An approximation to the Woods–Saxon potential based on a contact interaction

Abstract

We study a non-relativistic particle subject to a three-dimensional spherical potential consisting of a finite well and a radial $\delta -\delta '$ contact interaction at the well edge. This contact potential is defined by appropriate matching conditions for the radial functions, thereby fixing a self-adjoint extension of the non-singular Hamiltonian. Since this model admits exact solutions for the wave function, we are able to characterize and calculate the number of bound states. We also extend some well-known properties of certain spherically symmetric potentials and describe the resonances, defined as unstable quantum states. Based on the Woods–Saxon potential, this configuration is implemented as a first approximation for a mean-field nuclear model. The results derived are tested with experimental and numerical data in the double magic nuclei $^{132}$Sn and $^{208}$Pb with an extra neutron.

Appendices

On the self-adjointness of the Hamiltonian

The goal of the present appendix is a discussion on the self adjointness of the radial Hamiltonian (15). Setting the appropriate units such that $\hbar =1$ and $2 \mu =1$ this Hamiltonian may be written as

$$\begin{aligned} H=-\frac{\mathrm{d}^2}{\mathrm{d}r^2}+ [\theta (r-R)-1]V_0+a\delta (r-R) +b\delta '(r-R)+\frac{\ell (\ell +1)}{r^2}, \end{aligned}$$

(A.1)

$\ell \in {\mathbb {N}}_0$. Let us split it into $H=H_\ell +V(r)$, where

$$\begin{aligned} H_\ell =-\frac{\mathrm{d}^2}{\mathrm{d}r^2}+\frac{\ell (\ell +1)}{r^2}. \end{aligned}$$

(A.2)

For the sake of clarity, we first study $H_{\ell =0},$ which reduces to the one-dimensional Laplace operator in a given domain.

1.1 Zero angular momentum

We have to find a domain for $H_0$, which must be a subspace of $L^2[0,\infty )$. This domain must include all square integrable absolutely continuous functions, f(r), with absolutely continuous derivative and square integrable second derivative. Thus,

$$\begin{aligned} \int _0^\infty \{|f(r)|^2+|f''(r)|^2\}\,\mathrm{d}r<\infty . \end{aligned}$$

(A.3)

The boundary conditions at the origin should be specified in such a way that $H_0$ is Hermitian on its domain. In consequence, for any f(r), g(r) in the domain of $H_0$,

$$\begin{aligned} \langle h(r) | H_0 \, f(r) \rangle= & {} -\int _0^\infty h^*(r)\,f''(r)\,\mathrm{d}r =h^*(0)\,f'(0) - {h'}^*(0)\,f(0) - \int _0^\infty {h''}^*(r)\,f(r)\,\mathrm{d}r \nonumber \\= & {} h^*(0)\, f'(0) - {h'}^*(0)\, f(0) +\langle H_0 \, h(r) |f(r) \rangle . \end{aligned}$$

(A.4)

Then, $H_0$ is Hermitian in the given domain if, and only if, $h^*(0)\,f'(0)-{h'}^*(0)\,f(0)=0$, which happens if, and only if, $f(0)=cf'(0)$ for any function f(r) in this domain, where c is an arbitrary real constant. For $c=0$, we have that $f(0)=0$ with $f'(0)$ arbitrary. Since $c^{-1}f(0)=f'(0)$, another possible choice is $f'(0)=0$ with f(0) arbitrary. Here, we may say that $c=\infty$. All these possible choices select a domain, $\mathcal D$, in which $H_0$ is self-adjoint. We select any one of them.

After selecting a value of $c\in {\mathbb {R}}\cup \{\infty \}$, let us consider a subspace of ${\mathcal {D}}$, denoted by ${\mathcal {D}}(H_0)$. By definition, $f(r) \in {\mathcal {D}}(H_0)$ if, and only if,, $f(R)=f'(R)=0$. Choosing ${\mathcal {D}}(H_0)$ as the domain of $H_0$, we see that $H_0$ is symmetric (Hermitian), although not self adjoint, having deficiency indices (2, 2).

