Applying this simplified bound on $\beta$ to Lemma 3 gives a much-simplified
remainder bound.

$\displaystyle\norm{R_{n}(t_{0}+\Delta
t,t_{0})}<\left(4e\max_{\tau\in[t_{0},t_{0}+\Delta t]}\Lambda(\tau)\Delta
t\right)^{n+1}\sum_{j=1}^{M}\frac{\absolutevalue{a_{j}}}{(2\pi)^{k_{j}/2}}\binom{n+k_{j}}{k_{j}-1}$
(190)

For a fixed $n$, the binomial coefficient above grows with monotonically
$k_{j}$. This can be seen by writing

$\displaystyle\binom{n+k_{j}}{k_{j}-1}=\frac{(n+k_{j})!}{(n+1)!(k_{j}-1)!}=\frac{(n+k_{j})(n+k_{j}-1)\dots(k_{j})}{(n+1)!}$
(191)

On the right side, the numerator increases with $k_{j}$ while the denominator
remains fixed. However, the factor $(2\pi)^{-k_{j}/2}$ largely compensates
this growth. In order to see this, we first use Stirling’s formula to bound
the binomial.

$\begin{split}\binom{n+k}{k-1}&\leq\frac{1}{\sqrt{2\pi}}\sqrt{\frac{n+k}{(k-1)(n+1)}}\frac{(n+k)^{n+k}}{(n+1)^{n+1}(k-1)^{k-1}}\exp\left(\frac{1}{12(n+k)}-\frac{1}{12(n+1)+1}-\frac{1}{12(k-1)+1}\right)\\\
&<\sqrt{\frac{1}{2\pi}}\sqrt{\frac{n+k}{(k-1)(n+1)}}\left(\frac{n+k}{n+1}\right)^{n+1}\left(\frac{n+k}{k-1}\right)^{k-1}\exp\left(\frac{1}{12(n+k)}-\frac{1}{12(n+1)+1}\right)\end{split}$
(192)

The second line above uses the fact that $n>1$ and $k\geq 1$. Reintroducing
the $(2\pi)^{-k/2}$ factor and noticing that the sum of the first two factors
in the exponent is less than $1/600$, and $\exp(1/600)<1.002$, we have

$\frac{1}{(2\pi)^{k/2}}\binom{n+k_{j}}{k_{j}-1}\leq\frac{1.002}{2\pi}\sqrt{\frac{n+k}{(k-1)(n+1)}}\left(\frac{n+k}{n+1}\right)^{n+1}\left(\frac{n+k}{\sqrt{2\pi}(k-1)}\right)^{k-1}\;.$
(193)

In our application, we are interested in MPFs of order $2M$, hence we set
$n=2M$. From [27] we then have that

$\displaystyle
k_{j}=\left\lceil\frac{\sqrt{8}M}{\pi}\absolutevalue{\sin\left(\frac{\pi(2j-1)}{8M}\right)}^{-1}\right\rceil.$
(194)

We can now bound the coefficients using $\sin(x)\leq x$ and $\sin(x)\geq 4x/5$
for $x\in[0,1]$, which is satisfied in this case. This gives the lower bound

$k_{j}\geq\left\lceil\frac{8^{3/2}M^{2}}{\pi^{2}(2j-1)}\right\rceil\geq\left\lceil\frac{8^{3/2}M^{2}}{\pi^{2}(2M-1)}\right\rceil>\frac{\sqrt{128}M}{\pi^{2}}>M=\frac{n}{2}\;$
(195)

together with

$k_{j}\leq\left\lceil\frac{5\times
8\sqrt{8}M^{2}}{4(2j-1)\pi^{2}}\right\rceil\leq\left\lceil\frac{5\times
8\sqrt{8}M^{2}}{4\pi^{2}}\right\rceil<3M^{2}\;.$ (196)

Using the lowerbound, together with $n\geq 4$ (since $M>1$), we can get

$\sqrt{\frac{n+k}{(k-1)(n+1)}}=\sqrt{\frac{1}{k-1}+\frac{1}{n+1}}\leq\sqrt{\frac{2}{n-2}+\frac{1}{n+1}}<\sqrt{\frac{6}{5}}\;.$
(197)

Now we distinguish two cases. First let us assume that

$n+k\geq\sqrt{2\pi}(k-1).$ (198)

This implies that

$n+k\leq(n+1)\left(1+\frac{1}{\sqrt{2\pi}-1}\right)$ (199)

which combined with the lower bound on $k_{j}$ in (195) yields

$\begin{split}\left(\frac{n+k}{n+1}\right)^{n+1}\left(\frac{n+k}{\sqrt{2\pi}(k-1)}\right)^{k-1}&\leq\left(\frac{\sqrt{2\pi}}{\sqrt{2\pi}-1}\right)^{n+1}\left(\frac{n+k}{\sqrt{2\pi}(k-1)}\right)^{\frac{n+1}{\sqrt{2\pi}-1}}\\\
&\leq\left(\frac{\sqrt{2\pi}}{\sqrt{2\pi}-1}\right)^{n+1}\left(\frac{n+1}{(\sqrt{2\pi}-1)(k-1)}\right)^{\frac{n+1}{\sqrt{2\pi}-1}}\\\
&\leq\left(\frac{\sqrt{2\pi}}{\sqrt{2\pi}-1}\right)^{n+1}\left(\frac{n+1}{(\sqrt{2\pi}-1)(n/2-1)}\right)^{\frac{n+1}{\sqrt{2\pi}-1}}\\\
&\leq\left(\frac{\sqrt{2\pi}}{\sqrt{2\pi}-1}\right)^{n+1}\left(\frac{5}{\sqrt{2\pi}-1}\right)^{\frac{n+1}{\sqrt{2\pi}-1}}\\\
&\leq 3.7^{n+1}\end{split}$ (200)

where in the third line we used the lower bound $k>n/2$ and in the fourth the
lower bound $n\geq 4$. Now let us consider the second case wherein

$n+k<\sqrt{2\pi}(k-1)\Rightarrow(k-1)>(n+1)/(\sqrt{2\pi}-1)$ (201)

Defining $(k-1)=\eta(n+1)$ with $\eta\geq 1/(\sqrt{2\pi}-1)$ so that

$\begin{split}\left(\frac{n+k}{n+1}\right)^{n+1}\left(\frac{n+k}{\sqrt{2\pi}(k-1)}\right)^{k-1}&=\left(\left(1+\eta\right)\left(\frac{1+\eta}{\sqrt{2\pi}\eta}\right)^{\eta}\right)^{n+1}\\\
&\leq\left(\frac{\sqrt{2\pi}}{\sqrt{2\pi}-1}\right)^{n+1}<3.7^{n+1}\end{split}$
(202)

Then combining these two cases leads us to the following uniform upper bound

$\frac{1}{(2\pi)^{k/2}}\binom{n+k_{j}}{k_{j}-1}<\frac{1.002}{2\pi}\sqrt{\frac{6}{5}}3.7^{n+1}<0.16\times
3.7^{n+1}\;.$ (203)

Using this we can further upper bound the remainder as

$\displaystyle\begin{split}\norm{R_{2M}(t_{0}+\Delta
t,t_{0})}&<0.16\left(4e\times 3.7\max_{\tau\in[t_{0},t_{0}+\Delta
t]}\Lambda(\tau)\Delta t\right)^{2M+1}\norm{a}_{1}\\\
&<0.16\left(41\max_{\tau\in[t_{0},t_{0}+\Delta t]}\Lambda(\tau)\Delta
t\right)^{2M+1}\norm{a}_{1}\end{split}$ (204)

This last inequality, finally, has brought us to the sought after result of
Lemma 3. ∎

##  Appendix B Mathematical tools and notation

The purpose of this section is as reference for some of the mathematical
concepts and notation used throughout the paper. Subsection B.1 discusses
several combinatorial definitions and results invoked in the paper, while
Subsection B.2 discusses the vector, function, and operator norms used.

###  B.1 Combinatorics

Much of the technical work in this paper stems from computing high-order
derivatives of products and composite functions, which essentially becomes an
exercise in combinatorics. The purpose of this section is to introduce some of
the combinatorics tools used in this analysis.

The _multinomial coefficient_ is a generalization of the more-familiar
binomial coefficient, and it arises in several combinatorial situations. For
our purposes, it is defined by

$\displaystyle\binom{n}{s_{1},...,s_{k}}:=\frac{n!}{s_{1}!s_{2}!...s_{k}!}$
(205)

where $n\in\mathbb{Z}_{+}$ and the $(s_{q})_{q=1}^{k}$ are nonnegative
integers which sum to $n$. It is always a positive integer, corresponding to
the number of distinct ways of placing $n$ distinguishable items into $k$
boxes, where each box has a fixed number $s_{j}$ of items. In this work, we
will find occasion to make use of the multinomial when high-order derivatives
of a product.

$\displaystyle\left(\frac{d}{dt}\right)^{n}f_{1}(t)f_{2}(t)\dots f_{k}(t)$
(206)

Here, $(f_{q})_{q=1}^{k}$ are $n$-differentiable functions on an interval
$I\subseteq\mathbb{R}$. Employing the product rule, one is left to count all
the possible combinations of derivatives of each $f_{q}$. It turns out that
the multinomial is suited for performing this counting.

$\displaystyle\left(\frac{d}{dt}\right)^{n}\prod_{q=1}^{k}f_{q}(t)=\sum_{S}\binom{n}{s_{1},\dots,s_{k}}\prod_{q=1}^{k}\left(\frac{d}{dt}\right)^{s_{q}}f_{q}(t)$
(207)

The sum is taken over the set $S$ of sequences of nonnegative integers
$(s_{q})_{q=1}^{k}$ summing to $n$. As an aside, a similar multinomial sum to
the above arises when taking $n$ powers of a sum of $k$ distinct variables.
The same counting arguments are invoked.

On other occasions in the text below, we will be led to evaluate high-order
derivatives of the exponential of a function $a(t)$.

$\displaystyle\left(\frac{d}{dt}\right)^{n}e^{a(t)}$ (208)

This problem is also of a combinatorial nature. The solution we will rely on
is _Faà di Bruno’s formula_. In our specific context, this formula asserts
that

$\displaystyle\left(\frac{d}{dt}\right)^{n}e^{a(t)}=e^{a(t)}Y_{n}(a^{\prime}(t),a^{\prime\prime}(t),\dots,a^{(n)}(t))$
(209)

where $Y_{n}$ is the _complete exponential Bell polynomial_. [50] An explicit
formula is given by

$\displaystyle
Y_{n}(x_{1},x_{2},\dots,x_{n})=\sum_{C}\frac{n!}{c_{1}!c_{2}!\dots
c_{n}!}\prod_{j=1}^{n}\left(\frac{x_{j}}{j!}\right)^{c_{j}}$ (210)

where the sum is taken over the set $C$ of all sequences $(c_{j})_{j=1}^{n}$
such that $c_{j}\geq 0$ and

$\displaystyle c_{1}+2c_{2}+\dots+nc_{n}=n.$ (211)

Essentially, each coefficient in $Y_{n}$ counts the ways one can partition a
set of fixed size $n$ into subsets of given sizes and number. When one simply
wants to count the total number of possible partitions, one is led to the
_Bell numbers_ $b_{n}$. These are related to the $Y_{n}$ by evaluating all
arguments to one.

$\displaystyle b_{n}=Y_{n}(1,1,...1)$ (212)

More generally, for any $x\in\mathbb{R}$,

$\displaystyle Y_{n}(x,x^{2},\dots,x^{n})=x^{n}b_{n},$ (213)

which can be seen directly from (210) along with the sum rule (211). The Bell
numbers $b_{n}$ grow combinatorially, like a factorial. In fact, the following
upper bound is invoked in the paper [37].

$\displaystyle b_{n}<\left(\frac{.792n}{\log(n+1)}\right)^{n},\quad\forall
n\in\mathbb{Z}_{+}$ (214)

There is a natural way to form a single variable polynomial from $Y_{n}$ by
evaluating all arguments to the same value $x$. This is sometimes called
simply the _Bell polynomial_ or _Touchard polynomial_ $B_{n}(x)$.

$\displaystyle B_{n}(x)=Y_{n}(x,x,\dots,x)$ (215)

Of course, $b_{n}=B_{n}(1)$. The $n$th Bell polynomial $B_{n}(x)$ is also the
value of the $n$th moment of the Poisson distribution with mean $x$. From [49]
we have the following upper bound on $B_{n}$.

$\displaystyle B_{n}(x)\leq\left(\frac{n}{\log(1+\frac{n}{x})}\right)^{n}$
(216)

Notice that this bound for $x=1$ is very close to that for the Bell numbers in
(214). From their definitions, and intuitively from their combinatorial
natures, $Y_{n}$, $B_{n}$ and $b_{n}$ all grow monotonically, both in their
functional arguments and their index $n$.

Compared to these special functions, the reader is likely more familiar with
the simple factorial $n$! and it’s corresponding asymptotic expression given
by _Stirling’s approximation_. In the paper, we always make use of a version
of the approximation which gives strict bounds for $n>0$.

$\displaystyle\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}<n!<\sqrt{2\pi
n}\left(\frac{n}{e}\right)^{n}e^{1/(12n)}$ (217)

These bounds are extremely good, even for small $n$.

###  B.2 Norms

There are a couple different types of norms employed throughout the paper, on
a variety of mathematical objects. Here we define and discuss each of them.

The Schatten $p$-norm, is defined for any $v\in\mathbb{C}^{n}$, with any
$n\in\mathbb{Z}_{+}$. It is defined as

$\displaystyle\norm{v}_{p}:=\left(\sum_{j=1}^{n}\absolutevalue{v_{j}}^{p}\right)^{1/p}.$
(218)

In the paper, we will make use of the 1-norm several contexts, such as the
coefficients $a$ of the multiproduct formula.

$\displaystyle\norm{v}_{1}:=\sum_{j}\absolutevalue{v_{j}}$ (219)

In the main paper we claim our Hamiltonian simulation algorithm exhibits
$L^{1}$-norm scaling. Here we define what that means.

######  Definition 1.

Let $f:I\subseteq\mathbb{R}\rightarrow\mathbb{R}$ be a scalar function. The
$L^{1}$ norm of $f$ is defined as

$\displaystyle\norm{f}_{1}:=\int_{I}\absolutevalue{f(t)}dt$ (220)

assuming it exists and is finite. We say a Hamiltonian simulation algorithm
has $L^{1}$-norm scaling if it has complexity $O(\norm{f}_{1})$. Here, $f$ is
a function whose domain is the simulation interval $I$, and whose value at
$t\in I$ is some measure of the size of the Hamiltonian and its derivatives at
$t$.

In our presentation of the MPF algorithm, we take $f=\Lambda$ as defined in
Definition 1 of the main text. Although the same notation is used for both
vector (Schatten) and function ($L^{1}$) norms, the context should make clear
which of them is being used.

