Finite square well potential question Constants

[StephenD420]

I need to find B in terms of F in a finite square well potential

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

 

[anathema]

Dear Stephen,

To find B in terms of F, we can use the fact that the wavefunction must be continuous at the boundaries of the finite square well potential. This means that the wavefunction on the left side of the well (x < -a) must be equal to the wavefunction on the right side (x > a). This condition can be expressed mathematically as:

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

Solving for B, we get:

B = (Ae^(i*k2*a) - Ae^(-i*K*a)) / (e^(i*K*a) + e^(-i*K*a))

Now, we can substitute the expression for A from your equations:

A = Csin(k2*a) - Dcos(k2*a)

And then substitute the expressions for C and D:

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)

After some algebraic manipulation, we get:

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

Therefore, we have expressed B in terms of F, and we can see that B is a complex number. I hope this helps. Let me know if you have any further questions.