$\displaystyle{Lim \quad \Omega(\lambda',\mu,\lambda)\leq \Omega(0,\mu,\lambda)}$ as $\rho' \rightarrow + \infty$.

From the above property iv) we deduce that

$\forall \quad n \in \mathbb{N}$, $\displaystyle{\mid\mid K_{\lambda',\mu,\lambda}^{n}\mid\mid^{\frac{1}{n}}}$ $\rightarrow$  $\displaystyle{\mid\mid K_{0,\mu, \lambda}^{n}\mid\mid^{\frac{1}{n}}}$ as $\rho' \rightarrow + \infty$.

$\displaystyle{\forall \quad \epsilon > 0, \exists\quad \rho_{1}; \forall \quad \rho' \geq \rho_{1}}$ we have $\displaystyle{\mid\mid K_{0,\mu, \lambda}^{n}\mid\mid^{\frac{1}{n}} - \epsilon \leq \mid\mid K_{\lambda',\mu,\lambda}^{n}\mid\mid^{\frac{1}{n}} \leq \mid\mid K_{0,\mu, \lambda}^{n}\mid\mid^{\frac{1}{n}} + \epsilon}$.

set $\rho_{1} = \frac{\mu}{\lambda_{1}^{'}}$ then we get

$\displaystyle{\mid\mid K_{\lambda_{1}^{'},\mu,\lambda}^{n}\mid\mid^{\frac{1}{n}}\leq \mid\mid K_{0,\mu, \lambda}^{n}\mid\mid^{\frac{1}{n}} + \epsilon }$.

In particular, we get

\begin{equation}
\displaystyle{\Omega (\lambda_{1}^{'},\mu,\lambda)\leq \Omega(0,\mu, \lambda) + \epsilon }
\end{equation}

Now as

$\displaystyle{\Omega (\lambda',\mu, \lambda) = lim \quad \mid\mid K_{\lambda',\mu, \lambda}^{n}\mid\mid^{\frac{1}{n}}}$ as $n \rightarrow +\infty$

$\displaystyle{\forall \quad \epsilon > 0, \exists\quad n_{0}\in \mathbb{N}; \forall \quad n \geq n_{0}}$ we have $\displaystyle{\Omega (\lambda',\mu, \lambda) - \epsilon \leq \mid\mid K_{\lambda',\mu, \lambda}^{n}\mid\mid^{\frac{1}{n}} \leq \Omega_{(\lambda',\mu, \lambda)} + \epsilon}$

By using the above right inequality and the decreasing of the map $\rho' \rightarrow N_{\rho'}$ , we deduce that:

\begin{equation}
\displaystyle{\mid\mid K_{\lambda',\mu, \lambda}^{n}\mid\mid^{\frac{1}{n}} \leq \Omega(\lambda',\mu, \lambda) + \epsilon \leq  \Omega(\lambda_{1}^{'},\mu, \lambda) + \epsilon}
\end{equation}

and by (3.97) and (3.98) and for all $\epsilon > 0$, we get:

$\displaystyle{lim \quad\Omega_{\rho'}}$ $\leq$ $\displaystyle{\Omega(0,\mu, \lambda) + 2\epsilon}$ as $\rho'\rightarrow +\infty $

$\displaystyle{lim \quad\Omega(\lambda^{'},\mu, \lambda) = \Omega(0,\mu,\lambda)}$ as $\lambda' \rightarrow 0$.

vi) Let $\displaystyle{\tilde{N}_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda y_{1}\sqrt{r(y_{1})}} \int_{0}^{min(y,y_{1})}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$ where
$\delta = \rho'(\rho + \rho') - 1 = \rho'^{2} + \rho\rho' -1$

then  $\displaystyle{\tilde{N}_{\lambda',\mu,\lambda}(y, y_{1})}$ can be written as

\begin{equation}
\displaystyle{\tilde{N}_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda y_{1}}e^{\rho'y_{1}}(1 - \frac{y_{1}}{\rho'})^{\rho'^{2} -1} \int_{0}^{min(y,y_{1})}e^{-\rho's}(1 - \frac{s}{\rho'})^{-\rho'^{2}}[\frac{\rho' - y_{1}}{\rho' - s}]^{\rho\rho'} ds}
\end{equation}

As $0 \leq \leq y_{1}$, the map $\displaystyle{\mu \rightarrow [\frac{\rho' - y_{1}}{\rho' - s}]^{\rho\rho'}}$ is creasing with respect to $\mu$ and we get vi).

vii) Let $\displaystyle{\delta_{\beta} = \rho'(\frac{\beta}{\lambda} + \rho') - 1}$ and $\displaystyle{L_{2,\beta}([0, \rho'], r_{\beta}(y)dy)}$ be the space of square integrable  functions with respect to the measure $\displaystyle{r_{\beta}(y) = e^{2\rho'y}(1 - \frac{y}{\rho'})^{2\delta_{\beta}}}$.

To use the results of chapter VII of Kato's book [Kato] on the operators depending of a parameter, we shall consider the map:

\begin{equation}
\displaystyle{\mu \rightarrow <\phi, K_{\lambda', \mu, \lambda}\psi > = \int_{0}^{\rho'}\phi(-iy)\int_{0}^{\rho'}\tilde{N}_{\lambda', \mu, \lambda}(y,y_{1})\bar{\psi}(-iy_{1})dy_{1}r_{\beta}(y)dy}
\end{equation}

with $\displaystyle{\phi \in L_{2,\beta}([0, \rho'], r_{\beta}(y)dy)}$ and $\displaystyle{\psi \in L_{2,\beta}([0, \rho'], r_{\beta}(y)dy)}$

We being by showing that the map defined by (3.100) is continuous with respect to $\mu$ :

Let $\displaystyle{\tilde{N}_{\lambda', \beta}}$ be the expression of  $\displaystyle{\tilde{N}_{\lambda', \mu, \lambda}}$  where we have replaced  $\delta $ by $\delta_{\beta}$.

Now, we set $\displaystyle{f_{\mu}(y, y_{1}) = e^{2\rho'y}(1 - \frac{y}{\rho'})^{2\delta_{\beta}}\tilde{N}_{\lambda', \mu, \lambda}(y, y_{1})\phi(-iy)\bar{\psi}(-iy_{1})}$ then for $\mu \geq \beta$ we have:

$\displaystyle{\mid f_{\mu}(y, y_{1})\mid = e^{2\rho'y}(1 - \frac{y}{\rho'})^{2\delta_{\beta}}\tilde{N}_{\lambda', \beta}(y, y_{1})\mid \phi(-iy)\mid \mid\bar{\psi}(-iy_{1})\mid}$

then the continuity of (3.100) with respect to $\mu$ follows by applying the bounded convergence theorem.

Now to show that the map defined by (3.100) is analytic with respect to  $\mu$,  it suffices to observe that for all closed curve we have

\begin{equation}
\displaystyle{\int_{\gamma}\tilde{N}_{\lambda', \mu, \lambda}d\mu = 0}
\end{equation}

and to apply the following theorem:

(Reed-Simon  [Reed et al], theorem XII.8)

Let $\mathbb{E}$ be a complex Banach space and $L(\mathbb{E})$ be the set of bounded operators on $\mathbb{E}$ with respect to the operator norm.

Let $\mathbb{T}$ be an analytic function $(\mu \in \mathbb{C} \rightarrow \mathbb{T}(\mu) \in L(\mathbb{E}))$ and $\sigma_{0}$  be a simple eigenvalue of $T(\mu_{0})$.Then

There exists an analytic function $\sigma$ of $\mu$  for $\mu$ near $\mu_{0}$ such that  $\sigma(\mu)$ is simple eigenvalue of $\mathbb{T}(\mu)$ and $\sigma(\mu_{0}) = \sigma_{0}$.  $\square$

We conclude this section by noting that for $\lambda' \neq 0$, the operator $H_{\lambda', \mu, \lambda}$  has a rich set of spectral properties desired by physicists and for $\lambda' = 0$,  the results of Ando-Zerner  were a major contribution in the spectral study of the operator $H_{0, \mu, \lambda}$, however the density of its eigenvectors in Bargmann space is always open question.

§ STUDY OF THE SINGULARITY OF $\DISPLAYSTYLE{ H_{\LAMBDA',\MU, \LAMBDA}\VARPHI(Z) = \SIGMA(\LAMBDA', \MU, \LAMBDA) \VARPHI(Z)\, AT \, \INFTY}$

§.§ On bi-confluent Heun equation associated to eigenvalue problem of  $\displaystyle{ H_{\lambda',\mu, \lambda}}$

We consider the equation

$\displaystyle{(\lambda'z^{2} + i\lambda z)\varphi^{''}(z) + (i \lambda z^{2} + \mu z)\varphi^{'}(z) = \sigma(\lambda', \mu, \lambda) \varphi(z)}$       at   $\infty$   ($\star$)

In our differential equation ($\star$), if we put $\psi(z) = \varphi^{'}(z)$ then we get

\begin{equation}
\displaystyle{\psi^{'}(z) = \frac{\sigma(\lambda', \mu, \lambda)}{z(\lambda' z + i\lambda)}\varphi(z) - \frac{ i \lambda + \mu}{\lambda' z + i\lambda}\psi(z)}
\end{equation}

Now let $\phi(z) = \left (\begin{array} [c] {l} \varphi(z)\\
\quad\\
\psi(z)\\
\end{array} \right )$ then

\begin{equation}
\phi^{'}(z) = \left (\begin{matrix} \displaystyle{0}&\displaystyle{1}\\
\quad\\
\displaystyle{ \frac{\sigma(\lambda', \mu, \lambda)}{z(\lambda' z + i\lambda)}}&\displaystyle{- \frac{ i \lambda + \mu}{\lambda' z + i\lambda}}\\
\end{matrix} \right )\left (\begin{array} [c] {l} \varphi(z)\\
\quad\\
\psi(z)\\
\end{array} \right )
\end{equation}

If we put $\displaystyle{z = re^{i\theta}; r \in [0, +\infty[, \theta \in [0, 2\pi]}$ and $W(r) = \left (\begin{array} [c] {l} \varphi(re^{i\theta})\\
\quad\\
\psi(re^{i\theta})\\
\end{array} \right )$, we deduce that

\begin{equation}
\displaystyle{ W'(r) =e^{-i\theta}\mathbb{M}(re^{i\theta})W(r)}
\end{equation}

where  $\mathbb{M}(re^{i\theta})= \left (\begin{matrix} \displaystyle{0}&\displaystyle{1}\\
\quad\\
\displaystyle{ \frac{\sigma(\lambda', \mu, \lambda)}{re^{i\theta}(\lambda' re^{i\theta} + i\lambda)}}&\displaystyle{- \frac{ i \lambda + \mu}{\lambda' re^{i\theta} + i\lambda}}\\
\end{matrix} \right )$

In $\mathbb{C}^{2}$ we denote the norm of $W$ by  $\mid\mid W\mid\mid$ and the norm of matrix $\mathbb{M}$ by $\mid\mid\mid \mathbb{M}\mid\mid\mid$.

As  $\displaystyle{\mid\mid\mathbb{M}(re^{i\theta})W(re^{i\theta})\mid\mid \,\leq \,\mid\mid\mid \mathbb{M}\mid\mid\mid\,\mid\mid W\mid\mid}$, then we deduce that

\begin{equation}
\displaystyle{\mid\mid\mathbb{M}^{'}(r)\leq \mid\mid\mid \mathbb{M}\mid\mid\mid\,\mid\mid W(r)\mid\mid}
\end{equation}

Let $R > \rho'$ then the norm $\mid\mid\mid \mathbb{M}\mid\mid\mid$ is bounded outside the disk of radius $R$ and by applying the Gronwal Lemma, we deduce the following inequality

\begin{equation}
\displaystyle{ \forall \, r \in [R, + \infty[, \mid\mid \phi(z) \mid\mid \leq sup_{\theta \in [0, 2\pi]}\mid\mid W(Re^{i\theta}) \mid\mid e^{\mid\mid\mid \mathbb{M}\mid\mid\mid(\mid z\mid - R)}}
\end{equation}

In particular, we get

\begin{equation}
\displaystyle{\mid \varphi(z) \mid \leq sup_{\theta \in [0, 2\pi]}\mid\mid W(Re^{i\theta}) \mid\mid e^{\mid\mid\mid \mathbb{M}\mid\mid\mid(\mid z\mid - R)}}
\end{equation}

Therefore  $\varphi$ belongs to Bargmann space $\mathbb{B}_{0} $

§.§ On eigenvalues of $H_{\lambda', \mu, \lambda}$ acting on $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$

Let  $\rho' > 0, y \in [0, \rho']$ and $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$ be the Hilbert space of square integrable functions on $[0, \rho']$ with inner

\begin{equation}
\displaystyle{< u, v > = \int_{0}^{\rho'}u(y)\overline{v(y)}r(y)dy}
\end{equation}

with $r(y)$ called weight function.

We will construct the fonction $r(y)$ such that $H_{\lambda', \mu, \lambda}$ is symmetric on $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$.

For $\varphi \in \mathbb{B}_{0} $, consider  $u(y) = \varphi(-iy)$ and the eigenvalue problem associated to $H_{\lambda', \mu, \lambda}$:

\begin{equation}
\displaystyle{-y (\lambda - \lambda' y)u^{"}(y) + y(\lambda y +\mu y)u'(y) = \sigma(\lambda', \mu, \lambda)u(y)}
\end{equation}

By multiplying the differential equation (4.8) by $r(y)\overline{u(y)}$ and integrating it on $[0, \rho']$ to get

\begin{equation}
\left \{ \begin{array}{c} \displaystyle{\int_{0}^{\rho'}[-y (\lambda - \lambda' y)u^{"}(y)+y(\lambda y +\mu y)u'(y)] \overline{u(y)}r(y)dy =}\\
\displaystyle{\sigma(\lambda', \mu, \lambda)\int_{0}^{\rho'}\mid u(y)\mid^{2}r(y)dy \quad \quad \quad \quad \quad \quad \quad \quad \quad }
\end{array} \right.
\end{equation}

Integrating the first term  of integral equation by parts to get

$\displaystyle{[s(y)\overline{u(y)}u'(y)(\lambda' y - \lambda)]_{0}^{\rho'} - \int_{0}^{\rho'}s(y)(\lambda' y - \lambda)\mid u'(y)\mid^{2}dy}$

+ $\displaystyle{\int_{0}^{\rho'}[(\lambda y + \mu)s(y) - \frac{\partial}{\partial y}((\lambda' y - \lambda)s(y))]\mid u(y)\mid^{2}dy}$

\begin{equation}
\displaystyle{= \sigma(\lambda', \mu, \lambda)\int_{0}^{\rho'}\mid u(y)\mid^{2}r(y)dy}
\end{equation}

where $s(y) = yr(y)$

We choose $s(y)$ such that

\begin{equation}
\displaystyle{ \frac{\partial}{\partial y}((\lambda' y - \lambda)s(y)) = (\lambda y + \mu)s(y)}
\end{equation}

\begin{equation}
\displaystyle{ s(y) = ce^{\rho' y}\mid y - \rho'\mid^{\rho'(\rho + \rho') - 1}}\, and \,\displaystyle{ r(y) = \frac{s(y)}{y}} , c \, is \, const.
\end{equation}

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then $H_{\lambda', \mu, \lambda}$ is symmetric on $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$.

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\displaystyle{[s(y)\overline{u(y)}u'(y)(\lambda' y - \lambda)]_{0}^{\rho'} = 0}
\end{equation}

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\displaystyle{ - \int_{0}^{\rho'}s(y)(\lambda' y - \lambda)\mid u'(y)\mid^{2}dy} = \displaystyle{\sigma(\lambda', \mu, \lambda)\int_{0}^{\rho'}\mid u(y)\mid^{2}r(y)dy}
\end{equation}

i.e $H_{\lambda', \mu, \lambda}$ is symmetric on $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$.  $\square$

(i)  if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the the eigenvalues of $H_{\lambda', \mu, \lambda}$ are real.

(ii) On $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$, the operator $H_{\lambda', \mu, \lambda}$ takes the following form

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda}u(y) = \frac{1}{r(y)}\frac{\partial}{\partial y}(p(y)u'(y))}
\end{equation}

where $\displaystyle{p(y) = \lambda' y(\rho' - y) r(y)}$.

(i) It is a simple consequence of the above theorem and the fact that the space of the restrictions of the elements of the domain of the operator $ H_{\lambda', \mu, \lambda}$   as operator acting on Bargmann's space $\mathbb{B}_{0}$ is included in

$\displaystyle{\mathbb{B}_{2} = \{ u \in L_{2}((0, \rho'), r(y)dy); H_{\lambda', \mu, \lambda}u \in L_{2}((0, \rho'), r(y)dy); u(0) = 0\}}$

(ii) It is a simple consequence.  $\square$

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the eigenvalues of $H_{\lambda', \mu, \lambda}$ acting on $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$ are simple.

Let $u, v \in \displaystyle{\mathbb{B}_{2} }$ then $\displaystyle{vH_{\lambda', \mu, \lambda}u = \frac{v}{r(y)}\frac{\partial}{\partial y}(p(y)u'(y))}$, $\displaystyle{uH_{\lambda', \mu, \lambda}v = \frac{u}{r(y)}\frac{\partial}{\partial y}(p(y)v'(y))}$

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\displaystyle{r(y)(vH_{\lambda', \mu, \lambda}u - uH_{\lambda', \mu, \lambda}v = \frac{\partial}{\partial y}[p(y)(v(y)u'(y) - u(y)v'(y)]}
\end{equation}

If $\displaystyle{H_{\lambda', \mu, \lambda}u(y)  = \sigma(\lambda', \mu, \lambda)u(y) ; u \neq 0}$ and  $\displaystyle{H_{\lambda', \mu, \lambda}v(y)  = \sigma(\lambda', \mu, \lambda)v(y) ; v \neq 0}$  then

$\displaystyle{\frac{\partial}{\partial y}[p(y)(v(y)u'(y) - u(y)v'(y)] = 0}$, in particular $p(y)(v(y)u'(y) - u(y)v'(y)$ is a constant.

As $p(0)((v(0)u'(0) - u(0)v'(0)) = 0$ then  $v(y)u'(y) - u(y)v'(y) = 0 \, \forall \, y \in [0, \rho']$ and $ v $ is proportional to $u$ i.e. the geometric multiplicity  of $\sigma(\lambda', \mu, \lambda)$ is one.

Now, let $\displaystyle{Ker(H_{\lambda', \mu, \lambda}  - \sigma(\lambda', \mu, \lambda)I ) = \{u \in  \mathbb{B}_{2}; H_{\lambda', \mu, \lambda}u  = \sigma(\lambda', \mu, \lambda)u ; u \neq 0\}}$, showing that there are no Jordan blocks corresponding to  $\sigma(\lambda', \mu, \lambda)$ i.e;

$\displaystyle{Ker(H_{\lambda', \mu, \lambda}  - \sigma(\lambda', \mu, \lambda)I) = Ker(H_{\lambda', \mu, \lambda}  - \sigma(\lambda', \mu, \lambda)I)^{2} }$ $, \displaystyle{\forall\,  \sigma(\lambda', \mu, \lambda) \in \sigma(H_{\lambda', \mu, \lambda})}$

where $\displaystyle{\sigma(H_{\lambda', \mu, \lambda})}$ denotes the spectrum of $\displaystyle{H_{\lambda', \mu, \lambda}}$.

Let $\displaystyle{u \in Ker(H_{\lambda', \mu, \lambda}  - \sigma(\lambda', \mu, \lambda)I)}$ and $\displaystyle{v \in Ker(H_{\lambda', \mu, \lambda}  - \sigma(\lambda', \mu, \lambda)I)^{2} }$  such that if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\displaystyle{\frac{1}{r(y)}\frac{\partial}{\partial y}(p(y)u'(y))u(y) = \sigma(\lambda', \mu, \lambda)u(y), \quad y \in [0, \rho']}
\end{equation}

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\displaystyle{\frac{1}{r(y)}\frac{\partial}{\partial y}(p(y)v'(y))v(y) = \sigma(\lambda', \mu, \lambda)v(y) + u(y), \quad y \in [0, \rho']}
\end{equation}

Multiplyng  the differential equation by $r(y)\overline{v(y)}$ and  by $r(y)\overline{v(y)}$, integrating  by parts the obtained equations on $[0, \rho']$ to get

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\displaystyle{\int_{0}^{\rho'}p(y)u'(y)\overline{v}'(y)dy = \sigma(\lambda', \mu, \lambda)\int_{0}^{\rho'}r(y)u(y)\overline{v(y)}dy}
\end{equation}

if $\displaystyle{\rho'(\rho + \rho')  \geq 1}$ then the function $s(y)$ is bounded on $[0, \rho']$ this implies that

\begin{equation}
\left \{ \begin{array} {c} \displaystyle{\int_{0}^{\rho'}p(y)v'(y)\overline{u}'(y)dy = \sigma(\lambda', \mu, \lambda)\int_{0}^{\rho'}r(y)v(y)\overline{u(y)}dy \quad +}\\ \displaystyle{ \int_{0}^{\rho'}r(y)\mid u(y)\mid^{2}dy \quad \quad  \quad \quad  \quad \quad  \quad \quad  \quad  \quad \quad \quad  \quad \quad} \end{array} \right.
\end{equation}

Now (4.19) - (4.20) gives

\begin{equation}
\left \{ \begin{array} {c}  \displaystyle{\int_{0}^{\rho'}p(y)[ u'(y)\overline{v}'(y) -  v'(y)\overline{u}'(y)]dy = \quad \quad }\\
\displaystyle{\sigma(\lambda', \mu, \lambda)\int_{0}^{\rho'}r(y)[u(y)\overline{v(y)} - v(y)\overline{u(y)}]dy}\\
\displaystyle{ - \int_{0}^{\rho'}r(y)\mid u(y)\mid^{2}dy}    \quad  \quad  \quad  \quad \quad \quad  \quad \quad \end{array} \right.
\end{equation}

In particular we have

\begin{equation}
\left \{ \begin{array} {c} \displaystyle{2i\int_{0}^{\rho'}p(y)\Im m(u'(y)\overline{v}'(y))dy = }\\

\displaystyle{2i\sigma(\lambda', \mu, \lambda)\int_{0}^{\rho'}r(y)\Im m(u(y)\overline{v(y)}) - v(y)dy}\\

\displaystyle{ - \int_{0}^{\rho'}r(y)\mid u(y)\mid^{2}dy} \end{array} \right.
\end{equation}

As $p$ is real function and $ \sigma(\lambda', \mu, \lambda) \in \mathbb{R}$ then $\displaystyle{\int_{0}^{\rho'}r(y)\mid u(y)\mid^{2}dy = 0}$ which is contradictory with $u$ is eigenfunction associated to  $ \sigma(\lambda', \mu, \lambda)$.  $\square$

We end these first spectral properties on $H_{\lambda', \mu, \lambda}$ by the following observations

(1) The hamiltonian $H_{\lambda', \mu, \lambda}$ acting on Bargmann space (which is not selfadjoint) has a real spectrum.