In order to prove this latter statement, let us recall that the domain of the adjoint $H_0^\dagger$ is determined by

$$\begin{aligned} {{\mathcal {D}}}(H_0^\dagger )= \{ h(r)\in L^2[0,\infty );\; \exists \,g(r)\in L^2[0,\infty ); \langle h(r)|H_0\,f(r)\rangle =\langle g(r)|f(r)\rangle \}, \nonumber \\ \end{aligned}$$

(A.5)

for all f(r) in ${\mathcal {D}}(H_0)$. To obtain a basis of the deficiency subspaces [66], we have to solve the equations $h''(r)=\pm ih(r)$, where the solutions must be in $\mathcal D(H_0^\dagger )$. Let us choose the sign plus first. We obtain two linearly independent solutions, which are:

$$\begin{aligned} h_1(r)= & {} \left\{ \begin{array}{lll} C\,\mathrm{e}^{-\frac{\sqrt{2}}{2}\,r} \,\mathrm{e}^{-i\frac{\sqrt{2}}{2}\,r}, &{} &{} \quad r>R, \\ 0, &{} &{}\quad r<R, \end{array} \right. \\ h_2(r)= & {} \left\{ \begin{array}{ccc} 0, &{} &{}\quad r>R,\\ A\, \mathrm{e}^{-\frac{\sqrt{2}}{2}\,r} \,\mathrm{e}^{-i\frac{\sqrt{2}}{2}\,r}+ \frac{A(1-c)}{1+c}\, \mathrm{e}^{\frac{\sqrt{2}}{2}\,r} \,\mathrm{e}^{i\frac{\sqrt{2}}{2}\,r}, &{} &{}\quad r<R, \end{array} \right. \end{aligned}$$

where A and C are arbitrary constants. The linear independence of these two functions is obvious, so that they are a basis for the deficiency subspace corresponding to the plus sign. Similar analysis can be performed for the minus sign. This proves that the deficiency indices for $H_0$ with domain ${\mathcal {D}}(H_0)$ are precisely (2, 2). In this circumstance, $H_0$ admits an infinite number of self-adjoint extensions labelled by four independent real parameters. Domains for these self-adjoint extensions are determined by matching conditions at the point $r=R$ as usual [12, 13], where the exceptional cases $\beta =\pm 2$ are also included. The choice of the matching conditions (26) gives a two parametric family of self-adjoint extensions, which proves the self-adjointness of

$$\begin{aligned} H_r=-\frac{\mathrm{d}^2}{\mathrm{d}r^2}+ a\delta (r-R) +b\delta '(r-R), \end{aligned}$$

(A.6)

which is (A.1) with $\ell =0$ and without the term $V_0\,[\theta (r-R)-1]$. As we will explain at the end of the present appendix, adding this term to the potential does not change the self adjointness.

1.2 Higher angular momentum

For $\ell \ge 1$, we do not need to impose conditions at the origin of the type $f(0)=cf'(0)$, as the Hamiltonian (A.2) is essentially self-adjoint when its domain is given by the Schwartz space [67] supported on ${\mathbb {R}}^+:=[0,\infty )$. For these functions $f(0)=f'(0)=0$, so that $h^*(0)\,f'(0)-{h'}^*(0)\,f(0)$ is automatically zero. Then, we define ${\mathcal {D}}(H_\ell )$, $\ell \ne 0$, to be the space of functions $f(r)\in L^2[0,\infty )$ satisfying the following conditions [49]:

• 1. f(r) and $f'(r)$ are absolutely continuous.

• 2. $-f''(r)+[\ell (\ell +1)/r^2]f(r)$ is square integrable, i.e. it belongs to $L^2[0,\infty )$.

• 3. $f(0)=0$.

• 4. $f(R)=f'(R)=0$.

In order to obtain the deficiency subspaces for $H_\ell$, we have to find the square integrable solutions of the following pair of differential equations:

$$\begin{aligned} h''(r)-\frac{\ell (\ell +1)}{r^2}\,h(r)\mp ih(r)=0. \end{aligned}$$

(A.7)

For the minus sign in (A.7), the general solution is given by [49] (p. 478):

$$\begin{aligned} u(r)=A\,r^{1/2}\,J_{\ell +1/2}(r\sqrt{i})+B\,r^{1/2}\, Y_{\ell +1/2}(r\sqrt{i}), \end{aligned}$$

(A.8)

where $J_{\ell +1/2}$ and $Y_{\ell +1/2}$ are the Bessel and Neumann functions [68], respectively. Asymptotic properties of these functions show that [49, 68]:

$$\begin{aligned} r^{1/2}\,J_{\ell +1/2}(r\sqrt{i})\notin L^2[1,\infty ),&\quad&r^{1/2}\,J_{\ell +1/2}(r\sqrt{i}) \in L^2[0,1], \\ r^{1/2}\, Y_{\ell +1/2}(r\sqrt{i}) \in L^2[1,\infty ),&\quad&r^{1/2}\, Y_{\ell +1/2}(r\sqrt{i}) \notin L^2[0,1] . \end{aligned}$$

