1-D Schrödinger operators with Coulomb-like potentials

Function $Q$ of the form (1.2) near the origin is nonintegrable and therefore mapping $C^{\infty}_{0}(\mathbb{R})\ni\psi\mapsto\int_{\mathbb{R}}Q\psi\,dx$ is not a distribution. However we can find infinitely many functionals $q\in\mathcal{D}^{\prime}(\mathbb{R})$ which coincide with $Q$ outside the origin, i.e.,

Among such functionals there exists the family $\mathcal{F}_{Q}$ of distributions with the lowest order of singularity. It is easy to check that each $q\in\mathcal{F}_{Q}$ is continuous in space $C^{0,\gamma}_{0}(\mathbb{R})$ of Hölder continuous functions of compact support, but $q$ is not continuous in $C^{0}_{0}(\mathbb{R})$. In this sense, $q$ is more singular than Dirac’s $\delta$-function, but less singular than $\delta^{\prime}$-function. Moreover, if $q_{1}$ and $q_{2}$ belong to $\mathcal{F}_{Q}$, then $q_{2}-q_{1}=c\delta(x)$ for some complex constant $c$, and therefore $\mathcal{F}_{Q}=\{q_{1}+c\delta(x)\colon c\in\mathbb{C}\}$.

A word of explanation is necessary with regard to regularization of $Q$ given by (1.3). Here we use the analogy with formula $(\ln|x|)^{\prime}=\mathcal{P}\frac{1}{x}$. Suppose that $G$ is an antiderivative of $Q$ such that $G(x)=q_{-}\ln(-x)$ for $x\in(-a,0)$ and $G(x)=q_{+}\ln x$ for $x\in(0,a)$. Function $G$ specifies a regular distribution on the line, because it belongs to $L_{loc}^{1}(\mathbb{R})$. We set $g=G^{\prime}$, where $G^{\prime}$ is the derivative in the sense of distributions. Indeed, $g$ coincides with $Q$ outside the origin and $g\in\mathcal{F}_{Q}$. Let us now approximate $G$ in $\mathcal{D}^{\prime}(\mathbb{R})$ by the sequence of continuous functions

where $a$ is a $C^{1}$-function such that $a(-1)=q_{-}$ and $a(1)=q_{+}$ (see Fig. 1). Then distribution $g$ admits a regularization by $L_{loc}^{1}(\mathbb{R})$-functions having the form

The proof deals with the convergence of operators

and so we have the partial case of Theorem 1 when $U=0$. First of all, note that the trivial potential $U=0$ possesses a zero-energy resonance with half-bound state $h_{0}=1$. Since $\theta=1$, (1.7) and (1.8) become $\mu=\int_{\mathbb{R}}V\,dx=:\beta$ and

respectively. Hence, if the last condition holds, then operators $H_{\varepsilon}$, given by (2.1), converge to operator $\mathcal{H}$ associated with non-trivial point interactions (1.14), according to Theorem 1. In this case we obtain a partial transparency of the potentials $Q_{\varepsilon}+\varepsilon^{-1}V(\varepsilon^{-1}\,\cdot)$ in the limit. Next, in view of Lemma 1, the penetrability of these potentials is equivalent to the convergence of $Q_{\varepsilon}$ in the space of distributions.

Following are some examples of potentials $Q_{\varepsilon}$, illustrating the penetrability and impenetrability of the Coulomb-like potentials in the limit. Let us consider two regularizations of the classic Coulomb potential $Q(x)=-|x|^{-1}$:

(see Fig. 2). Both of the sequences $Q_{j,\varepsilon}$ converge to $-|x|^{-1}$ pointwise, but $Q_{1,\varepsilon}$ only converges in the sense of distributions. In the case of $Q_{1,\varepsilon}$, we have $q_{-}=1$, $q_{+}=-1$ and $\varkappa=-1$, and therefore condition (1.13) holds. In view of Theorem 2, potentials $Q_{0,\varepsilon}$ are asymptotically opaque; whereas $Q_{1,\varepsilon}$ are penetrable in the limit as $\varepsilon\to 0$. In other words, the transition probability $|T_{\varepsilon}(k)|^{2}$ calculated for $Q_{0,\varepsilon}$ tends to zero $\varepsilon\to 0$ for all $k$, but the corresponding probability for $Q_{1,\varepsilon}$ has a limit $|T(k)|^{2}$, which is a non-zero function of $k$.