(2) The PT-symmetry condition (space-time reflection) is not sufficient to provide that the spectrum  of  $H_{\lambda', \mu, \lambda}$ is real, there are some PT-symmetric hamiltonian with polynomial potentials producing non real eigenvalues, see  [Delabeare et al].

(3) The method used by  [Dorey et al] to give a rigorous proof of reality and positivity of the eigenvalues of some non-self-adjoint hamiltonian is not applicable to $H_{\lambda', \mu, \lambda}$ and even less to  $H_{\mu, \lambda}$

§ REVIEW SOME FONDAMENTAL SPECTRAL PROPERTIES OF $H_{\LAMBDA', \MU, \LAMBDA}$

We now turn to review  the fondamental spectral properties of $H_{\lambda', \mu, \lambda}$ that we will presented in above sections.

$\bullet$ We have restricted $H_{\lambda', \mu, \lambda}$ to the subspace of Bargmann space consisting of the functions that vanish at $0$. Thus we get rid of the vacuum state.

A central role is played by the following inequality (for $\mu > 0$):

\begin{equation}
\displaystyle{\Re e < H_{\lambda, \mu} \varphi , \varphi > = \mu \mid\mid A\varphi \mid\mid^{2} \geq  \mid\mid \varphi \mid\mid^{2}, \, \forall\,   \varphi \in D(H_{\lambda, \mu})}
\end{equation}

which obviously holds for polynomials ($\Re e$ means the real part).

$\bullet$ Now if $\varphi$ is in Bargmann space, $H_{\lambda, \mu}\varphi$ can be defined as an analytic function, so that it makes sense to take as domain of $H_{\lambda, \mu}$ the set of functions $\varphi$ such that $H_{\lambda, \mu}\varphi$ still is in Bargmann space. It turns out that so defined, $H_{\lambda, \mu}$
is the closure of $H_{\lambda, \mu_{\mid_{pol}}}$, has a compact inverse and $e^{-t H_{\lambda, \mu}}$  can be defined for positive $t$ (see  another section of this work for detailed proofs).

$\bullet$ Above, a rigorous mathematical analysis of domains of $H_{\mu, \lambda}$ is given , in particular, we have :

\begin{equation}
\displaystyle{D(H_{\mu, \lambda}) = D(H_{\mu, \lambda}^{min})}
\end{equation}

where $H_{\mu, \lambda}^{min}$ is the restriction of the closure of $H_{\mu, \lambda}$ on the polynomials.

$\bullet$ By using the following inequality :

$\displaystyle{\forall \, \epsilon > 0 , \exists \, C_{\epsilon, \lambda', \mu,  \lambda} > 0 ; \mid\mid H_{\mu, \lambda} \varphi \mid\mid^{2} \leq \epsilon \mid\mid A^{*^{2}}A^{2}\varphi \mid\mid  +  C_{\epsilon, \lambda', \mu,  \lambda} \mid\mid \varphi  \mid\mid \, \forall \, \varphi \in D(A^{*^{2}}A^{2})}$

we can  show   that for $\lambda' \neq 0$,  we have

\begin{equation}
\displaystyle{D(H_{\lambda', \mu, \lambda}) = D(A^{*^{2}}A^{2})}
\end{equation}

$\bullet$ Note that the domain of the adjoint and anti-adjoint parts of $H_{\mu, \lambda}$ are not included  in each other, nor is the domain of their commutator.

$\bullet$ The restriction of the closure of the anti-part $H_{I} = A^{*}(A + A^{*})A$ of $H_{\mu, \lambda}$  or of $H_{\lambda', \mu, \lambda}$ on the polynomials is symmetric but it is not self-adjoint, $D(H_{I}) \neq  D(H_{I}^{min})$ (where $H_{I}^{min}$  is the restriction of the closure of $H_{I}$ on the polynomials) and  that $H_{I}$ is chaotic in Devaney's sense. In particular, the spectrum  of $H_{I}$ is all $\mathbb{C}$.

$\bullet$ When we want to proof that the eigenvalues of $H_{\mu, \lambda}$ are real we define :

$\displaystyle{H_{\lambda'}: = H_{\lambda', \mu, \lambda} = \lambda' A^{*^{2}}A^{2} + H_{\mu, \lambda}}$

For positive $\lambda'$ and $\mu > 0$ we have,

\begin{equation}
\displaystyle{\Re e < H_{\lambda'}\varphi , \varphi > \geq \mu \mid\mid A \varphi \mid\mid^{2}, \, \forall\,   \varphi \in D(H_{\lambda'})}
\end{equation}

and $H_{\lambda'}$, which is also invertible satisfies:

\begin{equation}
\displaystyle{\mid\mid AH_{\lambda'}^{-1} \psi \mid\mid \leq \frac{1}{\mu}\mid\mid \psi\mid\mid}
\end{equation}

Now the set of $\phi$ satisfying $\displaystyle{\mid\mid A\phi \mid\mid^{2} \leq 1}$ is a compact set and we see that all $H_{\lambda'}$, map the unit ball into one compact set.

This enables us to prove that $\displaystyle{lim\, H_{\lambda'}^{-1} = H_{\mu, \lambda}^{-1}\, as \, \lambda' \longrightarrow 0}$ in the sense of the norm of operators and this implies the convergence of the eigenvalues.

(v) The operator $H_{\mu, \lambda}$  respectively $H_{\lambda',\mu, \lambda}$ has the form $\displaystyle{p(z)\frac{\partial^{2}}{\partial z^{2}} + q(z))\frac{\partial}{\partial z} }$ with $p(z) = i\lambda z$ (respectively $p(z) = \lambda'z^{2} + i\lambda z$) and $q(z) =  i\lambda z^{2} + \mu z$ of degree $2$ and $H_{\mu, \lambda}$  and $H_{\lambda',\mu, \lambda}$ are of Heun operator type respectively.

(vi) The eigenvalue problem associated to $H_{\mu, \lambda}$  (respectively $H_{\lambda',\mu, \lambda}$) does not satisfy the classical ordinary differential equation of the form:

\begin{equation}
\displaystyle{\sigma(z)\frac{\partial^{2}\varphi}{\partial z^{2}} + \tau(z))\frac{\partial \varphi}{\partial z}  = \alpha \varphi}
\end{equation}

where $\sigma(z)$ is a polynomial of degree  at most two, $\tau(z)$ is a polynomial of the degree exactly one and $\alpha$ is a constant.

(vii) If $\lambda \neq  0$, the eigenvalue problem associated to $H_{\mu, \lambda}$ can be written as follows :

\begin{equation}
\displaystyle{z\frac{\partial^{2}\varphi}{\partial z^{2}} +( z^{2} - i\rho z)\frac{\partial \varphi}{\partial z}  = - i\frac{\alpha}{\lambda} \varphi; \rho = \frac{\mu}{\lambda}}
\end{equation}

Let $ z = i\sqrt{2}\xi$ and $\varphi(\sqrt{2}\xi) = \psi(\xi)$ then (5.7) can be transformed to :

\begin{equation}
\displaystyle{\xi\frac{\partial^{2}\psi}{\partial \xi^{2}} +(- 2\xi^{2}  + \rho\sqrt{2} \xi)\frac{\partial \psi}{\partial\xi} - \frac{\alpha\sqrt{2}}{\lambda} \psi  = 0}
\end{equation}

(viii) If $\lambda' \neq  0$ and let $z = -i\rho'\xi$ and $\varphi(z) = \psi(\xi)$ then  the eigenvalue problem associated to $H_{\lambda', \mu, \lambda}$ can be transformed to follows :

\begin{equation}
\displaystyle{\xi(\xi - 1)\frac{\partial^{2}\psi}{\partial \xi^{2}} +(\rho'^{2}\xi^{2} + \rho'' \xi)\frac{\partial \psi}{\partial \psi}  = \frac{\alpha}{\lambda'} \psi; \rho' = \frac{\lambda}{\lambda'}, \rho'' = \frac{\mu}{\lambda'}}
\end{equation}

(ix)   Our equation  belongs to family of the bi-confluent  Heun equations of class $(0, 1, 1_{4})$

\begin{equation}
\displaystyle{H_{\alpha, \beta, \gamma, \delta}u(z) =  zu''(z) + (-2z^{2} - \beta z  + 1 + \alpha)u'(z) + [(\gamma - \alpha - 2)z -\delta] u(z) = 0}
\end{equation}

where  $\displaystyle{ \delta = \frac{1}{2}(\hat{\delta} + \beta(1+\alpha))}$

which have one  regular singular points $z_{0} = 0$ and one irregular singular point $z_{\infty} = \infty$.

This equation belongs to family of confluent Heun equations (CHE)

\begin{equation}
\displaystyle{H_{\alpha, \beta, \gamma, \delta,\epsilon}u(z) =  z(z - 1)u''(z) + (\alpha z^{2} + \beta z  + \gamma)u'(z) + (\delta z + \epsilon) u(z) = 0}
\end{equation}

which has two regular singular points $z_{0} = 0$ and $z_{1} = 1$ and one irregular singular point $z_{\infty} = \infty$.

These equations are more complicated than the equations for classical special functions such as Bessel functions, hypergeometric functions, and confluent hypergeometric functions (see [Betman et al.],  [Choun: [1-10)]]) and  [Abramovitz]).

Above equation includes, as a special (or confluent) case, the Mathieu equation, the equation for Coulomb spheroidal functions, and many other equations used in mathematical physics and applied mathematics. However, it cannot be solved in closed integral form, and analytic tools for investigation of its solutions are rather poor.

It turns out that integral relations are useful for analytic studies of equation and similar objects. Integral relations for equations of the Heun class were discussed in  [Erdelyi] and [Kasakov1 et al.]. In  [Kasakov2 et al.] such integral relations were used in order to obtain Fredholm integral equations for eigenfunctions of the boundary problems associated with equations of the Heun class and in [Kasakov1 et al.], integral relations between the eigenfunctions of different equations were established.

The aim result of this work is to specify the boundary conditions  of Schrodinger equations associated to “Gribov-Intissar” operators and to give an analysis of behavior of scattering amplitudes with respect the parameters.

As $ H_{\lambda', \mu, \lambda}$ and $H_{\mu, \lambda}$ are non-self-adjoint unbounded operators and that the space

$\displaystyle{ \mathbb{B}_{0} = \{ \varphi \in \mathbb{B}; \varphi (0) = 0 \}}$ is invariant under their actions then the first idea is to show  that their respectively resolvent sets  are non empty and to show that their respectively resolvent is  compact.

In  [Intissar2], it was showed that $ H_{\lambda', \mu, \lambda}^{-1}$ and $H_{ \mu, \lambda}^{-1}$  exist and they are compact using the compactness of the injection of their respective domains into Bargmann space and that their respective resolvent sets are non-empty.

Let $\psi \in \mathbb{B}_{0}$ and $\varphi \in D(H_{\lambda',\mu,\lambda})$ (respectively $\varphi \in D(H_{\mu,\lambda})$), we consider the equation  $H_{\lambda',\mu,\lambda}\varphi = \psi$  (respectively  $H_{\lambda',\mu,\lambda}\varphi = \psi$) which can be written under following form :

\begin{equation}
\displaystyle{(\lambda'z^{2} + i\lambda z)\varphi''(z) + (i\lambda z^{2}  + \mu z)\varphi'(z) = \psi(z)}
\end{equation}

respectively :

\begin{equation}
\displaystyle{ i\lambda z\varphi''(z) + (i\lambda z^{2}  + \mu z)\varphi'(z) = \psi(z)}
\end{equation}

Let $\psi \in \mathbb{B}_{0}$ and choosing the straight line connecting $-i\rho' , z \in \mathbb{C}$ parametrized  by

$\displaystyle{\gamma : [0, 1] \rightarrow \mathbb{C}, \gamma(t) = -i\rho' + t(z + i\rho') \quad \quad (\gamma(0) = -i\rho' , \gamma(1) = z)}$

If we define $\displaystyle{\int_{-i\rho'}^{z}\psi(\xi)d\xi:=\int_{\gamma}\psi(\xi)d\xi:= \int_{0}^{1}}\psi(\gamma(t))\gamma'(t)dt$ then our  equation can be transformed to following integral equation :

\begin{equation}
\displaystyle{\varphi(z) = \frac{1}{\lambda'}\int_{-i\rho'}^{z}e^{-i\rho'\eta}(\eta + i\rho')^{-(\delta +1)} [\int_{-i\rho'}^{\eta}e^{i\rho'\xi}(\xi + i\rho')^{\delta}\frac{\psi(\xi)}{\xi}d\xi]d\eta}
\end{equation}

In the same way we get on $]-i\infty, 0]$ an integral  equation with $\lambda' = 0$

However, the integral representation of $\varphi(z)$ in  the equations is hard to study in $\mathbb{C}$ for some existing results on eigenvalues and eigenfunctions of our operators. To overcome this difficulty, the study of integral equations associated to our equations are restricted on negative imaginary axis.Therefore led to put $\psi(y) = \varphi(-iy)$ ; $y \in [0, + \infty[$ to study the following problems :

\begin{equation}
\displaystyle{-\lambda y \psi_{\mu, \lambda}^{''}(y) + (\lambda y^{2} + \mu y)\psi_{\mu, \lambda}^{'}(y) = \sigma(\mu, \lambda)\psi_{\mu, \lambda}(y)}
\end{equation}

\begin{equation}
\displaystyle{(\lambda' y^{2} - \lambda y)\psi_{\lambda', \mu, \lambda}^{"}(y) + (\lambda y^{2} + \mu y)\psi_{\mu, \lambda}^{'}(y) = \sigma(\lambda', \mu, \lambda)\psi_{\lambda', \mu, \lambda}(y)}
\end{equation}

\begin{equation}
\displaystyle{  \mathbb{B}_{1} = \{\psi : [0, \infty[ \longrightarrow \mathbb{C}  \, entire \, ; \int_{0}^{\infty}\mid \psi(y)\mid^{2}e^{-\mid y \mid^{2}}dy < \infty \}}
\end{equation}

We note that above equations  are bi-confluent Heun equations and we observe that (5.17) possesses the origin as regular singular point and $\infty$ as irregular singular point but (5.17) possesses $ y = 0$ and $y = \displaystyle{\rho' = \frac{\lambda}{\lambda'}}$ as regular singular points  and $\infty$ as irregular singular point.

On negative imaginary axis an explicit inverse of $H_{\mu, \lambda}$  (respectively of $H_{\lambda', \mu, \lambda}$) is given by :

$\bullet$ For $ y \in [0, +\infty[$, we get:

\begin{equation}
\displaystyle{H_{\mu, \lambda}^{-1} \varphi(-iy) = \int_{0}^{\infty}\mathcal{N}_{\mu,\lambda}(y, s)\varphi(-is)ds}
\end{equation}

where  $\displaystyle{\mathcal{N}_{\mu,\lambda}(y, s) = \frac{1}{s}e^{-\frac{s^{2}}{2} - \frac{\mu}{\lambda}s}\int_{0}^{min(y, s)}e^{\frac{u^{2}}{2} +\frac{\mu}{\lambda}u}du}$

$\bullet$ The boundary conditions of the problem are given by:

\begin{equation}
\displaystyle{\frac{\partial u_{\mu, \lambda}}{\partial t}(t, y) -\lambda y \frac{\partial^{2} u_{\mu, \lambda}}{\partial y^{2}}(t, y) + (\lambda y^{2} + \mu y)\frac{\partial u_{\mu, \lambda}}{\partial y}(t, y) = 0}
\end{equation}

\begin{equation}
\displaystyle{ u_{\mu, \lambda}(t, 0) = 0 } \,and\,  \displaystyle{lim\, e^{-\frac{1}{2}(y - \frac{\mu}{\lambda})^{2}}\frac{\partial u_{\mu, \lambda}}{\partial y}(t, y) = 0}
\end{equation}

as $y \longrightarrow + \infty$ for all $t$

$\bullet$ For $ y \in [0, \rho^{'}]$, $\displaystyle{\rho' = \frac{\lambda}{\lambda'}}$ and $\displaystyle{\rho = \frac{\mu}{\lambda}}$, we get:

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda}^{-1} \varphi(-iy) = \int_{0}^{\rho'}\mathcal{N}_{\lambda', \mu,\lambda}(y, s)\varphi(-is)ds}
\end{equation}

where $\displaystyle{\mathcal{N}_{\mu,\lambda}(y, s) = \frac{1}{\lambda s}e^{\rho' s}(1 -  \frac{s}{\rho'})^{\delta}\int_{0}^{min(y, s)}e^{-\rho' u }(1 - \frac{u}{\rho'})^{-(\rho + 1)} du}$

And we have the following properties

(i) The boundary conditions on regular singular points $y = 0$ and $y = \rho'$ of the problem :

\begin{equation}
\displaystyle{\frac{\partial u_{\mu, \lambda}}{\partial t}(t, y) + (\lambda' y^{2} - \lambda y)\psi_{\lambda', \mu, \lambda}^{"}(y) + (\lambda y^{2} + \mu y)\psi_{\mu, \lambda}^{'}(y) = 0}
\end{equation}

\begin{equation}
\displaystyle{ u_{\mu, \lambda}(t, 0) = 0 } \mbox{ and }  \displaystyle{ u_{\mu, \lambda}(t, y)} \mbox{\,is\, an\, analytic\, function\, around\, }y = \rho'
\end{equation}

(ii) The space of the restrictions of the elements of the domain of the operator $ H_{\lambda', \mu, \lambda}$ is included in

\begin{equation}
\displaystyle{\mathbb{B}_{2} = \{u \in L_{2}([0, \rho'], r(y)dy); H_{\lambda', \mu, \lambda}u \in L_{2}([0, \rho'], r(y)dy) , u(0) = 0\}}
\end{equation}

where the expression of $H_{\lambda', \mu, \lambda}u$ is given by

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda}u = - \frac{1}{r(y)}\frac{\partial}{\partial y}[-\lambda'y(\rho' - y)r(y)\frac{\partial u}{\partial y}]} \, with\, \displaystyle{r(y) = \frac{1}{y}e^{\rho' y}\mid y - \rho'\mid^{\rho'(\rho' + \rho) - 1}}
\end{equation}

(iii) $\displaystyle{H_{\lambda', \mu, \lambda}}$ acting on $ L_{2}([0, \rho'], r(y)dy)$ with domain :

$\displaystyle{\mathbb{B}_{2} = \{u \in L_{2}([0, \rho'], r(y)dy); H_{\lambda', \mu, \lambda}u \in L_{2}([0, \rho'], r(y)dy) , u(0) = 0\}}$

is symmetric.

We will show that $H_{\lambda', \mu, \lambda}$ has one real eigenvector $\varphi_{0} \in \mathbb{B}_{2}$ such that $\varphi_{0}  \geq 0$ (almost everywhere with respect to $r(y)dy$ on  $[0, \rho']$).

If $\sigma_{0}$ stands for the eigenvalue of $H_{\lambda', \mu, \lambda}$ associated with $\varphi_{0}$, the symmetry of $H_{\lambda', \mu, \lambda}$ implies that, for all $\varphi \in \mathbb{B}_{2}$ we have $\displaystyle{\Re e[(H_{\lambda', \mu, \lambda} - \sigma_{0}I)\varphi \geq 0}$. The last results motivates the following definition:

(Local energy)

For any $\varphi \in \mathbb{B}_{2}$, the local energy is the  function $\mathcal{E} : [0, \rho'] \longrightarrow \overline{\mathbb{R}}$ defined by :

$\displaystyle{\mathcal{E}_{\varphi}(y) = \frac{\Re e (H_{\lambda', \mu, \lambda}\varphi(y))}{\Re e(\varphi(y))}}$.

