Did we establish that getting the second boundary condition for the attractive potential well requires integrating twice? Using epsilon for the bounds on the first integral (as in class) what would be the bounds on the second integral?
For the delta-function potential, I think we established that there is a discontinuity in dpsi(x)/dx, but that psi(x) is still continuous. The first result came from integrating once and the second condition, which is the conventional one, came from integrating twice. In either integration the limits were the same: -eps to +eps as eps goes to zero, i.e., centered on the singularity and shrinking.
I'm having trouble understanding...is there supposed to be a "help" between "you will get" and "on the HW"?
ReplyDeleteI think he means we literally get on top of the homework. :)
ReplyDeleteSounds fun!
ReplyDeleteDid we establish that getting the second boundary condition for the attractive potential well requires integrating twice? Using epsilon for the bounds on the first integral (as in class) what would be the bounds on the second integral?
ReplyDeleteFor the delta-function potential, I think we established that there is a discontinuity in dpsi(x)/dx, but that psi(x) is still continuous. The first result came from integrating once and the second condition, which is the conventional one, came from integrating twice. In either integration the limits were the same: -eps to +eps as eps goes to zero, i.e., centered on the singularity and shrinking.
ReplyDelete