Timeline for Justification for the matching condition for the wave function at potential jumps. Why is it both restrictive enough and sufficiently general?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Apr 27, 2011 at 3:47 | comment | added | Jess Riedel | There is a very clear physical reason why the wavefunction should be continuous: it's derivative is proportional to the momentum of the particle, so discontinuities imply that the state has an infinite-momentum component. | |
| Jul 27, 2010 at 11:09 | vote | accept | Rasmus | ||
| Jul 26, 2010 at 16:56 | comment | added | José Figueroa-O'Farrill | The idea is the following: suppose that $V$ has isolated discontinuities and let $x_0$ be the location of one such discontinuity. Replace $V$ on $[x_0-\epsilon, x_0+\epsilon]$ with another potential which is continuous and which tends to $V$ as $\epsilon\to 0$. Then you show that the wave-function which solves the Schrödinger equation for this new potential tends in the limit as $\epsilon\to0$ to the wave-function you want and that in this limit the first derivative remains continuous. This is not really proven in Cohen-Tannoudji et al. but only sketched. The details are not hard, though. | |
| Jul 26, 2010 at 15:23 | comment | added | Rasmus | Your first argument is not clear to me - I'll take a look at Cohen-Tannoudji. | |
| Jul 26, 2010 at 13:44 | history | answered | José Figueroa-O'Farrill | CC BY-SA 2.5 |