Finally, our paper makes use of a particular operator norm, namely the
_spectral norm_ , also known as the _induced 2-norm_ because it is induced
from the vector 2-norm. It is defined for any bounded operator $A$ on a
Hilbert space $\mathcal{H}$ by

$\displaystyle\norm{A}:=\sup_{v\in\mathcal{H}\setminus\\{0\\}}\frac{\norm{Av}_{2}}{\norm{v}_{2}}$
(221)

where $0\in\mathcal{H}$ is the zero vector. In our case, $A$ will always be
finite dimensional, and in such cases $\norm{A}$ is the largest singular value
of $A$. The spectral norm is invariant under left and right multiplication by
a unitary operator $U$, and $\norm{U}=1$. Like many (but not all) matrix
norms, the spectral norm is submultiplicative, a property we make frequent use
of.

$\displaystyle\norm{AB}\leq\norm{A}\norm{B}$ (222)

If $f$ is a function of a single variable, the notation $f^{(n)}$ will often
be used to denote the derivative of $f$ with respect to its variable. If
multiple variables are present, the proper variable will be specified. Note
that in the example $g^{(3)}(2x)$, the derivative is not being taken with
respect to $x$, but it is evaluated at $2x$.

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# Spectral analysis of self-adjoint first order differential operators
with nonlocal potentials

K. Debowska, I. L. Nizhnik

###### Abstract

The paper considers the general form of self-adjoint boundary value problems
for momentum operators with nonlocal potentials. We give an analysis of the
eigenvalue distribution as zeros of the characteristic functions, for which
their explicit representations are obtained in terms of nonlocal potentials.

2020 Mathematics Subject Classification: 34L05, 47E05.
Keywords and phrases: differential operator, nonlocal interaction, condition
for self-adjointness, resolvent structure, eigenvalues.

##  1 Introduction

The boundary values of solutions to boundary value problems for differential
equations are usually involved only in the boundary conditions. However, as
successful experience in studying important applied problems has shown, the
boundary characteristics of solutions are also involved into the equations.
For problems in quantum mechanics with interaction concentrated in the
neighborhood of a single point, or on a discrete set of such points, or even
in the neighborhood of a set $\Omega_{0}$ of a small dimension, models are
effective when the interaction is considered to be centered on such a set
$\Omega_{0}.$ From a mathematical point of view, this idea comes down to the
following: the indicated set $\Omega_{0}$ is attached to the boundary of the
domain, the equation is considered on functions that equal to zero in the
neighborhood of $\Omega_{0}$, and the resulting symmetric operator is extended
to self-adjoint operators. These operators describe interactions centered on
$\Omega_{0}$. A large number of works has been published in this direction, we
will only mention the important monograph [1]. In the same area of research,
[4] proposes a method in which the differential equation contains boundary
values on the set $\Omega_{0}$ and certain function factors. These function
factors are called nonlocal potentials in [4, 5]. For a number of models with
nonlocal potentials, a detailed spectral analysis [2, 4, 5, 6, 7, 8, 9, 10,
11] has been carried out. The very list of titles of these works indicates a
fairly wide range of research. We only note that the studied models are the
exact solvable models, which allowed to obtain a number of new efficient
algorithms for solving inverse spectral problems without using analogues of
the Gelfand–Levitan–Marchenko integral equations.

Let us demonstrate the above in more detail for the simplest self–adjoint
operator $\mathfrak{D}=i\frac{d}{dx}$ in the space $L_{2}(-\infty,+\infty)$ of
square–summable functions on the entire axis. The domain of the operator
$\mathfrak{D}$ is the Sobolev space $W_{2}^{1}(-\infty,+\infty)$ containing
functions $\psi,\,\psi^{\prime}\in L_{2}. $ We obtain an analogue of the
operator $\mathfrak{D}$, which describes the point interaction at the point
$x_{0}=0$, by considering the symmetric operator
$\mathfrak{D}_{\min}\psi(x)=i\frac{d\psi(x)}{dx}$, $x\neq 0$ on the set of
functions $W_{2}^{1}(\Omega),$ $\Omega=(-\infty,+\infty)\backslash\\{0\\}$
satisfying the boundary conditions $\psi(-0)=0,$ $\psi(+0)=0$. The operator
$\mathfrak{D}_{\min}^{*}=\mathfrak{D}_{\max}$, where the domain of
$\mathfrak{D}_{\max}$ consists of the entire space $W_{2}^{1}(\Omega).$ Every
self–adjoint extension $\mathfrak{D}_{\alpha}$ of the operator
$\mathfrak{D}_{\min}$ is a restriction of the operator $\mathfrak{D}_{\max}$
to the space $L_{2},$ and the domain of the operator $\mathfrak{D}_{\alpha}$
consists of the set $\\{\psi:\psi\in W_{2}^{1}(\Omega),\psi(+0)=e
^{i\alpha}\psi(-0)\\}$, where $\alpha$ is a real number characterizing the
operator $\mathfrak{D}_{\alpha}$. The operator $\mathfrak{D}_{\alpha}$ itself
describes the point interaction at the point $x_{0}=0$. Of course, the
operators $\mathfrak{D}$ and $\mathfrak{D}_{\alpha}$ can be perturbed by an
operator $V$ of the form $V\psi(x)=v_{1}(x)\psi(-0)+v_{2}(x)\psi(+0)$, where
the functions $v_{1}$ and $v_{2}$ belong to the space $L_{2}(\Omega),$ without
loss of the initial operators domain. However, the operators $\mathfrak{D}+V$
and $\mathfrak{D}_{\alpha}+V$ will be self-adjoint only for $V\equiv 0.$ To
preserve self-adjointness under such perturbations, it is necessary to change
the domain of the considered operators with nonlocal potentials, using
nonlocal boundary conditions, when information about the solution in the
entire region $\Omega$ is used in the conditions. For example, the operator is
self-adjoint

$\mathfrak{D}_{v,\alpha}=\mathfrak{D}_{\max}\psi+v(x)[\psi(-0)+\psi(+0)] $

on the domain

$ \\{\psi:\psi\in
W_{2}^{1}(\Omega),\psi(+0)-i\langle\psi,v\rangle_{L_{2}}=e^{i\alpha}[\psi(-0)
+i\langle\psi,v\rangle_{L_{2}}]\\}. $ (1.1)

In this case, any complex–valued function from the space $L_{2}$ can serve as
a nonlocal potential $v(x)$. The general case is considered in this paper. The
detailed proofs of the self–adjointness of the momentum operators with
nonlocal potentials on the entire axis with single–point perturbations and
operators on a finite interval with nonlocal potentials associated with
boundary points are given. The resolvents of such operators for nonreal $z$
are constructed explicitly as integral operators whose kernels have a finite
rank perturbation of the Green‘s function of operators without nonlocal
potentials. Spectral results for such models are considered. It is shown that
such models are exact solvable models.

##  2 Boundary value problem on the whole axis with one point interaction

Consider the simplest first-order differential operator
$\mathfrak{D}\psi=i\frac{d\psi(x)}{dx}$ on the whole axis $-\infty<x<+\infty$
in the space $L_{2}(-\infty,+\infty)$, whose domain is the Sobolev space
$W_{2}^{1}(-\infty,+\infty)$. This operator is self-adjoint with an absolutely
continuous spectrum on the whole axis $-\infty<\lambda<+\infty$ and with a
resolvent $(\mathfrak{D}-zI)^{-1}$, which for nonreal $z$ is an integral
operator:

$(\mathfrak{D}-zI)^{-1}h=\int_{-\infty}^{+\infty}\,g_{z}(x-y)\,h(y)\,dy,$
(2.1)

where

$g_{z}(x)=i\mathrm{sign}\,(\mathrm{Im}\,z)\theta(-x\mathrm{Im}\,z)e^{-izx}.$
(2.2)

Note that such a simple operator is the free momentum operator in quantum
mechanics. The Green‘s function $g_{z}(x)$ of the form $(\ref{2.2})$ has the
following simple properties:

$\begin{array}[]{c}\|g_{z}(\cdot)\|_{L_{2}}=\frac{1}{\sqrt{2|\mathrm{Im}\,z|}},\quad
g_{z}(-0)=i\theta(\mathrm{Im}\,z),\quad
g_{z}(+0)=-i\theta(-\mathrm{Im}\,z),\\\
g_{z}(+0)+g_{z}(-0)=i\mathrm{sign}\,(\mathrm{Im}\,z),\quad
g_{z}(+0)-g_{z}(-0)=-i,\quad\bar{g}_{\bar{z}}(x)=g_{z}(-x).\end{array}$ (2.3)

The boundary value problem with a one point interaction at the point $x_{0}=0$
for the momentum operator $\mathfrak{D}$ requires consideration of the
symmetric operator in the space $L_{2}$ of the form
$\mathfrak{D}_{\min}\psi(x)=i\frac{d\psi}{dx}$, $x\neq 0$,
$x\in\Omega=(-\infty,+\infty)\setminus\\{0\\}$ with domain $W_{2}^{1}(\Omega)$
containing functions $\psi(x)$ such that $\psi(-0)=0$ and $\psi(+0)=0$. The
adjoint operator of $\mathfrak{D}_{\min}$ is the operator
$\mathfrak{D}_{\max}\psi(x)=i\frac{d\psi}{dx},$ $x\neq 0$ and its domain is
the whole space $W_{2}^{1}(\Omega).$ The symmetry of the operator
$\mathfrak{D}_{\min}$ and the equality
$\mathfrak{D}_{\max}=\mathfrak{D}_{\min}^{*}$ immediately follows from Green‘s
formula:

$\langle\mathfrak{D}_{\max}\psi,\varphi\rangle_{L_{2}}-\langle\psi,\mathfrak{D}_{\max}\varphi\rangle_{L_{2}}=i\\{\psi(-0)\overline{\varphi(-0)}-\psi(+0)\overline{\varphi(+0)}\\},\quad\psi,\varphi\in
W_{2}^{1}(\Omega).$ (2.4)

######  Theorem 2.1.

For the self-adjoint operator $A$ to be an extension of the symmetric operator
$\mathfrak{D}_{\min}$, it is necessary and sufficient that the operator $A$ be
the restriction of the operator $\mathfrak{D}_{\max}$ to the set:

$\\{\psi:\psi\in W_{2}^{1}(\Omega),\,\,\psi(+0)=e^{i\alpha}\psi(-0)\\},$ (2.5)

where $\alpha$ is a real number characterizing the operator $A$. Each value of
$\alpha\in[0,2\pi)$ defines a self–adjoint extension of the operator
$\mathfrak{D}_{\min}$ and it is denoted by $\mathfrak{D}_{\alpha}$. For
nonreal $z$, the resolvent $(\mathfrak{D}_{\alpha}-zI)^{-1}$ is an integral
operator:

$(\mathfrak{D}_{\alpha}-zI)^{-1}h(x)=\int\,\mathcal{G}_{z}(x,y;\alpha)\,h(y)\,dy,$
(2.6)

where the kernel $\mathcal{G}_{z}(x,y;\alpha)$ is expressed in terms of the
Green‘s function $(\ref{2.2})$ as:

$\mathcal{G}_{z}(x,y,\alpha)=g_{z}(x-y)+g_{z}(x)\beta(z,\alpha)\bar{g}_{\bar{z}}(y)$
(2.7)

and the coefficient $\beta(z,\alpha)$ has the form

$\beta(z,\alpha)=i(1-e^{-i\alpha})\theta(\mathrm{Im}\,z)+i(e^{i\alpha}-1)\theta(-\mathrm{Im}\,z),\quad\overline{\beta}(\bar{z},\alpha)=\beta(z,\alpha).$
(2.8)

###### Proof.

The symmetry of the operator $\mathfrak{D}_{\alpha}$ follows from the Green‘s
formula $(\ref{2.4})$. To construct the kernel $\mathcal{G}_{z}(x,y,\alpha)$,
it suffices to solve the equation

$\psi(x)=g_{z}(x)i[\psi(+0)-\psi(-0)]+\int\,g_{z}(x-y)\,h(y)\,dy$ (2.9)

for any $h\in L_{2}$ satisfying the boundary condition
$\psi(+0)=e^{i\alpha}\psi(-0).$ From the equation $(\ref{2.9})$ and the
boundary conditions, we obtain a linear system for $\psi(+0)$ and $\psi(-0)$:

$\begin{array}[]{ll}\frac{1}{2}[\psi(+0)-\psi(-0)]=\frac{1}{2}\mathrm{sign}\,(\mathrm{Im}\,z)[\psi(+0)-\psi(-0)]+\langle
h,g_{\bar{z}}\rangle,\\\\[5.69054pt]
\psi(+0)-e^{i\alpha}\psi(-0)=0.\end{array}$ (2.10)

If we substitute the solution of the system $(\ref{2.10})$ into the equation
$(\ref{2.9})$, then we get

$\psi(x)=\int\,\mathcal{G}_{z}(x,y;\alpha)h(y)\,dy,$ (2.11)

where the kernel has the form $(\ref{2.7})$. It is easy to verify that the
solution $\psi$ of the form $(\ref{2.11})$, with the kernel
$\mathcal{G}_{z}(x,y,\alpha)$, satisfies the equation
$(\mathfrak{D}_{\alpha}-zI)^{-1}\psi=h$. ∎

######  Remark 2.1.

The Green‘s function $\mathcal{G}_{z}(x,y;\alpha)$ is related to the Green‘s
function $g_{z}(x)$ by the equality $(\ref{2.7})$ Therefore, due to (2.8) and
(2.3), the function

$E_{1}(x,z)=\mathcal{G}_{z}(x,-0;\alpha)=g_{z}(x)w(z,\alpha)$

where $w(z,\alpha)=\theta(\mathrm{Im}\,z)+e^{i\alpha}\theta(-\mathrm{Im}\,z)$.
Therefore, $E_{1}(-0,z)-e^{-i\alpha}E_{1}(+0,z)=i.$ Due to the boundary
conditions (2.5)

$\begin{array}[]{l} \displaystyle
\frac{1}{2}\Bigl{(}\mathcal{G}_{z}(-0,y;\alpha)+e^{-i\alpha}\mathcal{G}_{z}(+0,y;\alpha)\Bigr{)}=\mathcal{G}_{z}(-0,y;\alpha),\\\\[5.69054pt]
\displaystyle
\frac{1}{2}\Bigl{(}E_{1}(-0,z)+e^{-i\alpha}E_{1}(+0,z)\Bigr{)}=\frac{i}{2}\mathrm{sign}\,(\mathrm{Im}\,z).\end{array}$

The operators $\mathfrak{D}_{\alpha}$ in Theorem 2.1 are the models of point
interactions for momentum operators at the point $x_{0}=0$.

######  Remark 2.2.

Self–adjoint operators $\mathfrak{D}_{\alpha}$ with $\alpha\in[0,2\pi)$ are
unitary equivalent.