Therefore from what precedes, we get

Let $\sigma_{0}$ be the smallest eigenvalue of  $H_{\lambda', \mu, \lambda}$ then we have

($ \alpha$) For $\forall$ $\varphi \in \mathbb{B}_{2}$ such that $\displaystyle{\Re e(\varphi) \geq 0 , inf_{[0, \rho']}(\mathcal{E}_{\varphi}) \leq \sigma_{0} \leq sup_{[0, \rho']}(\mathcal{E}_{\varphi}) }$.

From above property, we deduce that:

($ \beta $) Let $\sigma_{0}$ be the smallest eigenvalue of  $H_{\lambda', \mu, \lambda}$  acting on Bargmann space then we have for
$\forall$ $\varphi \in D(H_{\lambda', \mu, \lambda}^{*})$ such that  $\Re e(\varphi) \geq 0 $,  $\displaystyle{inf_{[0, \rho']}\frac{[\Re e(H_{\lambda', \mu, \lambda}^{*}\varphi)}{\varphi}] \leq \sigma_{0} \leq sup_{[0, \rho']}\frac{[\Re e(H_{\lambda', \mu, \lambda}^{*}\varphi)}{\varphi}] }$.

By using Kato's inequalities, Thirring (1979) in [Thirring] have obtained the following theorem on Schrodinger operator acting on $L^{2}(\mathbb{R}^{n})$ (for this topic see also [Duffin] (1947), [Barnsley] (1978) and [Schmutz] (1985))

Let $\mathbb{H} = - \Delta + \mathbb{V}$ be a Schrodinger operator acting on $L^{2}(\mathbb{R}^{n})$ having an eigenvalue below the essential spectrum. Then the lowest eigenvalue $\epsilon_{0}$ of $\mathbb{H}$ is such that for any strictly positive $\varphi \in D(\mathbb{H})$,

$\displaystyle{inf_{\mathbb{R}^{n}}(\mathbb{V} - \frac{\Delta \varphi}{\varphi}) \leq \epsilon_{0} \leq sup_{\mathbb{R}^{n}}(\mathbb{V} - \frac{\Delta \varphi}{\varphi})}$.

In Gribov's Reggeon calculus, the Regge-poles are not the only singularities of the amplitude. There are also branch points which correspond to the exchange of several Reggeons. A Regge pole can be interpreted as corresponding to a single scattering. Regge cuts-multiple scatterings of hardons'$^{s }$ constituents.

We recall that the transition of a closed system of particles from an initial state $ \mid \varphi >$ to a final state $ \mid  \psi >$ is described in quantum theory by the $\mathbb{S}$ matrix (called scattering matrix) $\displaystyle{| \psi > = \mathbb{S} | \phi >}$ and the matrix elements of the $\mathbb{S}$  $\displaystyle{\mathbb{S}_{\psi, \varphi} = < \psi \mid \mathbb{S} \mid  \varphi >}$ can be represented in the form:

\begin{equation}
\displaystyle{\mathbb{S}_{\psi, \varphi} = \delta_{\psi, \varphi} + C\mathbb{T}_{\psi, \varphi}}
\end{equation}

where $\displaystyle{ \delta_{\psi, \varphi}  = 1}$  if the state does not change $(\mid \psi >  = \mid \varphi >)$ i.e. no interaction, $C$ denote the conservation of
energy and momentum and $\displaystyle{\mathbb{T}_{\psi, \varphi}}$ is called the transition (scattering) amplitude from the state $ \mid \varphi >$ to a final state $ \mid  \psi >$. We can consult the following references on the subject which summarizes the actual knowledge of Reggeon theory ( [Gribov2],   [Baker et al.], [Kaidalov] and [Boreskov et al.]).

§ GENERALIZED DIAGONALISATION  OF  SEMIGROUP $\DISPLAYSTYLE{E^{-TH_{\LAMBDA'', \LAMBDA', \MU, \LAMBDA}}}$, $T > 0$

§.§  Diagonalization of semigroup $e^{-tH_{\lambda'}}$,  $ t > 0$ where  $H_{\lambda'} = \lambda'A^{*^{2}}A^{2} + \mu A^{*}A + i\lambda (A^{*}(A + A^{*})A, \, \lambda' > 0$

In this under-section,  we consider the  following Cauchy problem  :

\begin{equation}
u'(t) = H_{\lambda'}u(t) ; u(0) = \phi \in B
\end{equation}

Let $\{\phi_{k}\}; k=1, 2,....$ are the eigenfunctions of  $H_{\lambda'}$ associated to eigenvalues $\{\sigma_{k}; k=1, 2,....\}$.

what conditions can be imposed on $H_{\lambda'}$ to write $u (t)$ as a limit of partial sums of expression $\displaystyle{\sum_{k=1}^{+\infty}C_{k}e^{-\sigma_{k}t}\phi_{k}}$ ?

Let $\lambda' \neq 0$, we will apply a fine result of Lidskii  [Lidskii] to show the existence of a total system of eigenfunctions of  $H_{\lambda'}$.

( [Intissar2])

Let $\lambda' > 0$ and $\lambda'^{2} \leq \mu\lambda' + \lambda^{2}$, we consider the Cauchy's problem :

$u'(t) = H_{\lambda'}u(t) ; t > 0 $ and $u(0) = \phi \in B$

$\displaystyle{H_{\lambda'} = (\lambda' z^{2} + i\lambda z)\frac{\partial^{2}}{\partial z^{2}} + (i \lambda z^{2} +\mu z)\frac{\partial }{\partial z}}$ acting on Bargmann space

$\displaystyle{\mathbb{B} = \{\varphi : \mathbb{C} \longrightarrow \mathbb{C} \, entire \, ; \int_{\mathbb{C}}\mid \varphi(z) \mid^{2} e^{-\mid z \mid^{2}}dxdy < + \infty \}}$

Then we have  $u(t) = \displaystyle{\sum_{k=1}^{+\infty}\frac{< \phi,\phi_{k}^{*} > }{< \phi_{k},\phi_{k}^{*} >}e^{-\sigma_{k}t}\phi_{k}}$  where

$ \phi_{k} \in D(H_{\lambda'})$ and $\phi_{k}^{*} \in D(H_{\lambda'}^{*}) = D(H_{\lambda'}) = D(S)$

$H_{\lambda'}\phi_{k} = \sigma_{k}\phi_{k}$ et $H_{\lambda'}^{*}\phi_{k}^{*} = \sigma_{k}\phi_{k}^{*}$

The proof of this proposition is based on a thin Lidskii theorem which will be given below in simplified form.

( [Lidskii], theorem 3, p:208)

Let $T$  be a linear operator with dense domain  $D(T)$ in Hilbert space  $\mathcal{H}$  compact resolvent.

We suppose that the eigenvalues of $T$ are simple and  $0 \in \rho(T)$ ($\rho(T)$ is the resolvent set. of $T$).

\begin{equation}
\left \{ \begin{array} {c} u'(t) + Tu(t) = 0 ; t > 0  \\
\quad\\
and \quad \quad \quad \quad \quad \quad \quad \quad\quad\\
\quad\\
u(0) = \phi \in \mathbb{B} \quad \quad \quad \quad.  \quad\\\
\end{array} \right.
\end{equation}

If $T$ satisfies the following two conditions

(i) $\exists \quad p ; 0 < p < 1$ and $\displaystyle{\sum_{k=1}^{+\infty}\mid \alpha_{k}\mid^{p}}$ converge

where $\{\alpha_{k}\}; k=1, 2, .....$ are the eigenvalues of $\sqrt{T^{*^{-1}}T^{-1}}$.

(ii) $\exists \quad \epsilon > 0 ; -\frac{\pi}{2} + \epsilon \leq < arg <T\phi, \phi> \leq \frac{\pi}{2} - \epsilon \quad \forall \phi \in D(T)$

(1) The problem (6.2)  admits an unique solution.

(2) The solution $u(t)$ of Cauchy's problem (6.2) is a limit of partial sums of expression

$\displaystyle{\sum_{k=1}^{+\infty}\frac{< \phi,\phi_{k}^{*} > }{< \phi_{k},\phi_{k}^{*} >}e^{-\beta_{k}t}\phi_{k}}$ o$\grave{u}$ $ \phi_{k} \in D(T)$ and $\phi_{k}^{*} \in D(T^{*})$

$T\phi_{k} = \beta_{k}\phi_{k}$ and $T^{*}\phi_{k}^{*} = \overline{\beta}_{k}\phi_{k}^{*}$

Let  $\beta_{0} > 0$ and  $H_{\lambda'} =  H_{1} + iH_{2}$ where $H_{1} = \lambda' S + \mu A^{*}A + \beta_{0}I$  and  $H_{2} = \lambda A^{*}(A + A^{*})A$ with $S = A^{*^{2}}A^{2}$

Then the verification of hypotheses of Lidskii's theorem is based on following lemma

Let $\lambda' > 0$ and $H_{\lambda'} = H_{1} + iH_{2}$, then we have

(i) $\exists \quad C > 0 ; \mid < H_{2}\phi, \phi >\mid \leq C \mid < H_{1}\phi, \phi >\mid \quad \forall \phi \in D(S)$.

(ii) the operator  $(I + H_{2}H_{1}^{-1})$ is invertible.

(iii) $\exists \quad \epsilon > 0 ; -\frac{\pi}{2} + \epsilon \leq <Arg <H_{\lambda'}\phi, \phi> \leq \frac{\pi}{2} - \epsilon \quad \forall \phi \in D(H_{\lambda'})$

(iv) the series $\displaystyle{\sum_{k=1}^{+\infty}\mid \alpha_{k}\mid^{p}}$ converges $\quad \forall p > \frac{1}{2}$

where $\{\alpha_{k}\}; k=1, 2, .....$ are the eigenvalues  $\sqrt{H_{\lambda'}^{*^{-1}}H_{\lambda'}^{-1}}$, (this operator is said to belong to Carleman's class $\mathcal{C}_{p}$)

(i) and (ii) follow from above section.

(iii) follow from (i).

(iv) As $H_{\lambda'}^{-1} =  H_{1}^{-1}(I + H_{2}H_{1}^{-1})^{-1}$ then we deduce that  $(I + H_{2}H_{1}^{-1})^{-1}$ is bounded by a constant $C$ and $H_{1}^{-1} \in \mathcal{C}_{p} \quad \forall p > \frac{1}{2}$ because its eigenvalues are $\gamma_{k} = \frac{1}{\lambda'k(k-1) + \mu k + \beta_{0}}\approx \frac{1}{k^{2}}$.

Let $\{\alpha_{k}\}$ be the eigenvalues of $\sqrt{H_{\lambda'}^{*^{-1}}H_{\lambda'}^{-1}}$,it is well known (by applying the minimax theorem, see  [Gohberg2 et al], p: 27)  there exist  $C > 0$ such that  $ \alpha_{k} \leq C \gamma_{k}$ in particular $\displaystyle{\sum_{k=1}^{+\infty} \alpha_{k}^{p} < + \infty \quad \forall p > \frac{1}{2}.}$

Now, as our operator checks the hypotheses of Lidskii's theorem then we deduce the proposition 6.1.

$\bullet$ For $\lambda' = 0$, above Lidskii's theorem is not applied then the question of diagonalisation of semigroup associated to  $H = \mu A^{*}A + i\lambda A^{*}(A + A^{*})A$  is open.

Nevertheless we have the following

§.§ Asymptotic behavior of $e^{-tH}$, $ t \rightarrow  + \infty$;$ H = \mu A^{*}A + i\lambda A^{*}(A + A^{*})A; \mu > 0$

We consider the Cauchy problem :

\begin{equation}
\left \{ \begin{array} {c} u'(t) = -Hu(t) , t > 0 \\
\quad\\
\n and \quad\quad  \quad\quad  \quad\quad  \quad\quad\\
\quad\\
\n u(0) = \phi \in B \quad\quad  \quad\quad\\
\end{array} \right.
\end{equation}

Then we have

(i) For $\mu > 0$, the semigroup $e^{-tH}$ associated to $H$ is compact in particular we have

\begin{equation}
\sigma (e^{-tH}) = e^{-t\sigma(H)} \cup \{0\} \end{equation}

(iii) Let $\sigma_{0}$ and $\sigma_{1}$ are respectively the smallest and the second eigenvalue of the operator $H$.

$\mid\mid e^{-tH}\mid\mid = e^{-\sigma_{0}t} + O(e^{-\epsilon t})\quad \forall \epsilon < \sigma_{1}$.

(iv) $< e^{-tH}\phi, \psi > = <\phi, \phi_{0} > <\phi_{0} , \psi >e^{-\sigma_{0}t} + O(e^{-\sigma_{0}t})\quad \forall \phi \in \mathbb{B}, \quad \forall \psi \in \mathbb{B}$ where $\phi_{0}$  is the eigenfunction of  $H$ associated to $\sigma_{0}$.

(i) This result was proven in [intissar2] and (ii) is a classical consequence of compact semigroups see  for example [Pazy].

(iii) Let $\sigma_{0}$ be the smallest eigenvalue of $H$. Since it is simple then there exist an eigenfunction $\phi_{0} \neq 0$ associated to $\sigma_{0}$, which generates an space  $\mathbb{F}_{0}$ of one dimension.

From now, we take $\mid\mid \phi_{0}\mid\mid = 1$,  then there exist $\mathbb{F}_{1} \subset \mathbb{B}$ such that

$\mathbb{B} = \mathbb{F}_{0} \oplus \mathbb{F}_{1}$, $\phi_{0} \, \notin \, \mathbb{F}_{1}$ and $e^{-tH}\mathbb{F}_{1}\subset \mathbb{F}_{1}.$

Let $P$ be the projection of  $B$ on $ \mathbb{F}_{0}$ and  $Q = I- P$ be the projection of $\mathbb{B}$ on $ \mathbb{F}_{1}$, then we can write

\begin{equation}
\phi = P\phi + Q\phi = c_{1}\phi + Q\phi, \quad \forall \phi \in \mathbb{B} \, \mbox{and} \, c_{1} \, \mbox{ is a constant }
\end{equation}

in particular

\begin{equation}
e^{-tH}\phi = c_{1}e^{-\sigma_{0}t}\phi + e^{-tH}Q\phi
\end{equation}

Let $r(T)$ be the spectral ray of an opérateur $T$, since $e^{-tH}$ is compact,  we can use a classical result  (see [Kato]) to obtain

\begin{equation}
r(e^{-tH}{_{\mid \mathbb{F}_{1}}}) = e^{-\sigma_{1}t}.
\end{equation}

$\sigma_{1}$ is the second eigenvalue of $H$.

Let's put it now, $t = n + s$ where $n$ is entire part of $t$  then  $0 \leq s \leq 1$ and we deduce that

$\displaystyle{\mid\mid e^{-tH}{_{\mid \mathbb{F}_{1}}}\mid\mid = \mid\mid e^{-nH}e^{-sH}{_{\mid \mathbb{F}_{1}}}\mid\mid \leq \mid\mid e^{-nH}{_{\mid \mathbb{F}_{1}}}\mid\mid}$.

As  $\displaystyle{lim \mid\mid e^{-nH}{_{\mid \mathbb{F}_{1}}}\mid\mid^{\frac{1}{n}} = e^{-\sigma_{1}}}$ as $n \longrightarrow +\infty$  then we  deduce that

$\displaystyle{\mid\mid e^{-nH}{_{\mid \mathbb{F}_{1}}}\mid\mid^{\frac{1}{n}} = e^{-\sigma_{1}}}$ +  $\displaystyle{\epsilon_{n}; \epsilon_{n} \rightarrow 0}$ as $n \rightarrow +\infty$.

$\displaystyle{\mid\mid e^{-nH}{_{\mid \mathbb{F}_{1}}}\mid\mid = (e^{-\sigma_{1}} + \epsilon_{n})^{n}}$

$\displaystyle{ = e^{-n\sigma_{1}}(1 + e^{\sigma_{1}}\epsilon_{n})^{n}}$

$\displaystyle{ =  e^{-n\sigma_{1}}(1 + \omega_{n})^{n}; \omega_{n}}$

$\displaystyle{ = e^{\sigma_{1}}\epsilon_{n}}$

Let $\epsilon < \sigma_{1}$, then

$\displaystyle{e^{-n\sigma_{1}}(1 + \omega_{n})^{n} = e^{-\epsilon t + \epsilon t-n\sigma_{1}}(1 + \omega_{n})^{n}}$

$\displaystyle{ = e^{-\epsilon t + n\epsilon + s\epsilon -n\sigma_{1}}(1 + \omega_{n})^{n}}$

$\displaystyle{= e^{-\epsilon t}e^{n(\epsilon-\sigma_{1})}e^{s\epsilon} (1 + \omega_{n})^{n}}.$

$\displaystyle{\mid\mid e^{-tH}{_{\mid \mathbb{F}_{1}}}\mid\mid  \leq \mid\mid e^{-nH}{_{\mid \mathbb{F}_{1}}}\mid\mid = e^{-\epsilon t}e^{n(\epsilon-\sigma_{1})}e^{s\epsilon} (1 + \omega_{n})^{n}}$.

it follows that

$\displaystyle{lim \quad Log (e^{\epsilon t}e^{-s\epsilon}\mid\mid e^{-tH}{_{\mid \mathbb{F}_{1}}}\mid\mid) = -\infty} $ as $t \rightarrow +\infty$

$\displaystyle{lim \quad e^{\epsilon t}\mid\mid e^{-tH}{_{\mid \mathbb{F}_{1}}}\mid\mid = 0 }$ as $t \rightarrow +\infty$.

(iv) Just write  $\displaystyle{ < e^{-tH}\phi, \psi > = <e^{-tH}P\phi, \psi > + < e^{-tH}Q\phi, \psi >}$.

and since  $\displaystyle{\mid < e^{-tH}Q\phi, \psi > \mid \leq c_{1}e^{-\epsilon t}}$ with $\epsilon < \sigma_{0}$, it follows that :

$\displaystyle{\mid < e^{-tH}\phi, \psi >\mid = Ce^{-\sigma_{0}t} + O(e^{-\sigma_{0}t}) \quad \forall \phi \in B, \quad \forall \phi \in \mathbb{B}}.$

where  $\displaystyle{C = < \phi, \phi_{0} >< \phi_{0}, \psi >}$.

§.§ Generalized diagonalization of semigroups $e^{-t H_{\lambda''}}$ and $e^{-t\sqrt{H}_{\lambda''}}$, $ t \geq 0$ , where $H_{\lambda"} = \lambda'' A^{*^{3}}A^{3} + \lambda' A^{*^{2}}A^{2} + \mu A^{*}A + i \lambda A^{*}(A+ A^{*})A$, $\lambda'' > 0$.

In this sub-section we prove that the system of generalized eigenvectors of the operator $H_{\lambda"} = \lambda'' A^{*^{3}}A^{3} + \lambda'A^{*^{2}}A^{2} + \mu A^{*}A + i \lambda A^{*}(A+ A^{*})A$, $\lambda'' > 0$ is an unconditional basis in Bargmann space $\mathbb{B}$.

We also give a generalized diagonalization in the Abel sense of the semigroups associated to this operator and to its square root, respectively. Finally, some open problems are pointed out.

$\bullet$ (A) Some preliminary results and completeness of system of generalized eigenvectors of the operator $H_{\lambda''}$

In the following, we adopt the notations,

$U\phi(z) =  A^{*}A \phi(z)$,  with domain $D(U) = \{\phi \in \mathbb{B}; U\phi \in \mathbb{B}\}$ (Harmonic Oscillator).

$S\phi(z) =  A^{*^{2}}A^{2}\phi(z)$,  with domain $D(S) = \{\phi \in \mathbb{B}; S\phi \in \mathbb{B}\}$

$\hat{G} = G + I$

$G\phi(z) = A^{*^{3}}A^{3} \phi(z)$,  with domain $D(G) = \{\phi \in \mathbb{B}; G\phi \in \mathbb{B}\}$

$H_{\lambda''} = H_{\lambda'', \lambda', \mu, \lambda}$

$H_{\lambda'} = H_{0, \lambda', \mu, \lambda}$

$H_{\mu} = H_{0, 0, \mu, \lambda}$

For $\lambda'' \neq 0$, we first begin to collect the basic results for the operators we defined previously.

1. In Bargmann space, the adjoint of the derivation operator $A$ is the multiplication operator $A^{*}$.

2. The injection from domain $D(A)$ of $A$ into the Bargmann space $\mathbb{B}$ is compact.

\begin{equation}\mid\mid \phi \mid\mid \leq \mid\mid A^{*}\phi \mid\mid ,  \phi \in D(A^{*})\end{equation}

4. $A^{*^{3}}A^{3} , A^{*^{2}}A^{2} $ and $A^{*}A $ are self-adjoint operators with compact resolvent.

5. $H_{\lambda''} $ is maximal accretive operator for $\lambda'' > 0$, $\lambda' \geq 0$ and $\mu \geq 0$.

6. For every $\epsilon > 0$ there exists $C_{\epsilon} > 0$ such that

\begin{equation}\displaystyle{\mid \mid H_{\lambda'}\phi \mid\mid \leq \epsilon \mid\mid G \phi \mid\mid + C_{\epsilon}\mid\mid \phi \mid\mid,  \, \forall \, \phi \in D(G)}\end{equation}

\begin{equation}\displaystyle{D_{max}(H_{\lambda''}) = D_{min}(H_{\lambda''}) = D(G) , \lambda'' \neq 0}\end{equation}

8. According to the perturbation theory of semigroups  [Kato]  and [Pazy], $- H_{\lambda''}$ generates an analytic semigroup $\displaystyle{e^{-tH_{\lambda''}}}$.

