Timeline for What happens with the energies of the boundstates in an infinite square well if we put a small potential step in the middel?
Current License: CC BY-SA 4.0
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| Dec 21, 2020 at 16:42 | comment | added | J. Murray | @NicolasSchmid Hi Nicolas. Remember that the way that I formulated this problem, the step height is contained within $\lambda$, not $V$. As a result, your correction to the energy should be linearly proportional to the step height. I haven't done the integral myself, but one way to check it is to simply let $a=L$; in that case, you've simply shifted the "floor" of the potential well up by $\lambda$, so all of the energy levels should be shifted up by the same amount. It looks like that's what happens with your result, which is good. | |
| Dec 21, 2020 at 13:59 | comment | added | Nicolas Schmid | Thank you! I calculated the first order energy correction with the advice you gave me and found that the energy is higher for a higher potential step (there is a factor height^2 in the first energy correction). For the influence of the length of the potential step it is a bit less clear because the energy correction is proportional to a/L - (2(-1)^n * sin(npia/L))/(n*pi). I plotted this function and it looks like sin(a) + a * const so it normally gets bigger when a is bigger, but sometimes it gets a bit smaller. | |
| Dec 21, 2020 at 13:47 | vote | accept | Nicolas Schmid | ||
| Dec 21, 2020 at 12:49 | history | answered | J. Murray | CC BY-SA 4.0 |