- #1

gfd43tg

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## Homework Statement

## Homework Equations

## The Attempt at a Solution

for ##r \le a## and ##l = 0##, the radial equation is

$$- \frac {\hbar^{2}}{2m} \frac {d^{2}u}{dr^{2}} - V_{0} = Eu $$

$$- \frac {\hbar^{2}}{2m} \frac {d^{2}u}{dr^{2}} - [V_{0} + E]u = 0$$

call ##k^{2} = \frac {2m}{\hbar^{2}}[V_{0} + E]##

The solution to this differential equation comes out to be

$$u(r) = A sin(kr) + B cos(kr)$$

Since ##u = R(r)r##, as ##r \rightarrow 0##, then ##\frac {cos(kr)}{r} \rightarrow \infty##, therefore ##B = 0##

Leaving ##u(r) = Asin(kr)##

For ##r > a##

$$- \frac {\hbar^{2}}{2m} \frac {d^{2}u}{dr^{2}} = Eu$$

##\kappa^{2} = - \frac {2mE}{\hbar^{2}}##

therefore, ##u(r) = Ce^{-\kappa r} + De^{\kappa r}##

As ##r \rightarrow \infty##, then the second term blows up, so ##D = 0##

At the ##r = a##, the ##u(r)## must be the same for both sides. Therefore,

##A sin(ka) = Ce^{- \kappa a}##

Also, the function must be continuous, so the derivative must be equal at ##r = a##

## kA cos(ka) = - \kappa C e^{-\kappa a}##

Dividing through,

$$\frac {1}{k} tan(ka) = - \frac {1}{\kappa}$$

From here, how am I supposed to find the bound states? I have always been confused about bound and unbound states.