A particle moves in the potential $$V(r)=\left\{\begin{aligned}\infty\ ,\ &0\leq r\leq a\ , \ r\geq b\\ 0 \ , \ & a<r<b\end{aligned}\right.$$ with $l=1$. We desire the energy eigenvalues.

The radial solution is $$R(r)=Aj_1(kr)+Bn_1(kr)$$ via spherical bessel functions and the continuity conditions demand $$B=-A\frac{j_1(ka)}{n_1(ka)}=-A\frac{j_1(kb)}{n_1(kb)}$$ or after some trigonometry $$\frac{\sin x-x\cos x}{\cos x-x\sin x}=\frac{\sin y-y\cos y}{\cos y-y\sin y}\Rightarrow ...\Rightarrow\tan(y-x)=\frac{y-x}{1-xy}$$ where $x=ka\ , \ y=kb$ which is a transcendental equation that i never seen before...does it have solutions,seems so...? Is there a book that i can find a similar problem, all textbooks i know have the solution for $l=0$.

  • $\begingroup$ The identity $\tan(y-x)=\frac{\tan(y)-\tan(x)}{1+\tan(y)\tan(x)}$ seems useful. Except that the sign in the denominator is "wrong"; otherwise it would have just been the usual $\tan(x)=x$ eigenvalues. $\endgroup$ Mar 6, 2018 at 16:29
  • $\begingroup$ I'll try the trigonometry again..maybe you are right.. $\endgroup$
    – Anastasios
    Mar 6, 2018 at 16:34
  • $\begingroup$ Damn..you are right the denominator is $1+xy$...but i think now i have two conditions $\tan(ka)=ka$ and $\tan(kb)=kb$,can i combine them to one as $\tan(k(b-a))=k(b-a)$ is this right?..Sorry for the extra question?...Anyway thanks! $\endgroup$
    – Anastasios
    Mar 6, 2018 at 17:01
  • $\begingroup$ Of course,what am i saying?...Thanks Anders, sometimes we need those extra eyes from someone else! $\endgroup$
    – Anastasios
    Mar 6, 2018 at 17:20
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    $\begingroup$ the correct transcendental equation is $\tan\xi=\frac{\xi}{c\xi^2+1}\ \ , \ \ \xi=k(b-a)\ , \ c=\frac{ba}{(b-a)^2}$ $\endgroup$
    – Anastasios
    Mar 18, 2018 at 12:55

1 Answer 1


The last transcendental equation is $$\tan(y-x)=\frac{y-x}{1+xy}$$ setting $$\xi=y-x=k(b-a)\ \ ,\ \ c=\frac{ba}{(b−a)^2}$$ we get $$\tan\xi=\frac{\xi}{c\xi^2+1}$$ and plotting both sides we can find solutions.


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