In the fall of 1964, rumors of mysterious circumstances arise regarding the behaviour of parish priest Father Flynn with one of the male students at St. Nicholas Catholic School. School principal Sister Aloysius, armed with her belief that sexual misconduct has occurred, looks to bring the priest to justice. But is he really innocent as he claims, or are her instincts right?
This emotional play was well directed by first time director Bob Kleinberg. Its pace was steady, with quick scene changes not disrupting the flow of the play. His set design allowed for four unique playing areas without requiring scenery to be moved during the play, assisted by Dennis Giblin's light design. Malca Giblin's costumes contributed to the somber mood of the production. One minor concern is the deep stage made the audience feel removed from several scenes, so we recommend sitting as close to the front row as possible.
Each of the four actors appeared to have definite goals for their characters, and each was successful in communicating their goals. Brendan Murray as Father Flynn was able to change between stern but caring teacher to friendly and gruff basketball coach, to kind and frightened teacher. Murray ably changed his emotions as the verbal barbs from Sister Aloysius (Michelle Trout) were thrown at him. Trout also was very effective, with her emotions running the gamut from her self-righteous attitude through to despair over the decisions she'd made. Compassionate Sister James (Catie Brown) struggled with complying to Sister Aloysius' counsels. Brown's expressions of trust and joy contrasted nicely with her realizations that the world is a dark place. Aurielia L. Spencer as Mrs. Muller, the mother of the student, was afraid and appeared to be beaten down by the world's constant attacks on her family.
Greenbelt Arts Center's solid production of Doubt: A Parable offers many opportunities for discussions about the motives of each of the show's characters. Did he, or didn't he? Should she, or shouldn't she? Will she or won't she? Check out this production and decide for yourself.
As I stepped into the director's role I sensed something. This was a new path for me so I'd start wondering. "Was the right course to take?" Could I convey the author's concepts?" Suddenly, I found that I was feeling doubt!
Webster's dictionary holds that it is "a feeling of uncertainty about the truth, reality or nature of something." But it seems to me that this is too simple, too anesthetic. The feeling of doubt is more guttural, more stealthier — it creeps in, like dark clouds of foreboding smothering your very certainty. What one day seems like the clearest, most sensible decision or path has slow1ly turned dim and murky. It is now up to you and your faith — in your reasoning, your beliefs, and/or yourself — to keep you moving forward.
Yet while doubt often seems like an emotionally overwhelming experience, its creation can also be a positive, useful tool in combating uncompromising resolute behavior. Carefully thought out logic can break down the barriers of unsupported arguments, allowing the fresh air of reason to bring a sensible perspective. This expression of "doubt" can be intellectually challenging and emotionally strengthening.
It is often said that Doubt, A Parable has two acts. The first act occurs at the theatre, as the audience members watch each character's emotional arc come to pass. The second act occurs afterwards as each audience member traverses his or her own arc from discussing the play's surface context (the innocence or guilt of Father Flynn) to exploring each character's doubts and how they dealt with them.
I hope you will enjoy tonights "first act" and I further hope that you walk away afterwards engrossed in your own "second act" as you discuss whose doubts were overwhelming and whose were strengthening.
As for me, as our talented actors became their characters, and as our imaginative set crew created St. Nicholas Church, my clouds of doubt began to disperse. Thanks for all your hard work.
This article can be linked to as: http://washingtondc.showbizradio.com/goto/6136.
I saw the national tour of "Doubt" when they played at the Hippodrome several years back. Since a week ago Friday, when I saw the GAC production for the first time, independent conversations I've had with other audience members have confirmed my opinion. I'm confident that it's no exaggeration when I say this production is one of the best I've seen anywhere. I may be GAC's GM, but I'm also a theatre lover. This show's a winner and more people should be seeing it. Only two weekends left. Don't miss out!