Therefore, the basis for the deficiency subspace with minus sign in (A.7) is given by the following pair of functions:

$$\begin{aligned} u_1(r)= & {} \left\{ \begin{array}{ccc} A\, r^{1/2}\, Y_{\ell +1/2}(r\sqrt{i}), &{} \mathrm{if} &{}\quad r>R, \\ 0, &{}\quad \mathrm{if} &{} r<R , \end{array} \right. \qquad \\ u_2(r)= & {} \left\{ \begin{array}{ccc} 0, &{}\quad \mathrm{if } &{}\quad r>R, \\ B\,r^{1/2}\,J_{\ell +1/2}(r\sqrt{i}), &{}\quad \mathrm{if} &{}\quad r<R, \end{array} \right. \end{aligned}$$

where A and B are constants. A similar result can be obtained for the plus sign in (A.7), so that the deficiency indices for $H_\ell$ with $\ell \ne 0$ are (2, 2). Self-adjoint extensions are obtained by suitable matching conditions at $r=R$ and depend on four real parameters. Again, the choice of matching conditions (26), where the exceptional cases $\beta =\pm 2$ are included, determines a self-adjoint Hamiltonian of the form,

$$\begin{aligned} H_r:=-\frac{\mathrm{d}^2}{\mathrm{d}r^2} +a\delta (r-R) +b\delta '(r-R)+\frac{\ell (\ell +1)}{r^2},\quad \ell \ne 0, \end{aligned}$$

(A.9)

which is (A.1) without the term $V_0\, [\theta (r-R)-1]$. Adding this term does not change anything in both cases ($\ell =0$ and $\ell \ne 0$). Once we have determined the domains for which (A.6) and (A.9) are self-adjoint, since the term $V_0\, [\theta (r-R)-1]$ is bounded and Hermitian, it is self-adjoint. Now, the Kato–Rellich theorem [66] says that if $H_r$ is self adjoint, so is $H_r+[\theta (r-R)-1]V_0$. We conclude that it is possible to determine domains such that (A.1) is self-adjoint for all values $\ell \in {\mathbb {N}}_0$.

Proofs of Theorems 1, 2 and 3

1.1 Proof of Theorem 1

In the first place, we show that the right-hand side of (27) is positive and strictly growing as a function of the relative energy. As will be proved in Theorem 2, there exists an upper bound for the angular momentum, $\ell _\text {max}$; hence, the term $8\beta (\ell +1)$ is always finite. From Theorem 6 in [69], there exits the following bounds for the following ratio of modified Bessel functions:

$$\begin{aligned} \sqrt{\sigma ^2+\ell ^2} +\ell +1 \le \frac{\sigma \,K_{\ell +3/2}(\sigma )}{K_{\ell +1/2}(\sigma )}< \sqrt{\sigma ^2+(\ell +1)^2}+\ell +1. \end{aligned}$$

(B.1)

Now, we can use the first inequality of (B.1) together with (31) to derive:

$$\begin{aligned} \phi _\ell (\sigma )= & {} \frac{(2+\beta )^2}{(2-\beta )^2}\, \frac{\sigma \,K_{\ell +3/2}(\sigma )}{K_{\ell +1/2}(\sigma )} - \frac{8\beta (\ell +1)}{(2-\beta )^2}+\frac{w_{\ell j}}{(2-\beta )^2} \\> & {} \frac{(\beta -2)^2+2 \ell \left( \beta ^2+4\right) +w_{\ell j}}{(\beta -2)^2} >0. \end{aligned}$$

In addition, using the Turan-type inequalities given in [70], we can prove the following relation:

$$\begin{aligned} \frac{\mathrm{d} \phi _\ell (\sigma )}{\mathrm{d}\sigma }= \frac{\sigma \,K_{\ell -1/2}(\sigma ) \,K_{\ell +3/2}(\sigma ) -\sigma \, K^2_{\ell +1/2}(\sigma )}{K^2_{\ell +1/2}(\sigma )} >0. \end{aligned}$$

(B.2)

This shows that $\phi _\ell (\sigma )$ is a strictly growing positive function on the variable $\sigma$ and, due to the definition of $\sigma (\varepsilon )$ (28), on $\varepsilon$.