Now we recall some useful definitions :

Let $L$ and $B$ be linear operators in a Hilbert space $\mathcal{H}$. We say that $B$ is $p$-subordinate $(p \in [0, 1])$  to $L$  if $D(L) \subset D(B)$. and there exists a strictly positive constant $C$ such that

\begin{equation}
\displaystyle{\mid\mid B \phi \mid\mid \leq C\mid\mid L \phi \mid\mid^{p}\mid\mid  \phi \mid\mid^{1-p}} \, \mbox{ for every} \, \phi \in D(L)
\end{equation}

For $p =1$, we say that $B$ is subordinate to $L$.

Let $L$ be a linear operator in a Hilbert space $\mathcal{H}$. The operator $B$ is said to be $L$-compact with order$ p \in [0, 1]$ if $D(L) \subset D(B)$
and for any $\epsilon > 0$, there exists a constant $C_{\epsilon} > 0$ such that

\begin{equation}
\displaystyle{\mid\mid B \phi \mid\mid \leq \epsilon \mid\mid L \phi \mid\mid^{p}\mid\mid  \phi \mid\mid^{1-p} + C_{\epsilon}\mid\mid \phi \mid\mid}  \mbox{for every}\, \phi \in D(L)
\end{equation}

The operator $B$ is called $L$-compact, if $B$ is $L$-compact with unit order.

From these definitions, we deduce the following results:

1. Let $L$ be an operator in $\mathcal{H}$ with a dense domain $D(L)$. and at least one regular point $\lambda$.

We suppose that:

($\alpha$) $D(L)) \subset D(B)$.

($\beta$)  For any $\epsilon > 0$, there exists a constant $C_{\epsilon} > 0$ such that  $\displaystyle{\mid\mid B \phi \mid\mid \leq \epsilon \mid\mid L \phi \mid\mid + C_{\epsilon} \mid\mid  \phi \mid\mid}$  for every $\phi \in D(L)$

($\gamma$) $\displaystyle{(L - \lambda I)^{-1}}$ is compact.

Then $\displaystyle{B(L - \lambda I)^{-1}}$ is a compact. operator

2. Let $L$ be an operator in $\mathcal{H}$ with a dense domain $D(L)$. and at least one regular point $\lambda$.

We suppose that:

($\alpha$) $ D(L) \subset D(B)$.

($\beta$)  $\displaystyle{B(L - \lambda I)^{-1}}$ is a compact. operator

Then for any $\epsilon > 0$ , there exists a constant $C_{\epsilon} > 0$ such that $\displaystyle{\mid\mid B \phi \mid\mid \leq \epsilon \mid\mid L \phi \mid\mid + C_{\epsilon} \mid\mid  \phi \mid\mid}$  for every $\phi \in D(L)$

$\bullet$ A sequence $\displaystyle{\{V_{n}\}_{n=1}^{\infty}}$  of subspaces of a Hilbert space $\mathcal{H}$ is called a basis of subspaces. if any vector $f$ belonging to $\mathcal{H}$ can be  uniquely represented as a series $\displaystyle{\phi = \sum_{ n= 1}^{\infty}\phi_{n}}$ such that $ \phi_{n} \in V_{n}$.

$\bullet$ A basis of subspaces is said to be unconditional if it remains a basis for $\mathcal{H}$  under any permutation of the subspaces appearing in it, i.e., if the preceding series converges unconditionally for any $\phi \in \mathcal{H}$.

A linearly independent sequence $\displaystyle{\{\phi_{n}\}_{n=1}^{\infty}}$  is called an unconditional basis with parentheses for $\mathcal{H}$ if there exists a
subsequence $\displaystyle{\{n_{k}\}_{k=1}^{\infty}}$. of the positive integers such that the subspaces spanned by the vectors $\displaystyle{\{\phi_{n}\}_{n_{k-1}}^{n_{k} - 1}}$ form an unconditional basis for $\mathcal{H}$.

Now we will prove  that the system of generalized eigenvectors of the Gribov operator $H _{\lambda''}$ is an unconditional basis in the Bargmann space $\mathbb{B}$. The analysis is essentially based on the results obtained by Markus in [Markus] and by Yacubov and Mamedov in [Yacubov et al], which we can apply to the operator of Reggeon field theory.

We begin by given some useful lemmas :

$H_{\mu}$ is $\frac{1}{2}$-subordinate to $\hat{G}$.

Let $\displaystyle{\hat{U} = \mu A^{*}A}$ then  $\displaystyle{\hat{U}}$ is a self-adjoint operator with compact resolvent in the Bargmann space $\mathbb{B}$, and $A^{*}A$ is a generator of an analytic semigroup.  Let $\displaystyle{\hat{U}\phi = \sum_{k=1}^{\infty} \lambda_{k} < \phi , e_{k} > e_{k}}$ be its spectral decomposition, where $\lambda_{k} = \mu k$ is the $kth$ eigenvalue of $\displaystyle{\hat{U}}$ associated to the eigenvector $\displaystyle{e_{k}(z) = \frac{z^{k}}{\sqrt{k!}}}$.

For $\beta > 0$, we define the operator $\displaystyle{\hat{U}^{\beta}}$ by

\begin{equation}
\displaystyle{\hat{U}^{\beta}\phi = \sum_{k=1}^{\infty} \lambda_{k}^{\beta} < \phi , e_{k} > e_{k}}
\end{equation}

with domain  $\displaystyle{D(\hat{U}^{\beta}) = \{\phi \in \mathbb{B}  ;  \sum_{k=1}^{\infty} \mid \lambda_{k}\mid^{2\beta} \mid < \phi , e_{k} >\mid^{2}  <  \infty \} }$.

From $\displaystyle{D(\hat{G}) = D(U^{3})}$ and $\displaystyle{\mid\mid \hat{G}\phi \mid\mid \sim \mid\mid U^{3}\phi \mid\mid }$, it is easy to deduce that

\begin{equation}
\displaystyle{D(\hat{G}^{\frac{3}{2}})  \subset D(H_{\mu})}
\end{equation}

There exists $C_{\mu, \lambda} > 0$  such that

\begin{equation}
\displaystyle{\mid\mid H_{\mu}\phi \mid\mid \leq C_{\mu, \lambda} \mid\mid U^{\frac{3}{2}}\phi \mid\mid }\mbox{ for any}  \displaystyle{\phi \in D(U^{\frac{3}{2}})}.
\end{equation}

Now as $\displaystyle{\mid\mid U^{\frac{3}{2}}\phi \mid\mid^{2} = < U^{\frac{3}{2}}\phi , U^{\frac{3}{2}}\phi >  = < U^{3}\phi , \phi > \leq \mid\mid U^{3}\phi \mid\mid \mid\mid \phi \mid\mid}$ we deduce that

\begin{equation}
\displaystyle{\mid\mid H_{\mu}\phi \mid\mid \leq C_{\mu, \lambda} \mid\mid U^{3}\phi \mid\mid^{\frac{1}{2}}\mid\mid \phi  \mid\mid^{\frac{1}{2} }} \mbox{ for any}  \displaystyle{\phi \in D(U^{3})}
\end{equation}

\begin{equation}
\displaystyle{\mid\mid H_{\mu}\phi \mid\mid \leq C_{\mu, \lambda} \mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}}\mid\mid \phi  \mid\mid^{\frac{1}{2} }}\mbox{ for any}  \displaystyle{\phi \in D(G)}
\end{equation}

Let $\displaystyle{\{e_{k}(z) = \frac{z^{k}}{\sqrt{k!}}\}_{k \in \mathbb{N}}}$  be the usuall basis of Bargmann space.

Let $(\lambda_{k}), k =1, 2, ...$ be the eigenvalues of $G$. Then for each $\lambda \neq \lambda_{k}$, we have

\begin{equation}
\displaystyle{(G - \lambda I)^{-1}\phi = \sum_{k=1}^{\infty}\frac{1}{\lambda_{k} -  \lambda} < \phi, e_{k} > e_{k}}
\end{equation}

Moreover, if $\Im m \lambda \neq 0$ and $\lambda$ belongs to a ray with origin zero and of angle $\theta$ with $\theta \neq 0$ and $\theta \neq \pi$ then we have:

\begin{equation}
\displaystyle{\mid\mid (G - \lambda I)^{-1}\mid\mid \leq \frac{1}{\mid \Im m \lambda \mid} = \frac{c(\theta)}{\mid \lambda \mid}}
\end{equation}

This follows immediately from the fact that G is a self-adjoint operator with compact resolvent.    $\square$

There exists a sequence of circles $\displaystyle{C(0, r_{k}), k = 1, 2, .... }$ with radii $r_{k}$  going to infinity such that

$\displaystyle{\mid\mid (\hat{G} - \lambda I)^{-1}\mid\mid \leq \frac{2}{\mid \lambda \mid^{\frac{2}{\beta}}}}$ for any $\beta \geq 3$ and $\mid \lambda \mid = r_{k}$.

First, we remark that the eigenvalues of the operator $\hat{G}$ are given by $\lambda_{k} = k(k-1)(k-2)$  and that $\lambda_{k+1} - \lambda_{k} \geq k^{2}$ for $ k \geq 2$.

We set $\displaystyle{ r_{k} = \frac{\lambda_{k+1} + \lambda_{k}}{2} = \lambda_{k} + \frac{\lambda_{k+1} - \lambda_{k}}{2}}$ then from the fact that $\hat{G}$ is a self-adjoint operator  and the equality  $\displaystyle{\mid\mid (\hat{G} - \lambda I)^{-1} \mid\mid = \frac{1}{d(\lambda , \sigma(\hat{G})}}$ where $\lambda \in \rho(\hat{G})$ and $d(\lambda , \sigma(\hat{G})$ denotes the distance between the point $\lambda$ and the spectrum of the operator $\hat{G}$ we get :

\begin{equation}
\displaystyle{\mid\mid (\hat{G} - \lambda I)^{-1} \mid\mid = \frac{1}{d(\lambda , \sigma(\hat{G})} \leq \frac{2}{\lambda_{k+1} - \lambda_{k}} \leq \frac{2}{k^{2}}}, \mbox{where} \mid \lambda \mid = r_{k}
\end{equation}

Now from $\displaystyle{ r_{k} = \frac{\lambda_{k+1} + \lambda_{k}}{2} = k^{3} - 2k^{2} + k \leq k^{3} \leq k^{\beta}}$ for any $\beta \geq 3$, we deduce that

$\displaystyle{ k^{-2} \leq (r_{k})^{-\frac{2}{\beta}}}$ and  $\displaystyle{\mid\mid (\hat{G} - \lambda I)^{-1} \mid\mid \leq 2(r_{k})^{-\frac{2}{\beta}} = \frac{2}{\mid\lambda\mid^{\frac{2}{\beta}}}}$ for any $\beta \geq 3$   $\square$

Let $\beta$ which satisfies $ 3 \leq \beta < 4$ then   $ \forall \, \epsilon > 0$ there exists  $C_{\epsilon} > 0$ such that

$\displaystyle{ \mid\mid H_{\mu}\phi \mid\mid \leq \epsilon \mid\mid \hat{G}\phi\mid\mid ^{\frac{2}{\beta}} \mid\mid \phi \mid\mid ^{1 - \frac{2}{\beta}}}$ $\forall \, \phi \in D(G)$.

Let us recall an interpolation inequality. For $a, b, \alpha $, and $\gamma$ positive numbers such that $\displaystyle{0 \leq \alpha \leq \gamma }$ then

\begin{equation}
\displaystyle{a^{\alpha}b^{\gamma-\alpha} \leq a^{\gamma} + b^{\gamma}}
\end{equation}

Hence, for each $\epsilon_{1} > 0$, there exists $C_{\epsilon_{1}} > 0$ such that

$\displaystyle{a^{\alpha}b^{\gamma-\alpha} \leq \epsilon_{1}a^{\gamma} +C_{\epsilon_{1}} b^{\gamma}}$

Writing $\displaystyle{\mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}} \mid\mid \phi \mid\mid^{\frac{1}{2}}}$ as   $\displaystyle{\mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}} \mid\mid \phi \mid\mid^{\frac{1}{2}} =  \mid\mid \phi \mid\mid^{1- \frac{2}{\beta}}\mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}} \mid\mid \phi \mid\mid^{\frac{2}{\beta}- \frac{1}{2}}}$  and as $\beta \leq 4$ we can apply the foregoing interpolation inequality to $\displaystyle{\mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}}}$ and $\displaystyle{\mid\mid \phi \mid\mid^{\frac{2}{\beta}- \frac{1}{2}}}$ with $\alpha = \frac{1}{2}$ and $\gamma = \frac{2}{\beta}$ to deduce that

\begin{equation}
\displaystyle{\mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}} \mid\mid \phi \mid\mid^{\frac{2}{\beta}- \frac{1}{2}} \leq \epsilon_{1}\mid\mid \hat{G}\phi \mid\mid^{\frac{2}{\beta}} + \, C_{\epsilon_{1}}\mid\mid \phi \mid\mid^{\frac{2}{\beta}} }
\end{equation}

for every $ \phi \in D(G)$

\begin{equation}
\displaystyle{\mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}} \mid\mid \phi \mid\mid^{\frac{1}{2}} \leq \epsilon_{1}\mid\mid \hat{G}\phi \mid\mid^{\frac{2}{\beta}}  \mid\mid \phi \mid\mid^{1 - \frac{2}{\beta}} + \, C_{\epsilon_{1}}\mid\mid \phi \mid\mid}
\end{equation}

for every $ \phi \in D(G)$

Finally, from this last inequality and from Lemma 6.9, we obtain

\begin{equation}
\displaystyle{ \mid\mid H_{\mu} \phi \mid\mid \leq C_{\mu,\lambda}\epsilon_{1}\mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}} \mid\mid \phi \mid\mid^{1- \frac{2}{\beta}} + \quad C_{\mu,\lambda}C_{\epsilon_{1}}\mid\mid \phi \mid\mid }
\end{equation}

for every $\phi \in D(G)$

That is, for all $\epsilon > 0$  there exists $C_{\epsilon} > 0 $ such that

\begin{equation}
\displaystyle{ \mid\mid H_{\mu} \phi \mid\mid \leq \epsilon \mid\mid \hat{G}\phi \mid\mid^{\frac{1}{2}} \mid\mid \phi \mid\mid^{1- \frac{2}{\beta}} + \quad C_{\epsilon}\mid\mid \phi \mid\mid }
\end{equation}

for every $ \phi \in D(G)$

.   $\square$

Let us now recall two results due to Yacubov and Mamedov [Yacubov et al]  and Markus [Markus] :

(Yakubov and Mamedov in [Yakubov et al])

Let $T$  be a closed densely defined linear operator in a Hilbert $\mathcal{H}$ with compact resolvent and assume that there exist a sequence of circles $C(0, r_{k}), k = 1, . . . , $ with radii $r_{k}$ going to infinity, a constant $C > 0$, and an integer $m \geq -1$ such that

\begin{equation}
\displaystyle{ \mid\mid (T - \lambda I)^{-1} \mid\mid  \leq \mid \lambda \mid^{m}} \mbox{ for } \mid \lambda \mid = r_{k}
\end{equation}

Then the spectrum of the operator $T$ is discrete and for any   $\phi \in D(T)$, there exists a subsequence of partial sums of the series $\displaystyle{\sum_{k}^{ }\mathbb{P}_{k}\phi }$ converging to $\phi$ in the sense of $\mathcal{H}$ where

$\displaystyle{\mathbb{P}_{k} = \frac{1}{2i\pi}\int_{C(0, r_{k})}(T - \lambda I)^{-1} d\lambda}$

( [Markus])

Let $T$ be a closed densely defined linear operator in a Hilbert $\mathcal{H}$ with compact resolvent, let $L$ be a normal operator
with compact resolvent and its spectrum lying on a finite number of rays, and. assume that,

\begin{equation}
\displaystyle{\overline{lim}\frac{\mathcal{N}(r, L)}{r^{1-p}} < \infty }\, as \,r \longrightarrow \infty \mbox{ for some} \,\, p \in [0, 1)
\end{equation}

where $\mathcal{N}(r, L)$ denotes the sum of the multiplicities of all eigenvalues of $L$ contained in the disk $\displaystyle{\{ \lambda \in \mathbb{C} ; \mid \lambda \mid \leq r\}}$.

If the operator $B =T- L $ is subordinate to $L$ with order $p$, then the system of root vectors of $T$ forms an unconditional basis with parentheses  in $\mathcal{H}$.

Theorems 6.13 and 6.14 are used in the proof of the next theorem, which forms the main part of this sub-section.

For $\lambda'' \neq 0$  and  $\lambda' = 0$, we have the following statements:

(i) The spectrum of the operator $H_{\lambda''}$ is discrete and for any   $\phi \in D(H_{\lambda''}^{2})$  there exists a subsequence of partial sums of series $\displaystyle{\sum_{n}^{}\mathbb{P}_{n}\phi}$ converging to $\phi$ in $\mathbb{B}$

(ii) The system of generalized eigenvectors of the operator $H_{\lambda''}$ is an  unconditional basis with parentheses in $\mathbb{B}$.

(i) Let $\lambda \in \rho(\hat{G})$. Due to the equality $\displaystyle{\hat{G}(\hat{G} - \lambda I)^{-1} = I + \lambda (\hat{G} - \lambda I)^{-1}}$ and from the inequality of Lemma 6.11, we obtain

\begin{equation}
\displaystyle{\mid\mid \hat{G}(\hat{G} - \lambda I)^{-1}\mid\mid \leq  1 + 2\mid \lambda\mid^{1 - \frac{2}{\beta}}}\mbox{ for any} \beta \geq 3\mbox{ and} \displaystyle{\mid \lambda \mid = r_{k} \longrightarrow +\infty}.
\end{equation}

Therefore,  for any $\beta \geq 3$ and $\displaystyle{\mid \lambda = r_{k}}$ sufficiently large, we have:

\begin{equation}
\n \displaystyle{\mid\mid \hat{G}(\hat{G} - \lambda I)^{-1}\mid\mid \leq  3\mid \lambda\mid^{1 - \frac{2}{\beta}}}
\end{equation}

Now, it follows from Lemma 6.12 that for each $\epsilon > 0$, there exists $C_{\epsilon} > 0$ such that:

\begin{equation}
\displaystyle{\mid\mid H_{\mu}(\hat{G} - \lambda I)^{-1}\phi \mid\mid \leq \epsilon \mid\mid \hat{G}(\hat{G} - \lambda I)^{-1}\phi \mid\mid^{\frac{2}{\beta}} \mid\mid (\hat{G} - \lambda I)^{-1}\phi \mid\mid^{1 - \frac{2}{\beta}} + C_{\epsilon} \mid\mid (\hat{G} - \lambda I)^{-1}\phi \mid\mid}
\end{equation}

From (6.29), (6.30) and lemma 6.11, it follows that

$\displaystyle{\mid\mid H_{\mu}(\hat{G} - \lambda I)^{-1}\phi \mid\mid \leq 3\epsilon \mid\mid \phi \mid\mid + 2C_{\epsilon}\mid \lambda \mid^{-\frac{2}{\beta}} = (3\epsilon  + 2C_{\epsilon}\mid \lambda \mid^{-\frac{2}{\beta}} )\mid\mid \phi \mid\mid}$

Hence, for $\epsilon < \frac{1}{3}$ and $\displaystyle{\mid \lambda \mid = r_{k} \longrightarrow +\infty}$ we get :

\begin{equation}
\displaystyle{\mid\mid H_{\mu}(\hat{G} - \lambda I)^{-1}\phi \mid\mid \leq q < 1.}
\end{equation}

Then, using the Neuman identity $\displaystyle{(\hat{G} + H_{\mu} - \lambda I)^{-1} = (\hat{G}  - \lambda I)^{-1}\sum_{k=1}^{\infty}[H_{\mu}(\hat{G} - \lambda I)^{-1}]^{k}}$,

we obtain:

\begin{equation}
\displaystyle{\mid\mid (\hat{G} + H_{\mu} - \lambda I)^{-1}\mid\mid \leq  C_{\mu}\mid\mid(\hat{G}  - \lambda I)^{-1}\mid\mid \leq \hat{C}_{\mu} \mid \lambda \mid^{-\frac{2}{\beta}}}
\end{equation}

as $\mid \lambda \mid = r_{k} \longrightarrow \infty$.

\begin{equation}
\displaystyle{\mid\mid (\lambda''\hat{G} + H_{\mu} - \lambda I)^{-1}\mid\mid \leq  C_{\lambda''}\mid\mid(\hat{G}  - \lambda I)^{-1}\mid\mid \leq \hat{C}_{\lambda''} \mid \lambda \mid^{-\frac{2}{\beta}}}
\end{equation}

as $\mid \lambda \mid = r_{k} \longrightarrow \infty$.

\begin{equation}
\displaystyle{\mid\mid ( H_{\lambda''} - \lambda I)^{-1}\mid\mid \leq \hat{C}_{\lambda''} \mid \lambda \mid^{-\frac{2}{\beta}}}
\end{equation}

as $\mid \lambda \mid = r_{k} \longrightarrow \infty$.

