#  Time evolution operator for a $h(1)\oplus
h(1)\mathbin{\mathchoice{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\displaystyle\displaystyle
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Hamiltonian; a self-consistent resolution method based on Feynman’s
disentangling rules

Nibaldo Edmundo Alvarez Moraga111Email address<EMAIL_ADDRESS>Autonomous Center of Theoretical Physics and Applied Mathematics,
2805 Place de Darlington, Montreal (Quebec), H3S1L4, Canada

###### Abstract

In this article the time evolution operator of two interacting quantum
oscillators, whose Hamiltonian is an element of the complex
$\left\\{h(1)\oplus
h(1)\right\\}\mathbin{\mathchoice{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\displaystyle\displaystyle
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\put(0.5,0.5){\arc[-90,90]{0.5}} \end{picture}}}}}u(2)$ algebra, is analyzed
using the Feynman time ordering operator techniques. This method is
consistently used to generate the conditions and to formally find explicit
disentangled expressions for such operator. In this way, it is shown that all
the problem reduces to solve a complex Riccati-type differential equation.
Some closed solutions to this differential equation are found and then
concrete disentangling expressions for the time-ordered evolution operator are
given. Finally, the time evolution of the coherent states linked to the
isotropic 2D quantum oscillator are analyzed under alternative time-
independent an time-dependent Hamiltonian systems.

##  1 Introduction

In quantum mechanics the time evolution of the state of a system is
determined, depending on the formulation we are working with, by solving the
Schrödinger equation, by constructing a unitary time evolution operator or by
determining the so-called propagator[1]–[7]. Certainly, all these formulations
are equivalent and the use of one or the other to describe the behavior or a
certain property of a quantum system will give the same results, the only
difference could be in the fact of weighting in which of them the calculation
of a certain property of the system we are studying is easier to realize,
although there is no single answer here either.

In the present article, we will disentangle the unitary time evolution
operator which associated time-dependent Hamiltonian is an element of the
complex Lie algebra $\left\\{h(1)\oplus
h(1)\right\\}\mathbin{\mathchoice{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\displaystyle\displaystyle
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\end{picture}}}}{\makebox[7.66663pt]{\raisebox{-0.58333pt}{\begin{picture}(1.0,1.0)\scriptstyle\displaystyle
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\put(0.5,0.5){\arc[-90,90]{0.5}} \end{picture}}}}}u(2).$ This algebra is a
composed algebra formed as the semi-direct sum of the direct sum of two
Heisenberg-Weyl Lie algebras $h(1)$ and the oscillator algebra $u(2).$ The
Hamiltonian this way constructed represents, in the two boson representation
space, two interacting oscillators with linear external coupling added. In
this sense, this article is a special case of the case of $N$ interacting
oscillators already treated in the literature [8]. However, here we will use a
self-consistent technique to disentangle completely the time evolution
operator, that is, the Feynman’s operator disentangling rules [9], and we will
construct some explicit expressions of it.

This article is organized as follows : In section 2 we briefly recall the
concept of time evolution of a physical state of a quantum system based on the
action of a time ordered unitary operator on the initial state of such a
system and at the same time we present the Feynman’s disentangling operator
technique, which allows to isolate an exponential operator from a unitary
operator of the time evolutiontype. There we also show the equivalence between
this technique and that based on the application of a unitary similarity
transformation. In section 3, we begin by expressing the time evolution
operator as the product of two exponential factors, the first having the
structure of a unitary displacement operator of the two dimensional quantum
oscillatortype, which acts absolutely at right of the second, which in turn
has the structure of a unitary time evolution operator whose Hamiltonian
belongs to the algebra $u(2).$ Then, we proceed to disentangle the second
exponential factor and establish the conditions under which a complete
disentangling of the time evolution operator is possible. These conditions
assume the form of a complex Riccati-type differential equation. In section 4
a closed solution of the complex Riccatitype equation for special choices of
the time dependent parameters is given, some of them suggested by the form of
the standard structure of a disentangled time-independent element of the
$SU(2)$ group. In section 5 we compute the time evolution of a class of
coherent states linked to the 2 D isotropic harmonic oscillator[10]. We show
that these states are also linked to a class of time-independent $u(2)$
Hamiltonian that can be reduced by means of a unitary squeezing mixed operator
to a one mode harmonic oscillator. In section 6, we repeat that we did in the
previous section but now with a class of coherent states linked to a 2 D time-
dependent $u(2)$-type of interacting oscillators. In section 7 we try another
option of disentangling ordering for the unitary time evolution operator.
There, we show the equivalence of the resulting equations with those of the
previous choice. Finally, in appendix A, by choosing a suitable representation
space, we transform the problem of solving a system of linear differential
equations into the equivalent one of solving a time ordered non-homogeneous
Schrödinger-type equation[11]. This last fact will allow us to choose the
$j=\frac{1}{2}$ representation space of the $su(2)$ algebra to compute the
matrix elements which are necessary to finish the construction of the above
mentioned two-dimensional displacement unitary operator.

##  2 Unitary time evolution operator

In quantum mechanics, the connection between theory and experiment is given by
the expectation values of the dynamical variables. For a system represented by
the Hamiltonian $\hat{H}(t),$ if $\hat{\Lambda}$ denotes a dynamical variable,
its expectation value is given by

$\langle\Lambda\rangle(t)=\langle\Psi(t)\mid\hat{\Lambda}\mid\Psi(t)\rangle,$
(2.1)

where the state $\mid\Psi(t)\rangle$ is determined by the Schrödinger equation

$i\hbar\frac{d}{dt}\mid\Psi(t)\rangle=\hat{H}(t)\mid\Psi(t)\rangle$ (2.2)

and the initial condition, when $t=t_{0},$ $\mid|\psi(t_{0})\rangle.$ The
solution of (2.2) can be written in the form

$\mid\Psi(t)\rangle=\hat{U}(t,t_{0})\mid\Psi(t_{0})\rangle,$ (2.3)

where $\hat{U}(t,t_{0})$ is the time-ordered unitary operator of the system
which verifies the differential equation

$i\hbar\frac{d}{dt}\hat{U}(t,t_{0})=\hat{H}\hat{U}(t,t_{0}),$ (2.4)

with the initial condition $\hat{U}(t_{0},t_{0})=\hat{I}.$ The solution of
(2.4) can be formally written as

$\hat{U}(t,t_{0})=e^{-\;\frac{i}{\hbar}\int_{t_{0}}^{t}\left[\hat{H}(s)\right]_{+}ds},$
(2.5)

where the subscript $+$ in the integral factor indicates that, when expanding
the exponential operator, the order of action of the Hamiltonian $\hat{H}(t)$
on the states depends on the value of the parameter $s,$ that is, the
Hamiltonian evaluated at an earlier time acts first than a Hamiltonian
evaluated at a later time. There are several important properties this
operator verifies, among them we have

$\hat{U}(t,0)=\hat{U}(t,t_{0})\hat{U}(t_{0},0),\quad{\rm
which\;is\;equivalent\;to}\quad\hat{U}(t,t_{0})=\hat{U}(t,0)\hat{U}^{-1}(t_{0},0),$
(2.6)

where $t\geq t_{0}\geq 0.$

###  2.1 Feynman’s disentangling rule and unitary transformations

In this section we use the Feynman’s disentangling rule to factorize a the
time-ordered unitary operator in terms of exponential factors. To do that, let
us start with any generic time-ordered operator of the form

$\hat{V}(t,t_{0})=e^{-\frac{i}{\hbar}\int_{t_{0}}^{t}\left[\hat{W}(s)\right]_{+}ds}.$
(2.7)

Let us now rewrite this operator in the form

$\hat{V}(t,t_{0})=e^{-\frac{i}{\hbar}\int_{t_{0}}^{t}\left[\hat{W}(s)-i\hbar\hat{\gamma}(s)+i\hbar\gamma(s)\right]_{+}ds},$
(2.8)

where we only have added to $\hat{W}(s)$ the null operator
$i\hat{\gamma}(s)-i\hat{\gamma}(s).$ Now, applying the Feynman’s disentangling
rule to this last operator we get

$\hat{V}(t,t_{0})=\hat{S}_{t}e^{-\frac{i}{\hbar}\int_{t_{0}}^{t}\left[\hat{S}^{-1}_{s}\left(\hat{W}(s)-i\hbar\gamma(s)\right)\hat{S}_{s}\right]_{+}ds}\;\hat{S}^{-1}_{t_{0}},$
(2.9)

where the $\hat{S}_{s}$ operator verifies

$\frac{d\hat{S}_{s}}{ds}=\hat{\gamma}(s)\hat{S}_{s},$ (2.10)

i.e., $\hat{S}_{s}$ is the time-ordered operator

$\hat{S}_{s}=e^{\int_{t_{0}}^{s}[\hat{\gamma}(\tau)]_{+}d\tau}\;\hat{S}_{t_{0}}$
(2.11)

Finally, by combining equations (2.9 ) and (2.10), we obtain

$\hat{V}(t,t_{0})=\hat{S}_{t}e^{-\frac{i}{\hbar}\int_{t_{0}}^{t}\left[\hat{S}^{-1}_{s}\hat{W}(s)\hat{S}_{s}+i\hbar\left(\frac{d}{ds}\hat{S}^{-1}_{s}\right)\hat{S}_{s}\
\right]_{+}ds}\;\hat{S}_{t_{0}}.$ (2.12)

We note that if $i\hat{\gamma}_{s}$ is Hermitian then $\hat{S}_{s}$ is
unitary, that is, ${\hat{S}}_{s}^{-1}={\hat{S}}_{s}^{\dagger}.$ The
expressions (2.9) and (2.12) are certainly equivalent, but while the former is
more useful for practical purposes, to generate conditions under which a
determined exponential operator can be factorized, for example, the latter is
more indicated at the moment of analyzing the general structure of the quantum
system we are studying.

##  3 Time evolution operator for the time-dependent $h(1)\oplus
h(1)\mathbin{\mathchoice{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\displaystyle\displaystyle
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The starting point of the present study is the two mode time-dependent
Hamiltonian

$\hat{H}(t)=\sum_{\sigma,\lambda=1}^{2}\hbar
w_{\sigma\lambda}(t)\hat{a}^{\dagger}_{\sigma}\hat{a}_{\lambda}+\sum_{\sigma=1}^{2}[F_{\sigma}(t)\hat{a}^{\dagger}_{\sigma}+F^{\ast}_{\sigma}(t)\hat{a}_{\sigma}]+B(t)\hat{I},$
(3.1)

where we have used essentially the same notation and structure used by S.
Kumar and C. L. Mehta in [8], but now with the restriction that the $\sigma$
and $\lambda$ indexes run only from 1 to 2. As $\hat{H}(t)$ is Hermitian,
$w_{11}(t),w_{22}(t)$ and $B(t)$ are real functions of time while
$w_{12}(t)=w_{21}^{\ast}(t)$ and $F_{\sigma}(t)$ are arbitrary complex
functions of time. The set of creation and annihilation operators satisfy the
$h(1)\oplus h(1)$ commutation relations

$[\hat{a}_{\sigma},\hat{a}_{\lambda}]=[\hat{a}^{\dagger}_{\sigma},\hat{a}^{\dagger}_{\lambda}]=0,\quad[\hat{a}_{\sigma},\hat{a}^{\dagger}_{\lambda}]=\delta_{\sigma\lambda},\quad\sigma,\lambda=1,2.$
(3.2)

The time evolution operator for the Hamiltonian (3.1) formally writes

$\hat{U}(t,t_{0})=\exp{\left[-\;\frac{i}{\hbar}\;\int_{t_{0}}^{t}\left[\sum_{\sigma,\lambda=1}^{2}\hbar
w_{\sigma\lambda}(s)\;\hat{a}^{\dagger}_{\sigma}\hat{a}_{\lambda}+\sum_{\sigma=1}^{2}\;[F_{\sigma}(s)\;\hat{a}^{\dagger}_{\sigma}+F^{\ast}_{\sigma}(s)\;\hat{a}_{\sigma}]+\beta(s)\right]_{+}ds\right]}.$
(3.3)

This operator have already been formally disentangled in the case of arbitrary
$N$ interacting quantum harmonic oscillators [8] by means of two successive
unitary similarity transformations, the first of translation and the second of
rotation. There, some conditions on the time-dependent $w_{\sigma\rho}(t)$
parameters have been given for the time evolution could be disentangled. Here,
we will disentangle this operator for the case $N=2$ by isolating an
exponential operator from the time evolution operator, with the help of the
Feynman’s time ordered operator technique [9]. It has been shown in the
literature that these two techniques are completely equivalent [11], they lead
exactly to the same results, but Feynman’s technique gives us more flexibility
in analyzing the conditions that allow us to solve the differential equations
necessary to obtain an explicit disentangled expression for the time evolution
operator.

###  3.1 Disentangling with the help of a unitary displacement operator

In this section we factorize the two mode time evolution operator with the
help of a unitary displacement operator. In accordance with equation (2.6) we
only need to compute $\hat{U}(t,0).$ Thus, by using the Feynman disentangling
rules, we can write this time evolution operator in the form

$\displaystyle\hat{U}(t,0)$ $\displaystyle=$
$\displaystyle\hat{S}(t)\exp\left[-\;\frac{i}{\hbar}\;\int_{0}^{t}\hat{S}^{-1}(s)\left[\sum_{\sigma,\lambda=1}^{2}\hbar
w_{\sigma\lambda}(s)\;\hat{a}^{\dagger}_{\sigma}\hat{a}_{\lambda}\right.\right.$
(3.4) $\displaystyle+$
$\displaystyle\left.\left.\sum_{\sigma=1}^{2}\;[(F_{\sigma}(s)-i\hbar\gamma_{\sigma}(s))\;\hat{a}^{\dagger}_{\sigma}+(F^{\ast}_{\sigma}(s)+i\hbar\gamma^{\ast}_{\sigma}(s))\;\hat{a}_{\sigma}]+\beta(s)\right]\hat{S}(s)ds\right]_{+},$

where

$\hat{S}(s)=\exp{\int_{0}^{s}\sum_{\sigma=1}^{2}[\gamma_{\sigma}(\tau)\hat{a}^{\dagger}_{\sigma}-\gamma^{\ast}_{\sigma}(\tau)\hat{a}_{\sigma}]_{+}d\tau},$
(3.5)

with the choice $\hat{S}(0)=\hat{I}.$ The unitary displacement operator
$\hat{S}(s)$ is by itself an exponential operator that need to be disentangled
for operating on the states or for combining with the other operators. Then,
its disentangled form can be reached by using again the Feyman factorization
rules:

$\hat{S}(s)=\hat{V}(s)\exp{\left[-\int_{0}^{s}\sum_{\sigma=1}^{2}\hat{V}^{-1}(\tau)\gamma^{\ast}_{\sigma}(\tau)\hat{a}_{\sigma}\hat{V}(\tau)d\tau\right]}$
(3.6)

where

$\hat{V}(\tau)=\exp{\left[\int_{0}^{\tau}\sum_{\sigma=1}^{2}[\gamma_{\sigma}(\nu)\hat{a}^{\dagger}_{\sigma}d\nu\right]},$
(3.7)

with the choice $\hat{V}(0)=\hat{I}.$ We note that according to the
commutation relations (3.2), no subscript $+$ is needed in (3.6) and (3.7).
Indeed,

$\hat{V}(\tau)=\exp{\left[\sum_{\sigma=1}^{2}c_{\sigma}(\tau)\hat{a}^{\dagger}_{\sigma}\right]},$
(3.8)

where $c_{\sigma}(\tau)=\int_{0}^{\tau}\gamma_{\sigma}(\nu)d\nu,$ or
$\frac{d}{d\tau}c_{\sigma}(\tau)=\gamma_{\sigma}(\tau).$ Thus, by inserting
(3.8) into (3.6) and using the relation
$e^{-c_{\mu}(\tau)\hat{a}^{\dagger}}\hat{a}_{\sigma}e^{c_{\mu}(\tau)\hat{a}^{\dagger}}=\hat{a}_{\sigma}+c_{\sigma}(\tau)\delta_{\sigma\mu},$
we get

