Factorization of the Non-Normal Hamiltonian of Reggeon Field Theory in Bargmann Space

Abstract

In this paper, we present a “non-linear” factorization of a family of non-normal operators arising from Gribov’s theory of the following form: ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }={\lambda }^{\prime }{A}^{{*}^2}{A}^2+\mu A^{*}A+i\lambda A^{*}}$$(A+A^{*})A,$ where the quartic Pomeron coupling ${\lambda }^{\prime }$, the Pomeron intercept $\mu$ and the triple Pomeron coupling $\lambda$ are real parameters, and $i^2=-1$. $A^{*}$ and A are, respectively, the usual creation and annihilation operators of the one-dimensional harmonic oscillator obeying the canonical commutation relation $[A,A^{*}]=I$. In Bargmann representation, we have ${\displaystyle A⟷\frac{d}{dz}}$ and ${\displaystyle A^{*}⟷z}$, $z=x+iy$. It follows that ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ can be written in the following form: ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }=p\left(z\right)\frac{d^2}{d{z}^2}+q\left(z\right)\frac{d}{dz},}$ where $p\left(z\right)={\lambda }^{\prime }{z}^2+i\lambda z$ and $q\left(z\right)=i\lambda z^2+\mu z.$ This operator is an operator of the Heun type where the Heun operator is defined by $H=p\left(z\right){\displaystyle \frac{d^2}{d{z}^2}}+q\left(z\right){\displaystyle \frac{d}{dz}}+v\left(z\right),$ where $p\left(z\right)$ is a cubic complex polynomial, $q\left(z\right)$ and $v\left(z\right)$ are polynomials of degree at most 2 and 1, respectively, which are given. For $z=-iy$, $H_{{\lambda }^{\prime },\mu ,\lambda }$ takes the following form: $H_{{\lambda }^{\prime },\mu ,\lambda }=-a\left(y\right){\displaystyle \frac{d^2}{d{y}^2}}+b\left(y\right){\displaystyle \frac{d}{dz}},$ with $a\left(y\right)=y(\lambda -{\lambda }^{\prime }y)$ and ${\displaystyle b\left(y\right)=y(\lambda y+\mu ).}$ We introduce the change of variable $y={\displaystyle \frac{\lambda }{2{\lambda }^{\prime }}}(1-cos\left(\theta \right))$, $\theta \in [0,\pi ]$ to obtain the main result of transforming ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ into a product of two first-order operators: ${\displaystyle {\tilde{H}}_{{\lambda }^{\prime },\mu ,\lambda }={\lambda }^{\prime }(\frac{d}{d\theta }+\alpha \left(\theta \right))(-\frac{d}{d\theta }+\alpha \left(\theta \right)),}$ with $\alpha \left(\theta \right)$ being explicitly determined.

1. Introduction

We consider Bargmann space [1]

$${\displaystyle \mathbb{B}=\{\phi :\mathbb{C}⟶\mathbb{C}entire;{\int }_{\mathbb{C}}|\phi \left(z\right){|}^2{e}^{-{\left|z\right|}^2}dxdy<\infty .\},}$$

(1)

with scalar product

$${\displaystyle <\phi ,\psi >=\frac{1}{\pi }{\int }_{\mathbb{C}}\phi \left(z\right)\overline{\psi \left(z\right)}{e}^{-{\left|z\right|}^2}dxdy;z=x+iy,(x,y)\in {\mathbb{R}}^2,}$$

(2)

and basis

$${\displaystyle e_n\left(z\right)=\frac{z^n}{\sqrt{n!}},n\in \mathbb{N}.}$$

(3)

We present a “non-linear” factorization of a family of non-normal operators:

$${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }={\lambda }^{\prime }S+\mu \mathcal{N}+i\lambda (\mathcal{N}A+A^{*}\mathcal{N}),}$$

(4)

characterizing the Reggeon field theory, (see Gribov in [2,3]), where

${\lambda }^{\prime }$, $\mu$ and $\lambda$ are real parameters, and $i^2=-1$.

$A^{*}$ and A are the creation and annihilation operators. These operators satisfy the commutation relation:

$$[A,A^{*}]=I.$$

(5)

In Bargmann space, we have

$${\displaystyle A⟷\frac{d}{dz}\mathrm{with}\mathrm{domain}D\left(A\right)=\{\phi \in \mathbb{B};\frac{d}{dz}\phi \in \mathbb{B}\},}$$

(6)

$${\displaystyle A^{*}⟷z\mathrm{with}\mathrm{domain}D\left(A^{*}\right)=\{\phi \in \mathbb{B};z\phi \in \mathbb{B}\},}$$

(7)

$${\displaystyle \mathcal{N}=A^{*}A=z\frac{d}{dz}\mathrm{with}\mathrm{domain}D\left(\mathcal{N}\right)=\{\phi \in \mathbb{B};z\frac{d\phi }{dz}\in \mathbb{B}\},}$$

(8)

($\mathcal{N}$ is called the number operator).

Furthermore,

$${\displaystyle \mathcal{S}=A^{{*}^2}{A}^2=z^2\frac{d^2}{d{z}^2}\mathrm{with}\mathrm{domain}D\left(\mathcal{S}\right)=\{\phi \in \mathbb{B};z^2\frac{d^2\phi }{d{z}^2}\in \mathbb{B}\}.}$$

(9)

Then

$${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }=({\lambda }^{\prime }{z}^2+i\lambda z)\frac{d^2}{d{z}^2}+(i\lambda z^2+\mu z)\frac{d}{dz}.}$$

(10)

We see that $e_n$ is an eigenfunction of $\mathcal{N}$ with eigenvalue n and of $\mathcal{S}$ with eigenvalue $n(n-1)$; i.e.,

$$\mathcal{N}{e}_n\left(z\right)=n{e}_n\left(z\right)$$

(11)

and

$$\mathcal{S}{e}_n\left(z\right)=n(n-1)e_n\left(z\right).$$

(12)

Remark 1.

(i) It is well-known that the Heun equation has the following form:

$${\displaystyle [p\left(z\right)\frac{d^2}{d{z}^2}+q\left(z\right)\frac{d}{dz}+v\left(z\right)]u\left(z\right)=0,}$$

where $p\left(z\right)$ is a cubic complex polynomial, and $q\left(z\right)$ and $v\left(z\right)$ are polynomials of degree at most 2 and 1, respectively, which are given.

The singularities of this equation, when two or more of them merge to an irregular singularity, produce the confluent, double confluent, biconfluent and triconfluent Heun equations.

The general Heun equation and its confluent forms appear in many fields of modern physics, such as general relativity, astrophysics, hydrodynamics, atomic and particle physics, etc. We can consult [4,5,6].