From the compactness of $\displaystyle{ (\hat{G} - \lambda I)^{-1}}$ and the previous Neuman identity,it follows that the resolvent of the operator $H_{\lambda''}$ is compact in $\mathbb{B}$. Therefore, due to estimate  (6.34) , the operator $H_{\lambda''}$ satisfies the assumptions of Theorem 6.13 and the first assertion of the theorem is proved.

(ii) Let $\displaystyle{\mathcal{N}(C(0, r_{k} ., \hat{G})}$. be the sum of the multiplicities of the eigenvalues of the operator $G$ which are included in the circle $C(0, r_{k})$ . with radius $\displaystyle{r_{k} = \frac{k(k - 1)(2k -1)}{2}}$. Then, for $\beta \geq 3$ we get :

\begin{equation}
\displaystyle{\overline{lim}\, \frac{\mathcal{N}(C(0, r_{k} ., \hat{G})}{r_{k}^{1- \frac{2}{\beta}}} = lim \, k. k^{3(\frac{2}{\beta} - 1)} \leq 1 }
\end{equation}

as $ k \longrightarrow +\infty$.

Consequently, the second assertion of the theorem follows from theorem 6.14.   $\square$

(1) By adding the term $\lambda'' A^{*^{3}}A^{3}$ to $H_{\lambda'}$, we obtain a strong diagonalization of $H_{\lambda''}$;

(2) For $\lambda'' = 0$ and $\lambda' \neq 0$ the results of the preceding theorem remain open, although we hope to obtain a similar diagonalization.

(3) For $\lambda'' = 0$ and $\lambda' = 0$  and $\mu \neq 0$ notice that $H_{\mu}$ is very far from normal. Not only its self-adjoint and skew-adjoint parts do not commute
but there is no inclusion in either way between their domains or with the domain of their commutator.

$\bullet$ (B) Generalized diagonalization of the semigroup $\displaystyle{e^{-tH_{\lambda''}^{\frac{1}{2}}}, t > 0}$

The aim of the end of this sub-section is to study the semigroup $\displaystyle{e^{-tH_{\lambda''}^{\frac{1}{2}}}, t > 0}$ on Bargmann space $\mathbb{B}$. In particular, we prove that the series  $\displaystyle{\sum_{n=1}^{+\infty}e^{-tH_{\lambda''}^{\frac{1}{2}}}\mathbb{P}_{n}}$ converges strongly to $\displaystyle{e^{-tH_{\lambda''}^{\frac{1}{2}}}}$.

We begin by recalling the summability method in Abel sense and Lidskii's theorem to  give a generalized diagonalization of semigroup associated to $H_{\lambda''}$.

Let $T$  be an operator with a dense domain and a discrete spectrum in a Hilbert space $\mathcal{H}$. In the following, $\{\mathcal{H}_{n}\}$ $n = 1, 2, . . . ,$ are used to denote
the system of all root subspaces of the operator $T$  which are associated to the eigenvalues numbered in increasing order and repeated according multiplicities. $\{\phi_{n}\}$, $n =1, 2, . . . $, denote the system of root vectors of the operator $T$, obtained by consecutive numbering of the basis of the sub-spaces $\{\mathcal{H}_{n}\}$ $n = 1, 2, . . . ,$, composed from the Jordan sequences.

The Fourier coefficients with respect to the system $\{\phi_{n}\}$, $n =1, 2, . . . $ are defined for any vector $\phi $ by $\displaystyle{c_{n} = \frac{< \phi , \phi_{n}^{*} >}{< \phi_{n}, \phi_{n}^{*} >}}$,  where the system $\{\phi_{n}^{*}\}$, $n =1, 2, . . . $, denote the system of root vectors of the operator $T^{*}$.

Even if the system $\{\phi_{n}\}$, $n =1, 2, . . . $ is  is complete, it is well known that it is not possible to say anything about the convergence of the Fourier series $\displaystyle{\sum_{n = 1}^{\infty}c_{n}\phi_{n}}$ to $\phi$.

Let $\alpha$ be a strictly positive real number, we introduce the following polynomials with respect to the real parameter $t$,

\begin{equation}
\displaystyle{P_{m}^{\alpha}(\frac{1}{\xi}, t) = \frac{\xi^{-\alpha}}{m!}\frac{\partial^{m}}{\partial \xi^{m}}e^{-t \xi^{-\alpha}}} ( m = 0, 1, 2, ....
\end{equation}

and we compare the series $\displaystyle{\sum_{n=1}^{\infty}c_{n}\phi_{n}}$ with the series  $\displaystyle{\sum_{n=1}^{\infty}c_{n}(t)\phi_{n}}$ whose  coefficients are defined in the following way:

$\bullet$  If $\phi_{n}$ is an eigenvector of the operator $T$ associated to the eigenvalue $\lambda_{n}$  without a Jordan chain, we set :

\begin{equation}
\displaystyle{c_{n}(t) = e^{-t\lambda_{n}^{\alpha}}c_{n}}\, \mbox{ where} \,\, \displaystyle{c_{n} = \frac{< \phi , \phi_{n}^{*} >}{< \phi_{n}, \phi_{n}^{*} >}}
\end{equation}

$\bullet$  If $\{ \phi_{n}, \phi_{n}^{1}, \phi_{n}^{2}, ......, \phi_{n}^{s}\} , s \geq 0$ is a Jordan chain corresponding to the eigenvalue $\lambda_{n}$, then we set :

\begin{equation}
\displaystyle{c_{n+j}(t) = e^{-t\lambda_{n}^{\alpha}}\sum_{m=0}^{s-j}P_{m}^{\alpha}(\lambda_{n}, t), 0 \leq j \leq s}
\end{equation}

We say that the series $\displaystyle{\sum_{n=1}^{\infty}c_{n}\phi_{n}}$  is summable by Abel's method of order $\alpha$ if there exists a sequence of integers $m_{0}, m_{1}, ...., m_{n}$ with $m_{0}= 1$ such that, for any $t > 0$, the series $\displaystyle{ u(t) = \sum_{m_{n-1}}^{m_{n} -1}c_{n}(t)\phi_{n}}$ converges and $\displaystyle{ lim \, u(t) = \phi}$ as $t \longrightarrow 0^{+}$.

Now, we need the following definition before stating Lidskii's theorem which played an important role in the spectral study of our operators.

Let $K$ be a compact operator in the Hilbert space $\mathcal{H}$. $K$ is said to belong to the Carleman-class $\mathcal{C}_{p}, p > 0$, with order $p$, if the
series $\displaystyle{\sum_{n=1}^{+\infty}s_{n}^{p}(K)}$. converges, where $s_{n}(K)$. are the eigenvalues of the operator $\displaystyle{\sqrt{K^{*}K}}$.

In the particular case $p = 2$, $\mathcal{C}_{2}$ is exactly the Hilbert-Schmidt class and for $p = 1$, $\mathcal{C}_{1}$ is said class of nuclear operators or trace operators.

For a systematic treatment of the operators of Carleman-class, we refer to the Gohberg and Krein's book [Gohberg et al].

(Lidskii's Theorem  [Lidskii])

Let $K$ be a compact operator acting on a Hilbert space $\mathcal{H}$. Suppose that the series $\displaystyle{\sum_{n = 1}^{\infty}s_{n}^{p}(K)}$ converges for some $p$, where $s_{n}^{p}(K)$ are the eigenvalues of the operator $\displaystyle{\sqrt{K^{*}K}}$  and suppose that the values of the quadratic form $< K\phi; \phi >$ lie in a sector $\mathbb{S}_{\rho}$ the complex $z$-plane defined by

\begin{equation}
\displaystyle{\mathbb{S}_{\rho} = \{ z \in \mathbb{C}; \mid arg(z) \mid  \leq \frac{\pi}{2\rho} \}} \mbox{ with} \rho > max (p, \frac{1}{2})
\end{equation}

Then for every $\phi$ in the range of the operator $K$, the corresponding Fourier series with respect to the system of root vectors of $K$ is summable by Abel's
method of order $\alpha$ to the vector $\phi$  where $ p \leq \alpha < \rho $ .

For $\lambda'' \neq 0$, the resolvent of $H_{\lambda''}$ belongs to Carleman-class $\mathcal{C}_{p}$ for any $p > \frac{1}{3}$.

As the eigenvalues of the self-adjoint operator $G$ are $\lambda_{n} = n(n-1)(n-2)$ then the resolvent of $G$ belongs to Carleman-class $\mathcal{C}_{p}$ for any $p > \frac{1}{3}$.

Now, as $\displaystyle{\lambda'' G + H_{\mu} -\sigma I)^{-1} = (\lambda'' G  -\sigma I)^{-1}\sum_{k = 0}^{\infty}[-H_{\mu}\lambda'' G  -\sigma I)^{-1}]^{k}}$, by applying the minimax theorem, it follows that the resolvent operator of $H_{\lambda''}$ belongs to Carleman-class  $\mathcal{C}_{p}$ for any $p > \frac{1}{3}$.   $\square$

For $\lambda'' > 0$ and $\mu > 0$,  there exist $\beta_{0} > 0$ and $\delta > 0$ such that the values of the quadratic form  $ < (H_{\lambda''} + \beta_{0}I)\phi , \phi >$ lie in the sector of complex $z$-plane,

\begin{equation}
\displaystyle{\mathbb{S}_{\delta} = \{ z \in \mathbb{C};  - \frac{\pi}{2} + \delta \leq arg(z)   \leq\frac{\pi}{2} - \delta  \}} \mbox{  for any} \phi \in D(H_{\lambda''} ).
\end{equation}

We remark that

$\displaystyle{\mid < i\lambda A^{*}(A + A^{*})A\phi , \phi > \mid  \leq \mid \lambda \mid \mid <  (A + A^{*})A\phi , A\phi > \mid }$

$\displaystyle{ \leq \mid \lambda \mid [\mid < A^{2}\phi , \phi > \mid + \mid < A\phi , A^{2}\phi > \mid ]}$

$\displaystyle{ \leq 2\mid \lambda \mid. \mid\mid A^{2}\phi  \mid\mid. \mid\mid A \phi  \mid\mid}$

Using the estimate $\displaystyle{ \mid\mid  \phi  \mid\mid \leq \mid\mid A \phi  \mid\mid}$, we get

$\displaystyle{\mid < i\lambda A^{*}(A + A^{*})A\phi , \phi > \mid  \leq 2\mid \lambda \mid. \mid\mid A^{3}\phi  \mid\mid. \mid\mid A \phi  \mid\mid + \beta_{0}\mid\mid \phi \mid\mid^{2}}$

$\displaystyle{ \leq \mid \lambda \mid [ \mid\mid A^{3}\phi  \mid\mid^{2} +  \mid\mid A \phi  \mid\mid^{2} ]  + \beta_{0}\mid\mid \phi \mid\mid^{2}}$

$\displaystyle{ \leq  C < (\lambda'' G + \mu A^{*}A + \beta_{0}I)\phi , \phi > }$.

This last inequality implies that

\begin{equation}
-\displaystyle{tan(\frac{\Im m< H_{\lambda''} \phi, \phi >}{\Re e< (H_{\lambda''} +\beta_{0}I) \phi, \phi >})} \mbox{ is bounded}.
\end{equation}

Then, there exists $\delta > 0$ such that

\begin{equation}
\displaystyle{- \frac{\pi}{2} + \delta \leq arg < (H_{\lambda''} + \beta_{0}I) \phi , \phi > \leq  \frac{\pi}{2} - \delta}
\end{equation}

for any $\phi \in D(H_{\lambda''})$

For $\lambda'' > 0$ and $\mu > 0$, we have the following statements:

(i) The system of generalized eigenvectors of the Gribov operator $H_{\lambda''}$ is  complete in Bargmann space $\mathbb{B}$.

(ii) The corresponding Fourier series with respect to the system of generalized eigenvectors of the Gribov operator $H_{\lambda''}$ is summable by Abel's
method of unit order  and the series $\displaystyle{\sum_{n = 1}^{\infty}e^{-t H_{\lambda''}}\mathbb{P}_{n}}$  converges strongly to  $\displaystyle{e^{-t H_{\lambda''}}}$ for any $t > 0.$

By means of Lemmas 6.20 and 6.21, the “Gribov-Intissar” operator $H_{\lambda''}$ satisfies the assumptions of Lidskii's Theorem ; therefore, the earlier theorem follows directly from the Lidskii's theorem.

In particular, if we set $\displaystyle{u(t) = e^{-t H_{\lambda''}}\phi}$, then $u(t)$ is the solution of the Cauchy problem:

\begin{equation}
\left \{ \begin{array} {c} \displaystyle{u'(t) = - H_{\lambda''}u(t)} \\
\quad\\
\displaystyle{u(0) = \phi} \quad\quad\quad \quad\quad  \\
\end{array}\right.
\end{equation}

and the series $\displaystyle{\sum_{n = 1}^{\infty}e^{-t H_{\lambda''}}\mathbb{P}_{n}}$  converges strongly to  $\displaystyle{e^{-t H_{\lambda''}}}$ for any $t > 0.$   $\square$

For the study of the semigroup $\displaystyle{e^{-t H_{\lambda''}^{\frac{1}{2}}}}$ on Bargmann space, we need the following theorems.

( [Kato])

Let $T$ be a maximal accretive operator acting in a Hilbert space $\mathcal{H}$. Then there exists an unique maximal accretive square root $\sqrt{T}$ of $T$ such that $(\sqrt{T})^{2} = T$ and $\sqrt{T}$ has the following additional properties:

(1) $\sqrt{T}$ is $m$-sectorial with the numerical range contained in the sector :

$\displaystyle{\mathbb{S} = \{ z \in \mathbb{C} ; \mid arg z \mid \leq \frac{\pi}{4} \}}$

(2) $D(T)$  is a core of $\sqrt{T}$ .

(3) $T$ has a compact resolvent if and only if $\sqrt{T}$ has.

From the previous theorem and Krein's theorem 20, Chap. I, see [Krein], one can define a square-root $\sqrt{H_{\lambda''}}$ such that $-\sqrt{H_{\lambda''}}$ generates an analytic semigroup $\displaystyle{e^{-t H_{\lambda''}^{\frac{1}{2}}}}$.

The function $\displaystyle{u(t) = e^{-t H_{\lambda''}^{\frac{1}{2}}}\phi}$ is a solution of the abstract elliptic problem  :

\begin{equation}
\left \{ \begin{array} {c} \displaystyle{u''(t) =  H_{\lambda''}u(t)} \\
\quad\\
\displaystyle{u(0) = \phi} \quad\quad\quad \quad\quad  \\
\end{array}\right.
\end{equation}

( [Macaev et al , Intissar3 (2001)])

Let $K$ be a compact operator acting on a Hilbert space $\mathcal{H}$. We suppose that:

(i) $\displaystyle{\mid  Arg < K\phi, \phi > \mid \leq \frac{\pi}{2}}$ for any $\phi \in \mathcal{H}$.

(ii) $\displaystyle{K \in  \mathcal{C}_{p}}$.

Then $\displaystyle{{\color{black}\sqrt{K}} \in  \mathcal{C}_{2p}}$.

As the operator $H_{\lambda''}$ belongs to Carleman-class $ \mathcal{C}_{p}$  for any $p > \frac{1}{3}$  it follows from the preceding theorem that $\sqrt{H_{\lambda''}}$ belongs to Carleman-class $ \mathcal{C}_{p}$  for any $p > \frac{2}{3}$. Thus, by Lidskii's theorem, we get the following result:

The series $\displaystyle{\sum_{n=1}^{\infty}e^{-t H_{\lambda''}^{\frac{1}{2}}}\mathbb{P}_{n}}$ converges strongly to $\displaystyle{e^{-t H_{\lambda''}^{\frac{1}{2}}}}$ for any $t > 0$. In particular, if we set $\displaystyle{u(t) =  e^{-t H_{\lambda''}^{\frac{1}{2}}}\phi}$ with $\phi \in D(G^{\frac{3}{2}})$, then $u(t)$ is the solution of the second-order problem (6.44).

For $\lambda'' = 0$ and $\lambda' \neq 0$, the resolvent of $H_{\lambda''}$ belongs Carleman-class $ \mathcal{C}_{p}$  for any $p > \frac{1}{2}$ see [Aimar2 et al]  or  [Intissar4 (1994)]. and we can apply the Lidskii's theorem to give a generalized diagonalization of the semigroup  $\displaystyle{e^{-t H_{\lambda''}}}$ for any $t > 0$. But the problem of diagonalization of  $\displaystyle{e^{-t H_{\lambda''}^{\frac{1}{2}}}}$ is open.

For $\lambda'' = 0$ and $\lambda' = 0$ and $\mu \neq 0$, the resolvent of $ H_{\mu}$ belongs to Carleman-class $\mathcal{C}_{1+ \epsilon}$ for any $\epsilon > 0$ see [Aimar3 et al]  or [Aimar4 et al] .. However, the completeness of the system of generalized eigenvectors of $H_{\mu}$ and the diagonalization of the semigroup $\displaystyle{e^{-t H_{\mu}}}$ are open problems.

§ REGULARIZED TRACE FORMULA ASSOCIATED TO “MAGIC GRIBOV-INTISSAR” OPERATOR ON BARGMANN SPACE

In this section, we obtain a regularized trace formula for magic Gribov operator
$ H = \lambda{''}G + H_{\mu,\lambda}$ acting on Bargmann space where
$$G = A^{*3}A^{3} \quad \quad and \quad \quad  H_{\mu,\lambda} = \mu A^{*}A + i\lambda A^{*}( A + A^{*})A$$
Here $A$ and $A^{*}$ are the standard Bose annihilation and creation operators and in Reggeon field theory, the real parameters $\lambda{''}$ is the magic coupling of Pomeron, $\mu$  is Pomeron intercept, $\lambda$ is the triple coupling of Pomeron and $i^{2} = -1$.

By applying some abstract results of  Sadovnichii-Podolskii [traces of operators with relatively compact perturbations. Mat. Sb. 193 (2), (2002) 129-152], we give the number of corrections sufficient for the existence of finite formula of the trace of  concrete magic Gribov's operator.

§.§  Introduction and main result of this section

As is known, the trace of a finite-dimensional matrix is the sum of all the eigenvalues. But in an infinite dimensional space, in general, ordinary differential operators do not have a finite trace.

In 1953, Gelfand and Levitan considered the Sturm-Liouville operator

$\left\{\begin{array}[c]{l}-y''(x) + q(x)y(x) = \sigma y(x)\\ \quad \\ y'(0) = 0, y'(\pi) = 0\\

\n q(x) \in \mathcal{C}^{1} [0, \pi],\quad \displaystyle{\int_{0}^{\pi}q(x)dx} = 0 \\ \end{array}\right.\hfill { }  (*)$

and derived the formula

$\displaystyle{\sum_{n=1}^{\infty}(\sigma_{n} -\lambda_{n}) = \frac{1}{4}(q(0) + q(\pi))}$ $\hfill { }  (**)$

where $\sigma_{n}$ are the eigenvalues of the above operator and $\lambda_{n} = n^{2}$ are the eigenvalues of the same operator with $q(x)=0$.

The proof of this regularized trace formula for the Sturm-Liouville operator can been found in [Gelfand et al]

The same regularized trace formula for the same problem was obtained with different method by Dikii [Dikii1] .

For the scalar Sturm-Liouville problems, there is an enormous literature (see for example [Dikii2]  or  [Sadovnichii1 et al]) on estimates of large eigenvalues and regularized trace formulae which may often be computed explicitly in terms of the coefficients of operators and boundary conditions.

After these studies, several mathematicians were interested in developing regularized trace formulae for different differential operators. According Sadovnichii and  Podolskii, these formulae gave rise to a large and very important theory, which started from the investigation of specific operators and further embraced the analysis of regularized traces of discrete operators in general form.

Among the results of Sadovnichii and  Podolskii established for abstract operators, we can recall that following :

Let $A_{0}$ be a self-adjoint positive discrete operator of domain $D(A_{0})$ acting in a Hilbert space, we denote by $\{\lambda_{n}\}$ its eigenvalues arranged in ascending order, $\{\phi_{n}\}$ is an orthonormal basis formed by the eigenvectors of $A_{0}$ and $R_{0}(\sigma)$ is its resolvent . By $B$ we denote the perturbing
operator, by $\{\sigma_{n}\}$ we denote the eigenvalues of the operator $A_{0} + B$ numbered in increasing order of their real parts, and $R(\sigma)$ stands for its resolvent.

Also, assume that $BA_{0}^{-\delta}$, $\delta > 0$ is a compact operator belonging to some finite-order
Schatten von Neumann class, i.e. the set of compact operators whose singular numbers form a convergent series $\displaystyle{\sum_{n=1}^{\infty}s_{n}^{p}}$
for some $p > 0$ is traditionally denoted by $\mathcal{C}_{p}$.

For operators $A_{0}$ and $B$ in  [Sadovnichii2 et al], the following theorem is proved.