On the other hand, if we show that $\varphi _\ell (\chi )$, the left-hand side of (27), is one to one and onto as a function between the following intervals:

$$\begin{aligned} \varphi _\ell (\chi ) : (j_{\ell +3/2,s-1}, j_{\ell +1/2,s})\subset (0,v_0) \mapsto (0,\infty ), \ s\in {\mathbb {N}}, \end{aligned}$$

(B.3)

we guarantee the unique existence of the bound state $\varepsilon _s$ in (32). Thus, it will be enough to demonstrate that $\varphi _\ell (\chi )$ is strictly monotonic on $\chi$ and that it covers the whole $(0,\infty )$ as $\chi \in (j_{\ell +3/2,s-1}, j_{\ell +1/2,s})$. In fact, the first derivative of $\varphi _\ell (\chi )$ meets

$$\begin{aligned} \frac{\mathrm{d}\varphi _\ell (\chi ) }{\mathrm{d}\chi }\, = \chi \; \frac{J^2_{\ell +1/2}(\chi ) - J_{\ell -1/2}(\chi )\,J_{\ell +3/2}(\chi )}{J^2_{\ell +1/2}(\chi )}>0, \end{aligned}$$

where the equality follows from standard properties of the Bessel functions [59] and the second relation from the Turan-type inequalities [50, 70]:

$$\begin{aligned} J^2_n(\chi )-J_{n-1}(\chi )\,J_{n+1}(\chi )> \frac{J^2_n(\chi )}{n+1}>0, \end{aligned}$$

(B.4)

where $\chi$ is real and $n>0$, with $n=\ell +1/2$. Finally, in the given intervals, the function $\varphi _\ell (\chi )$ is positive and

$$\begin{aligned} \lim _{\chi \rightarrow j_{\ell +3/2,s-1}} \varphi _\ell (\chi )=\varphi _\ell (j_{\ell +3/2,s-1})=0,\qquad \lim _{\chi \rightarrow j_{\ell +1/2,s}^-} \varphi _\ell (\chi ) =\infty . \end{aligned}$$

(B.5)

Now, we focus on the second part of the theorem concerning the number of bound states (33). The key feature is the existence of an integer $s_0$ for which

$$\begin{aligned} (j_{\ell +3/2,s-1}, j_{\ell +1/2,s}) \cap (0,v_0)=\emptyset , \ \forall s\ge s_0. \end{aligned}$$

Let us examine this in greater detail. The largest integer M for which $(j_{\ell +3/2,M-1}, j_{\ell +1/2,M})\subset (0,v_0)$ still holds is obviously given by

$$\begin{aligned} j_{\ell +1/2,M}< v_0,\qquad j_{\ell +1/2,M+1}> v_0. \end{aligned}$$

(B.6)

Since $\varphi _\ell (\chi (\varepsilon ))$ is strictly decreasing and $\phi _\ell (\sigma (\varepsilon ))$ strictly increasing as functions of $\varepsilon$, the condition

$$\begin{aligned} \varphi _\ell (\chi (\varepsilon ))= & {} \varphi _\ell (v_0)={v_0 J_{\ell +\frac{3}{2}}(v_0 )}/{J_{\ell +\frac{1}{2}}(v_0 )}\\> & {} \dfrac{(\beta -2)^2+2 \ell \left( \beta ^2+4\right) +w_{\ell j}}{(\beta -2)^2}=\underset{\varepsilon \rightarrow 0^+}{\lim }\phi _\ell (\sigma (\varepsilon )). \end{aligned}$$

implies the existence of an additional bound state whose energy is the closest to $\varepsilon =0$. In the particular case $v_0=j_{\ell +1/2,M}$, no additional bound state should be added to M. Independently, if

$$\begin{aligned} \underset{\varepsilon \rightarrow 1^-}{\lim }\varphi _\ell (\chi (\varepsilon ))=0 >\underset{\varepsilon \rightarrow 1^-}{\lim }\phi _\ell (\sigma (\varepsilon )) \end{aligned}$$

(B.7)

no bound state satisfying $\varepsilon \in (1-j^2_{\ell +1/2,1}/v_0^2, 1)$ appears. Only if $j_{\ell +1/2,1}>v_0$ the functions m and $m'$ defined in the present theorem are not independent. Nevertheless, a bound state with relative energy $\varepsilon \in (0,1)$ appears if, and only if, $\varphi _\ell (v_0)>\underset{\varepsilon \rightarrow 0^+}{\lim }\phi _\ell (\sigma )$ and $0 <\underset{\varepsilon \rightarrow 1^-}{\lim }\phi _\ell (\sigma )$ so that $n_\ell$ is also given by (33) and the proof is concluded.