(ii) The Hamiltonian $H_{{\lambda }^{\prime },\mu ,\lambda }$ has the form

$${\displaystyle p\left(z\right)\frac{d^2}{d{z}^2}+q\left(z\right)\frac{d}{dz},}$$

with ${\displaystyle p\left(z\right)={\lambda }^{\prime }{z}^2+i\lambda z}$ and ${\displaystyle q\left(z\right)=i\lambda z^2+\mu z}$ of degree 2.

Then, $H_{{\lambda }^{\prime },\mu ,\lambda }$ is an operator of the Heun type.

(iii) The singular points of the eigenvalue problem associated with this operator are ${\displaystyle 0,-\frac{i\lambda }{{\lambda }^{\prime }};{\lambda }^{\prime }\ne 0}$ and ∞. They are regular at ${\displaystyle 0\mathrm{and}\mathrm{at}-\frac{i\lambda }{{\lambda }^{\prime }}}$ but the singularity at ∞ is irregular.

(iv) The factorization of $H_{{\lambda }^{\prime },\mu ,\lambda }$ that we consider in this paper has nothing to do with the well-known factorization methods of Infeld and Hull [7], which cover essentially the operators associated with second-order equations reducible to the hypergeometric equation.

(v) The factorization method of Heun’s differential operator given by Ronveaux in [8] is not applicable because the differential equation of the eigenvalue problem associated with $H_{{\lambda }^{\prime },\mu ,\lambda }$ is not a Fuchsian second-order differential equation.

Now, we recall some spectral properties of $H_{{\lambda }^{\prime },\mu ,\lambda }$:

In 1987, we have given in [9] many spectral properties of ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda };{\lambda }^{\prime }\ne 0}$. In particular,

(i) its domain ${\displaystyle D\left(H_{{\lambda }^{\prime },\mu ,\lambda }\right)}$ coincides with the domain $D\left(\mathcal{S}\right)$ of $\mathcal{S}$.

(ii) If ${\lambda }^{\prime }>0$,

• ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ generates strong semigroups ${\displaystyle e^{-t{H}_{{\lambda }^{\prime },\mu ,\lambda }}}$ and $\exists \beta \in \mathbb{R}$ such that ${\displaystyle e^{-t(H_{{\lambda }^{\prime },\mu ,\lambda }+\beta I)}}$ is compact for any $t>0$.
• ${\displaystyle \sigma (e^{-t{H}_{{\lambda }^{\prime },\mu ,\lambda }})}$ = ${\displaystyle e^{-t\sigma \left(H_{{\lambda }^{\prime },\mu ,\lambda }\right)}\bigcup \left\{0\right\}}$ where ${\displaystyle \sigma \left(H_{{\lambda }^{\prime },\mu ,\lambda }\right)}$ is the spectrum of ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$.

(iii) Let ${\lambda }^{\prime }>0$ and ${\lambda }^{\prime 2}\le \mu {\lambda }^{\prime }+{\lambda }^2$,

• The spectrum of ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ is real.
• If we consider Cauchy’s problem,

$u^{\prime }\left(t\right)+H_{{\lambda }^{\prime },\mu ,\lambda }u\left(t\right)=0;t>0$ and $u\left(0\right)=\phi \in \mathbb{B},$

then we have

$${\displaystyle u\left(t\right)=\sum _{k=1}^{+\infty }\frac{<\phi ,{\phi }_k^{*}>}{<{\phi }_k,{\phi }_k^{*}>}{e}^{-{\sigma }_{k}t}{\phi }_k,}$$

(13)

where

${\phi }_k\in D\left(H_{{\lambda }^{\prime },\mu ,\lambda }\right)$ and ${\phi }_k^{*}\in D\left(H_{{\lambda }^{\prime },\mu ,\lambda }^{*}\right)=D\left(H_{{\lambda }^{\prime },\mu ,\lambda }\right)=D\left(\mathcal{S}\right),$

$H_{{\lambda }^{\prime },\mu ,\lambda }{\phi }_k={\sigma }_k{\phi }_k$ and $H_{{\lambda }^{\prime },\mu ,\lambda }^{*}{\phi }_k^{*}={\sigma }_k{\phi }_k^{*}$,

${\displaystyle \left\{{\sigma }_k\right\},k=1,2,\dots }$ are the eigenvalues and ${\displaystyle \left\{{\phi }_k\right\},k=1,2,\dots }$ are the eigenfunctions of ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$.

In 2019, we have given in [10] a complete spectral analysis of the following operator:

$${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }={\lambda }^{\prime }{A}^{{*}^{p+1}}{A}^{p+1}+\mu A^{{*}^p}{A}^p+i\lambda A^{{*}^p}(A+A^{*})A^p,p\in \mathbb{N}.}$$

(14)

We invite the reader to visit the following papers [11,12,13,14], where we found other interesting spectral properties of $H_{{\lambda }^{\prime },\mu ,\lambda }$ for ${\lambda }^{\prime }\ne 0$.

Recently, in 2023, in the case ${\lambda }^{\prime }=0$ and $\mu \ne 0$, the completeness of the eigenfunctions of ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ (limit operator in resolvent sense as ${\lambda }^{\prime }⟶0$) was established in [15] (this property was still an open question since 1982).

For ${\lambda }^{\prime }=\mu =0$ and $\lambda \ne 0$, the spectrum of ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ is $\mathbb{C}$, see [14].

The objective of this work is to write the second-order differential operator $H_{{\lambda }^{\prime },\mu ,\lambda }$ as the product of two first-order differential operators in spirit of the factorization method, as introduced by Schrödinger in [16].

In this paper, one considerably reduces the difficulty by studying the factorization of ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ on an imaginary axis. In this case, if we take the restriction $z=-iy;y\in \mathbb{R}$, then ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ has the following form:

$${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }=y({\lambda }^{\prime }y-\lambda )\frac{d^2}{d{y}^2}+y(\lambda y+\mu )\frac{d}{dy}.}$$

(15)

In Section 2, we recall the factorization method on the wave equation and on the Schrödinger equation with some classical factorizable singular potentials, and in Section 3, we present our main result of this work by introducing a non-linear change in variable to factorize ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda }}$ on $L_2[0,{\rho }^{\prime }]$, where ${\displaystyle {\rho }^{\prime }=\frac{\lambda }{{\lambda }^{\prime }}}$.

2. Factorization Method on Wave Equation and on Schrödinger Equation with Some Classical Factorizable Singular Potentials

2.1. Factorization Method on Wave Equation

The “binomial factorization” ${\displaystyle \frac{{\mathsf{\partial }}^2}{\mathsf{\partial }{t}^2}-c^2\frac{{\mathsf{\partial }}^2}{\mathsf{\partial }{x}^2}}$$\underset{=}{?}$${\displaystyle (\frac{\mathsf{\partial }}{\mathsf{\partial }t}-c\frac{\mathsf{\partial }}{\mathsf{\partial }x})(\frac{\mathsf{\partial }}{\mathsf{\partial }t}+c\frac{\mathsf{\partial }}{\mathsf{\partial }x})}$ has been known for decades and describes the decomposition of the one-dimensional wave equation into two so-called “one-way” waves propagating in opposite directions [17].