text:
Previous Canadian Bulldog's World LJN Wrestling Figure Hall of Fame Inductees:
George "The Animal" Steele
Luscious Johnny Valiant
Corporal Kirchner
The Iron Sheik
LJN's Wrestling Superstars action figures are generally considered the first series of wrestling toys to be marketed to the mainstream internationally. And I'm lucky to have collected them all back in the 1980's! Each month, I'll look at one classic figure and explain what made them so special.
Since opening this prestigious Hall of Fame in April, we've inducted several action figures that deserved inclusion because of their unique or unusual qualities - why use a generic wrestling figure in your make-believe battles when you can have someone who looks like, say, King Kong Bundy or Corporal Kirchner?
And while we certainly plan on upholding that rich tradition here at the Canadian Bulldog's World LJN Wrestling Figure Hall of Fame (official motto: "It's better than your stupid Hall of Fame, wiseass!"), we've also realized that we have never inducted a legitimate World Champion.
It's time to respect the legend.
The Iron Sheik was one of the very first figures LJN released in its Wrestling Superstars line in 1984, along with an anorexic Hulk Hogan, a lion-haired Andre The Giant, Rowdy Roddy Piper, Superfly Jimmy Snuka, Junkyard Dog, Big John Studd, Hillbilly Jim and Sheikie's partner-in-crime Nikolai Volkoff.
Sure, The Sheik's pants have seen better days, but consider that this guy has been dragged around and punished by the biggest and baddest folks WWF has to offer - your pants would probably be a little worn out, too! And while his infamous pointy-toed boots and handlebar 'stache are both in bad need of a paint job, it's still quite obvious that this particular figure is one bad mutha.
First, there's his physique. It's far more impressive than, say, the Hogan figure, and honestly only rivals Snuka's in the initial series. This is one piece of plastic that has put in considerable time in the gym.
Then there's the arrogance factor. No, not arrogance as in "Rick Martel's cologne", but rather him looking at you as if to say "I don't care if we're here on a busy street corner (or a desk, as the case may be), I'm going to flex my muscles."
In addition to being muscular and arrogant, intelligent wrestling fans know that The Iron Sheik was capable of giving his opponents a wrestling LESSON, and his action figure counterpart was able to do pretty much the same.
Take the photo to the right, for example. To some, it appear as Sheik is delivering a simple running powerslam to that no good, bleached-blond jabroni The Hulk Hogan. But no - he's doing it with just ONE ARM!
Amazingly, the figure was able to do this to another figure with NO additional leverage! You would think holding a figure of equal weight would make the first figure tip over, but Sheikie is sturdy as anything here. Can you imagine the actual Sheik doing that? Hell, can you imagine Brock Lesnar doing that?
You can also prepare other figures for suplexes using The Sheik's arm positions, but.... well, that's not nearly as much fun as balancing an opponent on a single effing shoulder.
And as awesome as having The Iron Sheik of 1984 to do your evil bidding was.... imagine The Sheik of today, knowing everything we do now?
For example, I had a Sheik figure and a B. Brian Blair figure back in the day. I knew they competed at WrestleMania III sure, but that match was nothing special.
But if we knew the whole "break his back/make him humble" story? DAMN! That would be a main event in any arena in the country, as Gorilla Monsoon was so often fond of saying.
By the way, do you have any idea how difficult it is to stage a camel clutch picture with two slightly-bendable-but-not-actually-bendable wrestling toys?? Granted, this isn't a legit camel clutch or anything... in fact, it kind of looks like Sheik and Blair are at a pub after a few too many beers.
Although.... I'm not sure how, in that scenario, one would explain why both guys are topless.
The final thing I'll say about The Iron Sheik's LJN figure is how well it can cut a promo. Well.... virtual promo; it's not like the doll can talk or anything; but you know what I mean.
Just look at him telling Kennedy Mac-Mahon about how things are. Iran is number one. USA? Hock-ptooey! There's no doubt in my mind that he's already instructed Mister Cameraman to zoom in on those abs, too.
Maybe he's cutting a promo on how Hogan is a no-good (expletive deleted) like The Justin Bieber? Or perhaps he's following through on one of his many threats issued via Twitter on a daily basis. With The Sheik, you just never know.
One thing is for sure - your LJN collection wasn't quite complete without the addition of the Champion of the Olympics, used to be bodyguard for Shah of Iran, WWF World Champion, man who defeated Howdy Doody Bob Backlund in Mad'son Square Garden.... The Iron Sheik.
Welcome to the Hall of Fame, bubba.