$\hat{S}(s)=\exp{\left[\sum_{\sigma=1}^{2}c_{\sigma}(s)\hat{a}^{\dagger}_{\sigma}\right]}\exp{\left[-\sum_{\sigma=1}^{2}c^{\ast}_{\sigma}(s)\hat{a}_{\sigma}-\sum_{\sigma=1}^{2}\int_{0}^{s}\gamma^{\ast}_{\sigma}(\tau)c_{\sigma}(\tau)d\tau\right]},$
(3.9)

which after some manipulation becomes

$\hat{S}(s)=\exp\left[\sum_{\sigma=1}^{2}[c_{\sigma}(s)\hat{a}^{\dagger}_{\sigma}-c^{\ast}_{\sigma}(s)\hat{a}_{\sigma}]\right]\exp\left[\frac{1}{2}\sum_{\sigma=1}^{2}\int_{0}^{s}[\gamma_{\sigma}(\tau)c^{\ast}_{\sigma}(\tau)-\gamma^{\ast}_{\sigma}(\tau)c_{\sigma}(\tau)]d\tau\right].$
(3.10)

Inserting this last result into equation (3.4), then using the generic
displacement operator property
$\hat{D}^{\dagger}(c_{\mu}(s))\hat{a}_{\sigma}\hat{D}(c_{\mu}(s))=\hat{a}_{\sigma}+c_{\sigma}(s)\delta_{\sigma\mu,}$
and its corresponding conjugate relation, and requiring that the linear terms
in the creation and annihilation operators under the integral sign be
eliminated from the exponential factor, we get

$\displaystyle\hat{U}(t,0)$ $\displaystyle=$
$\displaystyle\exp\left[\sum_{\sigma=1}^{2}[c_{\sigma}(s)\hat{a}^{\dagger}_{\sigma}-c^{\ast}_{\sigma}(s)\hat{a}_{\sigma}]\right]$
(3.11) $\displaystyle\times$
$\displaystyle\exp\left[-\;\frac{i}{\hbar}\;\int_{0}^{t}\left[\sum_{\sigma,\lambda=1}^{2}\hbar
w_{\sigma\lambda}(s)\;\hat{a}^{\dagger}_{\sigma}\hat{a}_{\lambda}+\frac{1}{2}\sum_{\sigma=1}^{2}\;[(F_{\sigma}(s)c^{\ast}_{\sigma}(s))+F^{\ast}_{\sigma}(s)c_{\sigma}(s))]+\beta(s)\right]ds\right]_{+},$

where

$i\hbar\frac{dc_{\sigma}(t)}{dt}-\sum_{\lambda=1}^{2}\hbar
w_{\sigma\lambda}(t)\;c_{\lambda}(t)-F_{\sigma}(t)=0,\quad\sigma=1,2.$ (3.12)

By the way, these equations are the same found in [8] with a little difference
in the sign of $c_{\sigma}(t),\sigma=1,2.$ In the appendix A, we use the time
order operator method to obtain a formal set of solutions for this coupled
differential equations[11]. Explicit expressions of these solutions are given
in equation (A.11). The advantage of the method detailed there lies in the
fact that one can choose at will the space of representation of the $u(2)$
generators. As we will see below, in our case, a suitable irreducible
representation is the standard $\mid j,m\rangle$ space, where $j=\frac{1}{2},$
and $m=\pm\frac{1}{2}.$ On the other hand, we will see in the next section,
the same solutions will allow us to disentangle the remaining $u(2)$ part of
the total unitary time evolution operator. This last detail have been already
been pointed out by S. Kumar and C. L. Mehta in [8].

###  3.2  Two boson realization of the $u(2)$ generators

The remaining unitary operator we need to disentangle is quadratic in the
creation and annihilation operators, that is,

$\hat{U}_{0}(t,0)=\exp\left[-\;\frac{i}{\hbar}\;\int_{0}^{t}\left[\sum_{\sigma,\lambda=1}^{2}\hbar
w_{\sigma\lambda}(s)\;\hat{a}^{\dagger}_{\sigma}\hat{a}_{\lambda}\right]ds\right]_{+}.$
(3.13)

Here, it is convenient to express it in terms of the Schwinger two boson
realization [12] of the $su(2)$ Lie algebra generators222In this article, we
write the two-modes creation and annihilation operators in the following
simplified form: $\hat{a}_{1}\equiv\hat{a}_{1}\otimes\hat{I}_{2},$
$\hat{a}_{2}\equiv\hat{I}_{1}\otimes\hat{a}_{2},$
$\hat{a}_{1}^{\dagger}\equiv\hat{a}_{1}^{\dagger}\otimes\hat{I}_{2}$ and
$\hat{a}_{2}^{\dagger}\equiv\hat{I}_{1}\otimes\hat{a}_{2}^{\dagger},$ then,
par example,
$\hat{a}_{1}^{\dagger}\hat{a}_{2}=(\hat{a}_{1}^{\dagger}\otimes\hat{I}_{2})(\hat{I}_{1}\otimes\hat{a}_{2})=\hat{a}_{1}^{\dagger}\otimes\hat{a}_{2}.$
Sometimes, we will also write $\hat{a}_{2}^{\dagger}\hat{a}_{1}$ in place of
$\hat{a}_{1}\hat{a}_{2}^{\dagger}=\hat{a}_{1}\otimes\hat{a}_{2}^{\dagger},$
for example. :

$\hat{J}_{+}=\hat{a}_{1}^{\dagger}\hat{a}_{2},\quad\hat{J}_{-}=\hat{a}_{1}\hat{a}_{2}^{\dagger},\quad\hat{J}_{3}=\frac{1}{2}(\hat{a}_{1}^{\dagger}\hat{a}_{1}-\hat{a}_{2}^{\dagger}\hat{a}_{2}),$
(3.14)

which verify the commutation relations

$[\hat{J}_{3},\hat{J}_{\pm}]=\pm\;\hat{J}_{\pm},\quad[\hat{J}_{+},\hat{J}_{-}]=2\hat{J}_{3},$
(3.15)

and the Casimir operator

$\hat{N}=\frac{1}{2}(\hat{a}_{1}^{\dagger}\hat{a}_{1}+\hat{a}_{2}^{\dagger}\hat{a}_{2}),$
(3.16)

which commutes with all them. Thus the unitary operator assumes the form

$\hat{U}_{0}(t,0)=\exp\left[-\;i\;\int_{0}^{t}\left[\frac{d\alpha(s)}{ds}\hat{N}+\frac{d\rho(s)}{ds}\hat{J}_{3}+w_{12}(s)\hat{J}_{+}+w_{21}(s)\hat{J}_{-}\right]_{+}ds\right]$
(3.17)

where we have defined the auxiliary functions $\alpha(s)$ and $\rho(s)$ in
such a way that $\frac{d\alpha(s)}{ds}=w_{11}(s)+w_{22}(s)$ and
$\frac{d\rho(s)}{ds}=w_{11}(s)-w_{22}(s),$ with $\alpha(0)=\rho(0)=0.$

At this stage, to untangle (3.17) we have several choices. All these choices
are completely equivalent with respect to the complexity of the differential
equations that we finally have to solve to obtain the final untangled form of
the unitary time evolution operator. In fact, once such a unitary operator has
been completely untangled following a determined choice, we can move from one
to another simply by using appropriate similarity transformations formed from
the untangled exponential factors.

####  3.2.1 A simple and direct choice of disentangling

For example, we could start by isolating an exponential factor raised to a
factor proportional to $\hat{N},$ followed by another exponential factor
raised to a factor proportional to $\hat{J}_{3},$ to finally disentangle the
remaining exponential factor raised to a given linear combination of the
ladder operators $\hat{J}_{\pm}.$ As $\hat{N}$ commutes with all $su(2)$
generators and
$e^{i\rho(s)\hat{J}_{3}}\hat{J}_{\pm}e^{-i\rho(s)\hat{J}_{3}}=e^{\pm
i\rho(s)}\;\hat{J}_{\pm},$ a reiterated application of the Feynman disentangle
exponential factor‘s rule leads without any difficulty to

$\hat{U}_{0}(t,0)=e^{-i\alpha(t)\hat{N}}\;e^{-i\rho(t)\hat{J}_{3}}\exp\left[\;\int_{0}^{t}\left[\eta(s)\hat{J}_{+}-\eta^{\ast}(s)\hat{J}_{-}\right]_{+}ds\right],$
(3.18)

where $\alpha(s)=\int_{0}^{s}[w_{11}(s)+w_{22}(s)]ds,$
$\rho(s)=\int_{0}^{s}[w_{11}(s)-w_{22}(s)]ds$ and
$\eta(s)=-iw_{12}(s)e^{i\rho(s)}.$

The disentangling of the last factor

$\hat{T}(t)=\exp\left[\int_{0}^{t}\left[\eta(s)\hat{J}_{+}-\eta^{\ast}(s)\hat{J}_{-}\right]_{+}ds\right],$
(3.19)

is not always possible, it depend on the form of the functions
$w_{12}(s)=w^{\ast}_{21}(s)$ and of the real functions $w_{11}(s)$ and
$w_{22}(s).$ Indeed, a first factorization in terms of exponential operators
is

$\hat{T}(t)=e^{\Lambda(t)\hat{J}_{+}}\exp\left[\int_{0}^{t}e^{-\Lambda(s)\hat{J}_{+}}\left[(\eta(s)-\lambda(s))\hat{J}_{+}-\eta^{\ast}(s)\hat{J}_{-}\right]e^{\Lambda(s)J_{+}}ds\right],$
(3.20)

where $\Lambda(s)=\int_{0}^{s}\lambda(\tau)d\tau$ or
$\lambda(s)=\frac{d\Lambda(s)}{ds}.$ By using in (3.20) the relation
$e^{-\Lambda(s)\hat{J}_{+}}\;\hat{J}_{-}\;e^{\Lambda(s)\hat{J}_{+}}=\hat{J}_{-}-2\Lambda(s)\hat{J}_{3}-(\Lambda(s))^{2}\hat{J}_{+}$
and making a suitable regrouping of the terms we get

$\hat{T}(t)=e^{\Lambda(t)\hat{J}_{+}}\exp\left\\{\int_{0}^{t}\left[[\eta(s)-\lambda(s)+\eta^{\ast}(s)(\Lambda(s))^{2}]\hat{J}_{+}-\eta^{\ast}(s)[\hat{J}_{-}-2\Lambda(s)\hat{J}_{3}]\right]_{+}ds\right\\}.$
(3.21)

By forcing the coefficient of $\hat{J}_{+}$ in the second exponential factor
to be null, we get

$\hat{T}(t)=e^{\Lambda(t)\hat{J}_{+}}\exp\left\\{-\int_{0}^{t}\;\eta^{\ast}(s)\;\left[\hat{J}_{-}-2\Lambda(s)\hat{J}_{3}]\right]_{+}ds\right\\}.$
(3.22)

with the condition

$\frac{d\Lambda(s)}{ds}-\eta(s)-\eta^{\ast}(s)(\Lambda(s))^{2}=0,\quad\Lambda(0)=0,$
(3.23)

which a differential equation of the Riccati type.

Finally, by performing the disentangling of the remaining exponential terms we
get

$\hat{T}(t)=e^{\Lambda(t)\hat{J}_{+}}e^{\Omega(t)\hat{J}_{3}}e^{\Gamma(t)\hat{J}_{-}},$
(3.24)

where

$\Omega(t)=2\int_{0}^{t}\eta^{\ast}(s)\Lambda(s)\;ds\quad\text{
and}\quad\Gamma(t)=-\int_{0}^{t}\eta^{\ast}(s)e^{\Omega(s)}ds.$ (3.25)

##  4 A closed solution of the complex Riccati type equation

It is clear if we were able to solve equation (3.23) for any arbitrary value
of the time dependent functions $w_{\sigma\lambda}(s)$ we would have succeed
in our task, but a closed solution of this differential is not always
possible. For a formal discussion about it see [13]–[14], and references
therein, for example. Here we will generate some closed solutions of this
equation by appealing to some structural criteria of the operator algebra.

###  4.1 Standard $su(2)$ type structure

A first good criterion to consider is the unitary character the operator
$\hat{T}$ and its counterbalanced exponential structure in terms of the
$su(2)$ generators. Then, similarly to the case when all coefficients are
time-independent constants, we can choose $\Omega(t)$ to be real at any time,
which implies that, as we will see in the following development,
$\Gamma(t)=-\Lambda^{\ast}(t).$ A sufficient condition for $\Omega(t)$ to be
real is

$\eta^{\ast}(s)\Lambda(s)=\eta(s)\Lambda^{\ast}(s).$ (4.1)

Taking into account this condition in equation (3.23) we get

$\frac{d\Lambda(s)}{ds}=\eta(s)(1+\|\Lambda(s)\|^{2}).$ (4.2)

Now, if in this last equation we use the polar forms
$\Lambda(s)=R(s)e^{i\theta(s)}$ and $\eta(s)=\|\eta(s)\|e^{i\phi(s)}$, where
$R(s),\theta(s)$ and $\phi(s)$ are real functions, and we separate it into its
real and imaginary parts, we get

$\frac{dR(s)}{ds}=(-1)^{k}\;\|\eta(s)\|(1+R^{2}(s)),\quad k=0,\pm
1,\cdots,\quad$ (4.3)

and the condition $\theta(s)=\theta_{0},\forall s$ and
$\phi(s)=\phi_{0}=\theta_{0}-k\pi,\;k=0,\pm 1,\cdots,\;\forall s,$ where
$\phi_{0}$ and $\theta_{0}$ are real constants. Thus by integrating equation
(4.3) we obtain

$R(s)=(-1)^{k}\;\tan[\int_{0}^{s}\|\eta(\tau)\|d\tau],\quad k=0,\pm 1,\cdots.$
(4.4)

Finally, taking into account all these results, we find a solution of equation
( 3.23), that is,

$\Lambda(s)=\tan[\int_{0}^{s}\|\eta(\tau)\|d\tau]\;e^{i\phi_{0}},$ (4.5)

provided that

$\eta(s)=\|\eta(s)\|\;e^{i\phi_{0}}.$ (4.6)

Hence, the $\Omega(t)$ function writes

$\Omega(t)=2\int_{0}^{t}\|\eta(s)\|\tan[\int_{0}^{\tau}\|\eta(\tau)\|\;d\tau]ds=2\ln\left(\sec[\int_{0}^{s}\|\eta(\tau)\|d\tau]\right)$
(4.7)

and the $\Gamma(t)$ function assumes the form

$\Gamma(t)=-\int_{0}^{t}\|\eta(s)\|e^{-i\phi_{0}}\sec^{2}[\int_{0}^{t}\|\eta(\tau)\|d\tau]ds=-\;e^{-i\phi_{0}}\tan[\int_{0}^{t}\|\eta(s)\|\;ds]=-\Lambda^{\ast}(t).$
(4.8)

Then, if condition (4.1) is met, the disentangled structure of the unitary
operator (3.19) is

$\hat{T}(t,0)=e^{\tan[\int_{0}^{t}\|w_{12}(\tau)\|d\tau]\;e^{i\phi_{0}}\;\hat{J}_{+}}e^{2\ln\left(\sec[\int_{0}^{t}\|w_{12}(\tau)\|d\tau]\right)\hat{J}_{3}}e^{-\tan[\int_{0}^{t}\|w_{12}(\tau)\|d\tau]\;e^{-i\phi_{0}}\hat{J}_{-}},$
(4.9)

where we have used $\|\eta(\tau)\|=\|w_{12}(\tau)\|,\;\forall\;\tau,$
according to the relation
$\eta(\tau)=-iw_{12}(\tau)e^{i\int_{0}^{\tau}(w_{11}(s)-w_{22}(s))ds}.$ It is
easy now to rewrite this last operator in the usual form