###### Proof.

The operator $\mathfrak{D}_{0}$ for $\alpha=0$ is a self-adjoint operator with
the domain $W_{2}^{1}(R^{1})$. This operator is given by the differential
operator $\mathfrak{D}_{0}=i\frac{d}{dx}.$ The operator
$\mathfrak{D}_{\alpha}$ is related to the operator $\mathfrak{D}_{0}$ by the
equality $\mathfrak{D}_{\alpha}=U_{\alpha}^{*}\mathfrak{D}_{0}U_{\alpha}$
where $U_{\alpha}$ is a unitary operator in the space
$L_{2}(-\infty,+\infty)$. The operator $U_{\alpha}$ is an operator of
multiplication by the function $\chi_{\alpha}(x)=1$ for $x>0$ and
$\chi_{\alpha}(x)=e^{-i\alpha x}$ for $x<0$. If the function
$\psi_{\alpha}(x)$ belongs to the domain of the operator
$\mathfrak{D}_{\alpha}$, then $U_{\alpha}\psi_{\alpha}\in W_{2}^{1}(R^{1})$
and the equality $\mathfrak{D}_{0}U_{\alpha}\psi_{\alpha}$ makes sense. The
operator $U_{\alpha}^{*}$ maps the function
$\mathfrak{D}_{0}U_{\alpha}\psi_{\alpha}$ into
$\mathfrak{D}_{\alpha}\psi_{\alpha}$. ∎

Note that Remark 2.2 is valid only for the first-order differential operators.
For Schrödinger operators, similar operators are different for different
$\alpha$ and they describe point interactions [1].

##  3 Momentum operator with nonlocal potentials

Let us now extend the constructed model using nonlocal potentials.

######  Definition 3.1.

In the space $L_{2}(-\infty,+\infty)$, the momentum operator with a local
interaction in the point $x_{0}=0$ and two nonlocal potentials $v_{1}(x)$ and
$v_{2}(x)$, which are complex-valued functions from the space $L_{2}$, is
defined on all functions from the Sobolev space $W_{2}^{1}(\Omega)$ by the
expression:

$A_{\max}\psi=\mathfrak{D}_{\max}\psi(x)+v_{1}(x)[\psi(-0)+\frac{i}{2}\langle\psi,v_{1}\rangle_{L_{2}}]+v_{2}(x)[\psi(+0)-\frac{i}{2}\langle\psi,v_{2}\rangle_{L_{2}}],\,\,x\neq
0.$ (3.1)

The restriction of the operator $A_{\max}$ to the domain

$\mathfrak{D}_{(v_{1},v_{2},\alpha)}=\\{\psi:\psi\in
W_{2}^{1}(\Omega),\psi(+0)-i\langle\psi,v_{2}\rangle_{L_{2}}=e^{i\alpha}[\psi(-0)+i\langle\psi,v_{1}\rangle_{L_{2}}]\\}$
(3.2)

is denoted by $A(v_{1},v_{2},\alpha)$, where $\alpha$ is a real number.

######  Lemma 3.1.

The momentum operators $A(v_{1},v_{2},\alpha)$ with nonlocal potentials are
symmetric operators in the space $L_{2}(-\infty,+\infty)$.

###### Proof.

The operator $A_{\max}$ (3.1) satisfies the following Green‘s formula:

$\begin{array}[]{ll}\langle
A_{\max}\psi,\varphi\rangle_{L_{2}}-\langle\psi,A_{\max}\varphi\rangle_{L_{2}}=i\\{[\psi(-0)+i\langle\psi,v_{1}\rangle_{L_{2}}]\cdot[\varphi(-0)+i\langle\varphi,v_{1}\rangle_{L_{2}}]^{*}-\\\\[5.69054pt]
-[\psi(+0)-i\langle\psi,v_{2}\rangle_{L_{2}}]\cdot[\varphi(+0)-i\langle\varphi,v_{2}\rangle_{L_{2}}]^{*}\\}\end{array}$
(3.3)

for any functions $\psi,\,\varphi\in W_{2}^{1}(\Omega)$. If we assume in this
Green‘s formula that the functions $\psi$ and $\varphi$ belong to
$\mathfrak{D}_{(v_{1},v_{2},\alpha)}$ of $(\ref{2.210})$, then in the
left–hand side of the equation $(\ref{2.211})$, we can write
$A(v_{1},v_{2},\alpha)$ instead of the operator $A_{\max}$. Using the boundary
conditions of $(\ref{2.210})$, we obtain that the right–hand side of the
equation $(\ref{2.211})$ is equal to zero. Then the operator
$A(v_{1},v_{2},\alpha)$ is symmetric if the domain of this operator is dense
in the space $L_{2}(-\infty,+\infty)$. Let us show that even the domain of the
operator $A_{\min}:$

$\\{\psi:\psi\in
W_{2}^{1}(\Omega),\psi(+0)-i\langle\psi,v_{2}\rangle_{L_{2}}=0,\,\psi(-0)+i\langle\psi,v_{1}\rangle_{L_{2}}=0\\}\subset\mathfrak{D}_{(v_{1},v_{2},\alpha)}$
(3.4)

is dense in the space $L_{2}(-\infty,+\infty)$. It is well–known that the set
of all compactly supported continuously differentiable functions
$\\{\psi_{0}\\}\in W_{2}^{1}(-\infty,+\infty)$ such that $\psi_{0}(0)=0$ is
dense in the space $L_{2}(-\infty,+\infty)$. Let us show that such functions
$\psi_{0}(x)$ can be approximated as close as possible by functions from the
domain of the operator $A_{\min}:$

$||\psi_{0}-\psi_{n}||\rightarrow 0,\quad\,n\rightarrow\infty.$

This means that the domain of the operator $A_{\min}$ and, therefore, the
domain of the operator $A(v_{1},v_{2},\alpha)$ are dense in the space
$L_{2}(-\infty,+\infty)$.

In order to prove this we construct two functions:

$\delta^{(+)}(x)=\begin{cases}\begin{array}[]{l}(1-x)^{2},\,\,\,x\in[0,1],\\\\[5.69054pt]
0,\quad\quad\quad\,\,\,x\overline{\in}[0,1],\end{array}\end{cases}$

and

$\delta^{(-)}(x)=\begin{cases}\begin{array}[]{l}(1+x)^{2},\,\,\,x\in[-1,0],\\\\[5.69054pt]
0,\quad\quad\quad\,\,\,x\overline{\in}[-1,0].\end{array}\end{cases}$

Then we define functions $\delta_{n}^{(+)}(x)=\delta^{(+)}(nx)$ and
$\delta_{n}^{(-)}(x)=\delta^{(-)}(nx)$, $n\in N$. Now consider the following
functions:

$\psi_{n}(x)=\psi_{0}(x)+C_{n}^{(+)}\delta_{n}^{(+)}(x)+C_{n}^{(-)}\delta_{n}^{(-)}(x).$

These functions $\psi_{n}(x)$ belong to the space $W_{2}^{1}(\Omega)$, and
with appropriately chosen constants $C_{n}^{(+)}$ and $C_{n}^{(-)}$, they
belong to the domain of the operator $A_{\min}$. Thus, substituting the
function $\psi_{n}(x)$ into two conditions of the formula (3.4), we get a
linear system:

$\begin{cases}\begin{array}[]{l}C_{n}^{(+)}-i[\langle\psi_{0},v_{2}\rangle+C_{n}^{(+)}\langle\delta_{n}^{(+)},v_{2}\rangle+C_{n}^{(-)}\langle\delta_{n}^{(-)},v_{2}\rangle]=0,\\\\[5.69054pt]
C_{n}^{(-)}+i[\langle\psi_{0},v_{1}\rangle+C_{n}^{(+)}\langle\delta_{n}^{(+)},v_{1}\rangle+C_{n}^{(-)}\langle\delta_{n}^{(-)},v_{1}\rangle]=0.\end{array}\end{cases}$
(3.5)

Since
$\displaystyle|\langle\delta_{n}^{(+)},v_{j}\rangle|\leq\frac{1}{\sqrt{5n}}||v_{j}||$
and
$\displaystyle|\langle\delta_{n}^{(-)},v_{j}\rangle|\leq\frac{1}{\sqrt{5n}}||v_{j}||$,
$j=1,2$, then the system (3.5) is uniquely solvable for sufficiently large
values of $n$. In addition, $\displaystyle
C_{n}^{(+)}=i\langle\psi_{0},v_{2}\rangle+O\Bigl{(}\frac{1}{\sqrt{n}}\Bigr{)}$
and $\displaystyle
C_{n}^{(-)}=-i\langle\psi_{0},v_{1}\rangle+O\Bigl{(}\frac{1}{\sqrt{n}}\Bigr{)}$
as $n\rightarrow\infty$. Therefore, $|C_{n}^{(+)}|$ and $|C_{n}^{(-)}|$ are
bounded by some constant $C$ as $n\rightarrow\infty$. This shows that

$\displaystyle||\psi_{0}-\psi_{n}||\leq\frac{C(||v_{1}||+||v_{2}||)}{\sqrt{n}}\rightarrow
0,\quad\,n\rightarrow\infty.$

Thus, the domain of the operator $A_{\min}$ and, consequently, the domain of
the operator $A(v_{1},v_{2},\alpha)$ are dense in the space
$L_{2}(-\infty,+\infty)$. And these operators are symmetric in
$L_{2}(-\infty,+\infty)$. ∎

From Definition 3.1 of the operators $A(v_{1},v_{2},\alpha)$ it follows that
two operators with different nonlocal potentials can have the same domains and
even coincide.

######  Theorem 3.1.

A necessary and sufficient conditions for the operators
$A(v_{1},v_{2},\alpha)$ and $A(\hat{v}_{1},\hat{v}_{2},\alpha)$ to have the
same domains are the following:

$v_{1}(x)+e^{i\alpha}v_{2}(x)=\hat{v}_{1}(x)+e^{i\alpha}\hat{v}_{2}(x)\overset{def}{=}v(x).$
(3.6)

The difference of two operators with the same domains

$A(v_{1},v_{2},\alpha)-A(v,0,\alpha)=K(v_{1},v_{2},\alpha)$ (3.7)

is an integral operator of rank at most $2$

$K(v_{1},v_{2},\alpha)\psi=\frac{i}{2}e^{i\alpha}v_{2}(x)\langle\psi,v_{1}\rangle-\frac{i}{2}e^{-i\alpha}v_{1}(x)\langle\psi,v_{2}\rangle.$
(3.8)

The difference of two operators
$A(v_{1},v_{2},\alpha)-A(\hat{v}_{1},\hat{v}_{2},\alpha)$ with nonlocal
potentials under the condition $(\ref{2.12})$ is an integral operator of rank
at most $4$. These integral operators are bounded and symmetric.

###### Proof.

The domain of the operators $A(v_{1},v_{2},\alpha)$ is determined by the
boundary condition (3.2) and can be rewritten as follows:

$\psi(+0)-e^{i\alpha}\psi(-0)=i[e^{i\alpha}\langle\psi,v_{1}\rangle+\langle\psi,v_{2}\rangle]$
(3.9)

and respectively for the operator $A(\hat{v}_{1},\hat{v}_{2},\alpha)$. Let us
determine the value $\psi(+0)$ of the boundary condition $(\ref{2.14})$ and
substitute it into the definition of the operator $A(v_{1},v_{2},\alpha)$. We
obtain

$A(v_{1},v_{2},\alpha)\psi=A(v,0,\alpha)\psi+K(v_{1},v_{2},\alpha)\psi,$
(3.10)

where $K(v_{1},v_{2},\alpha)$ is a bounded symmetric operator of rank at most
$2$. Since this is also true for the operator
$A(\hat{v}_{1},\hat{v}_{2},\alpha)$, then
$A(v_{1},v_{2},\alpha)-A(\hat{v}_{1},\hat{v}_{2},\alpha)$ is a difference of
two symmetric operators. Therefore, this is a bounded symmetric operator of
rank at most $4$. ∎

######  Remark 3.1.

The fact that operators with nonlocal potentials and the same domain can
differ by a bounded symmetric operator shows that the proof of the self-
adjointness of an arbitrary operator $A(v_{1},v_{2},\alpha)$ can be reduced to
proving the self-adjointness of the operators $A(v,0,\alpha)$ or
$A(\frac{1}{2}v,\frac{e^{-i\alpha}}{2}v,\alpha)$, where
$v=v_{1}+e^{i\alpha}v_{2}$ and the operators $A(v,0,\alpha)$,
$A(\frac{1}{2}v,\frac{e^{-i\alpha}}{2}v,\alpha)$ coincide.

Indeed, due to Theorem 3.1 we have

$A(\frac{1}{2}v,\frac{e^{-i\alpha}}{2}v,\alpha)-A(v,0,\alpha)=K(\frac{1}{2}v,\frac{e^{-i\alpha}}{2}v,\alpha)\equiv
0.$

We will denote the operator $A(\frac{1}{2}v,\frac{e^{-i\alpha}}{2}v,\alpha)$
by $A(v,\alpha).$ Its action is expressed in the form

$A(v,\alpha)\psi=\mathfrak{D}_{\max}\psi+v(x)\left[\frac{1}{2}\psi(-0)+\frac{e^{-i\alpha}}{2}\psi(+0)\right]$
(3.11)

with the boundary condition

$i\psi(-0)-ie^{-i\alpha}\psi(+0)=\langle\psi,v\rangle.$ (3.12)

######  Theorem 3.2.

For any potential $v\in L_{2}$ and a real number $\alpha$, the operator
$A(v,\alpha)$ of the form $(\ref{2.16})$–$(\ref{2.17})$ is self-adjoint in the
space $L_{2}$. For nonreal $z$, the resolvent $(A(v,\alpha)-zI)^{-1}$ is an
integral operator with the kernel $\mathcal{G}_{z}(x,y;v,\alpha)$

$(A(v,\alpha)-zI)^{-1}h(z)=\int\,\mathcal{G}_{z}(x,y;v,\alpha)h(y)\,dy.$
(3.13)

The kernel $\mathcal{G}_{z}(x,y;v,\alpha)$ is a rank at most $2$ perturbation
of the kernel $\mathcal{G}_{z}(x,y;\alpha)$ of the free momentum operator in
(2.7) (see Theorem 2.1):

$\displaystyle\mathcal{G}_{z}(x,y;v,\alpha)=\mathcal{G}_{z}(x,y;\alpha)+\frac{1}{\gamma(z)}\sum\limits_{j,k=1}^{2}\,E_{j}(x,z)\gamma_{jk}(z)\bar{E}_{k}(y,\bar{z}),$
(3.14)

where $E_{1}(x,z)=\mathcal{G}_{z}(x,-0;\alpha),$
$E_{2}(x,z)=\int\,\mathcal{G}_{z}(x,y;\alpha)v(y)\,dy.$ The parameters
$\gamma_{jk}(z)$ and $\gamma(z)$ are defined by the formulas:

$\begin{array}[]{ll}\displaystyle\gamma_{11}(z)=-\langle
E_{2}(\cdot,z),v(\cdot)\rangle_{L_{2}}=-\int\int\,\mathcal{G}_{z}(x,y;\alpha)v(y)\overline{v(x)}\,dy\,dx,\\\\[5.69054pt]
\displaystyle\gamma_{12}(z)=1+\langle
v(\cdot),E_{1}(\cdot,\overline{z})\rangle_{L_{2}}=1+\int\,\overline{\mathcal{G}_{\bar{z}}}(-0,y;\alpha)v(y)\,dy,\\\\[5.69054pt]
\displaystyle\gamma_{21}(z)=1+\langle
E_{1}(\cdot,z),v(\cdot)\rangle_{L_{2}}=1+\int\,\mathcal{G}_{z}(x,-0;\alpha)\overline{v(x)}\,dx,\\\\[5.69054pt]
\displaystyle\gamma_{22}(z)=\displaystyle-\frac{i}{2}\mathrm{sign}\,(\mathrm{Im}\,z),\\\\[5.69054pt]
\displaystyle\gamma(z)=\gamma_{11}(z)\gamma_{22}(z)-\gamma_{12}(z)\gamma_{21}(z)=\\\\[5.69054pt]
\displaystyle=\displaystyle\frac{i}{2}\mathrm{sign}\,(\mathrm{Im}\,z)\cdot\langle
E_{2}(\cdot,z),v(\cdot)\rangle_{L_{2}}-(1+\langle
E_{1}(\cdot,z),v(\cdot)\rangle_{L_{2}})(1+\langle
v(\cdot),E_{1}(\cdot,\overline{z})\rangle_{L_{2}}).\end{array}$ (3.15)

The matrix $\Gamma(z)=\|\gamma_{jk}(z)\|$ has the property
$\Gamma^{*}(\overline{z})=\Gamma(z),$ and $\gamma(z)=\det\Gamma$ has the
property $\overline{\gamma}(\overline{z})=\gamma(z)\neq 0$.