redpajama_set_name: RedPajamaCommonCrawl


You can view the project code at the time of this blog post by checking out this tag.
I now wanted to start building out some client-side code. My initial plan is to test out lots of currently available UI frameworks (React, Aurelia, Ember, etc) to see how they compare... but I've become increasingly interested in Vue.js so wanted to initially start with that.
I installed Vue.js using yarn, but the the first gotcha I encountered was the issue with standalone vs runtime only versions which just goes to show that it pays to properly read the documentation! This was quick to resolve with the suggested approach of adding an alias - however this will mean that should I wish to ever deploy the application that I would almost certainly need a separate webpack configuration file for production deployment.
...although there are other source map values that can be used.
This code fragment defines a new component that before being "mounted" (see the life-cycle) uses axios to request the data from the Repository (going via the proxy set up in a previous post). When it mounts it then iterates over the items loaded from the repository and renders the name of each of them as an individual element in a list.
At the moment this obviously doesn't handle any kind of browsing of the data and it is not sensibly split out into separate files for the components - however, the essential building blocks are now in place - the ability to retrieve and render data.


#34780319 - Hands of a missing kidnapped, abused, hostage, victim woman ..
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#77793003 - women in handcuff, fashion cloth posing on pink background, pretty..
#65789558 - the boy holding the cage , imprisoned, retarded, Child Abuse..
#79568957 - Sexual abuse with a terrorist attacking a scared woman in a dark..
#20802738 - Native Americans and African slave reenactors posing as part..
#31695905 - Alcohol slave or Alcoholism ,social problem.
#84636074 - Sexual abuse with a terrorist attacking a scared woman in a dark..
#76175331 - Violence, woman with long red curly hair gagged. Red-haired girl..
#56484623 - vintage tone of Hands of a missing kidnapped, abused, hostage,..
#99041454 - A woman is working on the computer. In the open air.
#10770194 - SONAKHALI, INDIA - JANUARY 17 child workers carry bricks carrying..
#49718229 - Trafficking or the concept of human rights violations.
#61923344 - Raider with mohawk hairstyle, young pretty slave in iron mask..


Today, we continue Spider-Verse week with yet another alternate universe Spider-Man, this time from Earth-1610 or (as it is more commonly known) the Marvel Ultimate Universe.
While not as immediately popular as Spider-Gwen, Miles Morales was immediately controversial. I have to admit that I was one of those that originally didn't like the idea of a new Spider-Man (then again, I've never really cared for the Ultimate line of comics anyways), but Mr. Morales has begun to grow on me. I always love the exciting new possibilities that come with a new character as we watch them grow and develop into a hero. There's just something refreshing about new teenage heroes, which is why I think Miles, Spider-Gwen, and the new Ms. Marvel are so hot right now.
Miles Morales was just a normal 13-year old kid with an aptitude for science. While visiting his career-criminal uncle, Miles was bitten by a stolen radioactive spider. Miles began to manifest spider-influenced powers, but he decided to leave the crime fighting to the real Spider-Man. After Peter Parker died at the hands of the Sinister Six, Miles felt guilt over not using his powers to help Spider-Man and adopted Peter's crime-fighting philosophy – that with great power comes great responsibility. Despite being young and wildly inexperienced, Miles overcame his own self-doubt and proved to the world that he can do whatever a Spider-Man can do.
Below is the Ultimate Spider-Man.
Enjoy, excelsior, and have a MARVELOUS New Year!
PS – Thanks again to Mark for getting Miles all done up in his proper Marvel Plot Points duds.
This entry was posted on December 31, 2014 by David W in Datafiles, Spider-Verse.