For the odd Coulomb potential $Q(x)=x^{-1}$, we can also provide two different regularizations, plotted in Fig. 3, as follows:

In this case, $q_{-}=q_{+}=1$ and so condition (1.13) holds for potentials $Q_{3,\varepsilon}$ only, when $\varkappa(t)=-t$. Therefore $Q_{3,\varepsilon}$ are penetrable in the limit as $\varepsilon\to 0$, unlike the potentials $Q_{2,\varepsilon}$, which are asymptotically opaque.

Many authors have regularized the Coulomb potentials by so-called truncated ones of the form

where $|a_{\varepsilon}|\leq c\varepsilon^{-1}$ (see Fig. 4). From asymptotical point of view, $R_{\varepsilon}$ can be regarded as potentials $Q_{\varepsilon}$ with $\varkappa=0$. It follows from the proof of Lemma 1 that $R_{\varepsilon}$ can converge in $\mathcal{D}^{\prime}(\mathbb{R})$ if and only if $q_{-}=q_{+}$, i.e., $Q(x)$ is the odd Coulomb potential near the origin. Moshinsky [8] was the first who noticed the penetrability of potential $\gamma/x$. On the other hand, a regularization of the even Coulomb potential $\gamma/|x|$ by the truncated potentials is always asymptotically opaque and leads to the Dirichlet condition in the limit [1, 3, 2, 7]. The same assertion is also valid for modified Coulomb interactions having the form $M_{\varepsilon}(x)=-\dfrac{1}{|x|+\varepsilon}$; this potentials also diverge in $\mathcal{D}^{\prime}(\mathbb{R})$. Such regularizations were considered in [1, 2, 4, 5, 6].

It should be noted that the equivalence of penetrability in the limit and the convergence in the space of distributions for potentials $Q_{\varepsilon}+\varepsilon^{-1}V(\varepsilon^{-1}\,\cdot)$ is consistent with Kurasov’s results [12, 14]. Kurasov has interpreted the formal differential expression $-\frac{d^{2}}{dx^{2}}-\frac{\gamma}{x}$ in $\mathbb{R}$ as a map from some Hilbert space to the space of distributions. This operator has been defined in the principal value sense

on the whole line. As shown in [14], $H$ is the self-adjoint operator that is defined by $H\phi=-\phi^{\prime\prime}-\gamma\phi/x$ on functions from $W_{2}^{2}(\mathbb{R}\setminus(-\varepsilon,\varepsilon))$ for every positive $\varepsilon>0$ and satisfying the boundary conditions

where $b_{\pm}(\phi)=\lim_{x\to\pm 0}\big{(}\phi^{\prime}(x)-\gamma\phi(\pm 0)\ln|x|\big{)}$. Indeed, such considerations implicitly presupposed the existence of a regularization $Q_{\varepsilon}+\varepsilon^{-1}V(\varepsilon^{-1}\,\cdot)$ of the pseudopotential $-\mathcal{P}\frac{1}{x}+\beta\delta(x)$, which converges in $\mathcal{D}^{\prime}(\mathbb{R})$. These coupling conditions agree with (1.14), if $\gamma=q_{+}=q_{-}$.

Returning to the question of penetrability of the one-dimensional Coulomb potentials, it is probably worth considering that this question has no unambiguous answer. One should agree with the authors of [11] that mathematics alone cannot tell which boundary conditions for the wave function at the origin should be chosen to model a given experimental situation.

Let us consider the equation

for given $f\in L^{2}(\mathbb{R})$ and $\zeta\in\mathbb{C}$. In one-sided neighbourhoods of the origin the last equation becomes

The following proposition was proved in [14].

