Exam preparation question

In summary, when R=\pi/2, there are no bound states and the precise range of bound states is determined by the restriction -V_{0}<E<0.
  • #1
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Homework Statement



I was reviewing some homework problems and looking at the solutions. There is one problem with a tiny step I just cannot rationalize and I am hoping someone can point me in the right direction.

I have a spherical finite well:

[tex] V = {- V_{0}: 0 < r < a}[/tex],

[tex] = {0: r \geq a} [/tex]

[tex]- k_{2} = k_{1} cot (k_{1} a)[/tex] (1)

Refining the notation,

[tex]\alpha = a \sqrt{(2m(E + V_{0})}/hbar = k_{1} a[/tex]

[tex]R = a \sqrt{(2m(V_{0})}/hbar[/tex] and [tex]k_{2} = \sqrt{(2m(V_{0})}/hbar[/tex]

So (1) may be rewritten as [tex] \sqrt{R^{2} - \alpha^{2}} = - \alpha cot (\alpha)[/tex]

Homework Equations



From part 1.

The Attempt at a Solution



I don't understand how at [tex] R = \pi/2 [/tex] there are no bound states.

Also, I am given this restriction: [tex] -V_{0} < E < 0 [/tex]

How is this justified and how is the precise range of bound states determined?
 
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  • #2
When [itex]R=\pi/2[/itex], the only solutions are at [itex]\alpha=\pm\pi/2[/itex]. In these cases, you get k2=0.
 

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