# Finding the Boundary Conditions of a Potential Well

1. Oct 7, 2009

### jmtome2

1. The problem statement, all variables and given/known data

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?

2. Relevant 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 -$$\alpha$$$$^{2}$$=$$\frac{2mE}{h(bar)^{2}}$$...

3. 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?

Last edited: Oct 7, 2009
2. Oct 9, 2009

### gabbagabbahey

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

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

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

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

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

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

?

3. Oct 9, 2009

### jmtome2

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

4. Oct 9, 2009

### 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 $\tan(2qa)=$___?

5. Oct 10, 2009

### jmtome2

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

6. Oct 10, 2009

### gabbagabbahey

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

7. Oct 10, 2009

### 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