basis. In the PW basis the key step is that the Fourier transform of 1 is a delta function. In the SW basis, the norm of $Y_{lm}$ gives you a delta function in the angular coordinates, and the norm of $j_{lm}$ gives you the delta in radius... | |
| Dec 12, 2016 at 17:43 | comment | added | Antonio Fazzolari | I was totally wrong about boson, of course. Stupid error, unrelevant. The term "pulverized" was an attempt to depict the (eventual) non 1 to 1 correspondence, not to address some physical process. I was already aware of the Plane wave expansion in term of Spherical Bessel functions, which is in fact the inverse of the one I reported above. But the goal here was not to find the expansion of a Spherical Wave in terms of Plane Waves or vice-versa, which are both known, but to estabilish a relation between the "norm" (infinite) of an PW and the "norm" of a SW, sorry if I did't explain me well. | |
| Dec 10, 2016 at 15:39 | history | edited | Andrew | CC BY-SA 3.0 |
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| Dec 10, 2016 at 15:29 | history | answered | Andrew | CC BY-SA 3.0 |