Finite Square Well, Ψ[SUB]III[/SUB] const related too Ψ[SUB]II[/SUB]?

[The Floating Brain]

Homework Statement:

There are two ledges with of U potential energy, and a cavern in between them of 0 potential energy. Such that

U(x) = { x <= 0, U, x > 0, x >= L, U }

Find the probability distribution for all 3 regions

Relevant Equations:

(ħ[SUP]2[/SUP]/2m)((d[SUP]2[/SUP]Ψ)/(dx[SUP]2[/SUP]) + U(x)v = EΨ


∫ Ψ * Ψ dx = 1

I'm following Griffith's Modern Physics 2nd edition chapter 5.

I got to the part where we make Ψ_I(0) = Ψ_II(0) I get that

αCe^α(0) = QAsin(Q(0)) - QBsin(Q(0)) => C = QA/α

But when I try to graph it, the region I distribution doesn't seem to equal the region II distribution at 0.

The book goes on on to insert C into an equation to solve for G (-αGe^-α(L) for the other side of the well), but I don't understand why it does this.

-αGe^-α(L) = (αC/Qs)sin(Q(L)) + C/cos(Q(L))


Why does what happens on one side affect what happens on the other?

 

[vela]

Homework Statement:: There are two ledges with of U potential energy, and a cavern in between them of 0 potential energy. Such that

U(x) = { x <= 0, U, x > 0, x >= L, U }

Find the probability distribution for all 3 regions
Relevant Equations:: (ħ²/2m)((d²Ψ)/(dx²) + U(x)v = EΨ


∫ Ψ * Ψ dx = 1

I'm following Griffith's Modern Physics 2nd edition chapter 5.

I got to the part where we make Ψ_I(0) = Ψ_II(0) I get that

αCe^α(0) = QAsin(Q(0)) - QBsin(Q(0)) => C = QA/α

But when I try to graph it, the region I distribution doesn't seem to equal the region II distribution at 0.

Your graph doesn't correspond to your work above. Back up a bit and tell us what you're using for $\psi_I$ and $\psi_{II}$. And how did you get rid of $B$ to solve for $C$ from one equation?