The future of digital signage, particularly ahead of us in the near-term, will be driven by direct view LED technology, and was one of the strongest trends over the last year in the commercial display space.
All signs are pointing to wider adoption, as designers of commercial, retail, and public spaces consider display technology as an alternative to conventional design finishes we've grown accustom to, like metal, glass, tile and stone.
As we look forward while learning from historical trends, we learn that display technologies are maturing rapidly, converging and integrating in ways that are energizing environments by delivering better experiences and helping businesses improve operations through engaging visual communications.
Here's how Samsung has entrenched itself in that evolution.
Today, we're rolling out full digital signage networks, and it's rare to walk into a business or even public space and not see some form of display technology. Like smartphones, tablets, and even wearable technology, displays and digital signage are becoming common place; part of the overall digital transformation that is bettering the use of communication dissemination. Put simply, we're optimizing our spaces and time for quick and effective information sharing — and that's changing the way businesses operate.
It owes to many things, like the increasing affordability of commercial displays and dramatic advances in picture quality. Set-up, installation and operation are all easier these days. But increased industry consumption also owes to experience — with solutions companies and their end-user clients better understanding what works and what's needed to make digital signage, interactive displays and collaborative screens effective across a variety of use cases.
When you can make a whole feature wall a digital canvas, or all the walls of a designed space, that becomes transformational. One of the San Francisco office tower lobbies of a CRM software giant uses the full main wall for reception and elevator access as a digital canvas to reinforce its branding campaign. Same goes for a giant streaming media company with LA-based headquarters, where fine-pitch LED and audio are showcased to make visitors feel immersed in some of the company's most popular products.
More specifically for 2019, LED will continue to see increasing adoption because the module-based displays offer more design flexibility. As spaces and architecture evolve in the business setting, so will displays to accommodate cutting edge designs.
Another trend for 2019 born out of last year is the evolution of workspaces. From open-concept offices to collaborative conference rooms, spaces need now more than ever to be equipped with displays and content management systems to carry proper information to employees and connect companies to each other in the 24/7 digital environment.
Formalized, top-down management meetings and big conference rooms are being replaced with collaborative teams, impromptu meetings and huddle spaces that necessitate a different set of tools. Samsung developed and released in 2018 a product called The Flip that went right to the heart of those changing workplace dynamics — a digital flip chart that allowed quick and easy collaboration and sharing on something affordable and portable. It's been a big hit, and is something of an entry point for workplaces we think will use more and more digital. Additionally, the consumption and implementation of super large displays — in the 75 to 98 inch range — is growing due to lower price points and the desire to modernize conference room projector-based technology.
But also consider the need for display technology across industries. We're thinking of health care facilities, the brick-and-mortar piece of the banking industry, even the manufacturing shop floor – places where employees or customers are looking for directions, updates, or simply a status of an ongoing project that can be refreshed in real-time.
The new year is nothing less than an entry point on the path to a rapidly evolving need for workplace and public space display technology. As business needs and consumer desires grow and change, so will Samsung's technology to accommodate digital transformation taking place at break-neck speed.
We're excited about the road ahead and what the journey will bring in terms of display technology — the best is yet to come.
Planning and deploying digital signage technology while it constantly evolves can be difficult — learn more about the process and best practices in this free white paper.