$\hat{T}(t)=\exp\left[\left(\int_{0}^{t}\|w_{12}(\tau)\|d\tau\right)\left(e^{i\phi_{0}}\hat{J}_{+}-e^{-i\phi_{0}}\hat{J}_{-}\right)\right].$
(4.10)

Thus the completely disentangled form of (3.13) writes

$\displaystyle\hat{U}_{0}(t,0)$ $\displaystyle=$
$\displaystyle\exp\left[{-i\int_{0}^{t}(w_{11}(s)+w_{22}(s))ds\;\hat{N}}\right]\;\exp\left[{-i\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds\;\hat{J}_{3}}\right]$
(4.11) $\displaystyle\times$
$\displaystyle\exp\left[\left(\int_{0}^{t}\|w_{12}(\tau)\|d\tau\right)\left(e^{i\phi_{0}}\hat{J}_{+}-e^{-i\phi_{0}}\hat{J}_{-}\right)\right].$

####  4.1.1 An explicit untangled form of the time evolution operator

In our case, when the condition (4.1) is met, which according to the $\eta(s)$
definition and to ( 4.6) is equivalent to

$\theta_{12}(t)=-\theta_{21}(t)=\phi_{0}+\frac{\pi}{2}-\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds,$
(4.12)

with no other constraint on the $w_{\sigma\lambda}(t)$ functions, then the
time evolution operator (3.3) can be completely disentangled. In this case,
the solutions of the system (3.12), as it has been shown in the above section,
can be obtained by using the matrix elements of the unitary operator (A.5) in
a given orthonormal basis. As

$\hat{W}(s)=(w_{11}(s)+w_{22}(s))\hat{N}+(w_{11}(s)-w_{22}(s))\hat{J}_{3}+w_{12}(s)\hat{J}_{+}+w_{21}(s)\hat{J}_{-},$
(4.13)

a suitable basis to compute the matrix elements of (A.5) is the
$j=\frac{1}{2}$ basis representation of the $su(2)$ Lie algebra:
$\mid\pm\rangle=\mid\frac{1}{2},\pm\frac{1}{2}\rangle.$ On the other hand, as
the unitary operator (A.5) is identical to the unitary operator (3.13), its
untangled form is given by equation (4.11). Thus, the matrix elements of
$\hat{S}(t,0)$ are given by

$\displaystyle
S_{11}(t)=\langle+\mid\hat{S}(t,0)\mid+\rangle=e^{-i\int_{0}^{t}w_{11}(s)ds}\cos\left(\int_{0}^{t}\|w_{12}(s)\|ds\right),$
(4.14) $\displaystyle\quad
S_{12}(t)=\langle+\mid\hat{S}(t,0)\mid-\rangle=e^{-i\int_{0}^{t}w_{11}(s)ds}\;e^{i\phi_{0}}\;\sin\left(\int_{0}^{t}\|w_{12}(s)\|ds\right),$
(4.15) $\displaystyle\quad
S_{21}(t)=\langle-\mid\hat{S}(t,0)\mid+\rangle=-e^{-i\int_{0}^{t}w_{22}(s)ds}\;e^{-i\phi_{0}}\;\sin\left(\int_{0}^{t}\|w_{12}(s)\|ds\right),$
(4.16) $\displaystyle\quad
S_{22}(t)=\langle-\mid\hat{S}(t,0)\mid-\rangle=e^{-i\int_{0}^{t}w_{22}(s)ds}\cos\left(\int_{0}^{t}\|w_{12}(s)\|ds\right).$
(4.17)

By inserting these last matrix elements in equation (A.11), which gives us the
$c_{\sigma(t)},\;\sigma=1,2,$ functions that verify the differential equation
system (3.12), we get

$\displaystyle c_{1}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\int_{0}^{t}w_{11}(s)ds}\cos\left(\int_{0}^{t}\|w_{12}(s)\|ds\right)\tilde{c}_{1}(0)$
(4.18) $\displaystyle+$ $\displaystyle
e^{-i\int_{0}^{t}w_{11}(s)ds}\;e^{i\phi_{0}}\;\sin\left(\int_{0}^{t}\|w_{12}(s)\|ds\right)\tilde{c}_{2}(0)$
$\displaystyle-$
$\displaystyle\frac{i}{\hbar}e^{-i\int_{0}^{t}w_{11}(s)ds}\int_{0}^{t}e^{i\int_{0}^{s}w_{11}(\tau)d\tau}\cos\left(\int_{s}^{t}\|w_{12}(\tau)\|d\tau\right)F_{1}(s)$
$\displaystyle-$
$\displaystyle\frac{i}{\hbar}e^{-i\int_{0}^{t}w_{11}(s)ds}\;e^{i\phi_{0}}\;\int_{0}^{t}e^{i\int_{0}^{s}w_{22}(\tau)d\tau}\sin\left(\int_{s}^{t}\|w_{12}(\tau)\|d\tau\right)F_{2}(s)$

and

$\displaystyle c_{2}(t)$ $\displaystyle=$
$\displaystyle-\;e^{-i\int_{0}^{t}w_{22}(s)ds}\;e^{-i\phi_{0}}\;\sin\left(\int_{0}^{t}\|w_{12}(s)\|ds\right)\tilde{c}_{1}(0)$
(4.19) $\displaystyle+$ $\displaystyle
e^{-i\int_{0}^{t}w_{22}(s)ds}\cos\left(\int_{0}^{t}\|w_{12}(s)\|ds\right)\tilde{c}_{2}(0)$
$\displaystyle+$
$\displaystyle\frac{i}{\hbar}e^{-i\int_{0}^{t}w_{22}(s)ds}\;e^{-i\phi_{0}}\;\int_{0}^{t}e^{i\int_{0}^{s}w_{11}(\tau)d\tau}\sin\left(\int_{s}^{t}\|w_{12}(\tau)\|d\tau\right)F_{1}(s)$
$\displaystyle-$
$\displaystyle\frac{i}{\hbar}e^{-i\int_{0}^{t}w_{22}(s)ds}\int_{0}^{t}e^{i\int_{0}^{s}w_{22}(\tau)d\tau}\cos\left(\int_{s}^{t}\|w_{12}(\tau)\|d\tau\right)F_{2}(s).$

With these results, we now are sure that a closed disentangled form of the
unitary time evolution operator (3.3) can be reached, provided that the phase
condition (4.12) be satisfied.

###  4.2 The case $\eta(s)=\eta_{0}e^{i(\phi_{0}+w_{0}s)},\;\eta_{0},\phi_{0}$
and $w_{0}$ constant

Let us now consider the case when $\eta(s)=\eta_{0}e^{i(\phi_{0}+w_{0}s)},$
where $\eta_{0},\phi_{0}$ and $w_{0}$ are given constants. As
$\eta(s)=-iw_{12}e^{i\rho(s)},$ this implies that $\|w_{12}(s)\|=\eta_{0}$ and

$\theta_{12}(s)=\phi_{0}+\frac{\pi}{2}+w_{0}\;s-\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau.$
(4.20)

Let us now note that here this condition can be interpreted in both
directions, that is, for given values of the time-dependent parameters
$w_{11}(s)$ and $w_{22}(s)$ we must choose $\theta_{12}(s)$ such that the
condition is met, and vice versa, or force all them to meet the condition and
thus obtain fixed values of the constants $\phi_{0}$ and $w_{0}.$ So, for
example, in the second case, we can choose $\theta_{12}(s)=\nu_{12}\;s,$ and
$w_{11}(s)-w_{22}(s)=\nu_{0},$ where $\nu_{12}$ and $\nu_{0}$ are constants,
and obtain the fixed values $\phi_{0}=-\frac{\pi}{2}$ and
$w_{0}=\nu_{0}+\nu_{12.}$

Then, taking into account condition (4.20) in equation (3.23), we get

$e^{-i(\phi_{0}+w_{0}s)}\frac{d}{ds}\Lambda(s)=\eta_{0}\left[1+\left(e^{-i(\phi_{0}+w_{0}s)}\Lambda(s)\right)^{2}\right],\quad\Lambda(0)=0.$
(4.21)

Now, by performing the change of variables

$\Lambda(s)=e^{i(\phi_{0}+w_{0}s)}\tilde{\Lambda}(s),$ (4.22)

this last equation can be put in the form:

$\frac{2\eta_{0}}{\delta}\;\frac{d}{ds}\tilde{\Lambda}(s)=\frac{\delta}{2}\left[1+\left(\frac{2\eta_{0}}{\delta}\right)^{2}\left(\tilde{\Lambda}(s)-\frac{iw_{0}}{2\eta_{0}}\right)^{2}\right],\quad\tilde{\Lambda}(0)=0,\quad$
(4.23)

where $\delta=\sqrt{4\eta_{0}^{2}+w_{0}^{2}}.$ By performing again a new
change of variables:

$\tilde{\Upsilon}(s)=\frac{2\eta_{0}}{\delta}\left(\tilde{\Lambda}(s)-\frac{iw_{0}}{2\eta_{0}}\right),$
(4.24)

equation (4.23) take the form

$\frac{d}{ds}\tilde{\Upsilon}(s)=\frac{\delta}{2}\left[1+\left(\tilde{\Upsilon}(s)\right)^{2}\right],\quad\tilde{\Upsilon}(0)=-\frac{iw_{0}}{\delta},$
(4.25)

which is easily separable and leads to the solution

$\tilde{\Upsilon}(s)=\frac{\delta\tan{\left(\frac{\delta
s}{2}\right)}-iw_{0}}{\delta+iw_{0}\tan\left(\frac{\delta s}{2}\right)}.$
(4.26)

Returning to the original variable with the help of (4.22) and (4.24), after
some manipulations, we get

$\Lambda(s)=\frac{2\eta_{0}e^{i(\phi_{0}+w_{0}s)}\tan{\left(\frac{\delta
s}{2}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}{\left(\frac{\delta
s}{2}\right)}}}\exp\left[-i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;s}{2}\right)\right]\right]$
(4.27)

Finally, by inserting this last result into (3.25), then integrating and
manipulating the resulting expressions we obtain

$\Omega(s)=\ln\left[\frac{\delta^{2}\sec^{2}\left(\frac{\delta\,s}{2}\right)}{\delta^{2}+w_{0}^{2}\tan^{2}\left(\frac{\delta\;s}{2}\right)}\right]+i\left[w_{0}\;s-2\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;s}{2}\right)\right]\right]$
(4.28)

and

$\Gamma(s)=-\frac{2\eta_{0}e^{-i\phi_{0}}\tan\left(\frac{\delta\;s}{2}\right)}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}\left(\frac{\delta\;s}{2}\right)}}\exp\left[-i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;s}{2}\right)\right]\right].$
(4.29)

Finally, by inserting these last results into (3.24), and using the general
relation
$e^{\alpha\hat{J}_{3}}\hat{J}_{-}e^{-\alpha\hat{J}_{3}}=e^{-\alpha}\hat{J}_{-},$
we obtain the untangle form of the $\hat{T}(t)$ unitary operator, that is

$\displaystyle\hat{T}(t)$ $\displaystyle=$
$\displaystyle\exp\left[\frac{2\eta_{0}e^{i(\phi_{0}+w_{0}t)}\tan{\left(\frac{\delta
t}{2}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}{\left(\frac{\delta
t}{2}\right)}}}\exp\left[-i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;t}{2}\right)\right]\right]\;\hat{J}_{+}\right]$
(4.30) $\displaystyle\times$
$\displaystyle\exp\left[\ln\left[\frac{\delta^{2}\sec^{2}\left(\frac{\delta\,t}{2}\right)}{\delta^{2}+w_{0}^{2}\tan^{2}\left(\frac{\delta\;t}{2}\right)}\right]\;\hat{J}_{3}\right]\exp\left[\frac{-2\eta_{0}e^{-i(\phi_{0}+w_{0}t)}\tan{\left(\frac{\delta
t}{2}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan{\left(\frac{\delta
t}{2}\right)}}}\exp\left[i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;t}{2}\right)\right]\right]\;\hat{J}_{-}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[i\left[w_{0}\;t-2\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;t}{2}\right)\right]\right]\;\hat{J}_{3}\right],$

which written in the standard form look like

$\displaystyle\hat{T}(t)$ $\displaystyle=$
$\displaystyle\exp\left\\{\arctan\left(\frac{2\eta_{0}\tan{\left(\frac{\delta
t}{2}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}{\left(\frac{\delta
t}{2}\right)}}}\right)\right.$ (4.31) $\displaystyle\times$
$\displaystyle\left.\left[e^{i\left(\phi_{0}+w_{0}\;t-\arctan\left(\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;t}{2}\right)\right)\right)}\;\hat{J}_{+}-e^{-i\left(\phi_{0}+w_{0}\;t-\arctan\left(\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;t}{2}\right)\right)\right)}\;\hat{J}_{-}\right]\right\\}$
$\displaystyle\times$ $\displaystyle
e^{i\left[w_{0}\;t-2\arctan\left(\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;t}{2}\right)\right)\right]\;\hat{J}_{3}}.$

As before, so that untangled form of the time evolution is reached, we need to
compute the $S$ matrix elements of unitary operator (3.13). For this we can
use again the vector basis of the $j=\frac{1}{2}$ irreducible representation
of su(2) algebra, doing that we obtain

$\displaystyle S_{11}(t)$ $\displaystyle=$
$\displaystyle\langle+\mid\hat{S}(t,0)\mid+\rangle=e^{-i\int_{0}^{t}w_{11}(s)ds}\;e^{i\frac{w_{0}\;t}{2}}\left[\cos\left(\frac{\delta
t}{2}\right)-i\frac{w_{0}}{\delta}\sin\left(\frac{\delta\;t}{2}\right)\right]$
$\displaystyle S_{12}(t)$ $\displaystyle=$
$\displaystyle\langle+\mid\hat{S}(t,0)\mid-\rangle=\frac{2\eta_{0}}{\delta}e^{-i\int_{0}^{t}w_{11}(s)ds}\;e^{i(\phi_{0}+\frac{w_{0}\;t}{2})}\;\sin\left(\frac{\delta\;t}{2}\right),$
$\displaystyle S_{21}(t)$ $\displaystyle=$
$\displaystyle\langle-\mid\hat{S}(t,0)\mid+\rangle=-\frac{2\eta_{0}}{\delta}e^{-i\int_{0}^{t}w_{22}(s)ds}\;e^{-i(\phi_{0}+\frac{w_{0}\;t}{2})}\;\sin\left(\frac{\delta\;t}{2}\right),$
$\displaystyle S_{22}(t)$ $\displaystyle=$
$\displaystyle\langle+\mid\hat{S}(t,0)\mid+\rangle=e^{-i\int_{0}^{t}w_{22}(s)ds}\;e^{-i\frac{w_{0}\;t}{2}}\left[\cos\left(\frac{\delta
t}{2}\right)+i\frac{w_{0}}{\delta}\sin\left(\frac{\delta\;t}{2}\right)\right].$

###  4.3 The case $\eta(s)=r(s)e^{i\phi(s)}$ with
$r(s)=\epsilon\frac{\eta_{0}}{w_{0}}\frac{d\phi}{ds}(s)>0\;\forall s,$ where
$\epsilon=\pm 1$ and $\eta_{0}$ and $w_{0}$ are constants

The last case treated above can be generalized if the norm of $\eta(s)$ and
its phase verify the following condition

$r(s)=\epsilon\;\frac{\eta_{0}}{w_{0}}\frac{d\phi}{ds}(s)=\begin{cases}+\eta_{0}\frac{d\phi(s)}{ds},&\text{if
$\frac{d\phi(s)}{ds}>0$}.\\\
-\frac{\eta_{0}}{w_{0}}\frac{d\phi(s)}{ds},&\text{if
$\frac{d\phi(s)}{ds}<0$}.\end{cases}$ (4.33)

where $\eta_{0}$ and $w_{0}$ are given non-negative real constants greater
than zero which have been written here in this way in order to make compatible
the physical quantities. According the definition of the parameter $\eta(s),$
this condition is equivalent to