Let’s consider the wave equation (with fundamental velocity set to one):

$${\displaystyle (\frac{d^2}{d{t}^2}-\frac{d^2}{d{x}^2})\phi (t,x)=0,}$$

(16)

for some function $\phi (t;x)$.

If $\phi$ satisfies ${\displaystyle \frac{{\mathsf{\partial }}^2}{\mathsf{\partial }t\mathsf{\partial }x}\phi (t,x)=\frac{{\mathsf{\partial }}^2}{\mathsf{\partial }x\mathsf{\partial }t}\phi (t,x)}$, it is easy to deduce the following identity:

$${\displaystyle \frac{{\mathsf{\partial }}^2}{\mathsf{\partial }{t}^2}-\frac{{\mathsf{\partial }}^2}{\mathsf{\partial }{x}^2}=(\frac{\mathsf{\partial }}{\mathsf{\partial }t}-\frac{\mathsf{\partial }}{\mathsf{\partial }x})(\frac{\mathsf{\partial }}{\mathsf{\partial }t}+\frac{\mathsf{\partial }}{\mathsf{\partial }x}).}$$

(17)

It follows that we have a valid solution for Equation (16):

$${\displaystyle (\frac{d}{dt}-\frac{d}{dx})\phi (x,t)=0or(\frac{d}{dt}+\frac{d}{dx})\phi (x,t)=0,}$$

(18)

where the first and second equation have solutions $\phi (x,t)=\varphi (t+x)$ and $\phi (t,x)=\varphi (t-x)$ respectively, for any function $\varphi$, So we conclude that the solutions to the wave equation are of the form $\phi (x,t)=\varphi (t\pm x)$.

If we consider the $1D$ wave equation with a constant speed c and focus on the problem:

$$\left\{\begin{array}{c}{\displaystyle \frac{{\mathsf{\partial }}^2}{\mathsf{\partial }{t}^2}-c^2\frac{{\mathsf{\partial }}^2}{\mathsf{\partial }{x}^2}=f}\hfill \\ \hfill \\ u(x,0)=u_0\left(x\right)\hfill \\ \hfill \\ {\displaystyle \frac{\mathsf{\partial }}{\mathsf{\partial }t}}u(x,0)=u_1\left(x\right),\hfill \end{array}\right.$$

(19)

then the solution to problem (19) is given by d’Alembert’s formula:

$$\left\{\begin{array}{c}{\displaystyle u(x,t)=\frac{u_0(x+ct)+u_0(x-ct)}{2}+\frac{1}{2c}{\int }_{x-ct}^{x+ct}{u}_1\left(s\right)ds}\hfill \\ \hfill \\ {\displaystyle +\frac{1}{2c}{\int }_0^t{\int }_{|y-x|<c(t-s)}f(y,s)ds}\hfill \end{array}\right.$$

(20)

Idea to proof this classic formula of d’Alembert is to use the above identity of the factorization.

2.2. Factorization Method on Schrödinger Equation

The Shrödinger equation is given by the usual equation [16]:

$${\displaystyle -\frac{d^2}{d{x}^2}+V\left[\left(x\right)\right]\phi \left(x\right)=E\phi \left(x\right),}$$

(21)

where

$${\displaystyle \phi \in L^2\left(\mathbb{R}\right)\mathrm{such}\mathrm{that}{\int }_{\mathbb{R}}{\left|\phi \right|}^{2}dx=1,}$$

(22)

$V:\mathbb{R}⟶\mathbb{R}$ is the multiplication operator associated with the potential $V=V\left(x\right)$ in ${\displaystyle L^2\left(\mathbb{R}\right),}$ and ${\displaystyle E}$ is the energy of the system.

Let us consider the Schrödenger operator [16]:

$${\displaystyle H=-\frac{d^2}{d{x}^2}+V.}$$

(23)

The factorization method introduced by Schrödinger [16], aims to transform the second order differential operator $H.$ into product of two first order operators a and $a^{+}$ plus a constant, i.e., to find two operators:

$${\displaystyle a=\frac{d}{dx}+\beta \left(x\right),}$$

(24)

and

$${\displaystyle a^{+}=-\frac{d}{dx}+\beta \left(x\right),}$$

(25)

such that the Hamiltonian H can be written as

$${\displaystyle H=a^{+}a.}$$

(26)

Let’s develop $a^{+}a$ to obtain

${\displaystyle a^{+}a\phi \left(x\right)=(-\frac{d}{dx}+\beta \left(x\right))(\frac{d}{dx}+\beta \left(x\right))\phi \left(x\right)}$

${\displaystyle =(-\frac{d}{dx}+\beta \left(x\right))({\phi }^{\prime }\left(x\right)+\beta \left(x\right)\phi \left(x\right)))}$

${\displaystyle =-{\phi }^{″}\left(x\right)-{\beta }^{\prime }\left(x\right)\phi \left(x\right)-\beta \left(x\right){\phi }^{\prime }\left(x\right)+\beta \left(x\right){\phi }^{\prime }\left(x\right)+{\beta }^2\left(x\right)\phi \left(x\right)}$

${\displaystyle =-{\phi }^{″}\left(x\right)+({\beta }^2\left(x\right)-{\beta }^{\prime }\left(x\right))\phi \left(x\right).}$

It follows that $\beta \left(x\right)$ must satisfy the following Ricatti equation:

$${\displaystyle -{\beta }^{\prime }\left(x\right)+{\beta }^2\left(x\right)=V\left(x\right).}$$

(27)

If we choose $\beta \left(x\right)=x$ then $V\left(x\right)=x^2-1.$

Let’s develop $a{a}^{+}$ to obtain

${\displaystyle a{a}^{+}\phi \left(x\right)}$${\displaystyle =-{\phi }^{″}\left(x\right)+({\beta }^2\left(x\right)+{\beta }^{\prime }\left(x\right))\phi \left(x\right)}$

and

${\displaystyle [a,a^{+}]\phi \left(x\right)=(a{a}^{+}-a{a}^{+})\phi }$${\displaystyle =2{\beta }^{\prime }\left(x\right)\phi \left(x\right).}$

Then for $\beta \left(x\right)=x$, we get ${\displaystyle [\frac{1}{\sqrt{2}}a,\frac{1}{\sqrt{2}}{a}^{+}]=1}$.

