Algebra in potential barrier problem

[Summer95]

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 Schrödinger 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.

 

[andrewkirk]

The complex exponential in the numerator of the first expression disappears when you take the modulus in the second, because $\forall x\in\mathbb{R}:\ |e^{ix}|=1$.
In the denominator of the second expression, square and add the imaginary and real parts to get the modulus.

 

[Summer95]

The complex exponential in the numerator of the first expression disappears when you take the modulus in the second, because $\forall x\in\mathbb{R}:\ |e^{ix}|=1$.
In the denominator of the second expression, square and add the imaginary and real parts to get the modulus.

Thank you! I also just realized I need to re-square F/A because I did not take the complex conjugate.

 

[Summer95]

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.

So the transmission probability is actually:
$T=\left | \frac{F}{A} \right |^{2}=\frac{F*F}{A*A}$ (where F* means the complex conjugate of F)
so doing that I get:

$\frac{4k^{2}k'^{2}}{4k^{2}k'^{2}cos^{2}k'L+(k^{2}+k'^{2})^{2}sin^{2}k'L}$

I guess I am still struggling with the algebra here. This should look like:
$T=\frac{4\frac{k'^{2}k^{2}}{(k^{2}-k'^{2})^{2}}}{sin^{2}k'L+4\frac{k'^{2}k^{2}}{(k^{2}-k'^{2})^{2}}}$

What am I missing?