$\theta_{12}(t)=-\theta_{21}(t)=\phi(t)+\frac{\pi}{2}-\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds,$
(4.34)

and

$\|w_{12}(s)\|=\epsilon\frac{\eta_{0}}{w_{0}}\frac{d\phi}{ds}(s).$ (4.35)

Assuming that holds, then (3.23) can be analytically solved. Indeed, by
performing the change of variables:

$\tilde{\Upsilon}(s)=\frac{2\eta_{0}}{\delta}\left(e^{-i\phi(s)}\;{\Lambda}(s)-i\frac{w_{0}}{2\epsilon\;\eta_{0}}\right),$
(4.36)

where $\delta=\sqrt{4\eta_{0}^{2}+w_{0}^{2}},$ equation (3.23) take the form

$\frac{d}{ds}\tilde{\Upsilon}(s)=\frac{\delta\;\epsilon}{2w_{0}}\frac{d\phi}{ds}(s)\left[1+\left(\tilde{\Upsilon}(s)\right)^{2}\right],\quad\tilde{\Upsilon}(0)=-i\frac{w_{0}}{\delta\;\epsilon},$
(4.37)

which is easily separable and leads to the solution

$\tilde{\Upsilon}(s)=\epsilon\;\frac{\delta\tan{\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)}-iw_{0}}{\delta+iw_{0}\tan\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)},$
(4.38)

where $\tilde{\phi}(s)=\phi(s)-\phi(0)$ and the $\epsilon$ parameter value
stands for the region where $\epsilon\frac{d\phi}{ds}(s)>0.$

Returning to the original variable with the aid of (4.36) and (4.38), after
some manipulations, we get

$\Lambda(s)=\frac{2\;\epsilon\eta_{0}e^{i\phi(s)}\tan{\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}{\left(\frac{\delta\;\tilde{\phi(s)}}{2w_{0}}\right)}}}\exp\left[-i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)\right]\right],$
(4.39)

Thus, by inserting this last result into (3.25) and then integrating and
manipulating the resulting expressions we obtain

$\Omega(s)=\ln\left[\frac{\delta^{2}\sec^{2}\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)}{\delta^{2}+w_{0}^{2}\tan^{2}\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)}\right]+i\left[\tilde{\phi}(s)-2\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)\right]\right]$
(4.40)

and

$\Gamma(s)=-\frac{2\;\epsilon\eta_{0}e^{-i\phi(0)}\tan\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)}}\exp\left[-i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(s)}{2w_{0}}\right)\right]\right].$
(4.41)

Finally, with the help of all these expression we can write the unitary
operator $\hat{T}(t)$ in the form

$\displaystyle\hat{T}(t)$ $\displaystyle=$
$\displaystyle\exp\left[\frac{2\;\epsilon\;\eta_{0}e^{i\phi(t)}\tan{\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}{\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}}}\exp\left[-i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right]\right]\;\hat{J}_{+}\right]$
(4.42) $\displaystyle\times$
$\displaystyle\exp\left[\ln\left[\frac{\delta^{2}\sec^{2}\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}{\delta^{2}+w_{0}^{2}\tan^{2}\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}\right]\;\hat{J}_{3}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[\frac{-\;2\;\epsilon\;\eta_{0}e^{-i\phi(t)}\tan{\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}{\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}}}\exp\left[i\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right]\right]\;\hat{J}_{-}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[i\left[\tilde{\phi}(t)-2\arctan\left[\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right]\right]\;\hat{J}_{3}\right],$

or in the the standard form, as we have done in the above sections:

$\displaystyle\hat{T}(t)$ $\displaystyle=$
$\displaystyle\exp\left\\{\arctan\left(\frac{2\;\epsilon\;\eta_{0}\tan{\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}}{\sqrt{\delta^{2}+w_{0}^{2}\tan^{2}{\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)}}}\right)\right.$
(4.43) $\displaystyle\times$
$\displaystyle\left.\left[e^{i\left(\phi(t)-\arctan\left(\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right)\right)}\;\hat{J}_{+}-e^{-i\left(\phi(t)-\arctan\left(\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right)\right)}\;\hat{J}_{-}\right]\right\\}$
$\displaystyle\times$ $\displaystyle
e^{i\left[\tilde{\phi}(t)-2\arctan\left(\frac{w_{0}}{\delta}\tan\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right)\right]\;\hat{J}_{3}},$

which reduces to (4.31) when $\phi(s)=\phi_{0}+w_{0}\;s.$

As before we can compute the $S$ matrix elements of (3.13) in the
$j=\frac{1}{2}$ representation space of the $su(2)$ Lie algebra. They are
given by

$\displaystyle S_{11}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\int_{0}^{t}w_{11}(s)ds}\;e^{i\frac{\tilde{\phi}(t)}{2}}\;\left[\cos\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)-i\frac{w_{0}}{\delta}\;\sin\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right]$
$\displaystyle S_{12}(t)$ $\displaystyle=$
$\displaystyle\;\frac{2\;\epsilon\;\eta_{0}}{\delta}e^{-i\int_{0}^{t}w_{11}(s)ds}\;e^{i\frac{(\phi(t)+\phi(0))}{2}}\;\sin\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right),$
$\displaystyle S_{21}(t)$ $\displaystyle=$
$\displaystyle-\;\frac{2\;\epsilon\;\eta_{0}}{\delta}e^{-i\int_{0}^{t}w_{22}(s)ds}\;e^{-i\frac{(\phi(t)+\phi(0))}{2}}\;\sin\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right),$
$\displaystyle S_{22}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\int_{0}^{t}w_{22}(s)ds}\;e^{-i\frac{\tilde{\phi}(t)}{2}}\;\left[\cos\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)+i\frac{w_{0}}{\delta}\;\sin\left(\frac{\delta\;\tilde{\phi}(t)}{2w_{0}}\right)\right],$

where in all the above expressions the sing of $\epsilon$ accords with the
sign of $\frac{d\phi}{ds}(s).$

###  4.4 The standard case when all parameters $w_{\sigma\lambda}$ are time-
independent

When all parameters $w_{\sigma\lambda}$ are time-independent, i.e.,
$w_{11}(s)=w_{11},$ $w_{22}(s)=w_{22},$ and
$w_{12}(s)=w_{21}^{\ast}(s)=\|w_{12}\|\;e^{i\theta^{0}_{12}},$ where
$w_{11}\neq w_{22}$ and $\theta^{0}_{12}\in[0,2\pi],$ the condition (4.34)
implies

$\theta^{0}_{12}=-\theta^{0}_{21}=\phi(s)+\frac{\pi}{2}-(w_{11}-w_{22})s,$
(4.45)

from were

$\phi(s)=\theta^{0}_{12}-\frac{\pi}{2}+(w_{11}-w_{22})s,\quad{\rm
and}\quad\tilde{\phi}(s)=(w_{11}-w_{22})s.$ (4.46)

Thus the different combinations of parameters are

$\frac{\eta_{0}}{w_{0}}=\frac{\epsilon\;\|w_{12}\|}{(w_{11}-w_{22})},\quad\delta=\frac{\epsilon\;w_{0}\;b}{(w_{11}-w_{22})}\quad{\rm
and}\quad\frac{\delta\tilde{\phi}(s)}{2w_{0}}=\frac{\epsilon\;b\;s}{2},$
(4.47)

where $b=\sqrt{4\|w_{12}\|^{2}+(w_{11}-w_{22})^{2}}.$ Hence, the unitary
operator $\hat{T}(t)$ takes the form

$\displaystyle\hat{T}(t)$ $\displaystyle=$
$\displaystyle\exp\left[\frac{2w_{12}e^{i(w_{11}-w_{22})t-i\frac{\pi}{2}}\tan{\left(\frac{b\;t}{2}\right)}}{\sqrt{b^{2}+(w_{11}-w_{22})^{2}\tan^{2}{\left(\frac{b\;t}{2}\right)}}}\exp\left[-i\arctan\left[\frac{(w_{11}-w_{22})}{b}\tan\left(\frac{b\;t}{2}\right)\right]\right]\;\hat{J}_{+}\right]$
(4.48) $\displaystyle\times$
$\displaystyle\exp\left[\ln\left[\frac{b^{2}\sec^{2}\left(\frac{b\;t}{2}\right)}{b^{2}+(w_{11}-w_{22})^{2}\tan^{2}\left(\frac{b\;t}{2}\right)}\right]\;\hat{J}_{3}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[\frac{-2w_{21}e^{-i(w_{11}-w_{22})t+i\frac{\pi}{2}}\tan{\left(\frac{b\;t}{2}\right)}}{\sqrt{b^{2}+(w_{11}-w_{22})^{2}\tan^{2}{\left(\frac{b\;t}{2}\right)}}}\exp\left[i\arctan\left[\frac{(w_{11}-w_{22})}{b}\tan\left(\frac{b\;t}{2}\right)\right]\right]\;\hat{J}_{-}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[i\left[(w_{11}-w_{22})(t)-2\arctan\left[\frac{(w_{11}-w_{22})}{b}\tan\left(\frac{b\;t}{2}\right)\right]\right]\;\hat{J}_{3}\right],$

Thus the operator (4.11) becomes

$\displaystyle\hat{U}_{0}(t,0)$ $\displaystyle=$
$\displaystyle\exp\left[-i(w_{11}+w_{22})\hat{N}\right]$ (4.49)
$\displaystyle\times$
$\displaystyle\exp\left[\frac{2w_{12}\tan{\left(\frac{b\;t}{2}\right)}}{\sqrt{b^{2}+(w_{11}-w_{22})^{2}\tan^{2}{\left(\frac{b\;t}{2}\right)}}}\exp\left[-i\frac{\pi}{2}-i\arctan\left[\arctan\left(\frac{(w_{11}-w_{22})}{b}\tan\left(\frac{b\;t}{2}\right)\right)\right]\right]\;\hat{J}_{+}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[\ln\left[\frac{b^{2}\sec^{2}\left(\frac{b\;t}{2}\right)}{b^{2}+(w_{11}-w_{22})^{2}\tan^{2}\left(\frac{b\;t}{2}\right)}\right]\;\hat{J}_{3}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[\frac{-2w_{21}\tan{\left(\frac{b\;t}{2}\right)}}{\sqrt{b^{2}+(w_{11}-w_{22})^{2}\tan^{2}{\left(\frac{b\;t}{2}\right)}}}\exp\left[i\frac{\pi}{2}+i\arctan\left[\frac{(w_{11}-w_{22})}{b}\tan\left(\frac{b\;t}{2}\right)\right]\right]\;\hat{J}_{-}\right]$
$\displaystyle\times$
$\displaystyle\exp\left[-2i\arctan\left[\frac{(w_{11}-w_{22})}{b}\tan\left(\frac{b\;t}{2}\right)\right]\;\hat{J}_{3}\right],$

where we can recognize the standard Bigolovuv disentangled form of an $su(2)$
exponential operator, and the $\hat{S}(t,0)$ matrix elements take the form

$\displaystyle S_{11}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\frac{(w_{11}+w_{22})\;t}{2}}\;\left[\cos\left(\frac{bt}{2}\right)-i\frac{(w_{11}-w_{22})}{b}\;\sin\left(\frac{bt}{2}\right)\right]$
$\displaystyle S_{12}(t)$ $\displaystyle=$
$\displaystyle-\frac{2iw_{12}}{b}e^{-i\frac{(w_{11}+w_{22})\;t}{2}}\;\sin\left(\frac{bt}{2}\right),$
$\displaystyle S_{21}(t)$ $\displaystyle=$
$\displaystyle-\frac{2iw_{21}}{b}e^{-i\frac{(w_{11}+w_{22})\;t}{2}}\;\sin\left(\frac{bt}{2}\right),$
$\displaystyle S_{22}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\frac{(w_{11}+w_{22})\;t}{2}}\;\left[\cos\left(\frac{bt}{2}\right)+i\frac{(w_{11}-w_{22})}{b}\;\sin\left(\frac{bt}{2}\right)\right].$
(4.50)

##  5 Time evolution of the coherent states linked to the commensurate 2 D
isotropic harmonic oscillator

The Schrödinger type coherent states of the two-dimensional isotropic harmonic
oscillator

$H_{0}=\hbar\hat{a_{1}}^{\dagger}\hat{a}_{1}+\hbar\hat{a_{2}}^{\dagger}\hat{a}_{2}$
(5.1)

can be defined as the eigentates of the two-mode annihilation operator

$\hat{A}_{\alpha\beta}=\alpha\hat{a}_{1}+\beta\hat{a}_{2},$ (5.2)

where $\alpha$ and $\beta$ are complex numbers which verify
$\|\alpha\|^{2}+\|\beta\|^{2}=1.$ Indeed, due to this last fact, the operators
$\hat{H}_{0},$ $\hat{A}_{\alpha\beta}$ and its adjoint
$\hat{A}^{\dagger}_{\alpha\beta}$ and the identity $\hat{I}$ satisfy
commutation relations of the oscillator algebra

$\left[\hat{A}_{\alpha\beta},\hat{A}^{\dagger}_{\alpha\beta}\right]=\hat{I},$
(5.3)

and

$[\hat{H}_{0},\hat{A}_{\alpha\beta}]=-\hbar\;\hat{A}_{\alpha\beta}\quad{\rm
and}\quad[\hat{H}_{0},\hat{A}^{\dagger}_{\alpha\beta}]=\hbar\hat{A}^{\dagger}_{\alpha\beta}.$
(5.4)

The Hamiltonian (5.1) is not the only $h(1)\oplus
h(1)\mathbin{\mathchoice{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\displaystyle\displaystyle
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\end{picture}}}}{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\textstyle\displaystyle
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\end{picture}}}}{\makebox[7.66663pt]{\raisebox{-0.58333pt}{\begin{picture}(1.0,1.0)\scriptstyle\displaystyle
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\end{picture}}}}{\makebox[6.11107pt]{\raisebox{-0.41666pt}{\begin{picture}(1.0,1.0)\scriptscriptstyle\displaystyle
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\put(0.5,0.5){\arc[-90,90]{0.5}} \end{picture}}}}}u(2)$-type Hamiltonian that
possess these characteristics [15], but also, for example, the following
Hamiltonian formed from the product of the ladder operators[16]
$\hat{A}^{\dagger}_{\alpha\beta}$ and $\hat{A}_{\alpha\beta}:$

$\hat{H}_{\alpha\beta}=\hbar\hat{A}^{\dagger}_{\alpha\beta}\hat{A}_{\alpha\beta}=\hbar\;[\|\alpha\|^{2}\hat{a}^{\dagger}_{1}\hat{a}_{1}+\|\beta\|^{2}\hat{a}^{\dagger}_{2}\hat{a}_{2}+\alpha^{\ast}\beta\hat{a}^{\dagger}_{1}\hat{a}_{2}+\alpha\beta^{\ast}\hat{a}^{\dagger}_{2}\hat{a}_{1}].$
(5.5)

Certainly, this Hamiltonian also verify

$[\hat{H}_{\alpha\beta},\hat{A}_{\alpha\beta}]=-\hbar\;\hat{A}_{\alpha\beta}\quad{\rm
and}\quad[\hat{H}_{\alpha\beta},\hat{A}^{\dagger}_{\alpha\beta}]=\hbar\hat{A}^{\dagger}_{\alpha\beta}.$
(5.6)

Since the set of operators
$\left\\{\hat{A}_{\alpha\beta},\hat{A}^{\dagger}_{\alpha\beta},\hat{I},\hat{H}_{\alpha\beta}\right\\}$
generate the one dimensional oscillator algebra, we can associate to this
system the one parameter Schrödinger type coherent states333Indeed, the
coherent states in (5.7) represent only a special case of a family of algebra
eigenstates of the operator (5.2) whose coefficients verify the relation
$\|\alpha\|^{2}+\|\beta\|^{2}=1.$ In this special case, the fundamental state
$\mid 0,0\rangle,$ coincides with the fundamental state of the Hamiltonian
(5.5) [15] .