Now let ${\displaystyle b=\frac{1}{\sqrt{2}}a}$ and ${\displaystyle b^{+}=\frac{1}{\sqrt{2}}{a}^{+}}$ which are respectively annihilation and creation operators on $L^2\left(\mathbb{R}\right)$ satisfying:

$$[b,b^{+}]=1.$$

(28)

The Bargmann transform ${\displaystyle \mathcal{B}:L^2\left(\mathbb{R}\right)⟶\mathbb{B},f⟶\phi }$ is defined by:

$$\left\{\begin{array}{c}{\displaystyle \mathcal{B}\left[f\right]\left(z\right)=\phi \left(z\right)={\int }_{\mathbb{R}}\mathcal{A}(z,q)f\left(q\right)dq}\hfill \\ \hfill \\ \mathrm{such}\mathrm{that}\hfill \\ \hfill \\ {\displaystyle \mathcal{A}(z,q)=\frac{1}{{\pi }^{\frac{1}{4}}}{e}^{-\frac{q^2}{2}+\sqrt{2}zq-\frac{z^2}{2}},z\in \mathbb{C}\mathrm{and}q\in \mathbb{R}.}\hfill \end{array}\right.$$

(29)

This transformation was introduced in 1960 by Bargmann in [1], page 198.

In $L^2\left(\mathbb{R}\right)$ representation, we recall that:

${•}_1$ The operators b and $b^{+}$ are adjoint with respect to the $L^2\left(\mathbb{R}\right)$-inner product.

${•}_2$ The Bargmann transform $\mathcal{B}$ is a holomorphic function and it is injective.

${•}_3$ $\mathcal{B}$ is an unitary transform from $L^2\left(\mathbb{R}\right)$ to $\mathbb{B}$ i.e., an isomorphism between $L^2\left(\mathbb{R}\right)$ and $\mathbb{B}$ that preserves the inner product.

${•}_4$ By the Bargmann transform, the operators b and $b^{*}$ defined on $L^2\left(\mathbb{R}\right)$, are transformed into the operators A and $A^{*}$ defined on Bargmann space $\mathbb{B}$ by:

$${\displaystyle A=\frac{\mathsf{\partial }}{\mathsf{\partial }z}\mathrm{and}{A}^{*}=z\mathrm{on}\mathrm{Bargmann}\mathrm{space}.}$$

${•}_5$ The Bargmann transform of the normalised Hermite functions

$${\displaystyle h_n\left(q\right)=\frac{1}{2^{n}n!\sqrt{\pi }}{(q-\frac{\mathsf{\partial }}{\mathsf{\partial }q})}^n{e}^{-\frac{q^2}{2}}}$$

(30)

is given by:

$${\displaystyle \mathcal{B}[\frac{1}{2^{n}n!\sqrt{\pi }}(q-\frac{\mathsf{\partial }}{\mathsf{\partial }q}{)}^n{e}^{-\frac{q^2}{2}}]\left(z\right)=\frac{z^n}{\sqrt{n!\sqrt{\pi }}}\mathcal{B}\left[e^{-\frac{q^2}{2}}\right]=\frac{z^n}{\sqrt{n!}},}$$

(31)

because ${\displaystyle \mathcal{B}\left[e^{-\frac{q^2}{2}}\right]={\pi }^{\frac{1}{4}}}$.

${•}_6$ These are exactly the normalized basis functions of the Bargmann space. We end this sub-section by giving the Bargmann transform inverse:

${•}_7$ ${\displaystyle {\mathcal{B}}^{-1}:\mathbb{B}⟶L^2\left(\mathbb{R}\right)}$

$${\displaystyle {\mathcal{B}}^{-1}\left[\phi \right]\left(z\right)=c{\int }_{\mathbb{C}}\phi \left(z\right)e^{-\frac{1}{2}(q^2+\overline{z})+\sqrt{2}q\overline{z}}{e}^{{\left|z\right|}^2}dxdy,c>0.}$$

(32)

2.3. Some Classical Factorizable Singular Potential

(i) The ${\displaystyle x^2+\frac{1}{x^2}}$ potential

Consider the super-potential ${\displaystyle \mathbb{W}=\frac{\alpha }{x}-x}$. The operators associated with this potential are

${\displaystyle a=-\frac{d}{dx}+\frac{\alpha }{x}-x}$ and ${\displaystyle a^{+}=\frac{d}{dx}+\frac{\alpha }{x}-x}$; these produce the Hamiltonians

$${\displaystyle H_{+}=a^{+}a=-\frac{d^2}{d{x}^2}+x^2+\frac{\alpha (\alpha -1)}{x^2}-2\alpha -1,}$$

(33)

$${\displaystyle H_{-}=a{a}^{+}=-\frac{d^2}{d{x}^2}+x^2+\frac{\alpha (\alpha +1)}{x^2}-2\alpha -1,}$$

(34)

with ${\displaystyle [a,a^{+}]=\frac{2\alpha }{x^2}+2}$. If $\alpha ={\displaystyle \frac{1}{2}}$, we obtain

$${\displaystyle H_{-}=a{a}^{+}=-\frac{d^2}{d{x}^2}+x^2+\frac{3/4}{x^2}.}$$

(35)

(ii) The ${\displaystyle \frac{1}{x^2}}$ potential

We did not need the harmonic oscillator term to make the ${\displaystyle \frac{1}{x^2}}$ potential factorizable.

The relevant operators are

${\displaystyle a=-\frac{d}{dx}+\frac{\alpha }{x}}$ and ${\displaystyle a^{+}=\frac{d}{dx}+\frac{\alpha }{x}}$; these produce the Hamiltonians

$${\displaystyle H_{+}=a{a}^{+}=-\frac{d^2}{d{x}^2}+\frac{\alpha (\alpha +1)}{x^2},}$$

(36)

$${\displaystyle H_{-}=a^{+}a=-\frac{d^2}{d{x}^2}+\frac{\alpha (\alpha -1)}{x^2},}$$

(37)

and commute as follows:

$${\displaystyle [a,a^{+}]=\frac{2\alpha }{x^2}and[a^{+},a]=-\frac{2\alpha }{x^2}.}$$

(38)

These above potentials are discussed in [18].

Remark 2.

The examples of the factorization methods on the Schrödinger equations given above in Section 2.2 and Section 2.3 are recognized as supersymmetric (SUSY) quantum mechanics in physics. It is interesting to consult the Reference [19] for a global review on this topic.

3. On Non-Linear Change of Variable to Factorize ${\mathit{H}}_{{\mathit{\lambda }}^{\prime },\mathit{\mu },\mathit{\lambda }}$

Let us start by noticing that if we put $u\left(y\right)=\phi (-iy)$, then

$$H_{{\lambda }^{\prime },\mu ,\lambda }u\left(y\right)=-a\left(y\right)u^{″}\left(y\right)+b\left(y\right)u^{\prime }\left(y\right),$$

(39)

where $u^{\prime }$ denotes the first derivative of u with respect to y, and $u^{″}$ denotes the second derivative of u with respect to y, $a\left(y\right)=y(\lambda -{\lambda }^{\prime }y)$ and $b\left(y\right)=y(\lambda y+\mu ).$

If we interpret y as a coordinate, $a\left(y\right)$ is like a diffusion coefficient that is positive for $0<y\le {\rho }^{\prime }$, where ${\displaystyle {\rho }^{\prime }=\frac{\lambda }{{\lambda }^{\prime }}}$.