The layout of a lettering piece is fairly important because it sets the foundation for your message. The advantage of hand lettering is that you are able to create unique compositions to make your message stand out. However, if the composition isn't clear the message will likely be lost.
When it comes to lettering the last thing you want is for someone to not receive your message because it's not clear enough. Our job as letterers is to communicate, and even the simplest of lettering pieces communicate something. However, if a lettering piece is too complex your composition will begin to detract or even obscure your message.
Even though we have the ability to make our lettering stand out with complex compositions doesn't mean we have to. The balance is struck between providing a basic structure to make your message clear, and adding some complexity to them to make the composition interesting enough.
It's often better to err on the side of simplicity first. Often times, a basic and clear composition is just what a lettering piece needs for it to stand out. You can look to geometric shapes first—I usually start off with a square or rectangle to begin my piece—and create basic guidelines for borders, margins, cap height, x-height, and the baseline.
This allows for you to focus on clarity through hierarchy, which is arguably more important than a unique composition. Hierarchy is making sure that the right words are emphasized. When you're sketching out your concepts, discover the words that are most important in order to covey your main message. Make those words more prominent than other words and use them to grab people's initial attention.
Hierarchy is MAKING sure that the right words are emphasized. No, wait… Hierarchy is making sure that the RIGHT WORDS are emphasized.
Starting with a basic layout will place emphasis on the hierarchy and help you focus on making the lettering itself clear and interesting. Sometimes it doesn't take an off-the-wall composition to make the piece interesting, but rather just making the piece interesting through styles, hierarchy, the Rule of Thirds, syncopation, and letter interactions.
Once you have the basics down, you can extrapolate on them to make things a little more interesting. You can try more interesting forms like circles, polygons, and free-form shapes. Adding curvilinear guidelines to a basic rectangle can make a composition seem more dynamic than it actually is without adding too much difficulty.
The compositions above and below both started as simple rectangles, and then I decided to take some liberties in how I could make it more interesting. On the one above, I used the rectangle as a bounding box to define the edges of my composition, but I added a curvilinear composition inside of it. On the one below, I broke through the boundary of the rectangle to have the composition feel more cohesive by interacting with the other letters.
As you could see in the last lettering piece, I broke through the letter H's cap height and baseline in order to fill out the piece. However, you can also place a word or object in the composition and adapt the form of the letters, baseline, and cap height to other objects. This can be seen with the ampersand encroaching into the two rows of lettering in the piece below.
When I was figuring out the composition for this piece, the word "and" was an odd word and I was considering giving itself a single row to itself. Instead, I shortened it to just an ampersand in a circle and placed it between two rows of lettering. This made things a bit more interesting and forced the letters around it to conform to its shape.
At this point, you can incorporate any number or combination of these techniques. Many of the most interesting compositions have some elements of basic shapes. Use basic shapes as a foundation and layer in enhancements to making your piece interesting enough to attract attention, yet clear enough to get your message across. You can do so by syncopating your layout, involving letter interactions in your composition, and playing with the layout's boundaries.
It hard to say what an interesting composition looks like from lettering piece to lettering piece. Each piece has a unique number of words and character lengths that demand tailored compositions; this is one thing that makes a lettering artist's skills valuable. The most interesting compositions will come from beginning with these basic ideas, getting creative with the piece you're working with, and adding your unique touch.


