Finding the Boundary Conditions of a Potential Well

[jmtome2]

Homework Statement

Show that the conditions for a bound state, Eqn1 and Eqn2, can be obtained by requiring the vanquishing of the denominators in Eqn3 at k=i\kappa. Can you give the argument for why this is not an accident?

Homework Equations

Eqn1: \alpha=q*tan(qa)
Eqn2: \alpha=-q*cot(qa)

Eqn3:

R= i*e^{-2ika}\frac{(q^{2}-k^{2})*sin(2qa)}{2kq*cos(2qa)-i(q^{2}+k^{2})*sin2qa}
T=e^{-2ika}\frac{2kq}{2kq*cos(2qa)-i(q^{2}+k^{2})*sin2qa}

The problem concerns a potential well of V_{0} from -a to a.
I believe that k^{2}=\frac{2mE}{h(bar)^{2}} and that \kappa^{2}=-\frac{2m(E-V_{0})}{h(bar)^{2}}

I also know that in order to reach these boundaries through another method in the book, they let -\alpha^{2}=\frac{2mE}{h(bar)^{2}}...

The Attempt at a Solution

I removed the demoninators from Eqn3 and tried to substitute k=\sqrt{\frac{2mE}{h(bar)^{2}}} into Eqn3... but not only was this result messy but I couldn't get the boundary conditions, Eqn1 and Eqn2 to come out. Help?

 

[gabbagabbahey]

I'm not 100% clear on your problem statement...First, what is q? Second are these the equations you meant to write:

R= ie^{-2ika}\frac{(q^{2}-k^{2})\sin(2qa)}{2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)}

T=e^{-2ika}\frac{2kq}{2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)}

k=\left(\frac{2mE}{\hbar^2}\right)^{1/2}

\kappa=\left(\frac{2m(V_0-E)}{\hbar^2}\right)^{1/2}

\alpha=\left(-\frac{2mE}{\hbar^2}\right)^{1/2}

?

 

[jmtome2]

<br /> q=\left(\frac{2m(E+V_0)}{\hbar^2}\right)^{1/2}<br />

 

[gabbagabbahey]

Okay, let's look at your eqn 3..."the vanquishing of the denominators" is just a fancy way of saying that 2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)=0...so \tan(2qa)=___?

 

[jmtome2]

=\left(\frac{sin(2qa)}{cos(2qa)}\right)?

 

[gabbagabbahey]

Well of course, but what is that equal to when 2kq\cos(2qa)-i(q^{2}+k^{2})\sin(2qa)=0?

 

[jmtome2]

Alright, I'll take a closer look at this at a later time to make sure that I understand it but I think the missing link is that I initially overlooked q and \alpha. I'm sure that the solution will be clear once I make it through the alegbra, thanks