We will use this proposition for some formal considerations. To proof the norm resolvent convergence of $H_{\varepsilon}$ we do really need more subtle estimates of the remainder terms; a stronger version of these asymptotics is presented in Lemma 2. Set $y_{\varepsilon}=(H_{\varepsilon}-\zeta)^{-1}f$ for $f\in L^{2}(\mathbb{R})$ and $\zeta\in\mathbb{C}\setminus\mathbb{R}$. We look for the formal asymptotics of $y_{\varepsilon}$, as $\varepsilon\to 0$, having the form

We also assume that the coupling conditions

hold, where $[\,\cdot\,]_{x}$ is the jump of a function at point $x$. Function $y_{\varepsilon}$ is a unique solution of equation

belonging to the domain of $H_{\varepsilon}$. Since the interval on which the $(\alpha\delta^{\prime}+\beta\delta)$-like perturbation is localized shrinks to a point, $u$ must solve the equation

This solution can not be uniquely determined without additional conditions at the origin. One naturally expects that these conditions depend on the perturbation.

Suppose that $|x|<\varepsilon<a$. Then we can as follows rewrite equation (3.4) in the terms of new variable $t=x/\varepsilon$. If we set $v^{\varepsilon}(t)=y_{\varepsilon}(\varepsilon t)$, then

Furthermore, in view of Proposition 1, matching conditions (3.3) imply

In particular, we have

Substituting (3.2) for $|x|<\varepsilon$ into (3.5) and applying (3.6) yield

Let us first suppose that potential $U$ is resonant. Since the supports of $U$ is contained in $\mathcal{I}$, a half-bound state $h$ is constant outside $\mathcal{I}$ and its restriction to $\mathcal{I}$ is a non-trivial solution of the boundary value problem

Moreover $h(\pm\infty)=h(\pm 1)$ and hence $h_{0}(-1)=1$ and $h_{0}(1)=\theta$. Then the equation in (3.7) has a one-parameter family of solutions $v_{0}=ch_{0}$. But owing to $v_{0}(-1)=u(-0)$, we have

Hence $v_{0}(1)=u(-0)h_{0}(1)=\theta u(-0)$. On the other hand, $v_{0}(1)=u(+0)$ by (3.6). From this we deduce

Next, problem (3.8) is solvable if and only if

because the corresponding homogeneous problem possesses non-trivial solutions. The last condition can be easy obtained by multiplying the equation in (3.8) by $h_{0}$ and integrating by parts. Combining (3.12) and (3.13) gives us

The linear system admits a nonzero solution $(u(-0),u(+0))$ if and only if

If (3.13) holds, then (3.8) has a one-parameter family of solutions $v_{1}=v_{1}^{*}+c_{1}h_{0}$. Let us fix $v_{1}$ such that $v_{1}(-1)=0$; this is possible, because $h_{0}(-1)\neq 0$.

We at last turn to problem (3.9). Multiplying the equation in (3.9) by half-bound state $h_{0}$ and integrating by parts, we can similarly compute the solvability condition

We can choose $v_{2}$ to satisfy $v_{2}(-1)=0$. Recalling now (3.11), we can rewrite (3.16) in the form

Therefore if potential $U$ is resonant and (3.15) holds, then the leading term $u$ of asymptotics (3.2) must solve the problem

where $\mu$ is given by (1.7). The coupling conditions at the origin agree with (1.9) in view of Remark 2.

In the case when either $U$ has no zero-energy resonance or else $U$ is resonant, but (3.15) does not hold, both the values $u(-0)$ and $u(+0)$ equal zero. Indeed, if $U$ is not resonant, then problem (3.7) has only trivial solution $v_{0}$ and then the first condition in (3.6) implies $u(0)=0$. On the other hand, if $U$ is resonant, but (3.15) does not hold, then system (3.14) has a unique solution $u(-0)=u(+0)=0$. Hence $u$ should be a solution of the problem