###### Proof.

The operator $A(v,\alpha)$ is symmetric in the space $L_{2}$ according to
Theorem 3.1. Therefore, in order to prove its self–adjointness, it is
sufficient to show that for any $h$ and some $z_{1}$, $z_{2}$ for which
$\mathrm{Im}\,z_{1}>0,$ $\mathrm{Im}\,z_{2}<0$ there exists a solution of the
problem

$(A(v,\alpha)-z_{j}I)^{-1}\psi(x)=h,\quad j=1,2.$ (3.16)

This problem is equivalent to the problem for finding a solution of the
equation:

$\mathfrak{D}_{\max}\psi-
z_{j}\psi=h-\frac{1}{2}v(x)(\psi(-0)+e^{-i\alpha}\psi(+0)),\quad j=1,2,$

satisfying the boundary condition $(\ref{2.17})$. This problem can be reduced
to the problem

$\psi(x)=-\mathcal{G}_{z}(x,-0;\alpha)\cdot\langle\psi,v\rangle+\int\,\mathcal{G}_{z}(x,y;\alpha)\left[h(y)-v(y)\frac{1}{2}(\psi(-0)+e^{-i\alpha}\psi(+0))\right]\,dy$
(3.17)

with the boundary condition $(\ref{2.17})$. Indeed, if the equation (3.17) has
a solution $\psi(x)\in L_{2}$, then it satisfies the equation

$\left(i\frac{d}{dx}-zI\right)\psi(x)=h(x)-v(x)\frac{1}{2}\left(\psi(-0)+e^{-i\alpha}\psi(+0)\right)$

and the boundary conditions (3.12). This follows from the properties of the
Green‘s function $\mathcal{G}_{z}(x,y,\alpha)$. A function

$E_{1}(x,z)=\mathcal{G}_{z}(x,-0;\alpha)=g_{z}(x)w(z,\alpha)$

satisfies an equality

$E_{1}(-0,z)-e^{-i\alpha}E_{1}(+0,z)=i$

in view of properties (2.3). The function representing the integrals in (3.17)
satisfies homogeneous conditions due to the boundary conditions (2.5).
Therefore, the solution $\psi(x)$ of the equation (3.17) satisfies the
condition

$\psi(-0)-e^{-i\alpha}\psi(+0)=-i\langle\psi,v\rangle,$

that is the boundary conditions for elements of the domain of the operator
$A(v,\alpha).$ The equation (3.17) can be reduced to a linear system of
equations for two numbers $\langle\psi,v\rangle$ and
$\psi_{s}=\frac{1}{2}\Bigl{(}\psi(-0)+e^{-i\alpha}\psi(+0)\Bigr{)}$. In order
to do this, one need to represent the quantities $\psi_{s}$ and
$\langle\psi,v\rangle$ on the left-hand side. Since any function of the form
$\displaystyle\varphi(x)=\int\,\mathcal{G}_{z}(x,y,\alpha)f(y)\,dy$, $f\in
L_{2}$ has the property

$\varphi_{s}=\frac{1}{2}\Bigl{(}\varphi(-0)+e^{-i\alpha}\varphi(+0)\Bigr{)}=\int\,\mathcal{G}_{z}(-0,y;\alpha)f(y)\,dy$

by virtue of boundary properties for the Green‘s function $\mathcal{G}_{z}$.
The function $E_{1}(x,z)$ satisfies the condition

$\frac{1}{2}\Bigl{(}E_{1}(-0,z)+e^{-i\alpha}E_{1}(+0,\alpha)\Bigr{)}=\frac{i}{2}\mathrm{sign}\,(\mathrm{Im}\,z).$

Then we obtain from (3.17) the following equality:

$\psi_{s}=-\frac{i}{2}\mathrm{sign}\,(\mathrm{Im}\,z)\langle\psi,v\rangle+\int\,\mathcal{G}(-0,y;\alpha)(h(y)-v(y)\psi_{s})\,ds.$
(3.18)

If the equation (3.17) is scalar multiplied on the right-hand side by the
function $v$ in $L_{2}$, then we obtain

$\langle\psi,v\rangle=-\langle
E_{1}(\cdot,z),v\rangle\langle\psi,v\rangle+\langle
h,E_{2}(\cdot,\bar{z})\rangle-\langle v,E_{2}(\cdot,x)\rangle\psi_{s}.$ (3.19)

The equations $(\ref{2.27d})$–(3.19) represent a linear system of the form

$\begin{pmatrix}-\gamma_{22}&\gamma_{12}\\\
\gamma_{21}&-\gamma_{11}\end{pmatrix}\begin{pmatrix}\langle\psi,v\rangle\\\
\psi_{s}\end{pmatrix}=\begin{pmatrix}\langle h,E_{1}\rangle\\\ \langle
h,E_{2}\rangle\end{pmatrix}.$ (3.20)

A matrix determinant

$\begin{vmatrix}-\gamma_{22}&\gamma_{12}\\\
\gamma_{21}&-\gamma_{11}\end{vmatrix}=\gamma(z)=\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21}$

for $|\mathrm{Im}\,z|\rightarrow\infty,$ $\gamma(z)\rightarrow-1.$ Therefore,
there are numbers $z_{1}$ and $z_{2}$ such that $\mathrm{Im}\,z_{1}>0$,
$\mathrm{Im}\,z_{2}<0$ with sufficiently large $|\mathrm{Im}\,z_{k}|$ that the
system (3.20) has a unique solution. Therefore, substituting the solution from
the system (3.20) into (3.17), we obtain that the equation (3.17) has a
solution for any $h\in L_{2}$ and for such $z=z_{k},\,\,k=1,2$. This means
that the symmetric operator $A(v,\alpha)$ has a deficiency index (0,0), and,
therefore, this operator is self-adjoint in the space $L_{2}$. Since the
solution to the system (3.20) can be represented as

$\begin{pmatrix}\langle\psi,v\rangle\\\
\psi_{s}\end{pmatrix}=\frac{1}{\gamma(z)}\begin{pmatrix}-\gamma_{11}&-\gamma_{12}\\\
-\gamma_{21}&-\gamma_{22}\end{pmatrix}\begin{pmatrix}\langle
h,E_{1}(\cdot,\bar{z})\rangle\\\ \langle
h,E_{2}(\cdot,\bar{z})\rangle\end{pmatrix},$ (3.21)

then substituting these solutions into (3.17), we obtain a representation of
the solution $\psi(x)$ of equation (3.17) for $h$ in the form (3.13) where the
Green‘s function $\mathcal{G}_{z}(x,y,v,\alpha)$ has the form (3.14) with the
specified parameter values in (3.16). ∎

Since the Green‘s function $\mathcal{G}_{z}(x,y,\alpha)$ is explicitly
expressed through the Green‘s function $g_{z}(x)$ according to formula (2.7),
then the Green‘s function $\mathcal{G}_{z}(x,y,v,\alpha)$ is explicitly
expressed through $\mathcal{G}_{z}(x,y,\alpha)$ according to Theorem 3.2.
Therefore, this function can be expressed explicitly in terms of $g_{z}(x)$.

######  Theorem 3.3.

For any potential $v\in L_{2}$ and a real number $\alpha$, an operator
$A(v,\alpha)$ of the form

$A(v,\alpha)\psi=i\frac{d\psi(x)}{dx}+v(x)\left[\frac{1}{2}\psi(-0)+\frac{e^{-i\alpha}}{2}\psi(+0)\right],\quad
x\neq 0$ (3.22)

with the boundary condition

$i\psi(-0)-ie^{-i\alpha}\psi(+0)=\langle\psi,v\rangle$ (3.23)

is self-adjoint in the space $L_{2}$. For nonreal $z$, the resolvent
$(A(v,\alpha)-zI)^{-1}$ is an integral operator with the kernel
$\mathcal{G}_{z}(x,y;v,\alpha)$

$(A(v,\alpha)-zI)^{-1}h(z)=\int\,\mathcal{G}_{z}(x,y;v,\alpha)h(y)\,dy.$
(3.24)

The kernel $\mathcal{G}_{z}(x,y;v,\alpha)$ is a perturbation of the kernel
$g_{z}(x-y)$ of the free momentum operator $(\ref{2.2})$ of rank at most $2$:

$\mathcal{G}_{z}(x,y;v,\alpha)=g_{z}(x-y)-\frac{1}{\gamma(z)}\sum_{j,k=1}^{2}\,e_{j}(x,z)\gamma_{j,k}(z)\bar{e}_{k}(y,\bar{z}),$
(3.25)

where

$\begin{array}[]{ll}\gamma(z)=\gamma_{11}(z)\gamma_{22}(z)-\gamma_{12}(z)\gamma_{21}(z),\\\\[5.69054pt]
\displaystyle\gamma_{11}(z)=\langle
e_{2},v\rangle-\frac{2i(1-e^{i\alpha})}{1+e^{i\alpha}},\\\\[5.69054pt]
\displaystyle\gamma_{12}(z)=-\frac{2}{1+e^{-i\alpha}}-\langle
v,g_{\overline{z}}\rangle,\\\\[5.69054pt]
\displaystyle\gamma_{21}(z)=-\frac{2}{1+e^{i\alpha}}-\langle
g_{z},v\rangle,\\\\[5.69054pt]
\gamma_{22}(z)=\frac{i}{1+e^{-i\alpha}}\left[\theta(\mathrm{Im}\,z)-e^{-i\alpha}\theta(-\mathrm{Im}\,z)\right],\\\\[5.69054pt]
e_{1}(x,z)=g_{z}(x),\,e_{2}(x,z)=\int\,g_{z}(x-y)v(y)\,dy.\end{array}$ (3.26)

###### Proof.

The operator $A(v,\alpha)$ is self-adjoint in the space $L_{2}$ according to
Theorem 3.2. Then, it is sufficient to show that for any $h\in L_{2}$ and some
nonreal $z$ there exists a solution of the problem

$(A(v,\alpha)-zI)^{-1}\psi(x)=h.$ (3.27)

This problem can be reduced to the problem

$\psi(x)=-g_{z}(x)\psi_{g}-e_{2}(x,z)\psi_{s}+\int\,g_{z}(x-y)h(y)\,dy$ (3.28)

with the boundary condition $(\ref{2.17d})$, where
$\psi_{g}=i\left[\psi(-0)-\psi(+0)\right]$,
$\psi_{s}=\frac{1}{2}\left[\psi(-0)+e^{-i\alpha}\psi(+0)\right]$. This is
equivalent to solving a linear system with respect to $\psi_{s}$ and
$\psi_{g}$. If we substitute $\psi_{s}$ from $(\ref{2.22d})$, and the solution
$\psi(x)$ from $(\ref{2.22d})$ to the boundary condition $(\ref{2.17d})$, we
get a linear system

$\begin{pmatrix}a_{11}(z)&a_{12}(z)\\\
a_{21}(z)&a_{22}(z)\end{pmatrix}\begin{pmatrix}\psi_{g}\\\
\psi_{s}\end{pmatrix}=\begin{pmatrix}\langle h,g_{\bar{z}}\rangle\\\ \langle
h,e_{2}(\cdot,\bar{z})\rangle\end{pmatrix},$ (3.29)

where

$\begin{array}[]{ll}\displaystyle
a_{11}(z)=\frac{i}{1+e^{-i\alpha}}\left[\theta(\mathrm{Im}\,z)-e^{-i\alpha}\theta(-\mathrm{Im}\,z)\right],\\\\[5.69054pt]
\displaystyle a_{12}(z)=\frac{2}{1+e^{-i\alpha}}+\langle
v,g_{\overline{z}}\rangle,\\\\[5.69054pt] \displaystyle
a_{21}(z)=\frac{2}{1+e^{i\alpha}}+\langle g_{z},v\rangle,\\\\[5.69054pt]
\displaystyle a_{22}(z)=\langle
e_{2},v\rangle-\frac{2i(1-e^{i\alpha})}{1+e^{i\alpha}}.\end{array}$

The determinant of the system $(\ref{2.23d})$ is equal to

$\displaystyle\gamma(z)$
$\displaystyle=\left[\frac{i}{1+e^{-i\alpha}}\left(\theta(\mathrm{Im}\,z)-e^{-i\alpha}\theta(-\mathrm{Im}\,z)\right)\right]\left[\langle
e_{2},v\rangle-\frac{2i(1-e^{i\alpha})}{1+e^{i\alpha}}\right]$
$\displaystyle-\left[\frac{2}{1+e^{-i\alpha}}+\langle
v,g_{\overline{z}}\rangle\right]\left[\frac{2}{1+e^{i\alpha}}+\langle
g_{z},v\rangle\right].$

It is easy to see that if $|\mathrm{Im}\,z_{j}|$, $j=1,2$ are large enough,
then $\gamma(z_{j})\neq 0$. Therefore, the system $(\ref{2.23d})$ has a unique
solution. Substituting the obtained values of $\psi_{g}$ and $\psi_{s}$ into
$(\ref{2.22d})$, we get a representation of the solution $\psi(x)$ in terms of
$h$ as follows:

$\psi(x)=\int\,\mathcal{G}_{z}(x,y;v,\alpha)h(y)\,dy,$ (3.30)

where $\mathcal{G}_{z}(x,y;v,\alpha)$ has the form
$(\ref{2.19d})$–$(\ref{2.20d})$ with

$\begin{array}[]{ll}\gamma_{11}(z)=a_{22}(z),\\\\[5.69054pt]
\gamma_{12}(z)=-a_{12}(z),\\\\[5.69054pt]
\gamma_{21}(z)=-a_{21}(z),\\\\[5.69054pt]
\gamma_{22}(z)=a_{11}(z).\\\\[5.69054pt] \end{array}$ (3.31)

The matrix $\Gamma(z)=\|\gamma_{jk}(z)\|$ has a property
$\Gamma^{*}(\overline{z})=\Gamma(z)$ and

$\gamma(z)=a_{11}(z)a_{22}(z)-a_{21}(z)a_{12}(z)=\gamma_{11}(z)\gamma_{22}(z)-\gamma_{12}(z)\gamma_{21}(z).$

∎

######  Remark 3.2.

Since the operator $A(v_{1},v_{2},\alpha)$ differs from the operator
$A(v,\alpha)$ by a symmetric operator of rank at most 2 according to Lemma
3.1, then its resolvent differs from the resolvent
$\mathcal{G}_{z}(x,y,v,\alpha)$ by an operator of rank 2.

