Bound states in finite spherical well

[gfd43tg]

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.

 

[Noctisdark]

Hi Maylis,
Try to expand tan(ka)
Note that the ground state have E~0
but no zero, we can give it a great approximation !

 

[PeroK]

Homework Statement

View attachment 83427

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.

Bound states are defined by E < 0, which you assumed to get the solution outside the well. So, this is the solution for bound states you have.

To solve the transcendental equation in $tan(ka)$ you need a bit of jiggery-pokery. See the earlier chapter on the 1D finite well for how it's done. Let $z = ka$ etc.