I need to find B in terms of F in a finite square well potential(adsbygoogle = window.adsbygoogle || []).push({});

I started with

-Ae^(-i*K*a) - Be^(i*K*a) = Csin(k2*a) - Dcos(k2*a)

and

Ae^(-i*K*a) - Be^(i*K*a) = i*K*k2 [C*cos(k2*a) - D*sin(k2*a)]

where

C = [sin(k2*a) + i*(K/k2)cos(k2*a)]*Fe^(i*K*a)

D = [cos(k2*a)- i*(K/k2)sin(k2*a)]*Fe^(i*K*a)

I added the first two and got

-2*B*e^(i*K*a) = {[i(k2*K +(K/k2)sin(k2*a) - (K^2 +1)cos(k2*a)]*Fe^(i*K*a)}*cos(k2*a) +

{[i(k2*K +(K/k2)cos(k2*a) + (K^2 +1)sin(k2*a)]*Fe^(i*K*a)}*sin(k2*a)

I do not know how to go from here.

Would B be equal to

(F/4)*e^(i*K*a)*{-4cos(k2*a) + 2i*[(k2/K)+(K/k2)]*sin(k2*a)}

what am I missing??

Thanks so much for your help.

Stephen

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# Homework Help: Finite square well potential question Constants

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