Recall that if the self-adjoint operator $A$ is perturbed in a Hilbert space
$H$ by the bounded operator

$V\psi=v_{1}\langle\psi,v_{2}\rangle+v_{2}\langle\psi,v_{1}\rangle,\quad
v_{1},v_{2}\in H,$

where $v_{1},v_{2}\in H,$ then the resolvent

$[(A+V-zI)^{-1}]h=R_{z}h-(R_{z}v_{1},R_{z}v_{2})\cdot\Gamma^{-1}(z)\cdot\begin{pmatrix}\langle
h,v_{1}\rangle\\\ \langle h,v_{2}\rangle\end{pmatrix}$

where $R_{z}=(A-zI)^{-1}$ and a numeric matrix has the following form:

$\Gamma(z)=\begin{pmatrix}\langle R_{z}v_{1},v_{1}\rangle&1+\langle
R_{z}v_{1},v_{2}\rangle\\\ 1+\langle R_{z}v_{2},v_{1}\rangle&\langle
R_{z}v_{2},v_{2}\rangle\end{pmatrix}.$

An explicit formula for the resolvent of the operator $A(v,\alpha)$ from
Theorem 3.2, Theorem 3.3 and Remark 3.1 allows one to get a representation of
the Green‘s function of operators $A(v_{1},v_{2},\alpha)$ in the form of rank
4 perturbations of the Green‘s function $g_{z}(x)$ of the free momentum
operator and the operator with one potential. However, it is possible to
obtain directly a representation of the resolvent by a rank 3 perturbation.

######  Theorem 3.4.

For potentials $v_{1},v_{2}\in L_{2}$ and a real number $\alpha\in[0,2\pi)$
describing the boundary conditions (3.2), the operator $A(v_{1},v_{2},\alpha)$
is self-adjoint in the space $L_{2}$. For nonreal $z$, its resolvent
$(A(v_{1},v_{2},\alpha)-zI)^{-1}$ is an integral operator with the kernel
$\mathcal{G}_{z}(x,y;v_{1},v_{2},\alpha)$

$(A(v_{1},v_{2},\alpha)-zI)^{-1}h(z)=\int\,\mathcal{G}_{z}(x,y;v_{1},v_{2},\alpha)h(y)\,dy.$
(3.32)

The kernel $\mathcal{G}_{z}(x,y;v,\alpha)$ has the following representation:

$\displaystyle\mathcal{G}_{z}(x,y;v_{1},v_{2},\alpha)=\mathcal{G}_{z}(x,y,\alpha)-\frac{1}{\gamma(z)}\sum\limits_{j,k=1}^{2}\mathcal{E}_{j}(x,z)\gamma_{jk}\bar{\mathcal{E}}_{k}(y,\bar{z}),$
(3.33)

where

$\begin{array}[]{ll}E_{0}(x,z)=\mathcal{G}_{z}(x,-0;\alpha),&\quad
E_{k}(x,z)=\int\,\mathcal{G}_{z}(x,y;\alpha)v_{k}(y)\,dy,\quad
k=1,2,\\\\[5.69054pt]
\mathcal{E}_{1}(x,z)=E_{1}+2iE_{0},&\quad\mathcal{E}_{2}(x,z)=E_{2}-2ie^{-i\alpha}E_{0},\end{array}$
(3.34)

and

$\begin{array}[]{l}\gamma(z)=\gamma_{11}(z)\gamma_{22}(z)-\gamma_{12}(z)\gamma_{21}(z),\\\\[5.69054pt]
\gamma_{11}(z)=2i\left(\mathrm{sign}\,(\mathrm{Im}\,z)+e^{i\alpha}\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle-e^{-i\alpha}\langle
E_{0},v_{2}\rangle-\frac{i}{2}\langle
E_{2},v_{2}\rangle\right),\\\\[5.69054pt]
\gamma_{12}(z)=2i\left(2e^{-i\alpha}\theta(\mathrm{Im}\,z)+\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle+e^{-i\alpha}\langle
E_{0},v_{1}\rangle+\frac{i}{2}\langle
E_{2},v_{1}\rangle\right),\\\\[5.69054pt]
\gamma_{21}(z)=-2i\left(2e^{i\alpha}\theta(-\mathrm{Im}\,z)+e^{i\alpha}\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle+\langle
E_{0},v_{2}\rangle-\frac{i}{2}\langle
E_{1},v_{2}\rangle\right),\\\\[5.69054pt]
\gamma_{22}(z)=-2i\left(-\mathrm{sign}\,(\mathrm{Im}\,z)+\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle-\langle
E_{0},v_{1}\rangle+\frac{i}{2}\langle
E_{1},v_{1}\rangle\right).\\\\[5.69054pt] \end{array}$ (3.35)

The matrix $\Gamma(z)=\|\gamma_{jk}(z)\|$ has the property
$\Gamma^{*}(\overline{z})=\Gamma(z),$ and $\gamma(z)=\det\Gamma$ has the
property $\overline{\gamma}(\overline{z})=\gamma(z)\neq 0$.

###### Proof.

For the symmetric operator $A(v_{1},v_{2},\alpha)$, the construction of the
Green‘s function is reduced to the equation

$\psi(x)=-\mathcal{E}_{1}(x,z)\psi_{1}-\mathcal{E}_{2}(x,z)\psi_{2}+\int\,\mathcal{G}_{z}(x,y;\alpha)h(y)dy,$
(3.36)

where

$\begin{array}[]{ll}E_{0}(x,z)=\mathcal{G}_{z}(x,-0;\alpha),&\quad
E_{k}(x,z)=\int\,\mathcal{G}_{z}(x,y;\alpha)v_{k}(y)\,dy,\quad
k=1,2,\\\\[5.69054pt]
\mathcal{E}_{1}(x,z)=E_{1}+2iE_{0},&\quad\mathcal{E}_{2}(x,z)=E_{2}-2ie^{-i\alpha}E_{0},\end{array}$
(3.37)

and

$\displaystyle\psi_{1}=\psi(-0)+\frac{i}{2}\langle\psi,v_{1}\rangle,\quad\psi_{2}=\psi(+0)-\frac{i}{2}\langle\psi,v_{2}\rangle.$
(3.38)

Indeed, the function $\psi(x)$ of $(\ref{3.30})$ satisfies the differential
equation $(A(v_{1},v_{2},\alpha)-z_{k}I)\psi=h$ for any $h\in L_{2}$ and
nonreal $z_{k}$ such that $\mathrm{Im}\,z_{1}>0$ and $\mathrm{Im}\,z_{2}<0$.
The equation $(\ref{3.30})$ has the following equivalent form:

$\psi(x)=E_{0}(x;z)[-2i\psi_{1}+2ie^{-i\alpha}\psi_{2}]-E_{1}(x,z)\psi_{1}-E_{2}(x,z)\psi_{2}+\int\,\mathcal{G}_{z}(x,y;\alpha)h(y)dy.$
(3.39)

From the equation $(\ref{3.31})$, using the fact that
$\mathcal{G}_{z}(+0,y;\alpha)=e^{i\alpha}\mathcal{G}_{z}(-0,y;\alpha)$, we
obtain

$\begin{array}[]{l}\psi(-0)=2\left(\psi_{1}-e^{-i\alpha}\psi_{2}\right)\theta(\mathrm{Im}\,z)-\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle\psi_{1}-\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle\psi_{2}+\langle
h,E_{0}(\cdot,{\bar{z}})\rangle,\\\\[5.69054pt]
\psi(+0)=-2e^{i\alpha}\left(\psi_{1}-e^{-i\alpha}\psi_{2}\right)\theta(-\mathrm{Im}\,z)-e^{i\alpha}\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle\psi_{1}-e^{i\alpha}\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle\psi_{2}+e^{i\alpha}\langle
h,E_{0}(\cdot,{\bar{z}})\rangle,\\\
\langle\psi,v_{k}\rangle=-2i\left(\psi_{1}-e^{-i\alpha}\psi_{2}\right)\langle
E_{0},v_{k}\rangle-\psi_{1}\langle E_{1},v_{k}\rangle-\psi_{2}\langle
E_{2},v_{k}\rangle+\langle h,E_{k}(\cdot,{\bar{z}})\rangle,\quad
k=1,2.\end{array}$ (3.40)

If we substitute the values $\psi(-0)$, $\psi(+0)$, $\langle\psi,v_{1}\rangle$
and $\langle\psi,v_{2}\rangle$ from $(\ref{3.32})$ to $(\ref{3.33})$, we get
the explicit form of the system for $\psi_{1}$ and $\psi_{2}$. It has the
following form:

$\begin{pmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{pmatrix}\begin{pmatrix}\psi_{1}\\\
\psi_{2}\end{pmatrix}=\begin{pmatrix}\langle
h,\mathcal{E}_{1}(\cdot,{\bar{z}})\rangle\\\ \langle
h,\mathcal{E}_{2}(\cdot,{\bar{z}})\rangle\end{pmatrix},$ (3.41)

where

$\begin{array}[]{ll}\gamma(z)=a_{11}a_{22}-a_{12}a_{21}=\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21}\\\
\gamma_{11}(z)=a_{22}(z)&\gamma_{12}(z)=-a_{12}(z)\\\
\gamma_{21}(z)=-a_{21}(z)&\gamma_{22}(z)=a_{11}(z),\end{array}$ (3.42)

and

$\begin{array}[]{l}a_{11}(z)=-2i\left(-\mathrm{sign}\,(\mathrm{Im}\,z)+\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle-\langle
E_{0},v_{1}\rangle+\frac{i}{2}\langle
E_{1},v_{1}\rangle\right),\\\\[5.69054pt]
a_{12}(z)=-2i\left(2e^{-i\alpha}\theta(\mathrm{Im}\,z)+\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle+e^{-i\alpha}\langle
E_{0},v_{1}\rangle+\frac{i}{2}\langle
E_{2},v_{1}\rangle\right),\\\\[5.69054pt]
a_{21}(z)=2i\left(2e^{i\alpha}\theta(-\mathrm{Im}\,z)+e^{i\alpha}\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle+\langle
E_{0},v_{2}\rangle-\frac{i}{2}\langle
E_{1},v_{2}\rangle\right),\\\\[5.69054pt]
a_{22}(z)=2i\left(\mathrm{sign}\,(\mathrm{Im}\,z)+e^{i\alpha}\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle-e^{-i\alpha}\langle
E_{0},v_{2}\rangle-\frac{i}{2}\langle
E_{2},v_{2}\rangle\right).\\\\[5.69054pt] \end{array}$

A determinant

$\gamma(z)=a_{11}a_{22}-a_{12}a_{21}\rightarrow-4\ ,\ \textrm{for}\ \
|\mathrm{Im}\,z|\rightarrow\infty,$

since

$\begin{array}[]{ll}a_{11}(z)=2i\mathrm{sign}\,(\mathrm{Im}\,z)+o(1),&a_{12}(z)=-4ie^{-i\alpha}\theta(\mathrm{Im}\,z)+o(1),\\\\[5.69054pt]
a_{21}(z)=4ie^{i\alpha}\theta(-\mathrm{Im}\,z)+o(1),&a_{22}(z)=2i\mathrm{sign}\,(\mathrm{Im}\,z)+o(1),\,\,|\mathrm{Im}\,z|\rightarrow\infty.\end{array}$

Therefore, there exist nonreal $z_{1}$, $z_{2}$ such that
$\mathrm{Im}\,z_{1}>0,$ $\mathrm{Im}\,z_{2}<0$ and $|\mathrm{Im}\,z_{k}|$,
$j=1,2$ large enough that $\gamma(z)\neq 0$ and the system $(\ref{eq3a})$ has
a unique solution. This gives us the self-adjointness of the operator
$A(v_{1},v_{2},\alpha)$. Moreover, we can show that
$\bar{a}_{jk}(\bar{z})=a_{kj}(z)$, then
$\bar{\gamma}_{jk}(\bar{z})=\gamma_{kj}(z)$. Therefore the matrix
$\Gamma(z)=\|\gamma_{jk}(z)\|$ has the property
$\Gamma^{*}(\overline{z})=\Gamma(z),$ and $\gamma(z)=\det\Gamma$ has the
property $\overline{\gamma}(\overline{z})=\gamma(z)$. A substitution of
$\psi_{1}$ and $\psi_{2}$ to the equation $(\ref{3.30})$ gives the explicit
form of the Green‘s function $(\ref{3.31d})$. ∎

######  Remark 3.3.

If we substitute $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ from $(\ref{3.32d})$
to $(\ref{3.31d})$, we obtain the explicit form of Green‘s function that
depends on $E_{0},E_{1},E_{2}$:

$\mathcal{G}_{z}(x,y;v_{1},v_{2},\alpha)=\mathcal{G}_{z}(x,y;\alpha)-\frac{1}{\gamma(z)}\sum_{j,k=0}^{2}\,E_{j}(x,z)c_{j,k}(z)\bar{E}_{k}(y,\bar{z}),$
(3.43)

where

$E_{0}(x,z)=\mathcal{G}_{z}(x,-0;\alpha),\quad
E_{k}(x,z)=\int\,\mathcal{G}_{z}(x,y;\alpha)v_{k}(y)\,dy,\ \ k=1,2$ (3.44)

and

$\begin{array}[]{lll}c_{00}(z)=-2i\gamma_{22}+2ie^{i\alpha}\gamma_{12}+2i\gamma_{11}-2ie^{-i\alpha}\gamma_{21}&c_{01}(z)=\gamma_{22}+e^{-i\alpha}\gamma_{21}&c_{02}(z)=\gamma_{12}+e^{-i\alpha}\gamma_{11}\\\
c_{10}(z)=-\gamma_{22}+e^{i\alpha}\gamma_{12}&c_{11}(z)=-\frac{i}{2}\gamma_{22}&c_{12}(z)=-\frac{i}{2}\gamma_{12}\\\
c_{20}(z)=-e^{i\alpha}\gamma_{11}+\gamma_{21}&c_{21}(z)=\frac{i}{2}\gamma_{21}&c_{22}(z)=\frac{i}{2}\gamma_{11}.\end{array}$
(3.45)