Now, let ${\displaystyle y\in [0,{\rho }^{\prime }]}$, then we consider the Cauchy problem associated with $H_{{\lambda }^{\prime },\mu ,\lambda }$ on $[0,{\rho }^{\prime }]$:

$$\left\{\begin{array}{c}{\displaystyle \frac{\mathsf{\partial }u(t,y)}{\mathsf{\partial }t}=a\left(y\right)\frac{{\mathsf{\partial }}^{2}u(t,y)}{\mathsf{\partial }{y}^2}-b\left(y\right)\frac{\mathsf{\partial }u(t,y)}{\mathsf{\partial }y},t>0,y\in [0,{\rho }^{\prime }],}\hfill \\ \hfill \\ {\displaystyle a\left(y\right)=y(\lambda -{\lambda }^{\prime }y),}\hfill \\ \hfill \\ {\displaystyle b\left(y\right)=y(\lambda y+\mu ).}\hfill \end{array}\right.$$

(40)

We introduce the non-linear change of variable ${\displaystyle y=\frac{{\rho }^{\prime }}{2}(1-cos\left(\theta \right))}$ and eliminate the first order derivative.

Let us define $\tilde{u}(t,\theta )$ by ${\displaystyle u(t,y)=\sigma \left(\theta \right)\tilde{u}(t,\theta )}$ where

$${\displaystyle \sigma \left(\theta \right)={\left[sin\left(\frac{\theta }{2}\right)\right]}^{\frac{1}{2}}{\left[cos\left(\frac{\theta }{2}\right)\right]}^{-(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}{e}^{-\frac{{\lambda }^2}{2{\lambda }^{\prime 2}}{\left[sin\left(\frac{\theta }{2}\right)\right]}^2.}}$$

Then, we obtain the following theorem:

Theorem 1. 

(Main result of this paper)

The function $\tilde{u}$ is the solution of the following Cauchy problem:

$$\left\{{\displaystyle \begin{array}{c}{\displaystyle \frac{\mathsf{\partial }\tilde{u}(t,\theta )}{\mathsf{\partial }t}=-{\tilde{H}}_{{\lambda }^{\prime },\mu ,\lambda }\tilde{u}(t,\theta ),t>0,\theta \in [0,\pi ]}\hfill \\ \hfill \\ {\displaystyle {\tilde{H}}_{{\lambda }^{\prime },\mu ,\lambda }={\lambda }^{\prime }(\frac{d}{d\theta }+\alpha \left(\theta \right))(-\frac{d}{d\theta }+\alpha \left(\theta \right))}\hfill \\ \hfill \\ where\hfill \\ \hfill \\ \alpha \left(\theta \right)=\frac{1}{4}[\frac{{\lambda }^2}{{\lambda }^{\prime 2}}sin\left(\theta \right)-cot\left(\frac{\theta }{2}\right)-2(\frac{\mu }{{\lambda }^{\prime }}-\frac{{\lambda }^2}{{\lambda }^{\prime 2}})tan\left(\frac{\theta }{2}\right)].\hfill \end{array}}\right.$$

(41)

