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I'm trying to learn how to apply the WKB approximation. Given the following problem:

An electron, say, in the nuclear potential $$U(r)=\begin{cases} & -U_{0} \;\;\;\;\;\;\text{ if } r < r_{0} \\ & k/r \;\;\;\;\;\;\;\;\text{ if } r > r_{0} \end{cases}$$ 1. What is the radial Schrödinger equation for the $\ell=0$ state?

2. Assuming the energy of the barrier (i.e. $k/r_{0}$) to be high, how do you use the WKB approximation to estimate the bound state energies inside the well?


For the first question, I thought the radial part of the equation of motion was the following

$$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +{\hbar^2 \ell(\ell+1)\over 2mr^2}+V(r) \right \} R(r)=ER(r)$$

Do I simply just let $\ell=0$ and obtain the following? Which potential do I use?

$$\left \{ - {\hbar^2 \over 2m r^2} {d\over dr}\left(r^2{d\over dr}\right) +V(r) \right \} R(r)=ER(r)$$

For the other question, do I use $\int \sqrt{2m(E-V(r))}=(n+1/2)\hbar π$, where $n=0,1,2,...$ ? If so, what are the turning points? And again, which of the two potentials do I use?

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  1. For first question, you must simly take $l = 0$(in your notations, $l$ is curved). It's because angular momentum is conserved in radial-symmetry field problem, and you can simply take $L^2 = \hbar^2 l(l+1)$. So, for $l = 0$ states you simply ommit $\frac{L^2}{2mr^2}$ term.
    Potential is given in problem. It behaves like constant for some $r$ values, and like $\frac{1}{r}$ otherwise. Point is that Schrodinger equation says that in first region ($r<r_0$) wave-function satisfies equation with $U= -U_0$ potential, and in region of $r>r_0$ it satisfies same equation, but with respective potential $U = k/r$. In addition, radial $R$-function is continious, which means that solution of equations in two regions must "fit" to get physical-relevant solution.

  2. Yes, you should use $\int \sqrt{2m(E-U_{eff}(r))} dr = \pi \hbar (n+1/2)$. For arbitrary $l=0,1,2$, $U_{eff} = U(r) + \frac{\hbar^2 l(l+1)}{2mr^2}$.For Wentzel-Krammers-Brilloin approximation in this problem, we expect well to be high, so there are some bound states concentrated in region $0 < r < r_0$. Of course, real states must satisfy exact Schrodinger equation, that suggest that wave-function is nonzero in $r>r_0$ region, but for some large potentials, our approximation is physically correct. So you should take only $U = -U_0$(which is exact for $r<r_0$ region) and halt points $r = r_0$ and $r=0$. First halt point is argumented physically above. Why second is $r = 0$ is explained in quantum mechanics textbooks with WKB method explanation(for central field problem), likewise $\int \sqrt{2m(E-U_{eff}(r))} dr = \pi \hbar (n+1/2)$ equation.

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  • $\begingroup$ Thanks. For #2, do you mean that the form that you solve is $\int_{0}^{r_{0}} \sqrt{2m(E+U_{0})} dr = \pi \hbar (n+1/2)$ where $U_{eff}=-U_{0}+\frac{\hbar^{2}\ell (\ell+1)}{2mr^{2}} = -U_{0}$ since $\ell = 0$? Also, it seems strange that the Coulomb potential isn't used here at all. $\endgroup$ – user2561523 Dec 15 '13 at 3:25
  • $\begingroup$ Yes, I think so(with minus sign in the expression $E-U$, of course. Physical reason is that this phenomenological nuclear physics problem. Wave-function correspond to $\alpha$-particle, that has positive charge sign. Nucleus has same sign of charge, therefore Coulomb potential here is repulsive! But, there some extra nuclear forces. We can approximate them by simple $-U_0$ potential well. They attract particle to nucleus. $\endgroup$ – Vladimir Dec 15 '13 at 9:24

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