(Sadovnichii-Podol'skii  [Sadovnichii2 et al])

Consider operator $A_{0}$ and $B$ be such that $A_{0}^{-1} \in C_{1}$ and $D(A_{0}) \subset D(B)$, and  suppose that there exist $\delta \in [0, 1)$ such that the operator $BA_{0}^{-\delta}$ can be continued to a bounded operator. Further, suppose that there exists $\omega \in [0, 1)$ such that $ \omega + \delta < 1$ and  $A_{0}^{-(1-\delta -\omega)}$ is a trace class operator, i.e. in $\mathcal{C}_{1}$. Then, there exist a subsequence of natural numbers $\{n_{m}\}_{m=1}^{\infty}$ and a subsequence of contours $\Gamma_{m} \subset \mathbb{C}$, that  for $\omega \geq  \frac{\delta}{l}$ the following relation holds:

\begin{equation}
\lim \limits_{m \rightarrow \infty}(\displaystyle{\sum_{j=1}^{n_{m}}(\sigma_{j} - \lambda_{j}) +\frac{1}{2\pi i}\int_{\Gamma_{m}}\sum_{k=1}^{l}\frac{(-1)^{k-1}}{k} Tr((BR_{0}(\sigma))^{k}d\sigma) = 0},
\end{equation}

In particular, for $l = 1$ we have

\begin{equation}
\lim \limits_{m \rightarrow \infty} \displaystyle{\sum_{j=1}^{n_{m}}(\sigma_{j} - \lambda_{j} -  <B\phi_{j}, \phi_{j}>) = 0}
\end{equation}

1) This theorem has been successfully applied to concrete ordinary differential operators as well as to partial differential operators, we can see some examples given in  [Sadovnichii2 et al]  [Sadovnichii5 et al]).

2) We can found in  [Sadovnichii3 et al] or in  [Sadovnichii4 et al], an excellent survey dedicated by Sadovnichii and Podolskii to the history of the state of the art in the theory of regularized traces of linear differential operators with discrete spectrum and a detailed list of publications related to the present aspect.

Usually, quantum Hamiltonians are constructed as self-adjoint operators; for certain situations,however, non-self-adjoint Hamiltonians are also of importance. In particular, the reggeon field theory (as invented by V. Gribov  [Gribov1]) for the high energy behaviour of soft processes is governed by the magic non-self-adjoint Gribov operator

\begin{equation}
H_{\lambda'',\lambda',\mu,\lambda} = \lambda{''}A^{*3}A^{3}+\lambda{'}A^{*2}A^{2} + \mu A^{*}A +i\lambda A^{*}(A + A^{*})A
\end{equation}

where $a$ and its adjoint $a^{*}$ are annihilation and creation operators, respectively, satisfying the canonical commutation relation $[A, A^{*}] = I$.

In the case $\lambda{''} = 0$, Ando and Zerner in  [Ando et al], Aimar in    [Aimar1 et al to Aimar7 et al] and Intissar in  [Intissar1 to intissar 20] have given a complete spectral theory for the operator $H_{0,\lambda{'},\mu,\lambda}$. In particular, one can consult the list of spectral properties of this operator summarized in  [Intissar4 et al] (2019).

In the case $\lambda{''} \neq  0$ and $\lambda{'} \neq 0$, some assumptions of Sadovnichii-Podlskii's theorem recalled above are not verified for the operator $H_{\lambda{''},\lambda{'},\mu,\lambda}$.

In the case $\lambda{''} \neq  0$ and $\lambda{'} = 0$, Aimar et al in  [Aimar6 et al] have given some spectral properties of magic Gribov operator $H_{\lambda{''},0,\mu,\lambda}$, which is more regular than $H_{0,\lambda{'},\mu,\lambda}$.

In this section, we continue the spectral study of $H_{\lambda{''},0,\mu,\lambda}$ and we apply the results of Sadovnichii-Podlskii cited above to obtain a regularized trace formula of this operator.

We adopt the following notations:

$ H = H_{\lambda{''},0,\mu,\lambda} = \lambda{''}G + H_{\mu,\lambda}$ where $G = A^{*3}A^{3}$   and   $H_{\mu,\lambda} = \mu A^{*}A + i\lambda A^{*}( A + A^{*})A$

Here $A$ and $A^{*}$ are the standard Bose annihilation and creation operators and $\lambda{''}$, $\mu$, $\lambda$ are real parameters and $i^{2} = -1$.

It is convenient to regard the above operators as acting on Bargmann space $\mathbb{B}$  [Bargmann1].

$\mathbb{B}$ is defined as a subspace of the space O($\mathbb{C}$) of holomorphic functions on $\mathbb{C}$  given by

\begin{equation}
\mathbb{B} = \{\phi\in O(\mathbb{C}) ; < \phi, \phi > < \infty \}
\end{equation}

where the paring

\begin{equation}
<\phi,\psi> = \displaystyle{\frac{1}{\pi}\int_{\mathbb{C}}}\displaystyle{\phi(z)\overline{\psi(z)}e^{-\mid z\mid^{2}}dxdy}\,\,\,  \forall\, \phi,\psi \in O(\mathbb{C})
\end{equation}

and $dxdy$ is Lebesgue measure on $\mathbb{C}$.

The Bargmann space $\mathbb{B}$ with $\mid\mid \phi \mid\mid = \sqrt{<\phi, \phi>}$ is a Hilbert space and $\displaystyle{e_{n}(z) = \frac{z^n}{\sqrt{n!}}};\\ n = 0, 1, ....$ is an orthonormal basis in $\mathbb{B}$.

In this representation, the standard Bose annihilation and creation operators are defined by

$\displaystyle{A\phi(z) = \phi^{'}(z)}$ with  maximal  domain $\displaystyle{D(A) = \{\phi \in \mathbb{B} \,  \mbox{ such  that} \,  A\phi \in \mathbb{B}\}}$

$\displaystyle{A^{*}\phi(z) = z\phi(z)}$
$\displaystyle{D(A^{*}) = \{\phi \in \mathbb{B}  \,  \mbox{ such  that} \,  A^{*}\phi \in \mathbb{B}\}}$

Accordingly, for the operator  $ H := H_{\lambda^{''}, 0,\mu,\lambda}$ we have

\begin{equation}
\left\{\begin{array}[c]{l}H \phi(z)= \lambda^{''} z^{3}\phi^{'''}(z)+i\lambda z\phi^{''}(z) + (i\lambda z^{2} + \mu z)\phi^{'}(z) \quad \quad \quad  \quad \quad \quad\\ \quad\\ \mbox{with \ maximal domain } \\
\quad\\
D(H_{max}) = \{\phi \in \mathbb{B} ; H\phi \in\mathbb{B}\} \\ \end{array}\right.
\end{equation}

Tomin, in an interesting article [Tonin], have derived several formulas of Gelfand-Levitan type for the first regularized trace of discrete operators under various conditions convenient for verification. But, for $\mu \neq 0 $ and $\lambda \neq 0 $ the operator $H_{\mu,\lambda} = H_{0, 0,\mu,\lambda}$ is not in the classes of perturbing linear operators considered by him.

Notice that in this case the operator $H_{\mu,\lambda}$ is very far from normal and not only its self-adjoint and skew-adjoint parts do not commute but there is no inclusion in either way between their domains or with the domain of their commutator.

It may be noted also that if  $\mu = 0 $ and $\lambda \neq 0 $ the spectrum of $H_{\mu,\lambda}$ is $ \sigma(H_{\mu,\lambda}) = \mathbb{C}.$

Let us begin by reviewing the most important properties of the $ H := H_{\lambda^{''},0,\mu,\lambda}$

i) Let Pol be the space of polynomials, then Pol is dense in $\mathcal{B}$.

ii) We define $H_{min}$ as the closure of restriction operator $ H_{\mid_{ Pol}}$ on the polynomials acting in Bargmann space, i.e.

\begin{equation}
\left\{\begin{array}[c]{l}H_{min}\phi= \quad \lambda^{''} z^{3}\phi^{'''}(z)+i\lambda z\phi^{''}  + (i\lambda z^{2} + \mu z)\phi^{'}\\
\quad\\
\mbox{with \ maximal domain } \\
\quad\\
\n D(H_{min}) = \{\phi \in \mathcal{B} ; \, \exists p_{n} \in  Pol\,  and \, \psi \in \mathcal{B} \, ; p_{n}\rightarrow \phi \mbox{ and }\,\,  Hp_{n}\rightarrow \psi\} \end{array}\right.
\end{equation}

iii) $D(H_{min}) = D(H_{max}) = D(G)$

In Bargmann space, it may be noted also that

iv) The operator $ G = A^{*3}A^{3} $ is positive, self adjoint operator.

v) The functions $\displaystyle{e_{n}(z) = \frac{z^n}{\sqrt{n!}}}$ are  the orthonormal eigenvectors of $G$ corresponding to the eigenvalues $\lambda_{n} = n(n-1)(n-2)$ for $n \geq 3 $ and $\lambda_{n} = 0 $ for $  n \in \{0, 1,2\}$

vi) Let $\widetilde{G} = G + I$ then  $ <\widetilde{G}\phi, \phi> \geq <\phi, \phi>$  $\forall \phi \in D(G)$

vii) $G $ could be replaced by $\widetilde{G}$ or $G + \sigma I$ with a scalar $\sigma$ without changing the nature of the problem that we will study.

viii) in [Aimar6 et al] (see theorem 3.3, p: 595), Aimar et al have shown that the spectrum of the magic Gribov operator is discrete and that the system of generalized eigenvectors of this operator is an unconditional basis in Bargmann space $\mathbb{B}$.

The goal of this section consists in establishing news spectral properties of this operator and the number of corrections sufficient for the existence of finite formula of the regularized trace.

Then the main results to which is aimed this paper can be stated as follows for the magic Gribov operator

Let $\mathbb{B} $ be the Bargmann space, $ H = \lambda{''}G + H_{\mu,\lambda}$ acting on $\mathbb{B}$ where $ G = A^{*3}A^{3}$ and $ H_{\mu,\lambda} = \mu A^{*}A + i\lambda A^{*}(A + A^{*})A$ , $A$ and $A^{*}$ are the standard Bose annihilation and creation operators.

Then there exists an increasing sequence of radius $r_{m}$ such that $r_{m} \rightarrow \infty$ as $ m \rightarrow \infty$  and

\begin{equation}
\displaystyle{\lim\limits_{m \rightarrow} (\sum_{n=0}^{m}(\sigma_{n} - \lambda{''}\lambda_{n}) + \frac{1}{2i\pi}\int_{\gamma_{m}} Tr[\sum_{k=1}^{4}\frac{(-1)^{k-1}}{k}[H_{\mu,\lambda}(\lambda{''}G - \sigma I)^{-1}]^{k}]d\sigma) = 0}
\end{equation}

- $\sigma_{n}$ are the eigenvalues of the operator $ H = \lambda{''}G + H_{\mu,\lambda}$

- $\lambda_{n} = n(n-1)(n-2)$ are the eigenvalues of the operator $ G $

- $ (\lambda{''}G - \sigma I)^{-1}$ is the resolvent of the operator $\lambda{''}G$

- $\gamma_{m}$ is the circle of radius $r_{m}$ centered at zero in complex plane.

This section is organized as follows. In following subsection 7.2, we improve some basic results of the operator $G$ and some key results to apply theorem 7..1. In subsection 7.3, we show that all assumptions of theorem 7.1 are fulfilled for the operators $ A_{0} = \lambda'' (G + I) $ and $B = H_{\mu,\lambda}$ with $\delta = \frac{1}{2}$, $\forall \omega \in [0, \frac{1}{6})$ and for $\frac{1}{8} \leq \omega < \frac{1}{6}$, we  deduce the main result of this Note. This work is concluded by noting that if $\lambda' \neq 0$ and $\lambda \neq 0$, the existence of finite formula of the regularized trace for the operator  $H_{\lambda'',\lambda',\mu,\lambda}$ defined by (7.3) is an open problem.

§.§ Some auxiliary results

We begin this under-section by improving some basic results of  [Aimar6 et al ] on the operator $G$ and by recalling some useful definitions.

For the discreteness of spectrum of the operator $ \tilde{G} = G + I$, it suffices to use the following Rellich's theorem (see[Krein], p. 386)

Let $B$ be a self-adjoint operator in $\mathbb{B}$ satisfying $<B\phi, \phi> \geq <\phi, \phi>, \phi \in D(B)$, where $D(B)$ is a domain of $B$.

Then, the spectrum of $B$ is discrete if and only if the set of all vectors $\phi \in D(B)$, satisfying $ <B\phi, \phi > \leq 1 $ is a precompact set.

In  [Aimar6 et al ], it was shown the following basic spectral properties on the operator $G$

1) The operator $G$ has a compact resolvent.

2) Let  $\lambda_{n} = n(n-1)(n-2)$ for $ n \geq 0 $ the eigenvalues of $G$ associated to eigenvectors $\displaystyle{e_{n}(z) = \frac{z^{n}}{\sqrt{n!}}}$. Then for each $\sigma \in \mathbb{C}$ such that $\sigma \neq \lambda_{n}$, we have

\begin{equation}
(G - \sigma I)^{-1}\phi = \displaystyle{\sum_{n=0}^{\infty}\frac{1}{\lambda_{n} - \sigma}<\phi, e_{n}>e_{n}}.
\end{equation}

Moreover, if $Im \sigma \neq 0$ and if $\sigma$ belongs to a ray with origin zero and of angle $\theta$ with $\theta \neq 0 $ and $\theta \neq \pi $ , we have

\begin{equation}
\mid\mid(G - \sigma I)^{-1}\mid\mid \leq \frac{1}{Im \sigma} = \frac{c(\theta)}{\mid \sigma \mid}
\end{equation}

3) There exists a sequence of circles $C(0, r_{n}) , n =1, 2, . . . ,$ with radii $r_{n}$ going to infinity such that

\begin{equation}
\mid\mid (G - \sigma I)^{-1}\mid\mid \leq \frac{2}{\mid\sigma\mid^{2/\beta}}
\end{equation}

for any $\beta \geq 3$ and $ \mid \sigma \mid =  r_{n}$ where $ r_{n} = \frac{\lambda_{n} + \lambda_{n+1}}{2}$

1) It is well known that the injection from $D(A)$ into the Bargmann space $\mathbb{B}$ is compact (see for example (iii) of lemma 3.6) and as the injection from  $D(A^{*3}A^{3})$ into $D(A)$ is continuous, then the injection from  $D(A^{*3}A^{3})$ into Bargmann space is also compact.

Classically the operators of the form  $ (A^{*3}A^{3} + I) $ are invertible. Then the resolvent set of $G$ is not void. Consequently the self-adjoint operator $G$ has compact resolvent and this proves again the discreteness of its spectrum.

For the properties 2) and 3) of this lemma see the lemmas 3.1 and 3.2 in  [Aimar6 et al ].  $\square$

1) The resolvent of the operator $G$ belongs to the class Carlemann $ \mathcal{C}_{p}$,  $\forall p > \frac{1}{3}$.
In particulary the operator $G$ belongs to class of operators with trace resolvent.

2) Let $\lambda_{m} = m(m-1)(m-2)$ then

\begin{equation}
\forall m \geq 2, \lambda_{m+1} - \lambda_{m} \geq \frac{3}{2}m^{2}
\end{equation}

3) Let $\lambda_{n} = n(n-1)(n-2)$ for $ n = 2, 3, ....$ and $ \beta > \frac{1}{3}$ . Then there exists an increasing sequence of positive numbers $\tilde{r}_{m} = \frac{\lambda_{m}^{\beta} + \lambda_{m+1}^{\beta}}{2}$ such that $\tilde{r}_{m} \rightarrow \infty $ as $ m \rightarrow \infty $ and

\begin{equation}
\displaystyle{\sum_{n=3}^{+\infty}\frac{1}{\mid \lambda_{n}^{\beta} - \tilde{r}_{m}\mid} \leq C}
\end{equation}

where $C$ is a positive number.

1) As $\lambda_{n} \sim n^{3}$ then  the series of term general $\frac{1}{\lambda_{n}^{p}}$ converges for all $p > \frac{1}{3}$, i.e. the resolvent of the operator $G$ belongs to Carleman's class  $C_{p}$,  $\forall p > \frac{1}{3}$.

In particulary the operator $G$ belongs to class of operators with trace resolvent.

2) We note that $\lambda_{m+1} - \lambda_{m} = 3m(m-1)$ then $\lambda_{m+1} - \lambda_{m} \geq \frac{3}{2}m^{2}$.

3) Let $m \geq 3$ and $\tilde{r}_{m} = \frac{\lambda_{m}^{\beta} + \lambda_{m+1}^{\beta}}{2}$ with $\lambda_{m}^{\beta} = m^{\beta}(m-1)^{\beta}(m-2)^{\beta}$ and $\beta > \frac{1}{3}$. Then

$ \displaystyle{\sum_{n=3}^{+\infty}\frac{1}{\mid \lambda_{n}^{\beta} - \tilde{r}_{m}\mid}}$ = $ \displaystyle{\sum_{n=3}^{m}\frac{1}{\tilde{r}_{m} - \lambda_{n}^{\beta}}}$ + $ \displaystyle{\sum_{n=m+1}^{+\infty}\frac{1}{\lambda_{n}^{\beta} - \tilde{r}_{m}}}$

Let $ n-m = k$ then

$\displaystyle{\sum_{n=m+1}^{+\infty}\frac{1}{\lambda_{n}^{\beta} - \tilde{r}_{m}}}$ = $\displaystyle{\sum_{k =1}^{+\infty}\frac{1}{\lambda_{m + k}^{\beta} - \tilde{r}_{m}}}$ = $\displaystyle{\sum_{k =1}^{+\infty}\frac{2}{2\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta} - \lambda_{m + 1}^{\beta}}}$

Now to estimate $\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta}$, we consider the function

\begin{equation}
f_{\beta}(x) = x^{\beta}(x-1)^{\beta}(x-2)^{\beta} \mbox{ for } x \geq 3
\end{equation}

It follows that

$f^{'}_{\beta}(x) = \beta f_{\beta - 1}(x)(3x^{2} - 6x + 2)$

$f^{''}_{\beta}(x) = \beta f_{\beta - 2}(x){\Large[ }(3 (3\beta -1) x^{4} - 12(3\beta -1) x^{3} + 6 (8\beta -3)x^{2} - 12 (2\beta -1)x + 4 (\beta -1){\Large]}$

For $3\beta (3\beta -1) \geq 0$ in particulary for $ \beta > \frac{1}{3}$ and sufficiently large $x$ , it follows that  $f^{''}_{\beta}(x)$ take positive values for sufficiently large $x$ ($ x > x_{min} \geq 3$) then it follows that $f^{'}_{\beta}(x)$ increases and in particulary $f^{'}_{\beta}(x) < f^{'}_{\beta}(x + h)$ for all $ h > 0$  and for $x > x_{min}$ therefore $ f_{\beta}(x_{min} + h) - f_{\beta}(x_{min}) \leq f_{\beta}(x + h) - f_{\beta}(x)$ for $ x \geq x_{min}$.

Thus there exists a number $m_{0}$ depending only on $\beta$ such that, for $ m > m_{0} \geq 3$ we have :

$\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta} \geq \lambda_{m_{0} + k}^{\beta} - \lambda_{m_{0}}^{\beta}$

By noting that $\lambda_{m_{0} + k}^{\beta} - \lambda_{m_{0}}^{\beta}\sim k^{3\beta}$ and

$\displaystyle{\sum_{k =1}^{+\infty}\frac{2}{2\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta} - \lambda_{m + 1}^{\beta}}}$ = $\displaystyle{\sum_{k =1}^{+\infty}\frac{2}{\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta} + \lambda_{m + k}^{\beta} - \lambda_{m + 1}^{\beta}}}$

$\displaystyle{ = \frac{2}{\lambda_{m + 1}^{\beta} - \lambda_{m}^{\beta}}}$ + $\displaystyle{\sum_{k =2}^{+\infty}\frac{2}{\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta} + \lambda_{m + k}^{\beta} - \lambda_{m + 1}^{\beta}}}$

$\displaystyle{ \leq \frac{2}{\lambda_{m + 1}^{\beta} - \lambda_{m}^{\beta}}}$ + $\displaystyle{\sum_{k =2}^{+\infty}\frac{2}{\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta}}}$ + $\displaystyle{\sum_{k=2}^{+\infty}\frac{2}{\lambda_{m + k}^{\beta} - \lambda_{m + 1}^{\beta}}}$

$\displaystyle{ \leq \sum_{k =1}^{+\infty}\frac{2}{\lambda_{m + k}^{\beta} - \lambda_{m}^{\beta}}}$ + $\displaystyle{\sum_{k =2}^{+\infty}\frac{2}{\lambda_{m + k}^{\beta} - \lambda_{m + 1}^{\beta}}}$

$\leq C_{1}$ $\displaystyle{\sum_{k =1}^{+\infty}\frac{1}{k^{3\beta}} = C}$ ($C_{1}$ does not depend on $m$)   $\square$

The operator $ H_{\mu,\lambda}( G - \sigma I)^{-1}$ is nuclear on Bargmann space where $\sigma$ belongs a resolvent set of the operator $G$.

Consider $\phi(z) = \displaystyle{\sum_{n=0}^{\infty}\phi_{n}e_{n}(z)}$ in Bargmann space $\mathbb{B}$, The matrix of the operator $H_{\mu,\lambda}$ in basis  ${e_{n}(z)}$ has the form

\begin{equation}
(H_{\mu,\lambda}\phi)_{n} = \alpha_{n-1}\phi_{n-1} + q_{n}\phi_{n} +\alpha_{n}\phi_{n+1}, n\geq 1
\end{equation}

with $(H_{\mu,\lambda}\phi)_{0} = 0$

$q_{n} = \mu n$, and $\alpha_{n} = i\lambda n\sqrt{n+1}$, ( $\mu$ and $\lambda$ are real numbers and $i^{2} = -1$).