The values $\gamma_{jk}(z)$ are given in Theorem 3.4. The matrix
$||c_{jk}||_{jk=0}^{2}$ is degenerate. Its first column is a linear
combination of second and third ones:

$\mathrm{col}\,(c_{00},c_{10},c_{20})=-2i\mathrm{col}\,(c_{01},c_{11},c_{21})+2ie^{i\alpha}\mathrm{col}\,(c_{02},c_{12},c_{22}).$

The equalities $c_{jk}(z)=\bar{c}_{kj}(\bar{z}),$
$\gamma(z)=\bar{\gamma}(\bar{z})\neq 0$ are valid providing the self-adjoint
property of the operator $A_{(v_{1},v_{2},\alpha)}:$

${\bar{\mathcal{G}}}_{\bar{z}}(y,x;v_{1},v_{2},\alpha)=\mathcal{G}_{z}(x,y;v_{1},v_{2},\alpha).$

##  4 Spectral analysis of the momentum operator with a nonlocal potential

An operator $A(v,\alpha)$ of the form (3.11) with a nonlocal potential $v\in
L_{2}(-\infty,+\infty)$ is self-adjoint in the space $L_{2}(-\infty,+\infty)$.
Its resolvent $R_{z}(v,\alpha)=(A(v,\alpha)-zI)^{-1}$ for nonreal $z$ is an
integral operator with kernel $\mathcal{G}_{z}(x,y;v,\alpha)$ which is a
perturbation of rank at most 2 of the Green‘s function of the free operator
$A_{0}=i\frac{d}{dx}$ with a domain $W_{2}^{1}(-\infty,+\infty)$. The spectrum
of the self-adjoint operator $A_{0}$ is absolutely continuous and is composed
of the whole real axis. Therefore, the operator $A(v,\alpha)$, in addition to
the absolutely continuous part of the spectrum, also has ordinary eigenvalues
with eigenfunctions belonging to the space $L_{2}(-\infty,+\infty)$.

######  Theorem 4.1.

Let a self-adjoint operator $A(v,\alpha)$ of the form (3.11) have a nonlocal
potential $v\in L_{2}(-\infty,+\infty)$. Then the real number $\lambda$ is an
eigenvalue of operator $A(v,\alpha)$ if and only if the function

$\psi_{\lambda}(x)=\theta(x)\left(-i\int_{x}^{\infty}\,e^{-i\lambda(x-y)}v(y)\,dy\right)+\theta(-x)\left(i\int_{-\infty}^{x}\,e^{-i\lambda(x-y)}v(y)\,dy\right)$
(4.1)

belongs to the space $L_{2}(R^{1})$ and two conditions are satisfied

$\begin{array}[]{l}\psi_{\lambda}(-0)+e^{-i\alpha}\psi_{\lambda}(+0)=2,\\\
\psi_{\lambda}(-0)-e^{-i\alpha}\psi_{\lambda}(+0)=-i\langle\psi_{\lambda},v\rangle_{L_{2}}.\end{array}$
(4.2)

If these conditions are satisfied, then the function $\psi_{\lambda}(x)$ is an
eigenfunction of the operator $A(v,\alpha)$ with an eigenvalue $\lambda$

$A(v,\alpha)\psi_{\lambda}(x)=\lambda\psi_{\lambda}(x).$ (4.3)

Each eigenvalue always has a multiplicity of one.

###### Proof.

If $\lambda$ is an eigenvalue of the operator $A(v,\alpha)$, then its
eigenfunction is satisfied (4.3). Therefore, for $x\neq 0$ the equation holds

$i\frac{d\psi_{\lambda}(x)}{dx}+v(x)\frac{1}{2}\Bigl{(}\psi_{\lambda}(-0)+e^{-i\alpha}\psi_{\lambda}(+0)\Bigr{)}=\lambda\psi_{\lambda}(x),\quad
x\neq 0.$ (4.4)

The value $\psi_{\lambda}(-0)+e^{-i\lambda}\psi_{\lambda}(+0)\neq 0$ because
then the nontrivial solution (4.4) does not belong to the space $L_{2}$. Since
the eigenfunction $\psi_{\lambda}$ is determined up to a scalar non-zero
factor, then, without a loss of generality, we can assume that the first
condition in (4.2) is satisfied. Then the solution (4.4) takes the form (4.1).
This solution decreases as $|x|\rightarrow\infty$, but must belong to the
space $L_{2}.$ In addition, since $\psi_{\lambda}\in\mathcal{D}(A(v,\alpha))$,
then the second condition in (4.2) holds. Otherwise, if the conditions
(4.1)–(4.2) are satisfied, then the function $\psi_{\lambda}(x)$ of the form
(4.1) is a solution (4.4), i.e., it is an eigenfunction with an eigenvalue
$\lambda$. Since the eigenfunction for the eigenvalue $\lambda$ takes the form
(4.1), then the eigenvalue $\lambda$ has a multiplicity one. ∎

We will demonstrate the efficiency of the conditions in Theorem 4.1 using the
following examples.

######  Example 4.1.

Let the potential $v(x)$ in the operator $A(v,\alpha)$ be equal to the complex
constant $k$ in the interval $x\in(0,1)$ and $v(x)\equiv 0$ for
$x\notin[0,1].$ Then the operator $A(v,\alpha)$ has an eigenvalue $\lambda=0$
only with $k=2ie^{i\alpha}.$ In this case, the eigenvalue is unique.

###### Proof.

It is easy to check all the conditions in Theorem 4.1 with $\lambda=0$ and
$k=2ie^{i\alpha}.$ The eigenfunction $\psi_{0}(x)=2e^{i\alpha}(1-x)$ with
$x\in(0,1)$ and $\psi_{0}(x)\equiv 0$ with $x\notin[0,1].$ If we assume
$\lambda\neq 0$ with chosen $k$, then the first condition in (4.2) has the
form $e^{-i\alpha}(2ie^{i\alpha}\frac{1-e^{i\alpha}}{\lambda})=2.$ The modulus
of the left-hand side in this equality is equal to $\displaystyle
2\frac{\sin\frac{\lambda}{2}}{\frac{\lambda}{2}}$ and is strictly less than 2
for $\lambda\neq 0$. Therefore, there are no other eigenvalues for the chosen
potential, except $\lambda=0$ . ∎

######  Example 4.2.

Let the nonlocal potential be $v(x)=\theta(x)ke^{-(1+i\gamma)x}$, where
$\gamma$ is a real number. If $k=e^{i\alpha}2i$, then the number
$\lambda=\gamma$ is the only eigenvalue with the eigenfunction
$\psi_{\gamma}(x)=\theta(x)(-ike^{-x})$ of the self-adjoint operator
$A(v,\alpha)$ with such nonlocal potential.

###### Proof.

From Theorem 4.1, we have

$\psi_{\lambda}=\theta(x)\Bigl{(}-i\int_{x}^{\infty}\,e^{-i\alpha(\lambda-y)}v(y)\,dy\Bigr{)}=\theta(x)\frac{1}{1+i(\gamma-\lambda)}(-iv(x)).$

It is easy to check that conditions (4.2) are satisfied only for
$\lambda=\gamma$. ∎

######  Example 4.3.

An operator $A(v,0)$ with a nonlocal potential $v(x)=2i\mathrm{sign}\,x\cdot
e^{-|x|}$ is a self-adjoint operator in the space $L_{2}(-\infty,+\infty)$.
This operator has only two simple (non-multiple) eigenvalues $\lambda=1,$
$\lambda=-1$ and absolutely continuous spectrum which is the entire real line.
The eigenfunctions for the specified eigenvalues can be represented as

$\psi_{\lambda}(x)=\frac{2}{1+\lambda^{2}}e^{-|x|}(1+i\lambda\mathrm{sign}\,x).$
(4.5)

###### Proof.

The specified function $\psi_{\lambda}\in L_{2}(-\infty,+\infty)$ satisfies
equation

$i\frac{d\psi_{\lambda}}{dx}+v(x)=\lambda\psi_{\lambda},\quad x\neq 0$

and conditions

$\psi_{\lambda}(+0)=1+i\lambda,\quad\psi_{\lambda}(-0)=1-i\lambda,\quad\langle\psi_{\lambda},v\rangle=2\quad\text{with}\,\,\lambda=1\,\,\,\text{and}\,\,\,\lambda=-1.$

Therefore, the function $\psi_{\lambda}$ belongs to the domain of the operator
$A(v,0)$ and satisfies equality

$A(v,0)\psi_{\lambda}=\lambda\psi_{\lambda}\quad\text{with}\,\,\,\,\lambda=1\,\,\,\,\text{and}\,\,\,\,\lambda=-1.$

Therefore, $\psi_{\lambda}$ is an eigenfunction of the operator $A(v,0)$ with
eigenvalues $\lambda=1,$ $\lambda=-1$. Since the condition

$\frac{1}{2}(\psi_{\lambda}(-0)+\psi_{\lambda}(+0))=2$

is satisfied only for $\lambda=1$ and $\lambda=-1,$ then the operator $A(v,0)$
has no other eigenvalues. ∎

##  5 Operators on a finite interval

Consider the first order differential expression $A(v_{1},v_{2})$ on a finite
interval $\Omega=(0,1)$ with nonlocal potentials $v_{1}$ and $v_{2},$ where
$v_{1}$ and $v_{2}$ are complex-valued functions from the space
$L_{2}(\Omega)$. The action of the expression $A(v_{1},v_{2})$ on an arbitrary
$\psi$ function from the Sobolev space $W_{2}^{1}(\Omega)$ is given by

$A(v_{1},v_{2})\psi(x)=i\frac{d\psi(x)}{dx}+v_{1}(x)\left[\psi(0)-\frac{i}{2}\langle\psi,v_{1}\rangle_{L_{2}}\right]+v_{2}(x)\left[\psi(1)+\frac{i}{2}\langle\psi,v_{2}\rangle_{L_{2}}\right].$
(5.1)

The equality $(\ref{21.1})$ defines a maximal operator in the space
$L_{2}(\Omega)$, whose domain is the space $W_{2}^{1}(\Omega)$. This operator
will be denoted by $A(v_{1},v_{2})$.

######  Lemma 5.1.

The operator $A(v_{1},v_{2})$ satisfies the Green‘s formula, which for two
arbitrary functions $\psi,\varphi\in W_{2}^{1}(\Omega)$ has the form

$\begin{array}[]{rr}\langle
A\psi,\varphi\rangle_{L_{2}}-\langle\psi,A\varphi\rangle_{L_{2}}=i\\{\left[\psi(1)+i\langle\psi,v_{2}\rangle\right]\cdot\left[\varphi(1)+i\langle\varphi,v_{2}\rangle\right]^{*}-\\\
-\left[\psi(0)-i\langle\psi,v_{1}\rangle\right]\cdot\left[\varphi(0)-i\langle\varphi,v_{1}\rangle\right]^{*}\\}.\end{array}$
(5.2)

###### Proof.

Substituting $(\ref{21.1})$ into the left-hand side of $(\ref{21.2})$, it
shows the same as expanding the parentheses in the right-hand side of
$(\ref{21.2})$. ∎

######  Definition 5.1.

For the operator $A(v_{1},v_{2})$ and a real number $\alpha\in[0,2\pi),$
consider the operator $A(v_{1},v_{2},\alpha)$ in the space $L_{2}(\Omega)$ as
a restriction of operator $A(v_{1},v_{2})$ on all functions of space
$W_{2}^{1}(\Omega)$ that satisfy the boundary condition

$\psi(1)+i\langle\psi,v_{2}\rangle_{L_{2}}=e^{i\alpha}[\psi(0)-i\langle\psi,v_{1}\rangle_{L_{2}}].$
(5.3)

We will denote the operator $A(0,0,\alpha)$ by $A_{\alpha}$ in the case
$v_{1}\equiv 0$ and $v_{2}\equiv 0.$

######  Lemma 5.2.

The operator $A(v_{1},v_{2},\alpha)$ is symmetric in the space
$L_{2}(\Omega).$

This is a simple corollary of Lemma 5.1.