Proof 

• Let ${\displaystyle q=\frac{\lambda }{2{\lambda }^{\prime }}(1-cos\theta )}$ and ${\displaystyle a\left(q\right)=q(\lambda -{\lambda }^{\prime }q)}$; then we have
  ${\displaystyle a\left(q\right)=q(\lambda -{\lambda }^{\prime }q)=\frac{\lambda }{2{\lambda }^{\prime }}(1-cos\theta )(\lambda -{\lambda }^{\prime }\frac{\lambda }{2{\lambda }^{\prime }}(1-cos\theta ))}$
  ${\displaystyle =\frac{{\lambda }^2}{4{\lambda }^{\prime }}(1-cos\theta )(1+cos\theta )=\frac{{\lambda }^2}{4{\lambda }^{\prime }}(1-co{s}^2\theta )=\frac{{\lambda }^2}{4{\lambda }^{\prime }}si{n}^2\theta .}$
  As ${\displaystyle q^{\prime }\left(\theta \right)=\frac{\lambda }{2{\lambda }^{\prime }}sin\theta }$, then ${\displaystyle {\left[q^{\prime }\left(\theta \right)\right]}^2=\frac{{\lambda }^2}{4{\lambda }^{\prime 2}}si{n}^2\theta }$ and ${\displaystyle \frac{a\left(q\right)}{{\left[q^{\prime }\left(\theta \right)\right]}^2}={\lambda }^{\prime }}$.
• Let $u(t,q(\theta \left)\right)=\sigma \left(\theta \right)v(t,\theta )$; then we have
  (1) ${\displaystyle \frac{du}{d\theta }=\frac{d}{d\theta }\left[\sigma \left(\theta \right)v(t,\theta )\right]}$, i.e., ${\displaystyle \frac{du}{d\theta }=\frac{dq}{d\theta }\frac{du}{dq}={\sigma }^{\prime }v+\sigma v^{\prime }(\star )⟹}$
  ${\displaystyle \frac{du}{dq}=\frac{1}{q^{\prime }\left(\theta \right)}[{\sigma }^{\prime }v+\sigma v^{\prime }].}$
  By applying (⋆), we obtain
  (2) ${\displaystyle \frac{d^{2}u}{d{\theta }^2}={\sigma }^{″}v+{\sigma }^{\prime }{v}^{\prime }+{\sigma }^{\prime }{v}^{\prime }+\sigma v^{″}=\sigma v^{″}+2{\sigma }^{\prime }{v}^{\prime }+{\sigma }^{″}v.(\star \star )}$
  (3) ${\displaystyle \frac{d^{2}u}{d{\theta }^2}=\frac{d}{d\theta }\left[\frac{dq}{d\theta }\frac{du}{dq}\right]=\frac{d^{2}q}{d{\theta }^2}\frac{du}{dq}+\frac{dq}{d\theta }\frac{d^{2}u}{d\theta dq}=q^{″}\frac{du}{dq}+\frac{dq}{d\theta }\frac{dq}{d\theta }\frac{d^{2}u}{d{q}^2}=}$
  ${\displaystyle q^{″}\frac{du}{dq}+{\left[q^{\prime }\left(\theta \right)\right]}^2\frac{d^{2}u}{d{q}^2}=q^{″}\left[\frac{1}{q^{\prime }\left(\theta \right)}({\sigma }^{\prime }v+\sigma v^{\prime })\right]+{\left[q^{\prime }\left(\theta \right)\right]}^2\frac{d^{2}u}{d{q}^2}}$ = ${\displaystyle \sigma v^{″}+2{\sigma }^{\prime }{v}^{\prime }+{\sigma }^{″}v,⟹}$
  ${\displaystyle {\left[q^{\prime }\left(\theta \right)\right]}^2\frac{d^{2}u}{d{q}^2}=\sigma v^{″}+2{\sigma }^{\prime }{v}^{\prime }+{\sigma }^{″}v-q^{″}\left[\frac{1}{q^{\prime }\left(\theta \right)}({\sigma }^{\prime }v+\sigma v^{\prime })\right].⟹}$
  ${\displaystyle \frac{d^{2}u}{d{q}^2}=\frac{1}{{\left[q^{\prime }\left(\theta \right)\right]}^2}\{\sigma v^{″}+2{\sigma }^{\prime }{v}^{\prime }+{\sigma }^{″}v-q^{″}\left[\frac{1}{q^{\prime }\left(\theta \right)}({\sigma }^{\prime }v+\sigma v^{\prime })\right]\}}$.
• Let ${\displaystyle u(t,q)=\sigma \left(\theta \right)v(t,\theta )}$⟹${\displaystyle q^{\prime }\left(\theta \right)\frac{du}{dq}={\sigma }^{\prime }\left(\theta \right)v+\sigma \left(\theta \right)\frac{dv}{d\theta }}$⟹
  •${\displaystyle \frac{du}{dq}=\frac{1}{q^{\prime }\left(\theta \right)}[\sigma \left(\theta \right)\frac{dv}{d\theta }+{\sigma }^{\prime }\left(\theta \right)v],}$
  •${\displaystyle \frac{d^{2}u}{d{q}^2}=\frac{1}{{\left[q^{\prime }\left(\theta \right)\right]}^2}[\sigma \left(\theta \right)\frac{d^{2}v}{d{\theta }^2}+(2{\sigma }^{\prime }\left(\theta \right)-\frac{q^{″}\left(\theta \right)\sigma \left(\theta \right)}{q^{\prime }\left(\theta \right)})\frac{dv}{d\theta }+({\sigma }^{″}\left(\theta \right)-\frac{q^{″}\left(\theta \right){\sigma }^{\prime }\left(\theta \right)}{q^{\prime }\left(\theta \right)})v],}$
  •${\displaystyle a\left(q\right)\frac{d^{2}u}{d{q}^2}=\frac{a\left(q\right(\theta \left)\right)}{{\left[q^{\prime }\left(\theta \right)\right]}^2}[\sigma \left(\theta \right)\frac{d^{2}v}{d{\theta }^2}+(2{\sigma }^{\prime }\left(\theta \right)-\frac{q^{″}\left(\theta \right)\sigma \left(\theta \right)}{q^{\prime }\left(\theta \right)})\frac{dv}{d\theta }+({\sigma }^{″}\left(\theta \right)-\frac{q^{″}\left(\theta \right){\sigma }^{\prime }\left(\theta \right)}{q^{\prime }\left(\theta \right)})v].}$
  ⟹
  •${\displaystyle a\left(q\right)\frac{d^{2}u}{d{q}^2}={\lambda }^{\prime }[\sigma \left(\theta \right)\frac{d^{2}v}{d{\theta }^2}+(2{\sigma }^{\prime }\left(\theta \right)-\frac{q^{″}\left(\theta \right)\sigma \left(\theta \right)}{q^{\prime }\left(\theta \right)})\frac{dv}{d\theta }+({\sigma }^{″}\left(\theta \right)-\frac{q^{″}\left(\theta \right){\sigma }^{\prime }\left(\theta \right)}{q^{\prime }\left(\theta \right)})v].}$
  As ${\displaystyle q^{\prime }\left(\theta \right)=\frac{\lambda }{2{\lambda }^{\prime }}sin\theta }$, then ${\displaystyle q^{″}\left(\theta \right)=\frac{\lambda }{2{\lambda }^{\prime }}cos\theta }$.
  It follows that ${\displaystyle \frac{q^{″}\left(\theta \right)}{q^{\prime }\left(\theta \right)}=cot\left(\theta \right)}$
  and
  •${\displaystyle a\left(q\right)\frac{d^{2}u}{d{q}^2}={\lambda }^{\prime }[\sigma \left(\theta \right)\frac{d^{2}v}{d{\theta }^2}+(2{\sigma }^{\prime }\left(\theta \right)-cot\left(\theta \right)\sigma \left(\theta \right))\frac{dv}{d\theta }+({\sigma }^{″}\left(\theta \right)-cot\left(\theta \right){\sigma }^{\prime }\left(\theta \right))v].}$
• Let ${\displaystyle b\left(q\right)=q(\mu +\lambda q)}$, then
  As ${\displaystyle \frac{du}{dq}=\frac{1}{q^{\prime }\left(\theta \right)}[\sigma v^{\prime }+{\sigma }^{\prime }v]}$, then we have
  ${\displaystyle b\left(q\right)\frac{du}{dq}=\frac{b\left(q\right)}{q^{\prime }\left(\theta \right)}[\sigma v^{\prime }+{\sigma }^{\prime }v].}$
  As ${\displaystyle (1-cos\theta )=2{\left[sin\frac{\theta }{2}\right]}^2}$ and ${\displaystyle sin\theta =2sin\frac{\theta }{2}cos\frac{\theta }{2}}$, it follows that
  ${\displaystyle \frac{q}{q^{\prime }}=tan\frac{\theta }{2}}$ where ${\displaystyle q\left(\theta \right)=\frac{\lambda }{2{\lambda }^{\prime }}(1-cos\theta )}$ and ${\displaystyle q^{\prime }\left(\theta \right)=\frac{\lambda }{2{\lambda }^{\prime }}sin\theta ,}$ then
  ${\displaystyle \frac{b\left(q\right)}{q^{\prime }\left(\theta \right)}=tan\frac{\theta }{2}(\mu +\lambda q\left(\theta \right)).}$
  It follows that
  •${\displaystyle b\left(q\right)\frac{du}{dq}=tan\frac{\theta }{2}(\mu +\lambda q\left(\theta \right))[\sigma \left(\theta \right)\frac{dv}{d\theta }+{\sigma }^{\prime }\left(\theta \right)v]}$
  and
  •${\displaystyle -a\left(q\right)\frac{d^{2}u}{d{q}^2}+b\left(q\right)\frac{du}{dq}={\lambda }^{\prime }[-\sigma \left(\theta \right)\frac{d^{2}v}{d{\theta }^2}+(cot\left(\theta \right)\sigma \left(\theta \right)-2{\sigma }^{\prime }\left(\theta \right)+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))\sigma \left(\theta \right))\frac{dv}{d\theta }}$
  ${\displaystyle +(cot\left(\theta \right){\sigma }^{\prime }\left(\theta \right)-{\sigma }^{″}\left(\theta \right)+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right)){\sigma }^{\prime }\left(\theta \right))v]}$
  and
  •${\displaystyle \frac{1}{\sigma \left(\theta \right)}[-a\left(q\right)\frac{d^{2}u}{d{q}^2}+b\left(q\right)\frac{du}{dq}]={\lambda }^{\prime }[-\frac{d^{2}v}{d{\theta }^2}+(cot\left(\theta \right)-2\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))\frac{dv}{d\theta }+}$
  ${\displaystyle (cot\left(\theta \right)\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}-\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)})v].}$
  Choosing
  ${\displaystyle (cot\left(\theta \right)-2\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))=0}$, i.e.,
  ${\displaystyle 2\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}=cot\left(\theta \right)+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right)).}$
  Then we deduce that
  •${\displaystyle \frac{1}{\sigma \left(\theta \right)}[-a\left(q\right)\frac{d^{2}u}{d{q}^2}+b\left(q\right)\frac{du}{dq}]={\lambda }^{\prime }[-\frac{d^{2}v}{d{\theta }^2}+(cot\left(\theta \right)\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}-\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}}$
  ${\displaystyle +tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}\left)v\right].}$
  Now, we observe that
  ${\displaystyle cot\left(\theta \right)\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}-\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}}$
  ${\displaystyle +\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}}$${\displaystyle =[cot\left(\theta \right)+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))]\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}}$
  ${\displaystyle -\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}}$
  ${\displaystyle =\left[2\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}\right]\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}}$
  ${\displaystyle -\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}}$${\displaystyle =2{\left[\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}\right]}^2}$${\displaystyle -\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}}$${\displaystyle ={\left[\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}\right]}^2+{\left[\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}\right]}^2}$${\displaystyle -\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}}$.
  If we put