$\mid
Z\rangle=\exp\left[Z\hat{A}^{\dagger}_{\alpha\beta}-Z^{\ast}\hat{A}_{\alpha\beta}\right]\mid
0,0\rangle,$ (5.7)

where $\mid 0,0\rangle$ denotes the two mode quantum oscillator fundamental
state, which implies that $\hat{A}_{\alpha\beta}\mid 0,0\rangle=0.$

The time evolution of a state $\mid\psi(0)\rangle$ governed by the Hamiltonian
(5.5 ) is given by

$\mid\psi(t)\rangle=e^{-\frac{i}{\hbar}\int_{0}^{t}\hat{H}_{\alpha\beta}ds}\mid\psi(0)\rangle=e^{-\frac{i}{\hbar}\hat{H}_{\alpha\beta}t}\mid\psi(0)\rangle,$
(5.8)

because $\hat{H}_{\alpha\beta}$ is time-independent. Thus, the time evolution
of the coherent states 5.7 is given by

$\mid Z(t)\rangle=e^{-\frac{i}{\hbar}\hat{H}_{\alpha\beta}t}\mid
Z\rangle=\exp\left[-i\hat{A}^{\dagger}_{\alpha\beta}\hat{A}_{\alpha\beta}\;t\right]\exp\left[Z\hat{A}^{\dagger}_{\alpha\beta}-Z^{\ast}\hat{A}_{\alpha\beta}\right]\mid
0,0\rangle.$ (5.9)

Then, using the generic relations
$e^{-i\lambda\hat{a}^{\dagger}\hat{a}}\hat{a}^{\dagger}e^{i\lambda\hat{a}^{\dagger}\hat{a}}=e^{-i\lambda}\hat{a}^{\dagger}$
and
$e^{-i\lambda\hat{a}^{\dagger}\hat{a}}\hat{a}e^{i\lambda\hat{a}^{\dagger}\hat{a}}=e^{i\lambda}\hat{a},$
we get the time evolution 2 D Shrödinger type coherent states

$\mid
Z(t)\rangle=\exp\left[Ze^{-it}\hat{A}^{\dagger}_{\alpha\beta}-Z^{\ast}e^{it}\hat{A}_{\alpha\beta}\right]\mid
0,0\rangle.$ (5.10)

The same result can be reached if we use the general technique developed in
the previous sections. In fact, it has been well established [17] that the
most general time-dependent Hamiltonian that preserves in time the coherent
state character of a system which has initially been prepared as a coherent
state is precisely of the type (3.1), and that for $N$ interacting quantum
harmonic oscillators. Also, it has been proven [17] that the general structure
of these coherent states, in the schrödinger picture, is given by

$\exp\left[\sum_{\sigma=0}^{N}\left(c_{\sigma}(t)\hat{a}^{\dagger}_{\sigma}-c^{\ast}_{\sigma}(t)\hat{a}_{\sigma}\right)\right]\mid\tilde{0}\rangle,$
(5.11)

where the $c_{\sigma}(t)$ coefficients are given by (A.11) (where the sum is
extended to $N$) and $\mid\tilde{0}\rangle$ represents the fundamental state
of the system.

For the case we are trying now ($N=2,\;F_{\sigma}(t)=B(t)=0$), the parameters
can be chosen in the following way

$w_{11}=\|\alpha\|^{2},\quad w_{22}=\|\beta\|^{2}\quad{\rm and}\quad
w_{12}=w^{\ast}_{21}=\alpha^{\ast}\beta,$ (5.12)

from where $w_{11}+w_{22}=1$ and $b=1.$ Thus from (4.50) we get the
$\hat{S}(t,0)$ matrix elements

$\displaystyle S_{11}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\frac{t}{2}}\;\left[\cos\left(\frac{t}{2}\right)-i\left(\|\alpha\|^{2}-\|\beta\|^{2}\right)\;\sin\left(\frac{t}{2}\right)\right]$
$\displaystyle S_{12}(t)$ $\displaystyle=$ $\displaystyle-2i\alpha^{\ast}\beta
e^{-i\frac{t}{2}}\;\sin\left(\frac{bt}{2}\right),$ $\displaystyle S_{21}(t)$
$\displaystyle=$
$\displaystyle-2i\alpha\beta^{\ast}e^{-i\frac{t}{2}}\;\sin\left(\frac{bt}{2}\right),$
$\displaystyle S_{22}(t)$ $\displaystyle=$ $\displaystyle
e^{-i\frac{t}{2}}\;\left[\cos\left(\frac{t}{2}\right)+i\left(\|\alpha\|^{2}-\|\beta\|^{2}\right)\;\sin\left(\frac{t}{2}\right)\right]$
(5.13)

and also from there we obtain the following $c_{\sigma}(t),\;\sigma=1,2,$
coefficients:

$c_{1}(t)=e^{-i\frac{t}{2}}\;\left[\cos\left(\frac{t}{2}\right)-i\left(\|\alpha\|^{2}-\|\beta\|^{2}\right)\;\sin\left(\frac{t}{2}\right)\right]\tilde{c}_{1}(0)-2i\alpha^{\ast}\beta
e^{-i\frac{t}{2}}\;\sin\left(\frac{bt}{2}\right)\tilde{c}_{2}(0)$ (5.14)

and

$c_{2}(t)=-2i\alpha\beta^{\ast}e^{-i\frac{t}{2}}\;\sin\left(\frac{bt}{2}\right)\tilde{c}_{1}(0)+e^{-i\frac{t}{2}}\;\left[\cos\left(\frac{t}{2}\right)+i\left(\|\alpha\|^{2}-\|\beta\|^{2}\right)\;\sin\left(\frac{t}{2}\right)\right]\tilde{c}_{2}(0),$
(5.15)

which according to the initial condition (5.7) verify
$c_{1}(0)=Z\alpha^{\ast}$ and $c_{2}(0)=Z\beta^{\ast}.$ This last condition,
in this case, is equivalent to $\tilde{c}_{1}(0)=Z\alpha^{\ast}$ and
$\tilde{c}_{2}(0)=Z\beta^{\ast}.$ By inserting these last constants into
(5.14) and (5.15) and simplifying the resulting expressions we get

$c_{1}(t)=Z\alpha^{\ast}e^{-it}\quad{\rm and}\quad
c_{2}(t)=Z\beta^{\ast}e^{-it}.$ (5.16)

Finally, by inserting (5.16) into (5.11) (for the particular case $N=2$) and
regrouping suitably the terms, we regain (5.10).

Let us here make some comment on the energy eigenstates and energy spectrum of
the Hamiltonian (5.5). A chain of energy eigenstates for this Hamiltonian can
be constructed in the usual way, i.e.,

$\mid
E_{n}\rangle=\mathcal{N}_{n;\alpha\beta}^{-\frac{1}{2}}\left(\hat{A}^{\dagger}_{\alpha\beta}\right)^{n}\mid
0,0\rangle,$ (5.17)

where $\mathcal{N}_{n;\alpha\beta}=\langle E_{n}\mid E_{n}\rangle,$ is a
normalization constant. As $\hat{H}_{\alpha\beta}\mid 0,0\rangle=0,$ then
according to the commutation relations (5.6) the energy spectrum of this
system is given by $E_{n}=n,\;n=0,1,2,\ldots,$ i.e., this system is
isospectral to the canonical quantum oscillator.

On the other hand, it is interesting to remark that, by means of a unitary
similarity transformation, the Hamiltonian (5.5) can be brought to a simpler
form. Indeed, if we write $\hat{H}{\alpha\beta}$ in the standard form

$\hat{H}_{\alpha\beta}=\hbar[\gamma_{0}\hat{N}+\gamma_{3}\hat{J}_{3}+\gamma_{-}\hat{J}_{+}+\gamma_{+}\hat{J}_{+}],$
(5.18)

where $\gamma_{0}=\|\alpha\|^{2}+\|\beta\|^{2}=1,$
$\gamma_{3}=\|\alpha\|^{2}-\|\beta\|^{2},$ $\gamma_{-}=\alpha^{\ast}\beta$ and
$\gamma_{+}=\alpha\beta^{\ast},$ and define
$\alpha=\|\alpha\|e^{i\theta_{\alpha}}$ and
$\beta=\|\beta\|e^{i\theta_{\beta}},$ the $su(2)$ unitary mixing operator

$\hat{T}_{\epsilon}=\exp\left[-\arctan\left(\epsilon\sqrt{\frac{1-\epsilon\gamma_{3}}{1+\epsilon\gamma_{3}}}\right)\left(e^{-i(\theta_{\alpha}-\theta_{\beta})}\hat{J}_{+}-e^{i(\theta_{\alpha}-\theta_{\beta})}\hat{J}_{-}\right)\right],\quad\epsilon=\pm
1,$ (5.19)

which disentangled form is

$\hat{T}_{\epsilon}=\exp\left[-\epsilon\sqrt{\frac{1-\epsilon\gamma_{3}}{1+\epsilon\gamma_{3}}}e^{-i(\theta_{\alpha}-\theta_{\beta})}\hat{J}_{+}\right]\exp\left[\ln\left(\frac{2}{1+\epsilon\gamma_{3}}\right)\hat{J}_{3}\right]\exp\left[\epsilon\sqrt{\frac{1-\epsilon\gamma_{3}}{1+\epsilon\gamma_{3}}}e^{i(\theta_{\alpha}-\theta_{\beta})}\hat{J}_{-}\right],\quad\epsilon=\pm
1,$ (5.20)

and whose action on the one mode annihilation operators is given by

$\hat{T}_{\epsilon}^{\dagger}\hat{a}_{1}\hat{T}_{\epsilon}=\sqrt{\frac{1+\epsilon\gamma_{3}}{2}}\hat{a}_{1}-\epsilon
e^{-i(\theta_{\alpha}-\theta_{\beta})}\sqrt{\frac{1-\epsilon\gamma_{3}}{2}}\hat{a}_{2}$
(5.21)

and

$\hat{T}_{\epsilon}^{\dagger}\hat{a}_{2}\hat{T}_{\epsilon}=\sqrt{\frac{1+\epsilon\gamma_{3}}{2}}\hat{a}_{2}+\epsilon
e^{i(\theta_{\alpha}-\theta_{\beta})}\sqrt{\frac{1-\epsilon\gamma_{3}}{2}}\hat{a}_{1},$
(5.22)

then
$\hat{T}_{1}^{\dagger}\hat{H}_{\alpha\beta}\hat{T}_{1}=\hbar\;\hat{a}_{1}^{\dagger}\hat{a}_{1}$
and
$\hat{T}_{-1}^{\dagger}\hat{H}_{\alpha\beta}\hat{T}_{-1}=\hbar\;\hat{a}_{2}^{\dagger}\hat{a}_{2}.$

##  6 Time evolution of the coherent states linked to a class of 2 D time-
dependent isotropic harmonic oscillator

Let us now consider the 2 dimensional time-dependent isotropic harmonic
oscillator system governed by the Hamiltonian

$\hat{H}_{\alpha\beta}(t)=\hbar\hat{A}^{\dagger}_{\alpha\beta}(t)\hat{A}_{\alpha\beta}(t),$
(6.1)

where the two mode annihilation operator is given by

$\hat{A}_{\alpha\beta}(t)=\alpha(t)\hat{a}_{1}+\beta(t)\hat{a}_{2},$ (6.2)

where $\alpha(t)$ and $\beta(t)$ are complex functions which verify

$\|\alpha(t)\|^{2}+\|\beta(t)\|^{2}=1,\quad\forall\;t.$ (6.3)

Due to this last fact, the operator $\hat{A}_{\alpha\beta}(t)$ and its adjoint
$\hat{A}^{\dagger}_{\alpha\beta}(t)$ satisfy the commutation relation

$\left[\hat{A}_{\alpha\beta}(t),\hat{A}^{\dagger}_{\alpha\beta}(t)\right]=\hat{I},$
(6.4)

where $\hat{I}$ is the identity operator. The explicit form of the time-
dependent Hamiltonian (6.1) is given by

$\hat{H}_{\alpha\beta}(s)=\hbar[\|\alpha(s)\|^{2}\hat{a}^{\dagger}_{1}\hat{a}_{1}+\|\beta(s)\|^{2}\hat{a}^{\dagger}_{2}\hat{a}_{2}+\alpha^{\ast}(s)\beta(s)\hat{a}^{\dagger}_{1}\hat{a}_{2}+\alpha(s)\beta^{\ast}(s)\hat{a}^{\dagger}_{2}\hat{a}_{1}],$
(6.5)

and as we have done in the previous section, we can make the following
correspondence between the parameters

$w_{11}(s)=\|\alpha(s)\|^{2},\quad w_{22}(s)=\|\beta(s)\|^{2}\quad{\rm
and}\quad w_{12}(s)=w^{\ast}_{21}(s)=\alpha^{\ast}(s)\beta(s),$ (6.6)

or if we define

$\alpha(s)=\cos(\rho(s))e^{i\theta_{\alpha}(s)}\quad{\rm
and}\quad\beta(s)=\sin(\rho(s))e^{i\theta_{\beta}(s)},$ (6.7)

we get the following description of them

$w_{11}(s)=\cos^{2}(\rho(s)),\quad w_{22}(s)=\sin^{2}(\rho(s))\quad{\rm
and}\quad
w_{12}(s)=w^{\ast}_{21}(s)=\cos(\rho(s))\sin(\rho(s))e^{i(\theta_{\beta}(s)-\theta_{\alpha}(s))},$
(6.8)

where $\rho(s),\theta_{\alpha}(s)$ and $\theta_{\beta}(s)$ are time-dependent
real functions.

Since the set of operators
$\left\\{\hat{A}_{\alpha\beta}(t_{0}),\hat{A}^{\dagger}_{\alpha\beta}(t_{0}),\hat{I},\hat{H}_{\alpha\beta}(t_{0})\right\\},\;\forall
t_{0}\geq 0,$ generate the one dimensional oscillator algebra, at the instant
$t_{0}$ we can associate to this system one parameter Schrödinger type
coherent states

$\mid
Z_{(t_{0})}\rangle=\exp\left[Z_{(t_{0})}\hat{A}^{\dagger}_{\alpha\beta}(t_{0})-Z^{\ast}_{(t_{0})}\hat{A}_{\alpha\beta}(t_{0})\right]\mid
0,0\rangle.$ (6.9)

Thus, the time evolution of these coherent states, according to (6.9 ), is
given by

$\mid
Z_{(t_{0})}(t)\rangle=e^{-\frac{i}{\hbar}\int_{t_{0}}^{t}\left[\hat{H}_{\alpha\beta}(s)\right]_{+}}\mid
Z_{(t_{0})}\rangle,$ (6.10)

where in general
$\left[\hat{H}_{\alpha\beta}(s),\hat{H}_{\alpha\beta}(\tau)\right]\neq 0$ when
$s\neq\tau.$

We note that in principle, $\mid Z_{(0)}(t_{0})\rangle\neq\mid
Z_{(t_{0})}\rangle,$ or more generally, $\mid Z_{(t_{0})}(t)\rangle\neq\mid
Z_{(t)}\rangle.$ When $t_{0}=0,$ we can write $\mid Z_{0}\rangle$ in place of
$\mid Z_{(0)}\rangle$ and $\mid Z_{0}(t)\rangle$ in place of $\mid
Z_{(0)}(t).$ In the following subsections, without loos of generality, we will
suppose that $t_{0}=0.$

###  6.1 A special case

Let us now consider the special case in which the above parameters verify the
relations

$\|w_{12}(s)\|=\epsilon\frac{\eta_{0}}{w_{0}}\frac{d\phi}{ds}(s),$ (6.11)

where $\epsilon=1$ or $\epsilon=-1$ in the region where
$\frac{d\phi}{ds}(s)>0$ or $\frac{d\phi}{ds}(s)<0,$ respectively, and

$\theta_{12}(s)=-\theta_{21}(s)=\phi(s)+\frac{\pi}{2}-\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau.$
(6.12)

Under these conditions, as we have seen above, the unitary operator present in
(6.10) can be, in principle, completely disentangled. By combining equation
(6.8) with (6.11) and (6.12), we find the following constraints on the
parameters:

$\frac{d\phi}{ds}(s)=\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho(s))$ (6.13)

and

$\theta_{\beta}(s)-\theta_{\alpha}(s)=\phi(s)+\epsilon\frac{\pi}{2}-\int_{0}^{s}\cos(2\rho(\tau))d\tau,$
(6.14)

where we have used that
$\theta_{12}(s)=\theta_{\beta}(s)-\theta_{\alpha}(s)+(1-\epsilon)\frac{\pi}{2}.$
We note that by integrating the first of these two last equations and
performing the derivative of the second we get

$\phi(t)=\frac{w_{0}}{2\;\eta_{0}}\int_{0}^{t}\sin(2\rho(s))ds+\phi(0)$ (6.15)

and

$\frac{d\theta_{\beta}}{ds}(s)-\frac{d\theta_{\alpha}}{ds}(s)=\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho(s))-\cos(2\rho(s)),$
(6.16)

which, although it represents an important restriction on the choice of
parameters, it also allows us some freedom to accommodate the relative phases.

