- #1

genxhis

- 37

- 1

*wells*of strength alpha at x = -a and x = +a. To solve the problem I split the region into the three obvious intervals [-inf, -a], [-a, a], and [a, +inf]. For the first two regions I expressed the solutions as (A or C) exp(i k x) + (B or D) exp(-i k x) and for the last as E exp(i k x) where k = sqrt(2 m E)/hbar. I then applied the two contraints at the two boundary conditions to reduce the five unknowns to just one. Finally I did some more algebra to find the trasmission coefficient as the square of (E/A). But the entire process was lengthy and tedious. I was wondering if someone could validate this answer:

[tex] T = \frac{1}{1 + 2 \beta^2[ (1+\beta^2) + (1-\beta^2)\cos 4ka -2\beta \sin4ka ]} [/tex]

where [tex] \beta = (m \alpha)/(\hbar k) [/tex].