Simplifying Transmission Coefficient Calculation for a Potential Barrier

[Nylex]

Hi, I've been doing a QM problem to do with a potential barrier going from x = 0 to x = a. I need to work out an expression for the transmission coefficient T and have a horrible looking expression:

I can't use LaTeX properly, so:

A = (F/4k_{1}k_{2}).e^ik_{1}a[e^-k_{2}a (k_{2} + ik_{1})(k_{1} - ik_{2}) + e^k_{2}a (k_{2} - ik_{1})(k_{1} + ik_{2})]

where k_{1} = (1/hbar)(2mE)^1/2, k_{2} = (1/hbar)[2m(V_{0} - E)]^1/2

How on Earth do I simplify this further? Thanks.

 

[dextercioby]

Is this what u meant...?
A = (F/4k_{1}k_{2})\cdot e^{ik_{1}a}[e^{-k_{2}a} (k_{2} + ik_{1})(k_{1} - ik_{2}) + e^{k_{2}a} (k_{2} - ik_{1})(k_{1} + ik_{2})]

,where k_{1} = (1/\hbar)(2mE)^{1/2}, k_{2} = (1/\hbar)[2m(V_{0} - E)]^{1/2}

Daniel.

P.S.Won't u be needing the square modulus of this expression?

 

[Galileo]

Yikes. I've done this problem before. I know it's not pretty.

Anyway, if you work out the brackets:
e^{-k_2a}(k_{2} + ik_{1})(k_{1} - ik_{2})
and
e^{k_2a}(k_{2} - ik_{1})(k_{1} + ik_{2})]
you can rewrite the sum as:

2k_1k_2\left(e^{k_2a}+e^{-k_2a}\right)+i(k_2^2-k_1^2)(e^{k_2a}-e^{k_2a})
The exponential combinations form hyperbolic sinusoids. The expression is equal to:
4k_1k_2\cosh(k_2a)+2i(k_2^2-k_1^2)\sinh(k_2a)

So the original expression becomes (divide both sides by F, since you're interested in T=|F/A|^2, right?):

\frac{A}{F}=e^{ik_1a}\left(\cosh(k_2a)-\frac{i}{2}\left(\frac{k_2^2-k_1^2}{k_1k_2}\right)\sinh(k_2a)\right)

Take the modulus squared on both sides (the term in brackets is in the form a+bi with a and b real, so it's simply a^2+b^2):

T^{-1}=\frac{|A|^2}{|F|^2}=\cosh^2(k_2a)+\frac{1}{4}\left(\frac{k_2^2-k_1^2}{k_1k_2}\right)^2\sinh^2(k_2a)

Now use the expressions for k_1, k_2 to show that:

\left(\frac{k_2^2-k_1^2}{k_1k_2}\right)^2=-4+\frac{V_0^2}{E(V_0-E)}

Use the identity cosh^2(x)-sinh^2(x)=1 and write out k_2 in the argument to get the final answer.
..whew

 

[Nylex]

Ahh, cheers both of you.