- #1

physicsjock

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I've been trying to work out how, for a finite well of high Vo and width L, the interior solution has the form L Sin(kx + d),

I see that if d=0 then the solution resembles an infinite well, so that implies d depends inversely on the wells potential. But I can't work out what d comes from, and why the constant at the front is the length of the well (why is the amplitude of the wave the length of the well)

d is just the phase of the wave, so does it represent the reflection of any particles which moved past the well?

Ive also been trying to work out what the k is, when I go through the process of finding the wave function inside I end up with (s/k)Asinkx + Acoskx after applying the boundary conditions.

where k

^{2}=2mE/[itex]\hbar^{2}[/itex]

and s

^{2}=2m(Vo-E)/[itex]\hbar^{2}[/itex]

i've been trying to work this out because the question also says that the solution inside the well should also satisfy

ka = n[itex]\pi - 2Sin^{-1}(\frac{k\hbar}{\sqrt{2mV_{o}}})[/itex]

which makes me think the d resembles that arcsin the above equation

Anyone have any ideas?