- #1
Damascus Road
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Question:
Determine the transmission coefficient for a rectangular barrier. Treat separately the three cases E<0, E=Vo, E>Vo (note that the wave function inside the barrier is different in the three cases.
and
V(x) = Vo > 0 for -a < x < a
= 0 for |x| > a
I attempted to do this like the finite square well and do it per "zone", because I'm not sure how to go about dealing with the energy first.
If I call - infinity -> -a "zone 1" and the potential there is zero, the general solution should be:
[tex]\varphi[/tex] = A[tex]e^{-kx}[/tex] + B[tex]e^{kx}[/tex], but the first term blows up to -infinity, so we're left with
[tex]\varphi[/tex] = B[tex]e^{kx}[/tex].
Can someone tell me what is wrong with this logic?
For E<Vo and x < -a [tex]\varphi[/tex] should include both A and B and be imaginary, but I don't know why.
Determine the transmission coefficient for a rectangular barrier. Treat separately the three cases E<0, E=Vo, E>Vo (note that the wave function inside the barrier is different in the three cases.
and
V(x) = Vo > 0 for -a < x < a
= 0 for |x| > a
I attempted to do this like the finite square well and do it per "zone", because I'm not sure how to go about dealing with the energy first.
If I call - infinity -> -a "zone 1" and the potential there is zero, the general solution should be:
[tex]\varphi[/tex] = A[tex]e^{-kx}[/tex] + B[tex]e^{kx}[/tex], but the first term blows up to -infinity, so we're left with
[tex]\varphi[/tex] = B[tex]e^{kx}[/tex].
Can someone tell me what is wrong with this logic?
For E<Vo and x < -a [tex]\varphi[/tex] should include both A and B and be imaginary, but I don't know why.