1.2 Proof of Theorem 2

For this proof, we use the number of bound states given in Eq. (33) of Theorem 1 in order to obtain

$$\begin{aligned} n_\ell =0 \qquad \forall \ \ell >\ell _{\max }. \end{aligned}$$

(B.8)

Note that in the derivation of (33), we have not used the assumption of the existence of $\ell _\text {max}$ that appears at the beginning of “Appendix B.1” so there is no circular reasoning. Due to the properties of the zeros of the Bessel function (30) and their asymptotic expressions for large order [59], there exists an integer $\ell _0$ such that

$$\begin{aligned} j_{\ell +1/2,1}>v_0\, , \qquad \ell \ge \ell _0, \end{aligned}$$

(B.9)

and therefore $M=0$. Eventually, we shall reach a value $\ell _\text {max}\ge \ell _0$ such that

$$\begin{aligned} \underset{\varepsilon \rightarrow 0^+}{\lim }\varphi _\ell (\chi (\varepsilon ))= & {} \varphi _\ell (v_0)={v_0 J_{\ell +\frac{3}{2}}(v_0 )}/{J_{\ell +\frac{1}{2}}(v_0 )}\\< & {} \frac{(\beta -2)^2+2 \ell \left( \beta ^2+4\right) +w_{\ell j}}{(\beta -2)^2}=\underset{\varepsilon \rightarrow 0^+}{\lim }\phi _\ell (\sigma (\varepsilon )), \end{aligned}$$

for all $\ell >\ell _{\max }$, hence $m=m'=0$. In effect, the existence of $\ell _\text {max}$ is a consequence of the dependence on the angular momentum of both sides in the previous inequality. It is clear that the right-hand side is a strictly increasing function with respect to $\ell$. In addition, using Theorem 3 of [71] it can be easily proved that the left-hand side fulfils

$$\begin{aligned} \dfrac{\partial \varphi _\ell (v_0)}{\partial \ell } \le 0. \end{aligned}$$

(B.10)

In consequence, if none of the conditions $j_{\ell +1/2,1}< v_0$, $\varphi _\ell (v_0)>\phi _\ell (0^+)$ hold, there is no bound state.

In order to complete the proof, we should consider a configuration in which an integer $s_0 \in {\mathbb {N}}$ such that $v_0= j_{\ell +1/2,s_0}$ exists. In such a case, the condition $\varphi _\ell (v_0)>\phi _\ell (0^+)$ is ill defined. Nevertheless, if $s_0>1$ we know the existence of at least one bound state. Thus, we have to consider the next value of the angular momentum for which $v_0\ne j_{\ell +1+1/2,s_0}$. If $s_0=1$, the bound state always exists, although if $\phi _\ell (v_0)<0$, its energy can be below $-V_0$. Thus, we have to consider the next value of the angular momentum for which $v_0\ne j_{\ell +1+1/2,1}$.

1.3 Appendix B.3 Proof of Theorem 3

Throughout the proof, we bear in mind the monotonicity properties of $\varphi _\ell (\varepsilon )$ and $\phi _\ell (\varepsilon )$ with respect to $\varepsilon$ demonstrated in appendix B.1.

• (a) The inequality (a) in (37) is just a consequence of $j_{\lambda ,i}<j_{\lambda ,i+1}$ given in (30).

• (b) To prove the inequality (b) in (37), we first take into account that the bound states characterized by n are determined, for a given $\ell$, by the function $\varphi _\ell (\varepsilon )$ restricted to the interval $(a_{n\ell },b_{n\ell })$, where

  $$\begin{aligned} a_{n\ell }=1- \frac{j^2_{\ell +1/2,n+1}}{v_0^2} , \quad b_{n\ell }=1- \frac{j^2_{\ell +1/2,n}}{v_0^2} , \quad n \in {\mathbb {N}}_0. \end{aligned}$$

  We need to consider both functions $\varphi _\ell (\varepsilon )$, $\varphi _{\ell +1} (\varepsilon )$, and therefore both intervals $(a_{n\ell },b_{n\ell })$, and $(a_{n(\ell +1)},b_{n(\ell +1)})$. Due to the properties of the zeros $j_{\ell +1/2,n }$ given in (30), the following relations are fulfilled:

  $$\begin{aligned} a_{n(\ell +1)}< a_{n\ell }, \qquad b_{n(\ell +1)}< b_{n\ell }. \end{aligned}$$

  Therefore, either (i) $b_{n(\ell +1)}\le a_{n\ell }$, or (ii) $a_{n\ell }<b_{n(\ell +1)}$.