It is complex symmetric tri-diagonale matrix (but not Hermitian!) of the form

\begin{equation}
\left\{\begin{array}[c]{l}H_{\mu,\lambda} = (h_{m,n})_{m,n=0}^{\infty}\quad with \quad the \quad elements\\ h_{nn} = \mu n\cr h_{n,n+1}=h_{n+1,n}= i\lambda n\sqrt{n+1}; n = 0,1,2, \\ and\\
h_{mn} = 0 \quad for \quad \mid m - n \mid > 1  \\ \end{array}\right.
\end{equation}

All elements of the matrix $H_{\mu,\lambda}$ have order $O(n^{\frac{3}{2}})$ as $n \rightarrow \infty$ and as all elements of the matrix $(G -\sigma I)^{-1}$ have order $O(n^{-3})$ as $n \rightarrow \infty$ then the elements of the matrix $ H_{\mu,\lambda}(G -\sigma I)^{-1}$ have order $O(n^{\frac{-3}{2}}) $ as $ n \rightarrow \infty$ therefore $ H_{\mu,\lambda}(G -\sigma I)^{-1}$ is a nuclear operator.  $\square$

1) Let be $\omega \in \mathbb{R} $ such that $ \omega < \frac{1}{6} $ then the operator $\tilde{G}^{-(\frac{1}{2} - \omega)}$ is a trace class operator on the Hilbert space $E$.

2) Let be $\delta \in \mathbb{R} $ such that $ \frac{1}{2} \leq \delta < \frac{2}{3} $ then the operator $H_{\mu,\lambda}\tilde{G}^{-\delta}$ is bounded and the operator $\tilde{G}^{\delta}(\tilde{G} - \sigma I)^{-1}$ is nuclear; $\sigma  \in \rho (\tilde{G})$.

3) Let be $\delta \in \mathbb{R}$ such that $ \frac{5}{6} < \delta \leq 1$, the operator $H_{\mu,\lambda}\tilde{G}^{-\delta}$ is nuclear and the operator $\tilde{G}^{\delta}(\tilde{G} - \sigma I)^{-1}$ is bounded; $\sigma  \in \rho (\tilde{G})$.

1)  The matrix of the operator  $\tilde{G}^{-(\frac{1}{2} - \omega)}$ in the base $e_{n}(z)$ is diagonal and its elements have order  $O(n^{-3(1/2 - \omega)}$ as $ n \rightarrow \infty $ then $\tilde{G}^{-(\frac{1}{2} - \omega)}$ $\forall \omega < \frac{1}{6}$ is a trace class operator on the Hilbert space $\mathbb{B}$.

2)$H_{\mu,\lambda}\tilde{G}^{-\delta}$ in the base $e_{n}(z)$ is tridioagonal and its elements have order $O(n^{\frac{3}{2}-3\delta})$ as $ n \rightarrow \infty $ then it easy to see that,if $ \frac{1}{2} \leq \delta $ then we obtain the property 2)

3) From 2) if $ \frac{5}{6} < \delta \leq 1$, the operator $H_{\mu,\lambda}\tilde{G}^{-\delta}$ is nuclear and the operator $\tilde{G}^{\delta}(\tilde{G} - \sigma I)^{-1}$ is bounded where $\sigma $ belongs to $\rho(\tilde{G})$ resolvent set of $\tilde{G}$.   $\square$

All assumptions of theorem 1.6.1 are fulfilled for the operators $A_{0} = \lambda''\tilde{G}$ and $B = H_{\mu, \lambda}$ in $E$ with $\delta = \frac{1}{2}$ $\forall \omega < \frac{1}{6})$. The condition $\omega \geq \frac{\delta}{l} = \frac{1}{2l}$ is satisfied $\forall l > 3$ then for $l = 4$ and $\frac{1}{8}\leq \omega < \frac{1}{6}$, we get that there exist a subsequence of naturel numbers $\{n_{m}\}_{m =1}^{\infty}$ and a sequence of contours $\Gamma_{m} \subset \mathbb{C}$ that the formula

\begin{equation}
\displaystyle{\lim\limits_{m \rightarrow \infty} (\sum_{k=0}^{n_{m}}(\sigma_{k} - \lambda{''}\lambda_{k}) + \frac{1}{2i\pi}\int_{\Gamma_{m}} \sum_{j=1}^{4}\frac{(-1)^{j-1}}{j}Tr[[H_{\mu,\lambda}(\lambda{''}G - \sigma I)^{-1}]^{j}]d\sigma) = 0}
\end{equation}

is true

The operator $BA_{0}^{-\frac{1}{2}} $ is bounded then take $\delta = \frac{1}{2}$ in Theorem 7.1

By above Lemma,  assertion 1) implies that assumption on $A_{0}^{-(1 - \delta - \omega)}$ of Theorem 7.1 is satisfied for $\delta  =\frac{1}{2}$ and $\forall \omega < \frac{1}{6}$ .

Hence, taking  $\frac{1}{8}\leq \omega < \frac{1}{6}$ and $l > 3$ of assertion 2), we deduce that assumption $\omega > \frac{\delta}{l}$ of Theorem 7..1 is
satisfied. All the assumptions of Theorem 7.1 are satisfied. Hence formula in this proposition holds.   $\square$

1) the choice to use $\displaystyle{f_{\beta}(x)  = x^{\beta}(x - 1)^{\beta}(x - 2)^{\beta}}$ for $x \geq 3$ (see (2.7)) in this work instead of using $f_{1}(x)$  permits us to consider the function of order $m$, $\displaystyle{f_{\beta}(x)  = x^{\beta}(x - 1)^{\beta}... (x - m+1)^{\beta}}$ for $x \geq m$, and to give (in another paper) the number corrections sufficient for existence of finite formula of the regularized trace of the operator

\begin{equation}
\displaystyle{H = A^{*n}A^{n} + \sum_{i+j \leq m}c_{i,j}A^{*^{i}}A^{^{j}} }
\end{equation}

where $c_{i,j}$ are complex numbers. If $m < 2(n -1)$, we can found that the value of $l$ is $\displaystyle{[\frac{m}{2n - m - 2}] + 1}$ (the brackets denote the integer part).

2) The following of this section is devoted to proof of regularized formula  for $n_{m} = m$. The key point is the following theorem which gives an estimate of $\mid \mid H_{\mu,\lambda}(\tilde{G} - \sigma)^{-1} \mid\mid_{1}$.

(estimatation of $\mid \mid H_{\mu,\lambda}(\tilde{G} - \sigma)^{-1} \mid\mid_{1}$)

For all $\alpha$ ; $\displaystyle{ 0 \leq \alpha < \frac{1}{6} }$, then there exists a sequence of numbers $\displaystyle{\tilde{r}_{m} = [\frac{1}{2}(\lambda_{m}^{\frac{1}{2} - \alpha} + \lambda_{m+1}^{\frac{1}{2} - \alpha}]^{\frac{1}{\frac{1}{2} - \alpha}}}$ such that

$\displaystyle{\lambda_{m} < \tilde{r}_{m} < \lambda_{m+1}}$ and $\displaystyle{\mid \mid H_{\mu,\lambda}(G - \sigma I)^{-1}\mid\mid_{1} = o(\tilde{r}_{m})^{-\alpha}}$  where  $\displaystyle{\mid\mid . \mid\mid_{1}}$ is the nuclear norm.

To estimate $\mid \mid H_{\mu,\lambda}(G - \sigma I)^{-1}\mid\mid_{1}$, we note that

$\mid \mid H_{\mu,\lambda}(\tilde{G} - \sigma I)^{-1}\mid\mid_{1} \leq \mid \mid H_{\mu,\lambda}\tilde{G}^{-\frac{1}{2}}\mid\mid.\mid \mid \tilde{G}^{\frac{1}{2}}(\tilde{G} - \sigma I)^{-1}\mid\mid_{1} = \mid \mid H_{\mu,\lambda}\tilde{G}^{-\frac{1}{2}}\mid\mid \displaystyle{\sum_{n=1}^{\infty}\frac{\lambda_{n}^{\frac{1}{2}}}{\mid \lambda_{n} - \sigma\mid}}$

$\displaystyle{\leq\mid \mid H_{\mu,\lambda}\tilde{G}^{-\frac{1}{2}}\mid\mid \sum_{n=1}^{\infty}\frac{\lambda_{n}^{\frac{1}{2}}}{\mid \lambda_{n} - \mid\sigma\mid\mid}}$

By using the inequality
$$\mid \frac{a^{\frac{1}{2}}b^{\alpha}(a^{\frac{1}{2} -\alpha} - b^{\frac{1}{2} -\alpha})}{a - b}\mid \leq 1 $$
we get

$\mid \mid H_{\mu,\lambda}(\tilde{G} - \sigma I)^{-1}\mid\mid_{1} \leq \mid \mid H_{\mu,\lambda}G^{-\frac{1}{2}}\mid\mid $ $\displaystyle{\sum_{n=1}^{\infty}\frac{\lambda_{n}^{\frac{1}{2}}}{ \lambda_{n}^{\frac{1}{2}}\mid \sigma\mid^{\alpha}\mid \lambda_{n}^{\frac{1}{2}-\alpha} - \mid\sigma\mid^{\frac{1}{2}-\alpha}\mid}}$

$=\frac{\mid \mid H_{\mu,\lambda}\tilde{G}^{-\frac{1}{2}}\mid\mid }{\mid\sigma\mid^{\alpha}} \displaystyle{\sum_{n=1}^{\infty}\frac{1}{\mid \lambda_{n}^{\frac{1}{2}-\alpha} - \mid\sigma\mid^{\frac{1}{2}-\alpha}\mid}}$

From the relation $\lambda_{n}^{\frac{1}{2}-\alpha}\sim n^{3(\frac{1}{2}-\alpha)}$ then for $0 \leq \alpha < \frac{1}{6}$ the operator $\tilde{G}^{-(\frac{1}{2}-\alpha)}$  is nuclear and the series converges.

By choosing $\tilde{r}_{m} = \displaystyle{[\frac{\lambda_{m}^{\frac{1}{2}-\alpha} + \lambda_{m+1}^{\frac{1}{2}-\alpha}}{2}]^{\frac{1}{\frac{1}{2}-\alpha}}}$, we deduce that

\begin{equation}
\displaystyle{\sum_{n=1}^{\infty}\frac{1}{\mid \lambda_{n}^{\frac{1}{2} - \alpha} - \tilde{r}_{m}^{\frac{1}{2} - \alpha}\mid}} \leq C
\end{equation}

where $C$ does not depend on $m$

The arbitrariness of the choice of $\alpha$ allows us to deduce that for $\mid \sigma \mid = \tilde{r}_{m}$ we have

\begin{equation}
\mid\mid H_{\mu,\lambda}(\tilde{G} - \sigma I)^{-1}\mid\mid_{1} = o(\frac{1}{\tilde{r}_{m}^{\alpha}}) \mbox{ as } m \rightarrow \infty
\end{equation}

In following subsection, we recall the key points of proof of theorem 7.1 applied to our operator to deduce the theorem 7.4 of this section.

§.§ The key points of proof of Sadovnichii-Podol'skii's theorem applied to “magic Gribov-Intissar” operator

Suppose that $\gamma_{m}$ are the circles of raddii $\{\tilde{r}_{m}\}$ centered at zero. As  $\mid\mid H_{\mu,\lambda}(\lambda^{''}\tilde{G} - \sigma I)^{-1}\mid\mid < 1$(see theorem 3.3  [Aimar6 et al ]) and $ H_{\mu,\lambda}(\lambda{''}\tilde{G} - \sigma I)^{-1}$ is nuclear, then the perturbation determinant

\begin{equation}
\displaystyle{D_{H_{\lambda'',0,\mu,\lambda}/\lambda^{''}\tilde{G}}(\sigma) = det[(H_{\lambda'',0,\mu,\lambda} - \sigma I)(\lambda^{''}\tilde{G} - \sigma I)^{-1}] = det[I + H_{\mu,\lambda}(\lambda^{''}G - \sigma I)^{-1}].}
\end{equation}

is defined.

\begin{equation}
\displaystyle{Tr[(\lambda^{''}\tilde{G} - \sigma I)^{-1} - (H_{\lambda'',0,\mu,\lambda} - \sigma I)^{-1}] = \frac{d}{d\sigma}(ln D_{H_{\lambda'',0,\mu,\lambda}/\lambda^{''}\tilde{G}}(\sigma))}
\end{equation}

\begin{equation}
\displaystyle{ln(D_{H_{\lambda'',0,\mu,\lambda}/\lambda^{''}\tilde{G}}(\sigma) = Tr(ln[I + H_{\mu,\lambda}(\lambda{''}\tilde{G} - \sigma I)^{-1}}]
\end{equation}

Most of the results on infinite determinants of Hilbert space operators can be founded in ( [Gohberg1 et al ], [Gohberg2 et al ],  [Reed et al],  [Simon1] or  [Simon2])

We denote the eigenvalues  of $H$ by $\{\sigma_{n}\}$ and we continue to denote  the eigenvalues  of $\lambda{''}\tilde{G}$ by $\{\lambda_{n}\}$ then we have

\begin{equation}
\displaystyle{\sum_{n=1}^{m}(\sigma_{n}-\lambda_{n})  = -\frac{1}{2i\pi}\int_{\gamma_{m}}\sigma Tr((\lambda{''}\tilde{G} + H_{\mu,\lambda} - \sigma I)^{-1} - (\lambda{''}\tilde{G} - \sigma I)^{-1})d\sigma}
\end{equation}

As $\mid\mid  H_{\mu,\lambda}(\lambda{''}\tilde{G} - \sigma I)^{-1})\mid\mid < 1$ then the form of a power series for the logarithm is legitimate and we deduce that

\begin{equation}
\displaystyle{\sum_{n=1}^{m}(\sigma_{n}-\lambda_{n})  =  -\frac{1}{2i\pi}\int_{\gamma_{m}}[Tr(H_{\mu,\lambda}R_{\sigma}^{0}) + Tr(\displaystyle{\sum_{k=2}^{\infty }\frac{(-1)^{k-1}}{k}(H_{\mu,\lambda}R_{\sigma}^{0})^{k}})]d\sigma.}
\end{equation}

where $R_{\sigma}^{0} = (\lambda^{''}\tilde{G} -\sigma I)^{-1}$

Since $\displaystyle{H_{\mu,\lambda}R_{\sigma}^{0}}$ is nuclear operator, its trace can be calculated as the matrix trace in orthonormal basis $\displaystyle{e_{n}(z) = \frac{z^{n}}{\sqrt{n!}}, n= 0, 1, ...}$ then

$\displaystyle{\frac{1}{2i\pi}\int_{\gamma_{m}}Tr(H_{\mu,\lambda}R_{\sigma}^{0})d\sigma = - \sum_{n=0}^{m}<H_{\mu,\lambda}e_{n}, e_{n}> = -\sum_{n=1}^{m}n\mu}$  then

\begin{equation}
\displaystyle{\sum_{n=1}^{m}(\sigma_{n}-\lambda_{n} - n\mu)} = \displaystyle{-\frac{1}{2i\pi}\int_{\gamma_{m}}Tr(\displaystyle{\sum_{k=2}^{\infty }}\frac{(-1)^{k-1}}{k}(H_{\mu,\lambda}R_{\sigma}^{0})^{k})}d\sigma.
\end{equation}

Now let us estimate the terms of this series for $k \geq 3$. Let $\tilde{r}_{m} = \displaystyle{[\frac{\lambda_{m}^{\frac{1}{2}-\alpha} + \lambda_{m+1}^{\frac{1}{2}-\alpha}}{2}]^{\frac{1}{\frac{1}{2}-\alpha}}}$, we follows the techniques of proof in  [Sadovnichii2 et al] of Sadovnichii and  Podol'skii theorem 1.1 by taking $\delta = \frac{1}{2}$ then the following inequalities are valid

i) the norm of  the operator $H_{\mu,\lambda}R_{\sigma}^{0}$ is estimated as

\begin{equation}
\left \{ \begin{array}{c}\displaystyle{_{_{_{_{_{\sigma \in \gamma_{_{m}} }}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\! max \mid\mid H_{\mu,\lambda}(\tilde{G} -\sigma I)^{-1}\mid\mid \leq
\mid\mid H_{\mu,\lambda}\tilde{G}^{\frac{-1}{2}}\mid\mid _{_{_{_{_{\sigma \in \gamma_{_{m}} }}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\! max \quad (\quad _{_{_{_{_{n}}}}}\!\!\!\!\!\!\! max \frac{\lambda_{n}^{\frac{1}{2}}}{\mid \lambda_{n} - \sigma\mid})}\\

\displaystyle{ \leq c\quad _{_{_{_{_{n}}}}}\!\!\!\!\!\!\!\! max \quad \frac{\lambda_{n}^{\frac{1}{2}}}{\mid \lambda_{n} - \tilde{r}_{m}\mid} }\\

\displaystyle{\leq \frac{c}{\tilde{r}_{m}^{\frac{1}{2}} - \lambda_{m}^{\frac{1}{2}}}}\\

c > 0, c = \mbox{ const}.
\end{array} \right.
\end{equation}

ii) The estimate $\displaystyle{\mid Tr(H_{\mu,\lambda}R_{\sigma}^{0})^{k}\mid \leq \mid\mid H_{\mu,\lambda}R_{\sigma}^{0}\mid\mid^{k-1}\mid\mid H_{\mu,\lambda}R_{\sigma}^{0}\mid\mid_{1}}$  implies

\begin{equation}
\displaystyle{\int_{\gamma_{m}}\mid Tr( H_{\mu,\lambda}R_{\sigma}^{0})^{k}\mid \mid d\sigma \mid \leq max_{_{\sigma \in \gamma_{m}}}\mid\mid H_{\mu,\lambda}R_{\sigma}^{0}\mid\mid_{1} \int_{\gamma_{m}}\mid\mid H_{\mu,\lambda}R_{\sigma}^{0}\mid\mid^{k-1} \mid d\sigma \mid}
\end{equation}

\begin{equation}
\displaystyle{\int_{\gamma_{m}}\mid\mid H_{\mu,\lambda}R_{\sigma}^{0}\mid\mid^{k-1} \mid d\sigma \mid \leq \frac{c^{k}\lambda_{m}^{\frac{1}{2}}}{\mid \lambda_{m}^{\frac{1}{2}} - \tilde{r}_{m}^{\frac{1}{2}}\mid^{k-2}}}
\end{equation}

iv) the remainder of the series for $ l \geq 3$ satisfies the relation:

$\mid \displaystyle{-\frac{1}{2i\pi}\int_{\gamma_{m}}Tr(\displaystyle{\sum_{k=l}^{\infty }\frac{(-1)^{k-1}}{k}(H_{\mu,\lambda}R_{\sigma}^{0})^{k}})}d\sigma \mid \displaystyle{\leq \mid\mid H_{\mu,\lambda}R_{\tilde{r}_{m}}^{0}\mid\mid_{1}}.\displaystyle{\int_{\gamma_{m}}\sum_{k=l}^{\infty }\mid\mid H_{\mu,\lambda}R_{\sigma}^{0}\mid\mid^{k-1}  \mid d\sigma\mid}$

$\displaystyle{\leq \mid\mid H_{\mu,\lambda}R_{\tilde{r}_{m}}^{0}\mid\mid_{1}\lambda_{m}^{\frac{1}{2}}} \displaystyle{\sum_{k=l}^{\infty}(\frac{c}{ \tilde{r}_{m}^{\frac{1}{2}} - \lambda_{m}^{\frac{1}{2}}})^{k-2}}$

$\leq \displaystyle{C\frac{\tilde{r}_{m}^{\frac{1}{2}}\mid\mid H_{\mu,\lambda}R_{\tilde{r}_{m}}^{0}\mid\mid_{1}}{(\tilde{r}_{m}^{\frac{1}{2}} - \lambda_{m}^{\frac{1}{2}})^{l-2}}} $

\begin{equation}
\leq \displaystyle{C\frac{\tilde{r}_{m}^{\frac{1}{2} -\alpha(l-2)}\mid\mid H_{\mu,\lambda}R_{\tilde{r}_{m}}^{0}\mid\mid_{1}}{(\tilde{a}_{m}^{\frac{1}{2}} - \lambda_{m}^{\frac{1}{2}})^{l-2}}} = o\displaystyle{(\tilde{r}_{m}^{\frac{1}{2} - \alpha(l-1)})}, \quad \quad m \rightarrow +\infty \quad \quad \quad \quad
\end{equation}

Then for all $ l \geq 2$, we get

\begin{equation}
\displaystyle{\lim\limits_{m \rightarrow \infty}\sum_{n=0}^{m}(\sigma_{n} - \lambda_{n}) + \frac{1}{2i\pi}\int_{\gamma_{m}} Tr[\sum_{k=1}^{l}\frac{(-1)^{k-1}}{k}[H_{\mu,\lambda}(\lambda{''}\tilde{G} - \sigma I)^{-1}]^{k}]d\sigma = o(\tilde{r}_{m}^{\frac{1}{2} - \alpha l})}
\end{equation}

The right-hand side of the above equality tends to zero as $m \rightarrow +\infty$  provided that $ \frac{1}{2} - \alpha l \leq 0$ and as $\alpha$ is subject to the constraint $\frac{1}{8}\leq \alpha < \frac{1}{6}$ then  $l > 3$. By taking $\omega = \alpha$ and $l = 4$ then the formula of above proposition  is true for $n_{m} = m$, this completes the proof of theorem 7.4

This work is concluded  by noting that

(i) If we consider  the tri-diagonal matrices

\begin{equation}
\left \{ \begin{array} {c}(\tilde{H}_{\mu,\lambda}\phi)_{n} = \tilde{q}_{n}\phi_{n} + z(\beta_{n-1}\phi_{n-1} +\beta_{n}\phi_{n+1}), n\geq 1  \quad \quad \quad   \quad \quad \quad\\
(\tilde{H}_{\mu,\lambda}\phi)_{0} = 0   \quad \quad  \quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad  \quad \quad \quad \quad \quad  \quad \quad   \quad \quad  \\
where  \quad \quad   \quad \quad  \quad \quad \quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad  \quad \quad \quad \quad  \quad \quad \quad \quad \\
\tilde{q}_{n} = \lambda^{''}\lambda_{n} + \mu n, \lambda_{n} = n(n-1)(n-2), \beta_{n} =  n\sqrt{n+1} \, \mbox{and} \, z = i\lambda
\end{array} \right.
\end{equation}

then we observe that the diagonal matrix denoted by $D$ majories the off-diagonal one denoted by $B$ in the sense of the following condition

\begin{equation}
\mid \tilde{q}_{n}\mid \rightarrow \infty \, \mbox{and}\, \frac{\beta_{n}^{2}}{\mid \tilde{q}_{n}\tilde{q}_{n+1}\mid} \rightarrow 0\, \mbox{ as} \, n \rightarrow \infty
\end{equation}

under  above condition, well-know methods of Perturbation Theory  [Kato] give information about the spectra  $\sigma (D + zB)$ which is discrete and $\sigma(D + zB) = \{\sigma_{n}(z)\}_{1}^{\infty}$, where for each $n$ $\sigma_{n}(z)$ is analytic function at least for small $\mid z \mid$, i.e., in the disk $\mid z\mid < r_{n}$ for some $r_{n} > 0$

we can introduce the regularized trace

\begin{equation}
tr(\sigma_{n}(z)) = \sum_{n=1}^{\infty}(\sigma_{n}(z) - \tilde{q}_{n} )
\end{equation}

as an entire function therefore we can give an evaluation of this regularized trace by following the method of  [Janas5 et al].

