Solve coefficients for four equations in square well

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 1K views
dsdelavega

Homework Statement


Hello, I am stuck on four equations for which I must find the coefficients A,B,I,J.
I have tried using latex but the commands don't seem to work.

Homework Equations


Four equations:

[tex]A+B = I+J[/tex]
[tex]\frac{\alpha}{k}(J-I) = A - B[/tex]
[tex]D = Ie^{ia(\alpha-k)} + Je^{-ia(\alpha + k)}[/tex]
[tex]D = -\frac{\alpha}{k}Ie^{ia(\alpha-k)} - \frac{\alpha}{k}Je^{-ia(\alpha + k)}[/tex]

The Attempt at a Solution


I = [2B + J((alpha/k)-1)]/ (1+ (alpha/k))

i got this from setting 1 and 2 equal to A then solving for I but i don't know where to go from here. Any tips?
 
Last edited by a moderator:
Physics news on Phys.org
You have five unknowns, A, B, D, I, and J, but only four equations. You can solve for four of the coefficients in terms of the other one.

I'd solve the last two equations for I and J in terms of D. Then once you have those, plug the results into the first two equations and solve for A and B.
 
vela said:
You have five unknowns, A, B, D, I, and J, but only four equations. You can solve for four of the coefficients in terms of the other one.

I'd solve the last two equations for I and J in terms of D. Then once you have those, plug the results into the first two equations and solve for A and B.

Yes, we were told that we should solve in terms of our incident wave direction which is the value B in this case.
So if i solve for I and J in terms of D does that mean just set 3 = 4 and solve for I and J?
 
vela said:
These are linear systems of equations, like x+y=2 and x-y=1. You solve them the same way.

I understand, I am just getting lost in the algebra trying to match terms.
Also, any chance you can tell me how to implement the latex equations? I know how to use latex but i don't know how to do it on a post. thanks!

when setting 3 = 4 then I solve for I, then the result I get is the following:

[tex]I = J e^{-2ia\alpha} \frac {k-\alpha} {k+\alpha}[/tex]

I don't know what to do after this.
The four equations are throwing me off because of the many variables.
Additionally,I can't plug this into equation 1 because I wouldn't know what A is.