######  Remark 5.1.

In Definition 5.1 of operator $A(v_{1},v_{2},\alpha)$ with nonlocal potentials
$v_{1},\,v_{2}\in L_{2}(0,1),$ the domain of the operator is defined as the
set of all functions $\psi\in W_{2}^{1}(0,1)$ satisfying the boundary
condition

$\psi(1)-e^{i\alpha}\psi(0)=-i\Bigl{[}\langle\psi,v_{2}\rangle+e^{i\alpha}\langle\psi,v_{1}\rangle\Bigr{]}.$

This condition can be equivalently represented by the functionals

$\psi_{1}=\psi(0)-\frac{i}{2}\langle\psi,v_{1}\rangle,\quad\psi_{2}=\psi(1)+\frac{i}{2}\langle\psi,v_{2}\rangle$

included in the definition of the operator $A(v_{1},v_{2},\alpha)$ in the form

$\psi(1)-e^{i\alpha}\psi(0)=2[\psi_{2}-e^{i\alpha}\psi_{1}].$ (5.4)

######  Theorem 5.1.

The operator $A_{\alpha}$ is a self-adjoint operator in the space
$L_{2}(\Omega)$. The spectrum of this operator consists of eigenvalues, all
numbers of the form

$\lambda_{n}=-\alpha+2\pi n,\quad\textrm{where}\quad n=0,\pm 1,\pm 2,\ldots,$
(5.5)

and the eigenfunctions are $\psi_{n}(x)=e^{-i\lambda_{n}x}.$ For any values of
the complex number $z$ that differ from the indicated eigenvalues, there is a
resolvent $(A_{\alpha}-zI)^{-1}$, which is an integral operator with a kernel
$g(x,y;z)$, that is,

$(A_{\alpha}-zI)^{-1}h(x)=\int_{0}^{1}g(x,y;z)h(y)dy.$ (5.6)

The kernel $g(x,y;z)$ is explicitly expressed as

$g(x,y;z)=e^{-iz(x-y)}[-i\theta(x-y)+\beta],\quad\beta=\frac{ie^{-iz}}{e^{-iz}-e^{i\alpha}}.$
(5.7)

###### Proof.

Let us prove representations $(\ref{21.5})$, $(\ref{21.6})$. For this,
consider the boundary value problem

$i\frac{d\psi(x)}{dx}-z\psi(x)=h(x),\quad\psi(1)=e^{i\alpha}\psi(0).$ (5.8)

The general solution of equation $(\ref{21.7})$ has the form

$\psi(x)=Ce^{-izx}-i\int_{0}^{x}e^{-iz(x-y)}h(y)dy,$ (5.9)

where $C$ is an arbitrary constant. We choose it so that the solution
$(\ref{21.8})$ satisfies the boundary conditions $(\ref{21.7})$. This leads to
the following equation, where $C:$

$C[e^{-iz}-e^{i\alpha}]=ie^{-iz}\int_{0}^{1}e^{izy}h(y)dy.$ (5.10)

If $z\neq\lambda_{n},$ then there is a unique solution of equation
$(\ref{21.9})$. Substituting this value of $C$ in $(\ref{21.8})$, we obtain
$(\ref{21.5})$ and $(\ref{21.6})$. The self-adjointness of the operator
$A_{\alpha}$ follows from its symmetry due to the Green‘s formula and the
existence of bounded operators $(A_{\alpha}-zI)^{-1}$ for all non-real $z$.
Note that each eigenvalue $\lambda_{n}$ has its own eigenfunction

$\psi_{n}(x)=e^{-i\lambda_{n}x}.$ (5.11)

∎

######  Remark 5.2.

The free moment operator $L=i\frac{d}{dx}$ is self-adjoint on the space
$L_{2}(-\infty,\infty)$ with the domain
$\mathcal{D}(L)=W_{2}^{1}(-\infty,\infty).$ The resolvent $(L-zI)^{-1},$ if
$\mathrm{Im}\,z\neq 0$, is the integral operator

$(L-zI)^{-1}h=\int g_{z}(x-y)h(y)dy,$

where the kernel of the integral operator is Green‘s function

$g_{z}(x)=i\mathrm{sign}\,(\mathrm{Im}\,z)\cdot\theta(-\mathrm{Im}\,zx)e^{-izx}.$
(5.12)

The Green‘s function $g(x,y;z)$ of the form $(\ref{21.6})$ is a rank one
perturbation of function $g_{z}:$

$g(x,y;z)=g_{z}(x-y)+e^{-izx}(\beta-i\theta(\mathrm{Im}\,z))e^{izy}=g_{z}(x-y)+e(x,z)w(z,\alpha)\bar{e}(y,\bar{z}),$
(5.13)

where

$\begin{array}[]{l}\displaystyle e(x,z)=g_{z}(x)+g_{z}(x-1),\quad
w(z,\alpha)=i\left[\frac{1}{e^{iz}-e^{-i\alpha}}\theta(\mathrm{Im}\,z)-\frac{1}{e^{-iz}-e^{i\alpha}}\theta(-\mathrm{Im}\,z)\right],\\\\[5.69054pt]
\bar{w}(\bar{z},\alpha)=w(z,\alpha).\end{array}$

######  Remark 5.3.

From the explicit form $(\ref{21.6})$ of the resolvent kernel $g(x,y;z)$, we
have the following equality

$g(1,y;z)=e^{i\alpha}g(0,y;z)=e^{i\alpha}\beta(z)e^{izy},\quad\textrm{where}\quad\beta=\frac{ie^{-iz}}{e^{-iz}-e^{i\alpha}}$
(5.14) $\overline{g}(y,x;z)=g(x,y;\overline{z}).$ (5.15)

Note that the equality $(\ref{21.12})$ is equivalent to the equality of the
operators
$\left[(A_{\alpha}-zI)^{-1}\right]^{*}=(A_{\alpha}-\overline{z}I)^{-1}$, which
is equivalent to a self-adjointness of the operator $A_{\alpha}$.

######  Theorem 5.2.

Consider two operators with several potentials $A(v_{1},v_{2},\alpha)$ and
$A(v,0,\alpha)$, where $v(x)=v_{1}(x)+e^{i\alpha}v_{2}(x)$. The domains of
definition of these operators are equal and described by the appropriate
boundary conditions $(\ref{21.3})$

$\psi(1)+i\langle\psi,v_{2}\rangle=e^{i\alpha}[\psi(0)-i\langle\psi,v_{1}\rangle];\quad\psi(1)=e^{i\alpha}[\psi(0)-i\langle\psi,v\rangle].$
(5.16)

The difference between these operators is bounded self-adjoint operator of
rank $2$:

$[A(v_{1},v_{2},\alpha)-A(v,0,\alpha)]\psi=\frac{i}{2}e^{-i\alpha}v_{1}\langle\psi,v_{2}\rangle-\frac{i}{2}e^{i\alpha}v_{2}\langle\psi,v_{1}\rangle.$
(5.17)

###### Proof.

Two boundary conditions $(\ref{21.13})$ for the specified connection
$v=v_{1}+e^{i\alpha}v_{2}$ coincide. According to $(\ref{21.1})$ the following
equality is true

$[A(v_{1},v_{2},\alpha)-A(v,0,\alpha)]\psi=v_{1}[\psi(0)-\frac{i}{2}\langle\psi,v_{1}\rangle]+v_{2}[\psi(1)+\frac{i}{2}\langle\psi,v_{2}\rangle]-v[\psi(0)-\frac{i}{2}\langle\psi,v\rangle].$
(5.18)

If we substitute $\psi(1)$ from $(\ref{21.13})$ into this equality and take
into account the explicit form of $v=v_{1}+e^{i\alpha}v_{2}$ we get
$(\ref{21.14})$. The right side of $(\ref{21.14})$ is the sum of two one-
dimensional mutually adjoint operators. Therefore, it is a bounded self-
adjoint operator of rank at most $2$. ∎

######  Remark 5.4.

Due to Theorem 5.2 we have

$A\left(\frac{1}{2}v,\frac{e^{-i\alpha}}{2}v,\alpha\right)-A(v,0,\alpha)\equiv
0.$

We will denote the operator $A(\frac{1}{2}v,\frac{e^{-i\alpha}}{2}v,\alpha)$
by $A(v,\alpha).$ Its action is expressed as follows

$A(v,\alpha)\psi(x)=i\frac{d\psi(x)}{dx}+\frac{1}{2}v(x)\left[\psi(0)+e^{-i\alpha}\psi(1)\right]$
(5.19)

with the boundary condition

$ie^{-i\alpha}\psi(1)-i\psi(0)=\langle\psi,v\rangle.$ (5.20)

######  Theorem 5.3.

The operator $A(v_{1},v_{2},\alpha)$ from Definition 5.1 with nonlocal
potentials $v_{1},v_{2}\in L_{2}(0,1)$ is self–adjoint in $L_{2}(0,1)$. For
nonreal $z$, the resolvent $(A(v_{1},v_{2},\alpha)-zI)^{-1}$ is an integral
operator with the kernel

$g_{z}(x,y;v_{1},v_{2},\alpha)=g_{z}(x,y;\alpha)-\frac{1}{\gamma}\sum\limits_{j,k=1}^{2}\mathcal{E}_{j}(x,z)\gamma_{jk}\bar{\mathcal{E}}_{k}(y,\bar{z}).$
(5.21)

where $g_{z}(x,y;\alpha):=g(x,y;z)$ is the Green‘s function from Theorem 5.1,

$\begin{array}[]{ll}\displaystyle E_{0}(x,z)=g_{z}(x,1;\alpha),&\quad
E_{k}(x,z)=\int_{0}^{1}\,g_{z}(x,y;\alpha)v_{k}(y)\,dy,\ \
k=1,2,\\\\[5.69054pt]
\mathcal{E}_{1}(x,z)=E_{1}-2ie^{i\alpha}E_{0},&\quad\mathcal{E}_{2}(x,z)=E_{2}+2iE_{0}.\end{array}$
(5.22)

and

$\begin{array}[]{l}\gamma=\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21},\\\\[5.69054pt]
\gamma_{11}(z)=-2i\left(1+2i\beta(z)-\langle E_{0},v_{2}\rangle+\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle+\frac{i}{2}\langle
E_{2},v_{2}\rangle\right)\\\\[5.69054pt]
\gamma_{12}(z)=-2i\left(2ie^{iz}\beta(z)+\langle
E_{0},v_{1}\rangle+e^{-i\alpha}\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle-\frac{i}{2}\langle
E_{2},v_{1}\rangle\right),\\\\[5.69054pt]
\gamma_{21}(z)=2i\left(-2ie^{i\alpha}\beta(z)+e^{i\alpha}\langle
E_{0},v_{2}\rangle+\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle+\frac{i}{2}\langle
E_{1},v_{2}\rangle\right),\\\\[5.69054pt]
\gamma_{22}(z)=2i\left(1-2ie^{i\alpha}e^{iz}\beta(z)-e^{i\alpha}\langle
E_{0},v_{1}\rangle+e^{-i\alpha}\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle-\frac{i}{2}\langle
E_{1},v_{1}\rangle\right).\end{array}$ (5.23)

The explicit form of the function $\beta(z)$ is given in $(\ref{21.6})$. The
matrix $\Gamma(z)=\|\gamma_{jk}(z)\|$ has the property
$\Gamma^{*}(\overline{z})=\Gamma(z),$ and $\gamma(z)=\det\Gamma$ has the
property $\overline{\gamma}(\overline{z})=\gamma(z)\neq 0$.

###### Proof.

For the symmetric operator $A(v_{1},v_{2},\alpha)$, the construction of the
Green‘s function and proof of self-adjointness is reduced to solve the problem

$\left(A(v_{1},v_{2},\alpha)-z_{k}I\right)\psi=h,$

for any $h\in L_{2}(0,1)$ and nonreal $z_{k}$, $\mathrm{Im}\,z_{1}>0,$
$\mathrm{Im}\,z_{2}<0$. The solvability of this equation is equivalent to
solvability of the following equation:

$i\psi^{\prime}(x)-z_{k}\psi(x)=h(x)-v_{1}(x)\left[\psi(0)-\frac{i}{2}\langle\psi,v_{1}\rangle_{L_{2}}\right]-v_{2}(x)\left[\psi(1)+\frac{i}{2}\langle\psi,v_{2}\rangle_{L_{2}}\right]$
(5.24)

with boundary condition $(\ref{21.3})$. The solution of the equation
$(\ref{k1})$ has the following form

$\psi(x)=-E_{0}(x,z)w-E_{1}(x,z)\psi_{1}-E_{2}(x,z)\psi_{2}+\int_{0}^{1}\,g_{z}(x,y;\alpha)h(y)\,dy,$
(5.25)

where

$\displaystyle\psi_{1}=\psi(0)-\frac{i}{2}\langle\psi,v_{1}\rangle,\,\,\psi_{2}=\psi(1)+\frac{i}{2}\langle\psi,v_{2}\rangle$
(5.26)

with the boundary condition

$\psi(1)-e^{i\alpha}\psi(0)=-iw,\quad
w=\langle\psi,v_{2}\rangle+e^{i\alpha}\langle\psi,v_{1}\rangle.$

According to Remark 5.1, we have the following equality:

$\psi(1)-e^{i\alpha}\psi(0)=-iw=2(\psi_{2}-e^{i\alpha}\psi_{1}).$

Therefore, the equation (5.25) can be represented as following:

$\begin{array}[]{l}\displaystyle\psi(x)=-E_{0}\cdot
2i(\psi_{2}-e^{i\alpha}\psi_{1})-E_{1}\psi_{1}-E_{2}\psi_{2}+\int\limits_{0}^{1}\,g_{z}(x,y;\alpha)h(y)\,dy,\,\,\text{or}\\\\[5.69054pt]
\displaystyle\psi(x)=-\mathcal{E}_{1}(x,z)\psi_{1}-\mathcal{E}_{2}(x,z)\psi_{2}+\int\limits_{0}^{1}\,g_{z}(x,y,\alpha)h(y)\,dy.\end{array}$
(5.27)

The system for $\psi_{1}$ and $\psi_{2}$ has the form (5.26), where
substituting the values $\psi(0)$ and $\psi(1)$ for function $\psi(x)$ from
$(\ref{eq2a})$ in the right–hand side, and the function from $(\ref{eq2a})$ in
the scalar products $\langle\psi,v_{1}\rangle$, $\langle\psi,v_{2}\rangle$.
The explicit form of system $(\ref{eq1a})$ has the following form:

$\begin{pmatrix}a_{11}&a_{12}\\\
a_{21}&a_{22}\end{pmatrix}\begin{pmatrix}\psi_{1}\\\
\psi_{2}\end{pmatrix}=\begin{pmatrix}\langle
h,\mathcal{E}_{1}(\cdot,{\bar{z}})\rangle\\\ \langle
h,\mathcal{E}_{2}(\cdot,{\bar{z}})\rangle\end{pmatrix},$ (5.28)

where

$\begin{array}[]{ll}\gamma(z)=a_{11}a_{22}-a_{12}a_{21}=\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21}\\\
\gamma_{11}(z)=a_{22}(z)&\gamma_{12}(z)=-a_{12}(z)\\\
\gamma_{21}(z)=-a_{21}(z)&\gamma_{22}(z)=a_{11}(z)\end{array}$ (5.29)

and

$\begin{array}[]{l}a_{11}(z)=2i\left(1-2ie^{i\alpha}e^{iz}\beta(z)-e^{i\alpha}\langle
E_{0},v_{1}\rangle+e^{-i\alpha}\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle-\frac{i}{2}\langle
E_{1},v_{1}\rangle\right),\\\\[5.69054pt]
a_{12}(z)=2i\left(2ie^{iz}\beta(z)+\langle
E_{0},v_{1}\rangle+e^{-i\alpha}\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle-\frac{i}{2}\langle
E_{2},v_{1}\rangle\right),\\\\[5.69054pt]
a_{21}(z)=-2i\left(-2ie^{i\alpha}\beta(z)+e^{i\alpha}\langle
E_{0},v_{2}\rangle+\langle
v_{1},E_{0}(\cdot,{\bar{z}})\rangle+\frac{i}{2}\langle
E_{1},v_{2}\rangle\right),\\\\[5.69054pt]
a_{22}(z)=-2i\left(1+2i\beta(z)-\langle E_{0},v_{2}\rangle+\langle
v_{2},E_{0}(\cdot,{\bar{z}})\rangle+\frac{i}{2}\langle
E_{2},v_{2}\rangle\right).\\\\[5.69054pt] \end{array}$ (5.30)

A determinant

$\gamma(z)=a_{11}a_{22}-a_{12}a_{21}\rightarrow-4\ ,\ \textrm{for}\ \
|\mathrm{Im}\,z|\rightarrow\infty,$

since

$\begin{array}[]{l}a_{11}(z)=2i\left(1-2ie^{i\alpha}e^{iz}\beta(z)\right)+o(1),\\\\[5.69054pt]
a_{12}(z)=2i\left(2ie^{iz}\beta(z)\right)+o(1),\\\\[5.69054pt]
a_{21}(z)=-2i\left(-2ie^{i\alpha}\beta(z)\right)+o(1),\\\\[5.69054pt]
a_{22}(z)=-2i\left(1+2i\beta(z)\right)+o(1),\,\,|\mathrm{Im}\,z|\rightarrow\infty.\end{array}$