(ii) In [Aimar7 et al], we studied the trace of the semigroup $e^{-tH}$ where

$ H = H_{\lambda^{''},0,\mu,\lambda} = \lambda^{''}A^{*3}A^{3} + \mu A^{*}A +i\lambda A^{*}(A + A^{*})A$

Using the estimates obtained in  [Intissar13] see also the next section which give an approximation of this semigroup by the unperturbed semigroup $\displaystyle{e^{-t\lambda^{''}A^{*3}A^{3}}}$ in nuclear norm, we have given an asymptotic expansion of this trace as $t \rightarrow 0^{+} $. This study is main topic of the following section.

§  REGULARIZED TRACE FORMULA OF THE SEMIGROUPS ASSOCIATED TO GRIBOV-INTISSAR'S OPERATORS IN BARGMANN REPRESENTATION

In above section, we had regularized the operator  $\mathbb{H}_{\mu,\lambda}$ by $\lambda''\mathbb{G}$ where  $\mathbb{G} = A^{*3} A^{3}$, i.e we had considered  $\mathbb{H}_{\lambda''} = \lambda'' \mathbb{G} + \mathbb{H}_{\mu,\lambda}$ where $\lambda''$ is the magic coupling of Pomeron. In this case, we had established an exact relation between the degree of subordination of the non-self-adjoint perturbation operator $\mathbb{H}_{\mu,\lambda}$ to the unperturbed operator $\mathbb{G}$ and the number of corrections necessary for the existence of finite formula of the regularized trace.

The goal of work in this section consists to study the trace of the semigroup $e^{-t\mathbb{}H_{\lambda''}}$, in particular to give an asymptotic expansion of this trace as $t \rightarrow 0^{+} $.

§.§  Preliminary introduction

Usually, quantum Hamiltonians are constructed as self-adjoint operators; for certain situations, however, non-self-adjoint Hamiltonians are also of importance. In particular, the reggeon field theory (as invented (1967) by V. Gribov [Gribov1]) for the high energy behaviour of soft processes is governed by the non-self-adjoint Gribov operator

$\mathbb{H}_{\lambda'',\lambda',\mu,\lambda} = \lambda''A^{*3}A^{3} + \lambda'A^{*2}A^{2} + \mu A^{*}A +i\lambda A^{*}(A + A^{*})A $

where $A$ and its adjoint $A^{*}$ are annihilation and creation operators, respectively, satisfying the canonical commutation relations $[A, A^{*}] :=  AA^{*} - A^{*}A = \mathbb{I}$.

It is convenient to regard the above operators as acting on Bargmann space $\mathbb{B}$ [Bargmann1] :

$ \mathcal{B} = \{\phi: \mathbb{C} \rightarrow \mathbb{C} \, entire \, ;  \displaystyle{\int_{\mathbb{C}}}\displaystyle{\mid \phi(z)\mid^{2}e^{-\mid z \mid ^{2}} dxdy} < \infty \}$

The Bargmann space $\mathbb{\mathbb{B}}$ with the paring :

$ < \phi, \psi > = \displaystyle{\int_{\mathbb{C}}}\displaystyle{\phi(z)\bar{\psi(z)}e^{-\mid z \mid ^{2}} dxdy}$

is a Hilbert space and $\displaystyle{e_{n}(z) = \frac{z^n}{\sqrt{n!}}; n = 0, 1, ....}$ is an orthonormal basis in $\mathbb{B}$.

In this representation, the standard Bose annihilation and creation operators are defined by

\begin{array}{ c }
A\phi(z) = \phi^{'}(z) \quad \quad \quad \quad \quad \quad \quad \quad\\
\quad\\
\mbox{with  maximal  domain }\quad \quad \quad \quad\\
\quad\\
D(A) = \{\phi \in \mathbb{B} ; \quad A\phi \in \mathbb{B}\}\quad \quad\\
\end{array}\right.$

\begin{array}{ c }
A^{*}\phi(z) = z \phi(z) \quad \quad \quad \quad \quad \quad \quad \quad\\
\quad\\
\mbox{with  maximal  domain }\quad \quad \quad \quad\\
\quad\\
D(A^{*}) = \{\phi \in \mathbb{B} ; \quad A^{*}\phi \in \mathbb{B}\}\quad \quad\\
\end{array}\right.$

Notice that $D(A) = D(A^{*})$ and $D(A)\hookrightarrow \mathbb{B}$ is compact.

It has been established in [Intissar14] that $ \mathbb{T}_{p} = A^{*^{p}}A^{p+1}$, $p \in \mathbb{N}$ is a chaotic operator and in  [Intissar2 et al], some sufficient conditions have been given on the weight sequence of a weighted chaotic shift operator $\mathbb{T}$ on a Hilbert space such that $\mathbb{T} + \mathbb{T}^{*}$ to be chaotic (in the sense of Devaney) where $\mathbb{T}^{*}$ is its adjoint. In particular, in  [Intissar2 et al], it has been shown  that $\mathbb{T}_{1} + \mathbb{T}_{1}^{*} = A^{*}(A + A^{*})A =:\mathbb{H}_{I}$ is chaotic operator. In Bargmann representation, there not exists semigroup generated by the operator $\mathbb{H}_{I}$, because $\sigma(\mathbb{H}_{I}) = \mathbb{C}$ where $\sigma(\mathbb{H}_{I})$ is the spectrum of $\mathbb{H}_{I}$.

Notice also that

For $\mathbb{H}_{\mu,\lambda} =  \mu A^{*}A + i\lambda A^{*}( A + A^{*})A $ with domain $D(H_{\mu, \lambda} = \{\phi \in \mathbb{B}\,  \mbox{such that} \, H_{\mu,\lambda}\phi \in \mathbb{B}\}$, we have:

i) For $\mu \neq 0$ and $\lambda \neq 0$,  $\mathbb{H}_{\mu,\lambda}$ is very far from normal and not only its self-adjoint and skew-adjoint parts do not commute but there is no inclusion in either way between their domains or with the domain of their commutator.

ii) For $\mu > 0$ and $\lambda \in \mathbb{R}$, the resolvent of $\mathbb{H}_{\mu,\lambda}$ is compact and $e^{-t\mathbb{H}_{\mu,\lambda}}$ is compact (see [Intissar2]).

iii) For $\mu > 0$ and $\lambda \in \mathbb{R}$, it has been established in [Aimar3 et al] and  [Aimar4 et al] that the resolvent of $\mathbb{H}_{\mu,\lambda}$ belongs to the class $\mathcal{C}_{1 + \epsilon} \quad \forall$ $\epsilon > 0$.

We recalling that a compact operator $\mathbb{K}$ acting on a  complex Hilbert space $\mathcal{H}$ belongs to the Carlemen class $\mathcal{C}_{p}$ of order $p$ if $\displaystyle{\sum_{n=1}^{\infty} s_{n}^{p} < \infty}$, where $s_{n}$ are $s$-numbers of operator $\mathbb{K}$ i.e, the eigenvalues of the operator $\sqrt{\mathbb{K}^{*}\mathbb{K}}$.

In particular, the operator $\mathbb{K}$ is called nuclear operator if $\mathbb{K} \in \mathcal{C}_{1}$ and Hilbert-Shmidt operator if $\mathbb{K} \in \mathcal{C}_{2}$

For $ p \geq 1$ the value $\displaystyle{(\sum_{n=1}^{\infty} s_{n}^{p})^{\frac{1}{p}}}$ is a norm denoted by $\mid\mid . \mid\mid_{p} $ and for $ p = 1$ it is called nuclear norm or trace norm.

We can consult [Gogberg2 et al] for a systematic study of operators of Carleman class  $\mathcal{C}_{p}$ of order $p$.

Let $ \mathbb{H}_{\lambda'} = \lambda'\mathbb{S} +  \mathbb{H}_{\mu,\lambda}$ where $\mathbb{S} = A^{*^{2}}A^{2}$, this operator is more regular that $\mathbb{H}_{\mu,\lambda}$, its semigroup $e^{-t\mathbb{H}_{\lambda'}}$ is analytic and it has been established in [Intissar 13] that the convergence of the usual Trotter product formula for $\mathbb{H}_{\lambda'}$ is of classical type and can be lifted to trace-norm convergence.

Morever, there exist $t_{0} > 0 $ and $C > 0$ such that :

$\mid\mid (e^{-\frac{t}{n}\lambda'\mathbb{S}} e^{-\frac{t}{n}\mathbb{H}_{\mu,\lambda}})^{n} - e^{-t\mathbb{H}_{\lambda'}}\mid\mid_{1} \leq$ $C\frac{log n}{n}$ , $ n = 2, 3, ..... $ $\forall$ $t \geq t_{0}$.

Now, if $\mathbb{H}_{\mu,\lambda}$ is regularized by $\lambda''\mathbb{G}$ where $\mathbb{G} = A^{*^{3}}A^{3}$ where $\lambda''$ is the magic coupling of Pomeron, we can consider:

\begin{equation}
\mathbb{H}_{\lambda^{''}} = \lambda''A^{*^{3}}A^{3} + \mu A^{*}A + i\lambda A^{*}(A  + A^{*})A
\end{equation}

$ \left\{\begin{array}{ c }
\mathbb{H}_{\lambda^{''}}\phi(z)= \lambda^{''} z^{3}\phi^{'''}(z)+i\lambda z\phi^{''}(z) + (i\lambda z^{2} + \mu z)\phi^{'}(z)\quad \quad \\
\quad\\
\mbox{with maximal  domain }\quad\quad \quad \quad \quad\quad \quad \quad \quad \quad \quad\quad \quad \quad \quad\\
\quad\\
D(\mathbb{H}_{\lambda^{''}}) = \{\phi \in \mathbb{B}; \quad \mathbb{H}_{\lambda^{''}}\phi \in \mathbb{B}\}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\
\end{array}\right.$

In  [Aimar6 et al] (see theorem 3.3, p. 595), it was shown that the spectrum of $H_{\lambda''}$ is discrete and that the system of generalized eigenvectors of this operator is an unconditional basis in Bargmann space $\mathcal{B}$.

Recently in [Intissar 19], it was established a regularized trace formula for $\mathbb{H}_{\lambda''}$.

More precisely, it was shown the following result:

(see above section)

Let $\mathbb{\mathbb{B}}$ be the Bargmann space and $\mathbb{H}_{\lambda^{''}} = \lambda^{''}\mathbb{G} + \mathbb{H}_{\mu,\lambda}$ acting on $\mathbb{\mathbb{B}}$

- $ \mathbb{G} = A^{*3}A^{3}$ and $ \mathbb{H}_{\mu,\lambda} = \mu A^{*}A + i\lambda A^{*}(A + A^{*})A$

- $A$ and $A^{*}$ are the standard Bose annihilation and creation operators satisfying

the commutation relation $[A , A^{*}] = \mathbb{I}$.

There exists an increasing sequence of radius $r_{n}$ such that $r_{n} \rightarrow \infty$ as $ n \rightarrow \infty$

$\displaystyle{Lim \sum_{k=0}^{n}(\sigma_{k} - \lambda''\lambda_{k}) =}$

\begin{equation}
\displaystyle{\lim\limits_{n \rightarrow \infty}\frac{1}{2i\pi}\int_{\gamma_{n}} tr[\sum_{j=1}^{4}\frac{(-1)^{j-1}}{j}[\mathbb{H}_{\mu,\lambda}(\lambda{''}\mathbb{G} -\sigma \mathbb{I})^{-1}]^{j}]d\sigma}
\end{equation}

-  $tr(\mathbb{A})$ denote the trace of operator $\mathbb{A}$.

- $\sigma_{k}$ are the eigenvalues of the operator $ \mathbb{H}_{\lambda^{''}} = \lambda{''}\mathbb{G} + \mathbb{H}_{\mu,\lambda}$

- $\lambda_{k} = k(k-1)(k-2)$ are the eigenvalues of the operator of $ \mathbb{G} $ associated to eigenvectors $e_{k}(z)$

- $ (\lambda{''}\mathbb{G} - \sigma \mathbb{I})^{-1}$  is the resolvent of the operator $\lambda{''}\mathbb{G}$

- $\gamma_{n}$ is the circle of radius $r_{n}$ centered at zero in complex plane.

The goal of this section consists to study the trace of the semigroup $e^{-t\mathbb{H}_{\lambda''}}$, in particular, to

give an asymptotic expansion of this trace as $t \rightarrow 0^{+} $.

Our procedure consists to prove that

i) The semigroup $e^{-t\mathbb{G}}$ generated by the operator $\mathbb{G}$ is analytic and nuclear (Gibbs analytic semigroup).

\begin{equation} \left \{ \begin{array} {c} \forall \epsilon > 0, \exists C_{\epsilon} > 0  ; \mid\mid \mathbb{H}_{\mu,\lambda}\phi\mid\mid \leq \epsilon\mid\mid \mathbb{G}\phi\mid\mid + C_{\epsilon}\mid\mid\phi\mid\mid \forall \phi \in D(\mathbb{G})\\
\quad\\
\mbox{where} \, D(\mathbb{G}) = \{\phi \in \mathbb{B}, \mathbb{G}\phi \in \mathbb{B}\}
\end{array} \right.
\end{equation}

iii) The operator $\mathbb{H}_{\mu,\lambda}\mathbb{G}^{-\delta}$ is bounded   $\forall$ $\delta \geq \frac{1}{2}$.

Now, with the aid of the results of Angelscu et al [Angelscu et al ] or of Zagrebnov [Zagrebnov1, Zagrebnov2 ] with Ginibre-Gruber inequality [Ginibre], it easy to prove that the series of general term $S_{k}(t)_{k\in \mathbb{N}}$ defined by:

$S_{0}(t)\phi = e^{-t\lambda''\mathbb{G}}\phi$

$S_{k+1}(t)\phi = - \displaystyle{ \int_{0}^{t}e^{-(t-s)\lambda''\mathbb{G}} \mathbb{H}_{\mu,\lambda} S_{k}(t)\phi ds}$

converges to $e^{-t\mathbb{H}_{\lambda''}}$ (nuclear norm).

By using the properties i), ii) and iii) we get:

\begin{equation}
\n {\small \displaystyle{\mid\mid e^{-t\mathbb{H}_{\lambda''}} - e^{-t\lambda''\mathbb{G}} \mid\mid_{1} = t\mid\mid e^{-t\lambda''\mathbb{G}} \mathbb{H}_{\mu,\lambda}\mid\mid_{1} + } \displaystyle{\mid\mid (\lambda''\mathbb{G})^{\delta} e^{-\frac{t}{3}\lambda''\mathbb{G} }\mid\mid_{1}O(t^{2})}; \quad t \rightarrow 0^{+} }
\end{equation}

To establish the above results, we  begin by given :

- In  subsection 8.2,  some spectral properties of semigroups $e^{-t\lambda''\mathbb{G}}$ and $e^{-t(\lambda''\mathbb{G} + \mathbb{H}_{\mu,\lambda})} $ in $\mathcal{C}_{p}$.

- In subsection 8.3, we give the proof of the above formula (8.4) for the trace of $e^{-t\mathbb{H}_{\lambda''}}$ as $t \rightarrow 0^{+} $.

§.§ Some spectral properties of semigroups $e^{-t\lambda''\mathbb{G}}$ and $e^{-t(\lambda''\mathbb{G} + \mathbb{H}_{\mu,\lambda})}$ in  Carleman  spaces $\mathcal{C}_{p}$

We begin by given some elementary spectral properties of $\mathbb{G}$ and $\mathbb{H}_{\lambda{''}}$ :

1) The operator $\mathbb{G}$ is self adjoint with compact resolvent.

2) The eigenvalues of $\mathbb{G}$ are $\lambda_{n} = n(n-1)(n-2)$ for $ n \geq 0 $ associated to eigenvectors $\displaystyle{e_{n}(z) = \frac{z^{n}}{\sqrt{n!}}}$.

3) $Lim \mid\mid \mathbb{G}e^{-t\mathbb{G}}\mid\mid = \frac{1}{e}$ as $ t \rightarrow 0.$

4) the resolvent of $\mathbb{G}$ belongs to $ \mathcal{C}_{p}$   $\forall p > \frac{1}{3}$.

5) $ e^{-t\mathbb{G}}$ is nuclear semigroup and $\mid\mid e^{-t\mathbb{G}}\mid\mid_{1} \leq Ct^{-\frac{1}{3}}.$

where the constant $C$ does not depend on $t$

6) $ e^{-t\mathbb{G}} \in \mathcal{C}_{p} $   $\forall p > 0$.

1) It is evident that $\mathbb{G}$ is self-adjoint and its resolvent is compact follows easily from the following Rellich theorem :

- (see [Fréchet], p. 386)

Let $\mathbb{B}$ be a self adjoint operator in a complex Hilbert space $\mathcal{H}$ such that
$< \mathbb{B}\phi, \phi> \quad \geq\quad <\phi, \phi>, \phi \in D(\mathbb{B})$, where $D(\mathbb{B}) = \{\phi \in \mathcal{H}; \mathbb{B}\phi \in \mathcal{H}\}$.

Then $\mathbb{B}$ is discrete if and only if $\{\phi \in D(\mathbb{B})$; $ < \mathbb{B}\phi, \phi) \leq 1 \}$ is pre-compact.

that the operator $\mathbb{G}$ is discrete.   $\square$

Also, we can prove the above result by using the following observation:

Since $D(A)$ is compactly embedded in Bargmann space $\mathbb{B}$ and $D(A^{*3}A^{3}) \hookrightarrow D(A)$ is continuous then $D(A^{*3}A^{3})$ is compactly embedded in Bargmann space $\mathbb{B}$.

Since $ (A^{*3}A^{3} + \mathbb{I}) $ is invertible  then the operator $\mathbb{G}$ is discrete.

2) It is obvious.

3) As $\mathbb{G}e^{-t\mathbb{G}}e_{n} = n(n-1)(n-2)e^{-tn(n-1)(n-2)}e_{n}$ we get

$\mid\mid \mathbb{G}e^{-t\mathbb{G}}\mid\mid  = \frac{1}{t}\quad_{_{n}}\!\!\!\!\!Sup\quad tn(n-1)(n-2)e^{-tn(n-1)(n-2)} = \frac{1}{e}$ this implies

\begin{equation}
\n \lim\limits_{t \rightarrow 0} \mid\mid \mathbb{G}e^{-t\mathbb{G}}\mid\mid = \frac{1}{e}
\end{equation}

4) As $\mathbb{G}$ is self adjoint and its eigenvalues are :

\begin{equation}
\lambda_{n} = n(n-1)(n-2) \sim n^{3}
\end{equation}

this implies that the series of general term  $\frac{1}{n^{3p}}$ converges $\forall p > \frac{1}{3}$ and consequently, the resolvent of $\mathbb{G}$ belongs to Carleman class $ \mathcal{C}_{p}$ $\forall p > \frac{1}{3}$.

5) $e^{-t\mathbb{G}}$ is self adjoint and it is of trace class, because the series of general term $e^{-t\lambda_{n}}$ converges $\forall t > 0 $.