  $${\displaystyle \alpha \left(\theta \right)=-\frac{{\sigma }^{\prime }}{\sigma },}$$

  (42)

  then

  $${\displaystyle {\alpha }^{\prime }\left(\theta \right)=-\frac{{\sigma }^{″}}{\sigma }+{\alpha }^2\left(\theta \right).}$$

  (43)
  It follows that
  ${\displaystyle cot\left(\theta \right)\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}-\frac{{\sigma }^{″}\left(\theta \right)}{\sigma \left(\theta \right)}+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}={\alpha }^2\left(\theta \right)+{\alpha }^{\prime }\left(\theta \right).}$
  Consider the operator

  $${\displaystyle \tilde{H}v=(\frac{d}{d\theta }+\alpha \left(\theta \right))(-\frac{d}{d\theta }+\alpha \left(\theta \right))v=-\frac{d^{2}v}{d{\theta }^2}+({\alpha }^2\left(\theta \right)+{\alpha }^{\prime }\left(\theta \right))v.}$$

  (44)
  For ${\displaystyle \alpha \left(\theta \right)=-\frac{{\sigma }^{\prime }}{\sigma }}$, we obtain

  $${\displaystyle \tilde{H}v=-\frac{d^{2}v}{d{\theta }^2}+(2{\alpha }^2\left(\theta \right)-\frac{{\sigma }^{″}}{\sigma })v.}$$

  (45)
  Now, if we choose $\sigma \left(\theta \right)$ such that

  $${\displaystyle 2\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}=cot\left(\theta \right)+tan\frac{\theta }{2}(\frac{\mu }{{\lambda }^{\prime }}+\frac{\lambda }{{\lambda }^{\prime }}q\left(\theta \right))},$$

  (46)

  then from (46), we deduce that
  ${\displaystyle 2\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}=cot\theta +\frac{\mu }{{\lambda }^{\prime }}tan\frac{\theta }{2}+\frac{{\lambda }^2}{2{\lambda }^{\prime 2}}tan\frac{\theta }{2}(1-cos\theta )}$, where ${\displaystyle q\left(\theta \right)=\frac{\lambda }{2{\lambda }^{\prime }}(1-cos\theta )}$
  ${\displaystyle =\frac{cos\theta }{sin\theta }}$ + ${\displaystyle \frac{\mu }{{\lambda }^{\prime }}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}}$ + ${\displaystyle \frac{{\lambda }^2}{2{\lambda }^{\prime 2}}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}·2{\left[sin\frac{\theta }{2}\right]}^2}$
  ${\displaystyle =\frac{cos\theta }{sin\theta }}$ + ${\displaystyle \frac{\mu }{{\lambda }^{\prime }}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}}$ + ${\displaystyle \frac{{\lambda }^2}{{\lambda }^{\prime 2}}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}·(1-{\left[cos\frac{\theta }{2}\right]}^2)}$
  ${\displaystyle =\frac{cos\theta }{sin\theta }}$ + ${\displaystyle \frac{\mu }{{\lambda }^{\prime }}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}}$ + ${\displaystyle \frac{{\lambda }^2}{{\lambda }^{\prime 2}}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}-\frac{{\lambda }^2}{{\lambda }^{\prime 2}}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}·{\left[cos\frac{\theta }{2}\right]}^2)}$
  ${\displaystyle =\frac{cos\theta }{sin\theta }}$ + ${\displaystyle \frac{\mu }{{\lambda }^{\prime }}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}}$ + ${\displaystyle \frac{{\lambda }^2}{{\lambda }^{\prime 2}}\frac{sin\frac{\theta }{2}}{cos\frac{\theta }{2}}}$−${\displaystyle \frac{{\lambda }^2}{{\lambda }^{\prime 2}}sin\frac{\theta }{2}cos\frac{\theta }{2}}$
  It follows that expression of the function $\sigma \left(\theta \right)$ is given by
  •${\displaystyle ln\left|\sigma \right(\theta \left)\right|}$ = ${\displaystyle \frac{1}{2}ln\left|sin\left(\theta \right)\right|}$−${\displaystyle (\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}$${\displaystyle ln|cos\frac{\theta }{2}|}$−${\displaystyle \frac{{\lambda }^2}{2{\lambda }^{\prime 2}}{(sin\frac{\theta }{2})}^2}$
  = ${\displaystyle {ln\left|sin\left(\theta \right)\right|}^{\frac{1}{2}}}$−${\displaystyle ln|cos\frac{\theta }{2}{|}^{(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}}$−${\displaystyle \frac{{\lambda }^2}{2{\lambda }^{\prime 2}}{(sin\frac{\theta }{2})}^2}$
  = ${\displaystyle {ln\left|sin\left(\theta \right)\right|}^{\frac{1}{2}}{\left|cos\frac{\theta }{2}\right|}^{-(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}}$−${\displaystyle \frac{{\lambda }^2}{2{\lambda }^{\prime 2}}{(sin\frac{\theta }{2})}^2.}$
  Then we deduce that

  $${\displaystyle \sigma \left(\theta \right)={\left[sin\left(\frac{\theta }{2}\right)\right]}^{\frac{1}{2}}{\left[cos\left(\frac{\theta }{2}\right)\right]}^{-(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}{e}^{-\frac{{\lambda }^2}{2{\lambda }^{\prime 2}}{\left[sin\left(\frac{\theta }{2}\right)\right]}^2}.}$$