Thus, by taking into account all these results at the moment of computing
equation (LABEL:S-matrix-elements-phi-derivative) we get

$\displaystyle S_{11}(t)$ $\displaystyle=$ $\displaystyle
e^{-\frac{it}{2}}e^{\frac{i}{2}\tilde{\Theta}_{\beta\alpha}(t)}\;\;\left[\cos\left(\tilde{\Phi}(t)\right)-i\frac{w_{0}}{\delta}\;\sin\left(\tilde{\Phi}(t)\right)\right]$
$\displaystyle S_{12}(t)$ $\displaystyle=$
$\displaystyle\;\frac{2\;\epsilon\;\eta_{0}}{\delta}e^{-\frac{it}{2}}e^{i\phi(0)}\;e^{\frac{i}{2}\tilde{\Theta}_{\beta\alpha}(t)}\;\sin\left(\tilde{\Phi}(t)\right),$
$\displaystyle S_{21}(t)$ $\displaystyle=$
$\displaystyle-\;\frac{2\;\epsilon\;\eta_{0}}{\delta}e^{-\frac{it}{2}}e^{-i\phi(0)}\;e^{-\frac{i}{2}\tilde{\Theta}_{\beta\alpha}(t)}\;\sin\left(\tilde{\Phi}(t)\right),$
$\displaystyle S_{22}(t)$ $\displaystyle=$ $\displaystyle
e^{-\frac{it}{2}}e^{-\frac{i}{2}\tilde{\Theta}_{\beta\alpha}(t)}\;\;\left[\cos\left(\tilde{\Phi}(t)\right)+i\frac{w_{0}}{\delta}\;\sin\left(\tilde{\Phi}(t)\right)\right],$

where $\phi(0)=\theta_{\beta}(0)-\theta_{\alpha}(0)-\epsilon\frac{\pi}{2},$

$\tilde{\Phi}(t)=\frac{\delta\tilde{\phi}(t)}{2w_{0}}=\frac{\delta}{4\eta_{0}}\int_{0}^{t}\sin(2\rho(s))ds$
(6.18)

and

$\tilde{\Theta}_{\beta\alpha}(t)=\tilde{\theta}_{\beta}(t)-\tilde{\alpha}_{\alpha}(t)=\int_{0}^{t}\left[\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho(\tau))-\cos(2\rho(\tau))\right]d\tau,$
(6.19)

where $\tilde{\theta}_{\alpha}(t)=\theta_{\alpha}(t)-\theta_{\alpha}(0)$ and
$\tilde{\theta}_{\beta}(t)=\theta_{\alpha}(t)-\theta_{\beta}(0).$

Hence, with these last expressions in possession we can compute the time
evolution of the coherent states (6.9). In fact, they are given by

$\mid
Z_{0}(t)\rangle=\exp\left[\sum_{\sigma=1}^{2}c_{\sigma}(t)\hat{a}^{\dagger}_{\sigma}-c^{\ast}_{\sigma}(t)\hat{a}_{\sigma}\right]\mid
0,0\rangle$ (6.20)

where

$\displaystyle c_{1}(t)=S_{11}(t)\tilde{c}_{1}(0)+S_{12}(t)\tilde{c}_{2}(0)$
(6.21) $\displaystyle
c_{2}(t)=S_{21}(t)\tilde{c}_{1}(0)+S_{22}(t)\tilde{c}_{2}(0),$ (6.22)

which must verify the initial conditions
$c_{1}(0)=Z_{0}\cos(\rho(0))e^{-i\theta_{\alpha}(0)}$ and
$c_{2}(0)=Z_{0}\sin(\rho(0))e^{-i\theta_{\beta}(0)}.$ As according to
(LABEL:S-matrix-elements-phi-derivative-2-D-time-dependent),
$S_{11}(0)=S_{22}(0)=1$ and $S_{12}(0)=S_{21}(0)=0,$ then from these last two
equations we have
$\tilde{c}_{1}(0)=Z_{0}\cos(\rho(0))e^{-i\theta_{\alpha}(0)}$ and
$\tilde{c}_{2}(0)=Z_{0}\sin(\rho(0))e^{-i\theta_{\beta}(0)}.$

Therefore

$\displaystyle
c_{1}(t)=S_{11}(t)Z_{0}\cos(\rho(0))e^{-i\theta_{\alpha}(0)}+S_{12}(t)Z_{0}\sin(\rho(0))e^{-i\theta_{\beta}(0)}$
(6.23) $\displaystyle
c_{2}(t)=S_{21}(t)Z_{0}\cos(\rho(0))e^{-i\theta_{\alpha}(0)}+S_{22}(t)Z_{0}\sin(\rho(0))e^{-i\theta_{\beta}(0)},$
(6.24)

from where

$\displaystyle c_{1}(t)$ $\displaystyle=$ $\displaystyle
e^{-\frac{it}{2}}e^{\frac{i}{2}\tilde{\Theta}_{\beta\alpha}(t)}\;Z_{0}\cos(\rho(0))e^{-i\theta_{\alpha}(0)}$
(6.25) $\displaystyle\times$
$\displaystyle\left[\cos\left(\tilde{\Phi}(t)\right)-i\frac{w_{0}}{\delta}\;\left(1+\frac{2\eta_{0}}{w_{0}}\tan(\rho(0))\right)\;\sin\left(\tilde{\Phi}(t)\right)\right]$

and

$\displaystyle c_{2}(t)$ $\displaystyle=$ $\displaystyle
e^{-\frac{it}{2}}e^{-\frac{i}{2}\tilde{\Theta}_{\beta\alpha}(t)}\;Z_{0}\sin(\rho(0))e^{-i\theta_{\beta}(0)}$
(6.26) $\displaystyle\times$
$\displaystyle\left[\cos\left(\tilde{\Phi}(t)\right)+i\frac{w_{0}}{\delta}\;\left(1-\frac{2\eta_{0}}{w_{0}}\cot(\rho(0))\right)\;\sin\left(\tilde{\Phi}(t)\right)\right].$

Finally, inserting these last coefficients in (5.11) (for N=2), we obtain the
time evolution of the coherent states associated to the 2 D time-independent
isotropic quantum oscillator Hamiltonian under the action of a special class
of extended time-dependent 2 D isotropic quantum oscillator Hamiltonian:

$\mid
Z_{0}(t)\rangle=\exp\left[c_{1}(t)\hat{a}^{\dagger}_{1}-c^{\ast}_{1}(t)\hat{a}_{1}\right]\exp\left[c_{2}(t)\hat{a}^{\dagger}_{2}-c^{\ast}_{2}(t)\hat{a}_{2}\right]\mid
0,0\rangle.$ (6.27)

We note that these coherent states are eigenstates of
$\hat{A}_{\alpha\beta}(0)=\cos(\rho(0))e^{i\theta_{\alpha}(0)}\hat{a}_{1}+\sin(\rho(0))e^{i\theta_{\beta}(0)}\hat{a}_{2}.$
Indeed, its action on the time-dependent coherent states is given by

$\displaystyle\hat{A}_{\alpha\beta}(0)\mid Z_{0}(t)\rangle$ $\displaystyle=$
$\displaystyle\left[c_{1}(t)\cos(\rho(0))e^{i\theta_{\alpha}(0)}+c_{2}(t)\sin(\rho(0))e^{i\theta_{\beta}(0)}\right]\;\mid
Z_{0}(t)\rangle$ (6.28) $\displaystyle=$ $\displaystyle
Z_{0}e^{-\frac{it}{2}}\;\left\\{\left[\cos\left(\frac{\tilde{\Theta}_{\beta\alpha}(t)}{2}\right)\cos\left(\tilde{\Phi}(t)\right)+\frac{w_{0}}{\delta}\sin\left(\frac{\tilde{\Theta}_{\beta\alpha}(t)}{2}\right)\sin\left(\tilde{\Phi}(t)\right)\right]\right.$
$\displaystyle+$ $\displaystyle
i\left[\sin\left(\frac{\tilde{\Theta}_{\beta\alpha}(t)}{2}\right)\cos\left(\tilde{\Phi}(t)\right)\cos(2\rho(0))-\frac{w_{0}}{\delta}\left(\cos(2\rho(0))+\frac{2\eta_{0}}{w_{0}}\sin(2\rho(0))\right)\right.$
$\displaystyle\times$
$\displaystyle\left.\left.\cos\left(\frac{\tilde{\Theta}_{\beta\alpha}(t)}{2}\right)\sin\left(\tilde{\Phi}(t)\right)\right]\right\\}\;\mid
Z_{0}(t)\rangle.$

On the other hand, we also have that these coherent states are eigenstates of
the generalized time-dependent lowering operator

$\hat{A}_{\left(\frac{c^{\ast}_{1}}{Z_{0}^{\ast}}\;\frac{c^{\ast}_{2}}{Z_{0}^{\ast}}\right)}(t)=\frac{c^{\ast}_{1}(t)}{Z_{0}^{\ast}}\hat{a}_{1}+\frac{c^{\ast}_{2}(t)}{Z_{0}^{\ast}}\hat{a}_{2}$
(6.29)

with associated eigenvalue equal to $Z_{0}.$ Indeed, using the properties of
the $\hat{S}$ matrix elements, the definition of the $c$‘s coefficients in
terms of them and the initial conditions that these coefficients verify, or
calculating directly from (5.16) and (6.26), we deduce that
$\|c_{1}(t)\|^{2}+\|c_{2}(t)\|^{2}=\|Z_{0}\|^{2},$ then

$\hat{A}_{\left(\frac{c^{\ast}_{1}}{Z_{0}^{\ast}}\;\frac{c^{\ast}_{2}}{Z_{0}^{\ast}}\right)}(t)\mid
Z_{0}(t)\rangle=\left(\frac{c^{\ast}_{1}(t)}{Z_{0}^{\ast}}c_{1}(t)+\frac{c^{\ast}_{2}(t)}{Z_{0}^{\ast}}c_{2}(t)\right)\mid
Z_{0}(t)\rangle=Z_{0}\mid Z_{0}(t)\rangle.$ (6.30)

Let us notice that we would have reached the same conclusion if we had worked
on Heinsenberg‘s picture.

Moreover, the action of the ladder operator (6.2) on the coherent states $\mid
Z_{0}(t)\rangle$ is given by

$\hat{A}_{\alpha\beta}(t)\mid Z_{0}(t)\rangle=\tilde{Z}_{(t)}\mid
Z_{0}(t)\rangle,$ (6.31)

where

$\displaystyle\tilde{Z}_{(t)}$ $\displaystyle=$ $\displaystyle
Z_{0}e^{-\frac{it}{2}}e^{\frac{i}{2}\left(\tilde{\alpha}(t)+\tilde{\beta}(t)\right)}$
(6.32) $\displaystyle\times$
$\displaystyle\left\\{\cos\left(\tilde{\Phi}(t)\right)\cos\left(\rho(t)-\rho(0)\right)\right.$
$\displaystyle-$
$\displaystyle\left.\frac{iw_{0}}{\delta}\left[\cos\left(\rho(0)+\rho(t)\right)+\frac{2\eta_{0}}{w_{0}}\sin\left(\rho(0)+\rho(t)\right)\right]\sin\left(\tilde{\Phi}(t)\right)\right\\},$

which can be expressed in a simpler way if we choose
$\tilde{\beta}(t)=-\tilde{\alpha}(t).$

####  6.1.1 Case $\rho(s)=\rho_{0},$ linear phases

When $\rho(s)=\rho_{0},$ where $\frac{\pi}{2}>\rho_{0}>0,$ from (6.18) and
(6.19) we deduce

$\tilde{\Phi}(t)=\frac{\delta}{4\eta_{0}}\sin(2\rho_{0})\;t$ (6.33)

and

$\tilde{\Theta}_{\beta\alpha}(t)=\tilde{\theta}_{\beta}(t)-\tilde{\alpha}_{\alpha}(t)=\left[\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho_{0})-\cos(2\rho_{0})\right]\;t.$
(6.34)

Assuming that $\tilde{\theta}_{\alpha}(t)=-\tilde{\beta}_{\alpha}(t),$ then
the two-mode annihilator (6.2) takes the form

$\displaystyle\hat{A}_{\alpha\beta}(t)$ $\displaystyle=$
$\displaystyle\cos(\rho_{0})\exp\left[i\theta_{\alpha}(0)-\frac{it}{2}\left(\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho_{0})-\cos(2\rho_{0})\right)\right]\;\hat{a}_{1}$
(6.35) $\displaystyle+$
$\displaystyle\sin(\rho_{0})\exp\left[i\theta_{\beta}(0)+\frac{it}{2}\left(\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho_{0})-\cos(2\rho_{0})\right)\right]\;\hat{a}_{2},$

which evaluated in $t=0$ is equal to
$\hat{A}_{\alpha\beta}(0)=\cos(\rho_{0})e^{i\theta_{\alpha}(0)}\;\hat{a}_{1}+\sin(\rho_{0})e^{i\theta_{\beta}(0)}\;\hat{a}_{2}.$
Then the time-evolution of the coherent states $\mid
Z_{0}\rangle=\exp[Z_{0}\hat{A}^{\dagger}_{\alpha\beta}(0)-Z_{0}^{\ast}\hat{A}_{\alpha\beta}(0)]\mid
0,0\rangle,$ which are eigenstates of $\hat{A}_{\alpha\beta}(0)$ with
associated eigenvalue equal to $Z_{0},$ under the action of the time-ordered
unitary operator governed by the Hamiltonian
$\hat{H}_{\alpha\beta}(s)=\hat{A}^{\dagger}_{\alpha\beta}(s)\hat{A}_{\alpha\beta}(s),$
is represented by the time-dependent coherent states (5.11), where

