- #1
Summer95
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Homework Statement
It is a potential barrier with E of the incoming matter wave E>U (greater than the height of the barrier). I have already done a LOT of algebra to get to the point where
##\frac{F}{A}=\frac{2kk'e^{-ikL}}{2kk'cos(k'L)-i(k'^{2}+k^{2})sin(k'L)}##
which I won't bother writing out because its done and I can check that this is the correct expression.
k and k' are clearly the usually values corresponding to the time independent Schrodinger equation outside and inside of the barrier, respectively. F is the amplitude of the transmitted wave and A is the amplitude of the incident wave. L is the width of the barrier.
Homework Equations
The transmission probability *should* simplify to:
##T=\frac{4\frac{k'^{2}k^{2}}{(k^{2}-k'^{2})^{2}}}{sin^{2}k'L+4\frac{k'^{2}k^{2}}{(k^{2}-k'^{2})^{2}}}##
The Attempt at a Solution
So the transmission probability
##T=\left | \frac{F}{A} \right |^{2}=\frac{4k^{2}k'^{2}(coskL-isinkL)^{2}}{4k^{2}k'^{2}cos^{2}k'L+(k^{2}+k'^{2})^{2}sin^{2}k'L-2ikk'cosk'Lsink'L(k^{2}+k'^{2})}##
I have tried to simplify this twice - the second time I started by just multiplying the whole bottom out. I won't write out all of my work because nothing after this point lead to anything useful but if I do get somewhere I will add it. Mostly I don't understand how to get rid of the exponential in the numerator (or the trigs associated with it) and all of the i's.