  • $\bullet$ If (i) is true, $(a_{n(\ell +1)},b_{n(\ell +1)}) \cap (a_{n\ell },b_{n\ell })=\emptyset$ so $-\varepsilon _{n\ell _{j}}<-\varepsilon _{n(\ell +1)_{j}}$ holds trivially.

  • $\bullet$ If (ii) is true, then we have three disjoint intervals: $(a_{n(\ell +1)},a_{n\ell })$, $(a_{n\ell },b_{n(\ell +1)})$ and $(b_{n(\ell +1)},b_{n\ell })$. If $\varepsilon _{n\ell _{j}}\in (b_{n(\ell +1)},b_{n\ell })$ or $\varepsilon _{n(\ell +1)_{j}}\in (a_{n(\ell +1)},a_{n\ell })$, then it is obvious that $-\varepsilon _{n\ell _{j}}<-\varepsilon _{n(\ell +1)_{j}}$. However, if $\varepsilon _{n\ell _{j}}, \varepsilon _{n(\ell +1)_{j}}\in (a_{n\ell },b_{n(\ell +1)})$, the situation needs to be studied in detail. Let us prove first:

    $$\begin{aligned} \varphi _{\ell }(\varepsilon _0)\ge \varphi _{\ell +1}(\varepsilon _0),\quad \phi _{\ell }(\varepsilon _0)<\phi _{\ell +1}(\varepsilon _0), \end{aligned}$$

    (B.11)

    $\varepsilon _0\in (a_{n\ell },b_{n(\ell +1)})$. The first part results from (B.10), since it holds for $v_0\in {\mathbb {R}}$, excluding the singularities of $\varphi _\ell$ [71]. For the second case, the bounds in (B.1) ensure

    $$\begin{aligned}&\frac{\sigma \,K_{(\ell +1)+3/2}(\sigma )}{K_{(\ell +1)+1/2}(\sigma )}-\frac{\sigma \,K_{\ell +3/2}(\sigma )}{K_{\ell +1/2}(\sigma )} > 1. \end{aligned}$$

    Consequently, using (27), we reach

    $$\begin{aligned} \phi _{\ell +1}(\varepsilon _0) - \phi _{\ell }(\varepsilon _0) > 1+\frac{w_{(\ell +1) j}-w_{\ell j}}{(2-\beta )^2}. \end{aligned}$$

    (B.12)

    In addition, for $\ell +1$, $j=(\ell +1)-1/2$ so, bearing in mind (8), the parameter $w_{(\ell +1) j}>0$. In a similar way, for $\ell$, $j=\ell +1/2$ and $w_{\ell j}<0$. Consequently, the second inequality in (B.11) is proved.

    Now, we may prove $\varepsilon _{n(\ell +1)_j} < \varepsilon _{n\ell _j}$ by contradiction. Let us assume $\varepsilon _{n(\ell +1)_j} \ge \varepsilon _{n\ell _j}$. With (B.11) and the monotonicity with respect to $\varepsilon$ above-mentioned, we find

    $$\begin{aligned} \phi _\ell (\varepsilon _{n\ell _j})\le & {} \phi _\ell (\varepsilon _{n(\ell +1)_j}) < \phi _{\ell +1}(\varepsilon _{n(\ell +1)_j}) =\varphi _{\ell +1}(\varepsilon _{n(\ell +1)_j}) \\\le & {} \varphi _{\ell +1}(\varepsilon _{n\ell _j})\le \varphi _{\ell }(\varepsilon _{n\ell _j}). \end{aligned}$$

    From here, it follows $\varphi _\ell (\varepsilon _{n\ell _j})\ne \phi _{\ell }(\varepsilon _{n\ell _j})$, which is clearly absurd because $\varepsilon _{n\ell _j}$ is a bound state, and the equality, (27), must be satisfied.

• (c) The inequality (c) in (37) is proved taking into account that the only dependence on j in the secular equation (27) is through $w_{\ell j}$. As we have already pointed out, $w_{\ell (\ell -1/2)}>0 >w_{\ell (\ell +1/2)}$. In consequence, $\phi _\ell$ is greater for $j=\ell -1/2$. Since $\varphi _\ell$ is independent of j, we only need to consider the interval $(a_{n\ell },b_{n\ell })$, for which the inequality is proved, as has been done before, by contradiction.