$\displaystyle c_{1}(t)$ $\displaystyle=$ $\displaystyle
Z_{0}e^{-\frac{it}{2}}\cos(\rho(0))\exp\left[-i\theta_{\alpha}(0)+\frac{it}{2}\left(\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho_{0})-\cos(2\rho_{0})\right)\right]\;$
(6.36) $\displaystyle\times$
$\displaystyle\left[\cos\left(\frac{\delta}{4\eta_{0}}\sin(2\rho_{0})\;t\right)-i\frac{w_{0}}{\delta}\;\left(1+\frac{2\eta_{0}}{w_{0}}\tan(\rho_{0})\right)\;\sin\left(\frac{\delta}{4\eta_{0}}\sin(2\rho_{0})\;t\right)\right]$

and

$\displaystyle c_{2}(t)$ $\displaystyle=$ $\displaystyle
Z_{0}e^{-\frac{it}{2}}\sin(\rho(0))\exp\left[-i\theta_{\beta}(0)-\frac{it}{2}\left(\frac{w_{0}}{2\;\eta_{0}}\sin(2\rho_{0})-\cos(2\rho_{0})\right)\right]$
(6.37) $\displaystyle\times$
$\displaystyle\left[\cos\left(\frac{\delta}{4\eta_{0}}\sin(2\rho_{0})\;t\right)+i\frac{w_{0}}{\delta}\;\left(1-\frac{2\eta_{0}}{w_{0}}\cot(\rho_{0})\right)\;\sin\left(\frac{\delta}{4\eta_{0}}\sin(2\rho_{0})\;t\right)\right].$

Finally, the action of the operator (6.2) on the time-dependent coherent
states is given by

$\displaystyle\hat{A}_{\alpha\beta}(t)\mid Z_{0}(t)\rangle$ $\displaystyle=$
$\displaystyle
Z_{0}e^{-\frac{it}{2}}\left\\{\cos\left(\frac{\delta}{4\eta_{0}}\sin(2\rho_{0})\;t\right)\right.$
(6.38) $\displaystyle-$
$\displaystyle\left.\frac{iw_{0}}{\delta}\left[\cos\left(2\rho_{0}\right)+\frac{2\eta_{0}}{w_{0}}\sin\left(2\rho_{0}\right)\right]\sin\left(\frac{\delta}{4\eta_{0}}\sin(2\rho_{0})\;t\right)\right\\}\mid
Z_{0}(t)\rangle.$

We notice that if we choose $\frac{2\eta_{0}}{w_{0}}=\tan(2\rho_{0}),$ where
$\frac{\pi}{4}\geq\rho>0,$ then
$\hat{A}_{\alpha\beta}(t)=\hat{A}_{\alpha\beta}(0),$ i.e., the time-dependent
coherent states $\mid Z_{0}(t)\rangle$ are eigenstates of
$\hat{A}_{\alpha\beta}(0)$ with associated eigenvalue equal to $Z_{0}e^{-it},$
as we have seen in section 5.

####  6.1.2 Logarithmic phases

Let us now particularize the results of the previous subsection to the case
where $\rho(s)=\arctan(t_{0}+s),\;t_{0}>0,s>0.$ Then using (6.7) we get

$\alpha(s)=\frac{1}{\sqrt{1+(s+t_{0})^{2}}}e^{i\theta_{\alpha}(s)}\quad\text{
and}\quad\beta(s)=\frac{s+s_{0}}{\sqrt{1+(s+t_{0})^{2}}}e^{i\theta_{\beta}(s)},$
(6.39)

from where we deduce

$w_{11}(s)=\frac{1}{1+(s+t_{0})^{2}},\quad
w_{22}=\frac{(s+t_{0})^{2}}{1+(s+t_{0})^{2}}\quad\text{
and}\quad\|w_{12}\|=\|w_{21}\|=\frac{s+t_{0}}{1+(s+t_{0})^{2}}.$ (6.40)

Moreover, from (6.18) and (6.19) , we can compute the phases

$\tilde{\Phi}(s)=\frac{\delta}{4\eta_{0}}\ln\left(\frac{1+(s+t_{0})^{2}}{1+t^{2}_{0}}\right)$
(6.41)

and

$\tilde{\Theta}_{\beta\alpha}(s)=\frac{w_{0}}{2\eta_{0}}\ln\left(\frac{1+(s+t_{0})^{2}}{1+t^{2}_{0}}\right)+s+2\arctan(t_{0})-2\arctan(s+t_{0}).$
(6.42)

Then, again, if we choose
$\tilde{\theta}_{\alpha}(s)=-\tilde{\theta}_{\beta}(s),$ the time-dependent
ladder operator writes

$\displaystyle\hat{A}_{\alpha\beta}(s)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{1+t_{0}^{2}}}\exp\left[i\theta_{\alpha}(0)-\frac{i}{2}\left(\frac{w_{0}}{2\eta_{0}}\ln\left(\frac{1+(s+t_{0})^{2}}{1+t^{2}_{0}}\right)+s+2\arctan(t_{0})-2\arctan(s+t_{0})\right)\right]\;\hat{a}_{1}$
$\displaystyle+$
$\displaystyle\frac{t_{0}}{\sqrt{1+t_{0}^{2}}}\exp\left[i\theta_{\beta}(0)+\frac{i}{2}\left(\frac{w_{0}}{2\eta_{0}}\ln\left(\frac{1+(s+t_{0})^{2}}{1+t^{2}_{0}}\right)+s+2\arctan(t_{0})-2\arctan(s+t_{0})\right)\right]\;\hat{a}_{2}.$

The time-dependent coherent states are obtained from equation (5.11), with the
$c_{\sigma}(t),\sigma=1,2$ given by

$\displaystyle c_{1}(t)$ $\displaystyle=$ $\displaystyle
e^{-\frac{it}{2}}\exp\left[i\frac{t}{2}+\frac{i\delta}{4w_{0}}\ln\left(\frac{1+(t+t_{0})^{2}}{1+t_{0}^{2}}\right)-i\left(\arctan(t+t_{0})-\arctan(t_{0})\right)\right]\;Z_{0}\frac{1}{\sqrt{1+t_{0}^{2}}}e^{-i\theta_{\alpha}(0)}{\sqrt{1+t_{0}^{2}}}$
$\displaystyle\times$
$\displaystyle\left[\cos\left(\frac{\delta}{4w_{0}}\ln\left(\frac{1+(t+t_{0})^{2}}{1+t_{0}^{2}}\right)\right)-i\frac{w_{0}}{\delta}\;\left(1+\frac{2\eta_{0}\;t_{0}}{w_{0}}\right)\;\sin\left(\frac{\delta}{4w_{0}}\ln\left(\frac{1+(t+t_{0})^{2}}{1+t_{0}^{2}}\right)\right)\right]$

and

$\displaystyle c_{2}(t)$ $\displaystyle=$ $\displaystyle
e^{-\frac{it}{2}}\exp\left[-i\frac{t}{2}-\frac{i\delta}{4w_{0}}\ln\left(\frac{1+(t+t_{0})^{2}}{1+t_{0}^{2}}\right)+i\left(\arctan(t+t_{0})-\arctan(t_{0})\right)\right]\;Z_{0}\frac{t_{0}}{\sqrt{1+t_{0}^{2}}}e^{-i\theta_{\beta}(0)}$
$\displaystyle\times$
$\displaystyle\left[\cos\left(\frac{\delta}{4w_{0}}\ln\left(\frac{1+(t+t_{0})^{2}}{1+t_{0}^{2}}\right)\right)+i\frac{w_{0}}{\delta}\;\left(1-\frac{2\eta_{0}}{w_{0}\;t_{0}}\right)\;\sin\left(\frac{\delta}{4w_{0}}\ln\left(\frac{1+(t+t_{0})^{2}}{1+t_{0}^{2}}\right)\right)\right],$

where $\theta_{\beta}(0)$ and $\theta_{\alpha}(0)$ are linked by the relation
$\theta_{\beta}(0)=\theta_{\alpha}(0)+\phi(0)+\frac{\pi}{2}.$

##  7 Another equivalent disentangling order

Another equivalent disentangling order for the unitary operator (3.17) is,
first isolating an exponential factor raised to a factor proportional to
$\hat{N},$ then continuing with an exponential factor raised to a factor which
is proportional the raising operator $\hat{J}_{+},$ to continue with an
exponential operator raised to a factor which is proportional to
$\hat{J}_{3},$ and finally leave an exponential factor raised to a factor
which is proportional to the lowering operator $\hat{J}_{-}$ absolutely to the
right. Although, as we have already mention in previous section, by following
a new order of disentangling of the operators we will obtain the same results,
the equation we have to solve to reach them appear in a slightly different
form, which could allow to visualize more easily the possible resolution
methods.

Then applying the Feynman’s disentangling rule in that order we have

$\hat{U}_{0}(t,0)=\exp\left[-\;i\;\int_{0}^{t}(w_{11}(s)+w_{22}(s))ds\hat{N}\right]\exp\left[\tilde{\Lambda}(t)\hat{J}_{+}\right]\exp\left[\tilde{\Omega}(t)\hat{J}_{3}\right]\exp\left[\tilde{\Gamma}(t)\hat{J}_{-}\right],$
(7.1)

where $\tilde{\Lambda}(s),\tilde{\Omega}(s)$ and $\tilde{\Gamma}$(s) verify

$i\frac{d\tilde{\Lambda}}{ds}(s)=w_{12}(s)+(w_{11}(s)-w_{22}(s))\tilde{\Lambda}(s)-w_{21}(s)\tilde{\Lambda}^{2}(s),\quad\tilde{\Lambda}(0)=0,$
(7.2)
$i\frac{d\tilde{\Omega}}{ds}(s)=(w_{11}(s)-w_{22}(s))-2w_{21}(s)\tilde{\Lambda}(s),\quad\tilde{\Omega}(0)=0,$
(7.3)

and

$\tilde{\Gamma}(s)=-i\int_{0}^{s}w_{21}(\tau)e^{\tilde{\Omega}(\tau)}\;d\tau,$
(7.4)

respectively.

We note that by making
$\tilde{\Lambda}(s)=\exp\left[-i\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau\right]\Lambda(s),$
equations (7.3) and (7.2) become

$i\frac{d\tilde{\Omega}}{ds}(s)=(w_{11}(s)-w_{22}(s))+2i\eta^{\ast}(s)\Lambda(s),\quad\tilde{\Omega}(0)=0,$
(7.5)

and

$\frac{d\Lambda}{ds}(s)=\eta(s)+\eta^{\ast}(s)\Lambda^{2}(s),\quad\Lambda(0)=0,$
(7.6)

respectively, where
$\eta(s)=-iw_{12}(s)e^{i\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau},$ i.e.,
we have to solve the same differential equation as in the previous sections.

Also we observe that (7.6) can be transformed into a ordinary second order
linear differential equation by making

$\Lambda(s)=-\frac{\dot{u}(s)}{u(s)\eta^{\ast}(s)},$ (7.7)

that is,

$\ddot{u}(s)-\frac{\dot{\eta^{\ast}}(s)}{\eta^{\ast}(s)}\dot{u}(s)+\|\eta(s)\|^{2}u(s)=0,\quad\dot{u}(0)=0,$
(7.8)

where $\dot{u}(s)\equiv\frac{du}{ds}(s),$ etc. On the other hand, by inserting
(7.7) into equation (7.5) and then integrating, we get

$\tilde{\Omega}(s)=-2\ln u(s)-i\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau.$
(7.9)

In the same way, inserting this last expression into equation (7.4), after a
little handling of the equations, we get

$\tilde{\Gamma}(s)=-\int_{0}^{s}\frac{\eta^{\ast}(\tau)}{u^{2}(\tau)}d\tau.$
(7.10)

For example, if $\eta(s)=\eta_{0}e^{-i\theta_{0}s^{2}},$ where $\eta_{0}$ and
$\theta_{0}$ are real constants verifying $\eta_{0}>0,$ this last equation
takes the form

$\ddot{u}(s)-2i\theta_{0}s\dot{u}(s)+\eta_{0}^{2}u(s)=0,\quad\dot{u}(0)=0,$
(7.11)

which solution in terms of the Kummer confluent hypergeometric function is

$u(s)={}_{1}F_{1}[i\frac{\eta_{0}^{2}}{4\theta_{0}},\frac{1}{2},i\theta_{0}s^{2}].$
(7.12)

As

$\dot{u}(s)=-\eta_{0}^{2}\;s\;{}_{1}F_{1}[1+i\frac{\eta_{0}^{2}}{4\theta_{0}},\frac{3}{2},i\theta_{0}s^{2}],$
(7.13)

from the above equations, going back to the original functions we finally get

$\tilde{\Lambda}(t)=\eta_{0}\;t\;\frac{{}_{1}F_{1}[1+i\frac{\eta_{0}^{2}}{4\theta_{0}},\frac{3}{2},i\theta_{0}t^{2}]}{{}_{1}F_{1}[i\frac{\eta_{0}^{2}}{4\theta_{0}},\frac{1}{2},i\theta_{0}t^{2}]}\exp\left[-i\left(\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds+\theta_{0}t^{2}\right)\right],$
(7.14)

$\tilde{\Omega}(t)=-2\ln\left({}_{1}F_{1}[i\frac{\eta_{0}^{2}}{4\theta_{0}},\frac{1}{2},i\theta_{0}t^{2}]\right)-i\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds$
(7.15)

and

$\tilde{\Gamma}(t)=-\eta_{0}\int_{0}^{t}\frac{e^{i\theta_{0}s^{2}}\;ds}{\left({}_{1}F_{1}[i\frac{\eta_{0}^{2}}{4\theta_{0}},\frac{1}{2},i\theta_{0}s^{2}]\right)^{2}}.$
(7.16)

We note that, at first sight, a closed expression for the integral in (7.16)
seems not to be easy to perform, then at first instance, an expansion of it in
Taylor series would be recommended.

###  7.1 One more transformation

We observe that the second order linear differential equation (7.8) can be
expressed as a first order differential equation system by making the change
of variable $\dot{u}(s)=\eta^{\ast}(s)v(s),$ where $v(s)$ is an arbitrary
function of $s$ which verify $v(0)=0,$ i.e.,

$\dot{u}(s)-\eta^{\ast}(s)v(s)=0\quad\text{and}\quad\dot{v}(s)+\eta(s)u(s)=0,\quad\dot{u}(0)=v(0)=0.$
(7.17)

Thus, by comparing these last relations, assuming that
$\lim_{s\;\rightarrow\;0}\frac{\dot{u}(s)}{v(s)}$ exist and is equal to
$\eta^{\ast}(0),$ we can write

$\frac{\dot{v}(s)}{u(s)}=-\frac{\dot{u}^{\ast}(s)}{v^{\ast}(s)},\quad\leftrightarrow\quad
u(s)\dot{u}^{\ast}(s)+\dot{v}(s)v^{\ast}(s)=0.$ (7.18)

Then, if we express the functions $u(s)$ and $v(s)$ in the polar form, i.e.,
$u(s)=u_{0}(s)e^{i\theta_{u}(s)}$ and $v(s)=v_{0}(s)e^{i\theta_{v}(s)},$ and
insert them into this last equation, then by separating the real and imaginary
part of the resulting equation, we get the following system of differential
equations:

$\frac{1}{2}\frac{d}{ds}\left(u_{0}^{2}(s)+v_{0}^{2}(s)\right)=0,\quad\text{and}\quad
v_{0}^{2}(s)\dot{\theta_{v}}(s)-u_{0}^{2}(s)\dot{\theta_{u}}(s)=0.$ (7.19)

Thus, by integrating the first of these equations, and taking into account the
initial conditions, we obtain

$u_{0}(s)=\mid\cos[\varrho(s)]\mid\quad\text{and}\quad
v_{0}(s)=\mid\sin[\varrho(s)]\mid,\quad\varrho(0)=0,$ (7.20)

and then the second of these equations becomes

$\dot{\theta}_{u}(s)=\tan^{2}[\varrho(s)]\dot{\theta}_{v}(s),$ (7.21)

where we must add the constraint

$\eta(s)=-\left(\frac{\dot{v}_{0}(s)}{u_{0}(s)}+i\dot{\theta}_{v}(s)\frac{v_{0}(s)}{u_{0}(s)}\right)\exp\left[i\left(\theta_{v}(s)-\theta_{u}(s)\right)\right]=-iw_{12}(s)e^{i\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau},$
(7.22)

which is separated in two when comparing the norms and phases:

