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 A-B is given?
(Sorry if it seems stupid to ask but you said assume one thing and then you state something completely different).
May 20th 2013, 01:07 AM
Re: Annoying algebra
yes, there's always more . . . .
A steady beam of particles travels in the x-direction 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 stationary-state approach, the beam of particles is represented by an energy eigenfunction of the form
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(E0-V0)/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 / A-B = k1/k2 * C+D / C-D
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.