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Why does it when introducing u_0^2 make it proportional to L^2V_0 and later for the special case of the delta potential it is proportional u_0 without a square is proportional to L^2V_0? — Preceding unsigned comment added by Laserphysicsguy ( talk • contribs) 17:14, 14 December 2021 (UTC)
I just added the ref. to Griffiths, ie
{{
cite book}}
: |edition=
has extra text (
help)as there wasn't one. It's practically identical as whats in the book on the pg90-95 area.
- Lorand 11:21, 21 May 2006 (UTC)
Is it considered more standard to have the potential be 0 inside the box and a value outside or to have the potential be zero everywhere except inside the box, where it would have a value of -V? I'm tempted to re-write the section with the second convention, unless there are strong objections. KristinLee 01:18, 3 June 2006 (UTC)
I think it would be helpful if more information was added to the page in general. It seems to be all equations and therefore not especially helpful to a physics student like me. Maybe a couple pictures or figures?
Anyone else think this article seems rather incomplete? The final result for the wave functions is missing, along with a discussion of the allowed energies. Also, I agree that the derivation should probably be redone so that the potential is zero outside the box. GhostTrain 04:20, 1 May 2007 (UTC)
I have continued this page to cover graphical solutions and links to the infinite square well,the delta-function potential and the spherical cavity. (I've stuck with the current convention as it is closer to the infinite square well case, and also the variable k is closer to the actual energy. (The conversion from one to the other is trivial; essentially the variables and k interchange.) I think this page is fine now and can have the warning removed. As far as quantum tunnelling is concerned, I've changed "because of" to "cf" - they are closely related phenomena but with different observable consequences. Annafitzgerald ( talk) 16:46, 30 July 2009 (UTC)
These are two totally different cases; the articles should not be merged. Pfalstad 02:51, 4 August 2006 (UTC)
Yes, don't merge! Bamse 04:50, 4 August 2006 (UTC)
The first paragraph mentions Quantum Tunneling, but that effect is not measurable in this model, but rather at Finite potential barrier (QM). You could, however, conclude that something like Quantum Tunneling will happen at a Finite Potential Barrier-like situation, but, again, it is not something that can be observed in this model: so one should either link to that case and mention Quantum Tunneling as something that is apparent in this model, and measurable in that model, or one should delete that section altogether.
Maybe I ought to elaborate why I consider it unmeasurable, for it is true that the equation isn't zero at any point in the equation, but rather approaches zero as does for a negative x. However, I don't know if any observable phenomenons (maybe a lack of knowledge on my part) where one would have a finite potential and still observe 'particles' outside the well (mostly because the well is infinitely large), whereas there are many examples (such as alpha-decay, STM, etc.) that follow a more barrier-like potential. Boreras ( talk) 17:52, 29 February 2008 (UTC)
This should be enough to get you started. I'll watch the page for the next two weeks (maybe more), so if you are stuck, or have questions, just add them below. Headbomb { ταλκ κοντριβς – WP Physics} 14:36, 19 August 2009 (UTC)
Is the double finite potential covered anywhere? The Delta potencial page talks about the double delta case, but I don't see the double finite case. If not a section about the double well as a crude molocule model could be added here, Griffiths has a derivation. Timetraveler3.14 ( talk) 23:12, 14 November 2014 (UTC)
The treatment is the usual in text books on QM. So far ok. Has somebody ever tried to do this in e.g. 2D? Would be nice to see how it works here, because it is not simply an extension of the 1D case. Thanks, in advance. — Preceding unsigned comment added by 129.187.45.25 ( talk) 15:55, 4 March 2015 (UTC)
Thanks. — Preceding unsigned comment added by Koitus~nlwiki ( talk • contribs) 13:20, 11 August 2020 (UTC)
in the special case of the narrow well, the approximate equation becomes v^2 =v0^2 - v0^4. Where is the explanation for this statement? Why is this the case? 2A02:A45E:4B73:1:1DE8:5829:4A13:3C1A ( talk) 11:57, 6 May 2024 (UTC)