Therefore, there exist nonreal $z_{1}$, $z_{2}$ such that
$\mathrm{Im}\,z_{1}>0$, $\mathrm{Im}\,z_{2}<0$ and $|\mathrm{Im}\,z_{k}|$,
$j=1,2$ large enough that $\gamma(z)\neq 0$ and the system $(\ref{eq3a})$ has
a unique solution. This gives us the self-adjointness of the operator
$A(v_{1},v_{2},\alpha)$. We can show that $\bar{a}_{jk}(\bar{z})=a_{kj}(z)$,
then $\bar{\gamma}_{jk}(\bar{z})=\gamma_{kj}(z)$. Therefore, the matrix
$\Gamma(z)=\|\gamma_{jk}(z)\|$ has the property
$\Gamma^{*}(\overline{z})=\Gamma(z),$ and $\gamma(z)=\det\Gamma$ has the
property $\overline{\gamma}(\overline{z})=\gamma(z)$. Substituting the values
of $\psi_{1}$, $\psi_{2}$ in the right–hand side of $\psi(x)$ in the equation
$(\ref{eq2a}),$ we obtain the representation of Green‘s function
$(\ref{eq5a})$. ∎

######  Remark 5.5.

If we substitute $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ from $(\ref{eq4a})$
to $(\ref{eq5a})$, we can describe the Green‘s function explicitly using
$E_{0},E_{1},E_{2}$:

$g_{z}(x,y;v_{1},v_{2},\alpha)=g_{z}(x,y;\alpha)-\frac{1}{\gamma}\sum_{j,k=0}^{2}\,E_{j}(x,z)c_{j,k}(z)\bar{E}_{k}(y,\bar{z}),$
(5.31)

where

$E_{0}(x,z)=g_{z}(x,1;\alpha),\quad
E_{k}(x,z)=\int_{0}^{1}\,g_{z}(x,y;\alpha)v_{k}(y)\,dy,\ \ k=1,2$ (5.32)

and

$\begin{array}[]{lll}c_{00}=4\gamma_{11}-4e^{i\alpha}\gamma_{12}-4e^{-i\alpha}\gamma_{21}+4\gamma_{22}&c_{01}=-2ie^{i\alpha}\gamma_{11}+2i\gamma_{21}&c_{02}=-2ie^{i\alpha}\gamma_{12}+2i\gamma_{22}\\\
c_{10}=2ie^{-i\alpha}\gamma_{11}-2i\gamma_{12}&c_{11}=\gamma_{11}&c_{12}=\gamma_{12}\\\
c_{20}=2ie^{-i\alpha}\gamma_{21}-2i\gamma_{22}&c_{21}=\gamma_{21}&c_{22}=\gamma_{22}.\end{array}$
(5.33)

The values $\gamma_{jk}(z)$ are given in Theorem 5.3. The matrix
$||c_{jk}||_{jk=0}^{2}$ is degenerate. Its first column is a linear
combination of second and third ones:

$\mathrm{col}\,(c_{00},c_{10},c_{20})=2ie^{-i\alpha}\mathrm{col}\,(c_{01},c_{11},c_{21})-2i\mathrm{col}\,(c_{02},c_{12},c_{22}).$

The equalities $c_{jk}(z)=\bar{c}_{kj}(\bar{z}),$
$\gamma(z)=\bar{\gamma}(\bar{z})\neq 0$ are valid providing the self-adjoint
property of the operator $A(v_{1},v_{2},\alpha):$

${\bar{g}}_{\bar{z}}(y,x;v_{1},v_{2},\alpha)=g_{z}(x,y;v_{1},v_{2},\alpha).$

######  Remark 5.6.

If we substitute $v_{1}=\frac{1}{2}v$ and $v_{2}=\frac{e^{-i\alpha}}{2}v$ to
$(\ref{eq5a})$–$(\ref{k6})$ from Theorem 5.3, we obtain the explicit form of
the resolvent of the operator $A(v,\alpha)$ from Remark 5.4. Indeed,

$(A(v,\alpha)-zI)^{-1}h(z)=\int\,g_{z}(x,y;v,\alpha)h(y)\,dy$ (5.34)

with the kernel

$\displaystyle g_{z}(x,y;v,\alpha)=g_{z}(x,y;\alpha)-\frac{1}{\det
F}\sum\limits_{j,k=1}^{2}\,e_{j}(x,z)f_{jk}(z)\bar{e}_{k}(y,\bar{z}),$ (5.35)

where $e_{1}(x,z)=g_{z}(x,1;\alpha),$ $\displaystyle
e_{2}(x,z)=\int_{0}^{1}\,g_{z}(x,y;\alpha)v(y)\,dy$ and

$\begin{array}[]{ll}f_{11}(z)=\langle e_{2},v\rangle,\\\\[5.69054pt]
f_{12}(z)=-\left(e^{i\alpha}+\langle
v,e_{1}(\cdot,\overline{z})\rangle\right),\\\\[5.69054pt]
f_{21}(z)=-\left(e^{-i\alpha}+\langle e_{1},v\rangle\right),\\\\[5.69054pt]
f_{22}(z)=\displaystyle\frac{i}{2}\frac{e^{-iz}+e^{i\alpha}}{e^{-iz}-e^{i\alpha}},\\\\[5.69054pt]
\det F=f_{11}(z)f_{22}(z)-f_{12}(z)f_{21}(z).\end{array}$ (5.36)

##  6 Spectral analysis of momentum operator on a finite interval

The considered models of self-adjoint operators are exactly solvable models.
For the operator $A(v,0,\alpha)$, the eigenvalue problem is to find nontrivial
solutions of the boundary value problem

$i\frac{d\psi}{dx}+v(x)\left[\psi(0)-\frac{i}{2}\langle\psi,v\rangle_{L_{2}}\right]=\lambda\psi,$
(6.1) $\psi(1)=e^{i\alpha}\left[\psi(0)-i\langle\psi,v\rangle_{L_{2}}\right].$
(6.2)

The general solution of the equation $(\ref{3.1})$ has the following form

$\psi(x)=C_{1}e^{-i\lambda x}-C_{2}\cdot
i\cdot\int\limits_{0}^{x}e^{-i\lambda(x-y)}v(y)dy,$ (6.3)

where the constant $C_{1}$ is arbitrary, and the constant $C_{2}$ is related
to the solution by the equality

$C_{2}=-\left[\psi(0)-\frac{i}{2}\langle\psi,v\rangle_{L_{2}}\right].$ (6.4)

The condition that the solution $\psi(x)$ of the form $(\ref{3.3})$ satisfies
the boundary condition $(\ref{3.2})$ leads to the equation

$C_{1}(e^{-i\lambda}-e^{i\alpha})-C_{2}\cdot i\cdot
e^{-i\lambda}\tilde{v}=-ie^{i\alpha}\langle\psi,v\rangle,$ (6.5)

where

$\tilde{v}=\int\limits_{0}^{1}e^{i\lambda y}v(y)dy.$ (6.6)

If we substitute the solution $\psi$ of the form $(\ref{3.3})$ into
$\langle\psi,v\rangle$, we get

$\langle\psi,v\rangle=C_{1}\tilde{v}^{*}-C_{2}i\hat{v},$ (6.7)

where

$\hat{v}=\int\limits_{0}^{1}\int\limits_{0}^{x}e^{-i\lambda(x-y)}v(y)\overline{v(x)}dydx.$
(6.8)

Excluding from $(\ref{3.5})$ and $(\ref{3.6})$ the value
$\langle\psi,v\rangle$, we have

$C_{1}\left(e^{-i\lambda}-e^{i\alpha}+ie^{i\alpha}\tilde{v}^{*}\right)-iC_{2}\left(e^{-i\lambda}\tilde{v}+ie^{i\alpha}\hat{v}\right)=0.$
(6.9)

We get

$C_{1}(2i+\tilde{v}^{*})+iC_{2}(2-\tilde{v})=0,$ (6.10)

excluding $\psi(0)$ and $\langle\psi,v\rangle$ from $(\ref{3.3})$,
$(\ref{3.4})$ and $(\ref{3.6})$. Thus, the constants $C_{1}$ and $C_{2}$
satisfy the homogeneous system of equations $(\ref{3.8})$, $(\ref{3.9})$. The
condition of the existence of non-trivial solutions of this system is
equivalent to the determinant $\chi(\lambda)$ of this system is equal to zero.

This determinant $\chi(\lambda)$ has the following explicit form

$\chi(\lambda)=i(2i+\tilde{v}^{*})(e^{-i\lambda}+ie^{i\alpha}\hat{v})+i(2-\hat{v})(e^{-i\lambda}-e^{i\alpha}+ie^{i\alpha}\tilde{v}^{*}).$
(6.11)

Given above leads to the following theorem.

######  Theorem 6.1.

The real zeros of the function $\chi(\lambda)$ (6.3) and only them are the
eigenvalues of the operator $A(v,0,\alpha)$. The non-trivial solutions of the
system $(\ref{3.8})$–$(\ref{3.9})$ give an explicit form of the eigenfunctions
according to the formula $(\ref{3.3})$.

Let us consider in more detail a particular case of the operator
$A(v,0,\alpha)$, when the nonlocal potential is a constant $v(x)\equiv V$,
$\alpha=\pi$ and $e^{i\alpha}=-1$, that is the operator $A(V,0,\pi).$ Then,
using explicit form $(\ref{tildev})$ and $(\ref{3.7})$ for $\tilde{v}$ and
$\hat{v}$, the characteristic function $\chi(\lambda)$ can be written in the
following form

$\chi(\lambda)=4+2\frac{e^{-i\lambda}-1}{\lambda}\Bigl{[}\lambda+\overline{V}-V+\frac{i}{2}|V|^{2}\Bigr{]}.$
(6.12)

The equation $\chi(\lambda)=0$ can be rewritten as follows

$\frac{e^{-i\lambda}-1}{\lambda}=\frac{-2}{\lambda-2iS(V)},\quad\textrm{where}\quad
S(V)=\mathrm{Im}\,V-\frac{1}{4}|V|^{2}.$ (6.13)

Since the equation for the eigenvalues $\lambda$ includes only the quantity
$S(V)$ that depends on the potential $V$, then different potentials $V$ can
have the same eigenvalues. So the potentials $V=4i$ and $V=0$ have the same
values $S(4i)=S(0)=0$ and generate the same spectrum $(\ref{21.4})$:
$\lambda_{n}=(2n-1)\pi.$

######  Definition 6.1.

We will call the quantity $S(V)=\mathrm{Im}\,V-\frac{1}{4}|V|^{2}$ the
spectral characteristic of the constant nonlocal potential $V$.

######  Lemma 6.2.

Spectral characteristic $S(V)$ is less than or equal to $1$.

######  Lemma 6.3.

A general form of constant nonlocal potentials with $S(V)=0$ is
$V=2\sin{2\varphi}+i4\sin^{2}{\varphi}$, where $\varphi\in[0,\pi]$.

Passing to trigonometric functions in the equation $(\ref{3.12})$, we get an
equivalent equation for $\displaystyle\xi=\frac{2}{\pi}\lambda$,

$\displaystyle
F(\xi)\equiv\frac{\tan{\frac{\pi}{4}\xi}}{\frac{\pi}{4}\xi},\quad
F(\xi)=\frac{1}{S(V)},$ (6.14)

where $\displaystyle\lambda=\frac{\pi}{2}\xi$. Since the function $F(\xi)$ is
even for $\xi$, then the solution of the equation $(\ref{3.13})$ will be
symmetric with respect to the point $\xi=0$. The analysis of the equation
$(\ref{3.13})$ proves lemma.

######  Statement 6.4.

The equation $(\ref{3.13})$ has a solution $\xi=0$ and therefore $\lambda=0$
if and only if $V=2i$. In this case $\lambda=0$ has a multiplicity $2$. The
orthogonal eigenfunctions corresponding to the value $\lambda=0$ have the form
$\psi_{1}=1$, $\displaystyle\psi_{2}=x-\frac{1}{2}$.

######  Definition 6.5.

We will call the potential $V=2i$ resonant.

######  Theorem 6.2.

The spectrum of the self–adjoint operator $A(v,0,\pi)$ consists of eigenvalues
$\lambda$ which are solutions of the equation:

$\frac{\tan\frac{\lambda}{2}}{\frac{\lambda}{2}}=S^{-1}(v).$ (6.15)

If the resonant potential $V=2i$ generates the spectrum: a double eigenvalue
$\lambda=0$ and positive eigenvalues

$\lambda_{n}=(2n+1)\pi-\frac{2}{(2n+1)\pi}+O\Bigl{(}\frac{1}{n^{3}}\Bigr{)},\quad
n=1,2,\ldots$

with $S^{-1}(V)>1$, then all positive eigenvalues consist of
$\lambda_{0}\in(0,\pi)$, $\lambda_{n}\in(2n\pi,(2n+1)\pi)$ having the form

$\lambda_{n}=(2n+1)\pi-\frac{2}{(2n+1)\pi
S^{-1}(V)}+O\Bigl{(}\frac{1}{n^{3}}\Bigr{)},\quad n=1,2,\ldots$

If $S^{-1}(V)<0$, then all positive eigenvalues have the following form:

$\lambda_{n}=(2n-1)\pi-\frac{2}{(2n-1)\pi|S^{-1}(V)|}+O\Bigl{(}\frac{1}{n^{3}}\Bigr{)},\quad
n=1,2,\ldots$

All negative eigenvalues are symmetric to the positive ones having the forms
$-\lambda_{n}$.

###### Proof.

It follows from the solution of equation (6.15). The graph of $F(\xi)$
demonstrates these statements (see Figure 1). The intersection of the graph of
function $y=F(\xi)$ (see (6.14)) with horizontal lines $y=S^{-1}(V)$ gives the
points with the abscissas $\displaystyle\xi_{n}=\frac{2}{\pi}\lambda_{n}.$ ∎

Figure 1: Graph of F

Acknowledgments The authors wish to express their gratitude to Petru A.
Cojuhari and Leonid P. Nizhnik for problems posed and many useful discussions
during the preparation of the manuscript. I.N. is grateful to the Simons
Foundation for the financial support which has helped to accomplish this
manuscript.

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Kamila Debowska
<EMAIL_ADDRESS>
AGH University of Science and Technology
Faculty of Applied Mathematics
al. Mickiewicza 30, 30-059, Krakow, Poland

Irina L. Nizhnik
<EMAIL_ADDRESS>
Institute of Mathematics of NASU
Tereshchenkivska Str.,3, 01 601 Kyiv, Ukraine