
Annoying algebra
It's late and I really am at an end on this, I just can't seem to get it.
Note: A+B = C+D
So, I need to find B in terms of A.
A+B / AB = k1/k2 * C+D / CD
cross multiply
(A+B)k2 / (AB)k1 = C+D / CD
but A+B = C+D
so
(A+B)k2 / (AB)k1 = A+B / CD
=> (A+B)k2 / (AB)k1) / (A+B) = CD
(A+B)k2 / (AB)k1 * (A+B) = CD
Aagh!
Sorry, I can't do more, please help

Re: Annoying algebra
Hey froodles01.
Just for clarification can you tell us all the information you started off with? You said A+B=C+D which is one piece of information: is there any more?
Since you have four variables to start off with. it means you will need one more two get a result in two variables. Can we assume that your second result involving AB is given?
(Sorry if it seems stupid to ask but you said assume one thing and then you state something completely different).

Re: Annoying algebra
Hi.
yes, there's always more . . . .
A steady beam of particles travels in the xdirection and is incident on a finite square barrier of height V0, extending from x =0 to x = L. Each particle in the beam has mass m and total energy E0 =2V0. Outside the region of the barrier, the potential energy is equal to 0.
In the stationarystate approach, the beam of particles is represented by an energy eigenfunction of the form
ψ(x)=
Ae^ik1x + Be^−ik1x for x< 0
Ce^ik2x + De^−ik2x for 0 ≤ x ≤ L
Fe^ik1x for x>L
where A, B, C, D and F are constants and k1 and k2 are wave numbers appropriate for the different regions.
Express the wave numbers k1 and k2 in terms of V0, m and n. (which I have done  hooray [k1 = √2mE0/ħ and k2 = √2m(E0V0)/ħ])
Now consider the special case where k2L = pi/2 (which does not correspond to a transmission resonance). Use your answer to part (b) to show that
A+B / AB = k1/k2 * C+D / CD
Use your answer for part (a) and the above equation to find B in terms of A. Hence calculate the reflection coefficient, R, of the beam and deduce the value of the transmission coefficient, T.
Useful equations include
A + B = C + D and
k1A – k1B = k2C – k2D
Ce^ik2L + De^ik2L = Fe^ik1L
and k2Ce^ik2L – k2De^ik2L = k1Fe^ik1L
k2L = pi/2
. . . . .well, you did ask!
I should be able to do this, but left myself too little time & now I'm too thick to think enough.