$\|w_{12}(s)\|^{2}=\dot{\varrho}^{2}(s)+\tan^{2}\left(\varrho(s)\right)\dot{\theta}^{2}_{v}(s)$
(7.23)

and

$\arctan\left(\frac{\dot{\theta}_{v}(s)}{\dot{\varrho}(s)}\tan\left(\varrho(s)\right)\right)+\theta_{v}(s)-\theta_{u}(s)=\frac{\pi}{2}+\theta_{12}(s)+\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau,\quad\dot{\varrho}(s)\neq
0.$ (7.24)

In terms of the new variables we have

$\displaystyle\tilde{\Lambda}(t)$ $\displaystyle=$
$\displaystyle-\frac{v(t)}{u(t)}\exp\left[-i\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds\right]$
(7.25) $\displaystyle=$
$\displaystyle-\left|\tan\left((\varrho(t)\right)\right|\exp[i(\theta_{v}(t)-\theta_{u}(t))]\exp\left[-i\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds\right],$

$\tilde{\Omega}(t)=\ln\left(\sec^{2}({\varrho}(t))\right)-2i\theta_{u}(t)-i\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds,\quad\theta_{u}(0)=0,$
(7.26)

and

$\tilde{\Gamma}(t)=\int_{0}^{t}\|w_{12}(s)\|\left(1+\tan^{2}\left(\varrho(s)\right)\right)\exp\left[-i\left[\arctan\left(\frac{\dot{\theta}_{v}(s)}{\dot{\varrho}(s)}\tan\left(\varrho(s)\right)\right)+\theta_{v}(s)+\theta_{u}(s)\right]\right]ds.$
(7.27)

####  7.1.1 Case $\theta_{u}(s)=0$

When $\theta_{u}(s)=0,$ from (7.19) we deduce that
$\theta_{v}(s)=\theta_{v}^{0}$ is constant. Then, from (7.23) we
get444Assuming that $\varrho(s)$ is an increasing function of $s.$

$\dot{\varrho}(s)=\|w_{12}(s)\|,\quad\text{or}\quad{\varrho}(s)=\int_{0}^{s}\|w_{12}(\tau)\|d\tau.$
(7.28)

Thus, by inserting this last result into equations (7.25), (7.26) and (7.27),
we obtain

$\tilde{\Lambda}(t)=-\left|\tan\left(\int_{0}^{s}\|w_{12}(s)\|ds\right)\right|\exp[i\theta_{v}^{0}]\exp\left[-i\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds\right],$
(7.29)

$\tilde{\Omega}(t)=\ln\left(\sec^{2}({\varrho}(t))\right)-i\int_{0}^{t}(w_{11}(s)-w_{22}(s))ds,$
(7.30)

and

$\tilde{\Gamma}(t)=\left|\tan\left[\int_{0}^{t}\|w_{12}(s)\|ds\right]\right|\exp\left[-i\theta_{v}^{0}\right],$
(7.31)

respectively, where the following constraint must be respected

$\theta_{v}^{0}=\frac{\pi}{2}+\theta_{12}(s)+\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau.$
(7.32)

####  7.1.2 Case $\dot{\theta}_{v}(s)=\dot{\varrho}(s)$ and Fresnel type
solutions

When $\dot{\theta_{v}}(s)=\dot{\varrho}(s),$ i.e, when
$\theta_{v}(s)=\varrho(s)+\theta_{v}^{0},$ from (7.19) we deduce

$\dot{\theta}_{u}(s)=\tan^{2}\left(\varrho(s)\right)\dot{\varrho}(s),$ (7.33)

which implies that

$\theta_{u}(s)=\tan\left(\varrho(s)\right)-\varrho(s)+\theta_{u}^{0}.$ (7.34)

Thus, inserting the results into (7.23), we get

$\dot{\varrho}(s)\sec\left(\varrho(s)\right)=\pm\|w_{12}(s)\|,$ (7.35)

whose solution is given by

$\varrho(s)=2\arctan\left\\{\tanh\left[\pm\frac{1}{2}\left(\int_{1}^{s}\|w_{12}(\tau)\|d\tau-\int_{1}^{0}\|w_{12}(\tau)\|d\tau\right)\right]\right\\}.$
(7.36)

Therefore, in this case, the constraint (7.24) writes

$3\varrho(s)-\tan\left(\varrho(s)\right)+\theta_{v}^{0}-\theta_{u}^{0}=\frac{\pi}{2}+\theta_{12}(s)+\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau.$
(7.37)

For example, choosing $\|w_{12}(s)\|=w_{12}^{0}|\cos\left(\nu s^{2}\right)|,$
where $w_{12}^{0}$ and $\nu$ are non-negative real constants, we get

$\varrho(s)=2\arctan\left\\{\tanh\left[\frac{w_{12}^{0}\sqrt{\pi}\sqrt{1+\cos(2\nu
s^{2})}\mathbb{F}_{C}\left(\sqrt{\nu}\sqrt{\frac{2}{\pi}}s\right)\sec(\nu
s^{2})}{4\sqrt{\nu}}\right]\right\\},$ (7.38)

where

$\mathbb{F}_{C}(x)=\int_{0}^{x}\cos(\frac{\pi}{2}\tau^{2})d\tau=\sqrt{\frac{2}{\pi}}\sum_{n=0}^{\infty}(-1)^{n}\frac{(\sqrt{\frac{\pi}{2}}x)^{4n+1}}{(2n)!(4n+1)},$
(7.39)

is the Fresnel cosinus integral [18].

Finally, by inserting these results into equations (7.25), (7.26) and (7.27),
we can obtain the desired disentangling of the unitary operator (3.17), under
the restrictions imposed by equation (7.37).

####  7.1.3 An additional trigonometric reparameterization

Looking at equation (7.23), we observe that the additional trigonometric
reparameterization

$\dot{\varrho}(s)=\|w_{12}(s)\|\cos(\Theta(s))\quad\text{and}\quad\tan(\varrho(s)\dot{\theta}_{v}(s)=\|w_{12}(s)\|\sin(\Theta(s)),$
(7.40)

where $\Theta(s)$ is a real function555If we suppose that
$\dot{\varrho}(s)>0,$ then we can choose
$-\frac{\pi}{2}<\Theta(s)<\frac{\pi}{2}.$of $s,$ could help us to better
visualize the global structure of the final disentangled form of the unitary
time evolution operator. In terms of this new function $\Theta(s),$ equation
(7.23) verifies automatically, whereas equations (7.21) and (7.24) become

$\dot{\theta}_{u}(s)=\|w_{12}(s)\|\tan(\varrho(s))\sin(\Theta(s)),$ (7.41)

and

$\Theta(s)+\theta_{v}(s)-\theta_{u}(s)=\frac{\pi}{2}+\theta_{12}(s)+\int_{0}^{s}(w_{11}(\tau)-w_{22}(\tau))d\tau,\quad\dot{\varrho}(s)\neq
0,$ (7.42)

respectively. For example, when $\dot{\varrho}(s)\neq 0,$ the function
$\tilde{\Gamma}(t)$ in (7.27) assumes the form

$\tilde{\Gamma}(t)=\int_{0}^{t}\|w_{12}(s)\|\sec^{2}\left[\varrho(s)\right]\exp\left[-i\left[\Theta(s)+\theta_{v}(s)+\theta_{u}(s)\right]\right]ds,$
(7.43)

which can be directly integrated with the help of equations (7.40 ) and
(7.41), we get

$\tilde{\Gamma}(t)=\left|\tan\left(\varrho(t)\right)\right|\exp\left[-i\left(\theta_{v}(t)+\theta_{u}(t)\right)\right].$
(7.44)

So, proceeding in this way, we can generate disentangled expressions for the
original unitary time evolution operator. As we have mentioned before, we
cannot solve the original Riccati differential equation for any value of the
time-dependent $w$ functions, nor we can do it with the alternative
transformed versions of this equation. However, we have proven here that this
is possible when there are appropriated connections among these parameters.

##  8 Conclusions

In this article we have used the Feynman disentangling operator rules to
disentangle the time evolution operator of a system which Hamiltonian is an
element of the complex $\left\\{h(1)\oplus
h(1)\right\\}\mathbin{\mathchoice{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\displaystyle\displaystyle
3
\roundcap\put(0.5,0.5){\makebox(0.0,0.0)[]{\raisebox{0.83333pt}{$\displaystyle+$}}}
\put(0.5,0.5){\arc[-90,90]{0.5}}
\end{picture}}}}{\makebox[9.99997pt]{\raisebox{-0.83333pt}{\begin{picture}(1.0,1.0)\textstyle\displaystyle
3
\roundcap\put(0.5,0.5){\makebox(0.0,0.0)[]{\raisebox{0.83333pt}{$\textstyle+$}}}
\put(0.5,0.5){\arc[-90,90]{0.5}}
\end{picture}}}}{\makebox[7.66663pt]{\raisebox{-0.58333pt}{\begin{picture}(1.0,1.0)\scriptstyle\displaystyle
3
\roundcap\put(0.5,0.5){\makebox(0.0,0.0)[]{\raisebox{0.58333pt}{$\scriptstyle+$}}}
\put(0.5,0.5){\arc[-90,90]{0.5}}
\end{picture}}}}{\makebox[6.11107pt]{\raisebox{-0.41666pt}{\begin{picture}(1.0,1.0)\scriptscriptstyle\displaystyle
3
\roundcap\put(0.5,0.5){\makebox(0.0,0.0)[]{\raisebox{0.41666pt}{$\scriptscriptstyle+$}}}
\put(0.5,0.5){\arc[-90,90]{0.5}} \end{picture}}}}}u(2)$ composed Lie algebra.
For such a purpose, we have taken advantage of the two-boson representation
space of this algebra. In particular, we have written some generators in terms
of the generators of the $su(2)$ algebra in the two-boson Schwinger
representation. We have proven that the complete disentangling of that time
evolution operator depends exclusively on our capacity to solve a standard
Riccati-type equation with time dependent coefficients, which are related
directly with the time-dependent frequencies, or set of coefficients, of the
$u(2)$ expansion part of the original Hamiltonian. In other words, in the case
$N=2,$ we have proved that the conditions to disentangle the temporal
evolution operator are reduced to determining the appropriate combination of
frequency values that allow us to solve the Riccati-type differential
equation. All that because we know that not always is possible to have a
closed solution for such non-linear differential equation. Nevertheless, we
have found some relations among the frequencies that allow us to solve that
differential equation and disentangle completely the unitary time evolution
operator. By the way, these relations are not the only ones that work. Here,
in section 7, we have recalled some others that can lead us to solutions in
terms of hypergeometric or Fresnel functions, for example.

Another important fact we have illustrated in this article is that once that
the disentangling of the $u(2)$ sector of the unitary evolution operator has
been done, the matrix elements of this same disentangled factor can be used to
complete the disentangling of the $h(1)\oplus h(1)$ sector, no matter the
representation we use to compute these matrix elements. That because it have
been shown that the differential equation system, which serve to untangle the
exponential factor containing the linear terms in the unitary time evolution
operator, can be expressed as a time ordered operator equation. It is in this
sense that we say that the Feynman disentanglement method is self-consistent
in dealing with this problem.

By way of example, we have computed the time evolution of a class of coherent
states linked to the 2 D isotropic harmonic oscillator as well as to a class
of coherent states linked to a system of two time-dependent $u(2)$ interacting
oscillators. This study could be of some interest from the fact that here we
have shown that we can associate the same set of coherent states to two
quantum systems whose Hamiltonians are not connected by a unitary
transformation.

## Acknowledgments

This article is dedicated to the memory of Dr. Luciano Edmundo Laroze Barrios,
teacher at Department of Physics, Universidad Técnica Fedérico Santa María,
Valparaíso, Chile: Teacher, could you be my thesis supervisor? Neither a
solemn yes nor a resounding no for an answer, just a big smile and body
movements launched into space, imitating the gestures of a marine diver
submerged in the ocean and saying out loud, we will dive in time! July 1988.

## References

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##  Appendix A Solving a differential equation system by using the concept of
time ordering operator

The untangled closed form of the general time evolution operator of the system
we are dealing with will not be possible until the solution of the
differential equation system (3.12) is available. I have been shown in the
literature, see [11], that (3.12) can be solved for any number of interacting
oscillators using the concept of time ordering operator. Indeed, let us
consider a complete orthonormal set of states
$\left\\{\mid\psi_{\sigma}\rangle\right\\}$ where the functions
$c_{\sigma}(t),$ $w_{\sigma\lambda}(t)$ and $F_{\sigma}(t)$ be represented by

$c_{\sigma}(t)=\langle\psi_{\sigma}\mid\chi(t)\rangle,\quad
w_{\sigma\lambda}(t)=\langle\psi_{\sigma}\mid\hat{W(t)}(t)\mid\psi_{\lambda}\rangle,\quad{\rm
and}\quad F_{\sigma}(t)=\langle\psi_{\sigma}\mid F(t)\rangle,$ (A.1)

where $\hat{W}$ is an appropriate operator acting on these basis states. Then,
with these definitions (3.12) assumes the form

$i\hbar\frac{d}{dt}\mid\chi(t)\rangle-\hbar\hat{W}(t)\mid\chi(t)\rangle-\mid
F(t)\rangle=0.$ (A.2)

The associated homogeneous associated equation of this latter is given by

$i\frac{d}{dt}\mid\Phi(t)\rangle=\hat{W}(t)\mid\Phi(t)\rangle,$ (A.3)

which is a equation of the Schrödinger type, whose solution can be found using
the time ordering technique, that is

$\mid\Phi(t)\rangle=\hat{S}(t,t_{0})\mid\Phi(t_{0})\rangle,$ (A.4)

where

$\hat{S}(t,t_{0})=e^{-i\int_{t_{0}}^{t}\left[\hat{W}(s)\right]_{+}ds}.$ (A.5)

Now, if we try in (A.2) a particular solution $\mid\chi_{p}(t)\rangle$ of the
type

$\mid\chi_{p}(t)\rangle=\hat{S}(t,t_{0})\mid Z(t)\rangle,$ (A.6)

and then we perform some reductions to the resulting expressions, we get

$i\hbar\frac{d}{dt}\mid Z(t)\rangle=\hat{S}^{\dagger}(t,t_{0})\mid
F(t)\rangle,$ (A.7)

from where

$\mid
Z(t)\rangle=-\frac{i}{\hbar}\int_{t_{0}}^{t}\hat{S}^{\dagger}(s,t_{0})\mid
F(s)\rangle ds,$ (A.8)

and by consequence

$\mid\chi_{p}(t)\rangle=-\frac{i}{\hbar}\hat{S}(t,t_{0})\;\int_{t_{0}}^{t}\hat{S}^{\dagger}(s,t_{0})\mid
F(s)\rangle ds.$ (A.9)

In this way we obtain the general solution of (A.2) in the operator form

$\mid\chi(t)\rangle=\hat{S}(t,t_{0})\mid\chi(t_{0})\rangle-\frac{i}{\hbar}\hat{S}(t,t_{0})\int_{t_{0}}^{t}\hat{S}^{\dagger}(s,t_{0})\mid
F(s)\rangle ds.$ (A.10)

By projecting this general expression on the basis states
$\mid\psi_{\sigma}\rangle,$ we finally arrive to

$c_{\sigma}(t)=\sum_{\lambda}S_{\sigma\lambda}(t,t_{0})\tilde{c}_{\lambda}(t_{0})-\frac{i}{\hbar}\sum_{\lambda\nu}S_{\sigma\lambda}(t,t_{0})\int_{t_{0}}^{t}S^{\ast}_{\nu\lambda}F_{\nu}(s)ds,$
(A.11)

where the superscripts under the sum symbols run from 1 to the total number of
time-dependent oscillators, in our case it is equal to 2.