Timeline for Rectangular potential barrier
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2015 at 11:11 | vote | accept | Gennaro Marco Devincenzis | ||
| Sep 8, 2015 at 8:33 | comment | added | Victor Buendía | Because in fact there're no real boundaries. That last boundary condition on $E$ is imposed by us, not by the physical problem. We have to eliminate one constant if we want to solve the system (we have 6 constants, 5 equations!). So we assign a value to that last constant to have 5 constants - 5 equations and obtain the reflection/transmission coefficients without the constant $E$. I have one degree of freedom, so I can do with it whatever I want. And I don't want to have that annoying constant in my results. | |
| Sep 7, 2015 at 12:37 | comment | added | Gennaro Marco Devincenzis | I don't see why there are no correct boundaries. | |
| Sep 7, 2015 at 7:52 | comment | added | Victor Buendía | Of course, you can set $E=3$, but that doesn't simplify things. There's no "correct" boundaries. So you want an arbitrary one that fits your classical way of thinking. If the particle comes from left, how are you going to have a particle coming from right (represented by $Ee^{-ik_1x}$? That has no sense, and that's why we choose $E=0$. And of course, if you don't choose $E=0$, then transmission or reflection coefficients don't have sense -and they're the most important magnitude in this problem! | |
| Sep 6, 2015 at 15:20 | comment | added | Gennaro Marco Devincenzis | That's ok, it wasn't clear to me that eliminating a constant was completely arbitrary, and now things begin to be a little clearer. Still, who says that setting a constant to $0$ is the right boundary condition and not, for example, $E=3$? In the end the theoretical constructions have to match experiments and I see no physical reason to set $E=0$. | |
| Sep 6, 2015 at 15:03 | history | answered | Victor Buendía | CC BY-SA 3.0 |