  (47)
  Now, let
  ${\displaystyle f\left(\theta \right)={\left[sin\left(\frac{\theta }{2}\right)\right]}^{\frac{1}{2}}⟹}$${\displaystyle f^{\prime }\left(\theta \right)=\frac{1}{4}{\left[sin\left(\frac{\theta }{2}\right)\right]}^{-\frac{1}{2}}cos\frac{\theta }{2}}$, ⟹
• ${\displaystyle \frac{f^{\prime }\left(\theta \right)}{f\left(\theta \right)}=\frac{1}{4}{\left[sin\left(\frac{\theta }{2}\right)\right]}^{-1}cos\frac{\theta }{2}}$${\displaystyle =\frac{1}{4}cot\left(\frac{\theta }{2}\right)}$,
  ${\displaystyle g\left(\theta \right)={\left[cos\left(\frac{\theta }{2}\right)\right]}^{-(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}⟹}$${\displaystyle g^{\prime }\left(\theta \right)=\frac{(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}{2}{\left[cos\left(\frac{\theta }{2}\right)\right]}^{-(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}}+1)}sin\left(\frac{\theta }{2}\right)}$, ⟹
  •${\displaystyle \frac{g^{\prime }\left(\theta \right)}{g\left(\theta \right)}=}$${\displaystyle \frac{(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}{2}{\left[cos\left(\frac{\theta }{2}\right)\right]}^{-1}sin\left(\frac{\theta }{2}\right)}$${\displaystyle =\frac{(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})}{2}tan\left(\frac{\theta }{2}\right),}$
  ${\displaystyle h\left(\theta \right)=e^{-\frac{{\lambda }^2}{2{\lambda }^{\prime 2}}{\left[sin\left(\frac{\theta }{2}\right)\right]}^2}⟹}$${\displaystyle h^{\prime }\left(\theta \right)=-\frac{{\lambda }^2}{2{\lambda }^{\prime 2}}sin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)}$${\displaystyle e^{-\frac{{\lambda }^2}{2{\lambda }^{\prime 2}}{\left[sin\left(\frac{\theta }{2}\right)\right]}^2}}$, ⟹
  •${\displaystyle \frac{h^{\prime }\left(\theta \right)}{h\left(\theta \right)}=}$${\displaystyle -\frac{{\lambda }^2}{2{\lambda }^{\prime 2}}sin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right)=}$${\displaystyle -\frac{{\lambda }^2}{4{\lambda }^{\prime 2}}sin\left(\theta \right)}$
  and
  ${\displaystyle W\left(\theta \right)=f\left(\theta \right)g\left(\theta \right)h\left(\theta \right)}$.
  Then
  ${\displaystyle W^{\prime }\left(\theta \right)=f^{\prime }\left(\theta \right)g\left(\theta \right)h\left(\theta \right)+f\left(\theta \right)g^{\prime }\left(\theta \right)h\left(\theta \right)+f\left(\theta \right)g\left(\theta \right)h^{\prime }\left(\theta \right)}$
  and
  ${\displaystyle \frac{W^{\prime }\left(\theta \right)}{W\left(\theta \right)}=\frac{f^{\prime }\left(\theta \right)}{f\left(\theta \right)}+\frac{g^{\prime }\left(\theta \right)}{g\left(\theta \right)}+\frac{h^{\prime }\left(\theta \right)}{h\left(\theta \right)}}$${\displaystyle =\frac{1}{4}[cot\left(\frac{\theta }{2}\right)+2(\frac{\mu }{{\lambda }^{\prime }}+\frac{{\lambda }^2}{{\lambda }^{\prime 2}})tan\left(\frac{\theta }{2}\right)-\frac{{\lambda }^2}{{\lambda }^{\prime 2}}sin\left(\theta \right)]}$.
  As ${\displaystyle \alpha \left(\theta \right)=-\frac{{\sigma }^{\prime }\left(\theta \right)}{\sigma \left(\theta \right)}}$, we deduce that

  $${\displaystyle \alpha \left(\theta \right)=\frac{1}{4}[\frac{{\lambda }^2}{{\lambda }^{\prime 2}}sin\left(\theta \right)-cot\left(\frac{\theta }{2}\right)-2(\frac{\mu }{{\lambda }^{\prime }}-\frac{{\lambda }^2}{{\lambda }^{\prime 2}})tan\left(\frac{\theta }{2}\right)].}$$

  (48)

  □

4. Conclusions

A linear differential operator T admits factorization if it can be represented as a product of lower order operators of the same type (see [20,21]).

In this paper, we were focused on a non-normal linear differential operator acting in Bargmann space which characterizes the Reggeon field theory. We have introduced a change of variable as indicated in abstract of this paper, to transform the second-order differential operator $H_{{\lambda }^{\prime },\mu ,\lambda }$, ${\lambda }^{\prime }\ne 0$ into a product of two first-order operators.

The eigenvalue problem associated to $H_{{\lambda }^{\prime },\mu ,\lambda }$, ${\lambda }^{\prime }\ne 0$ is a confluent differential equation of second order of Heun type. This confluent equation having an irregular singularity at infinity, it is therefore not a Fuchsian second order differential equation as in [8]. The factorization of $H_{{\lambda }^{\prime },\mu ,\lambda }$, ${\lambda }^{\prime }\ne 0$ considered in this paper has nothing to do with the well-known factorization methods of Infeld and Hull [7] which cover essentially the operators associated to second order equations reducible to a hypergeometric equation.

For $p=1$ and ${\lambda }^{\prime }\ne 0$, the expression of Equation (14) is none other than $H_{{\lambda }^{\prime },\mu ,\lambda }$ and for $p=2$, it is none other than the infinitesimal generator of the semigroup studied in [11].

As in [22], an opportunity for my further research would be to assert the existence of a factorization of $(p+1)$ order linear differential operator of Equation (14) studied in [10], for $p\ge 2$.

Notice that the case ${\lambda }^{\prime }=0$ is entirely different, although the mathematical study of $H_{0,\mu ,\lambda }$ began in 1982 in [13], the question of its factorization is still open. Nevertheless, using the technique developed in [15] to show the completeness of its eigenfunctions, a factorization of $H_{0,\mu ,\lambda }$ is undertaken:

For $z=-iy,y\in \mathbb{R}$, we transformed $H_{0,\mu ,\lambda }$ into ${\displaystyle {\tilde{H}}_{0,\mu ,\lambda }=\frac{\lambda }{4}[-\frac{d^2}{d{y}^2}+V\left(y\right)]}$ where ${\displaystyle V\left(y\right)=\frac{3/4}{y^2}+y^2[{(y^2+\rho )}^2-2]}$, ${\displaystyle \rho =\frac{\mu }{\lambda };\lambda \ne 0}$ and $V(-y)=V\left(y\right)$.