text:
What the Netflix Documentary 'Bleeding Edge' Gets Right About the Dangers of Medical Devices in America
By Alexandra Sifferlin
Updated: July 27, 2018 10:20 AM ET | Originally published: July 27, 2018 9:00 AM EDT
A new Netflix Original documentary sheds light on a topic that's opaque to many: the medical device industry in America.
The film, called The Bleeding Edge, claims that the way medical devices make it to the U.S. market is lax and potentially dangerous. The filmmakers interview doctors and patients who say they've been personally harmed by an array of devices—including cobalt-based hip replacements, vaginal mesh and robotic surgery devices—and who feel that the guardian group of the medical device industry, the U.S. Food and Drug Administration (FDA), is not doing enough to protect them.
It's normal for medical devices to come with risks for adverse reactions, and it's a physician's responsibility to discuss those risks with their patients. However, medical devices often escape scrutiny on their path to the American market, the filmmakers say, and demand more research and testing than they currently have.
The FDA began regulating approvals for medical devices in 1976. Little has changed about the process since then, but the devices have: critics in The Bleeding Edge argue that medical devices today have become much more complex, and testing requirements haven't evolved along with them.
Their main concern is a loophole which allows many devices to reach the market without much clinical testing. New devices are supposed to go through a pre-market approval process similar to a new drug, which requires that the device be tested in humans to show it's both effective and safe. Yet because so many new devices are brought to market, including new iterations of already approved devices, some medical companies argue that it's too expensive to send each device through massive amounts of clinical testing.
To address that, a provision called the 510(k) pathway was created, whereby a manufacturer can also obtain approval if they can prove that their device is "substantially equivalent" to another device already on the market. The problem, the film argues, is that while that pathway was supposed to be an exception, it's become a common way devices are approved for sale, and sometimes, it creates an approval chain of devices that are on the market without recent testing.
"That provision, which was meant as an exception, in essence [is] a loophole," says Dr. David Kessler, a former FDA commissioner, in The Bleeding Edge. "[It] became the rule, so that the vast majority of devices today regrettably are regulated under this framework."
The topic is a timely one, and there are signs that oversight is weakening even further. In June, the Trump administration released its proposed plans for an FDA reorganization, which would limit the agency's purview, rather than expand it.
One of the devices targeted in the film for not undergoing enough pre-market testing or oversight has also recently come under fire. Just a week before Netflix released The Bleeding Edge, Bayer, the pharmaceutical company behind Essure—a controversial, permanent form of birth control that, when placed inside the fallopian tubes, creates scar tissue that prevents pregnancy—announced it was dropping the product.
More than 16,000 lawsuits have been filed against Bayer over Essure, with women citing complications like migraines, hair loss, organ perforation and even unintended or dangerous pregnancies, according to ConsumerSafety.org. Many of these women have banded together online in Facebook groups, and the documentary follows several of the women who have been affected. Some women said the coils came loose, causing severe bleeding and the need for surgeries, including hysterectomies.
On July 20, Bayer announced that Essure will no longer be available in the U.S. "This decision is based on a decline in U.S. sales of Essure in recent years and the conclusion that the Essure business is no longer sustainable," Bayer said in a statement. The company also says that the device is safe, writing in a statement on Thursday that, "Bayer is reminding women with Essure that the safety profile of Essure, the only Food and Drug Administration (FDA)-approved non-incisional form of permanent birth control, remains positive and unchanged."
"This is exactly why we do this work," said Kirby Dick, one of the filmmakers of The Bleeding Edge, in a statement about the announcement. "This film has already changed policies and changed history."
The FDA Approved the First New Endometriosis Treatment In a Decade


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text:
\section{Introduction}

\,\,\,\,The complete understanding of the high energy scattering processes
is impossible without the calculation of
the contributions of the different unitarization corrections
to the amplitude. In the QCD Pomeron framework, \cite{bfkl,bfklsum,nlbfkl},
these corrections are arose due the self Pomeron interactions via the
triple Pomeron interactions vertices \cite{vert1,vert2}.
These self Pomeron interactions lead to the very complicated  picture of the
amplitude's evolution with rapidity, see for example
\cite{eglla1,eglla2,eglla3,eglla4,jimwalk}.

  There are few main approaches which claim that they properly describe the
Pomeron self interactions.
The first one is presented as a chain of the evolution equations
with Balitsky-Kovchegov (BK) equation as a main field equation of the hierarchy, \cite{BK}.
The BK equation may be formulated in the framework of the Color Glass Condensate Approach
(CGK), \cite{jimwalk}, and may be properly analyzed in the terms
of s-channel interacting color dipoles, \cite{dipmod}. This equation
correctly describes the interaction of two non identical
objects, for example  such interactions as DIS on nuclei, see
\cite{bksols,bdepkov,jimsol,bksemi1,bksemi2,bk-pheno,kks,iim}.
Another approach is the QCD Reggeon Field Theory (RFT-QCD),
which is formulated in the momentum space and based on the
standard diagrammatic calculus,
\cite{eglla1,eglla2,eglla3,eglla4,braun1,braun2,braun3}. In this approach the BK equation
describes a resummation of the ''fan'' diagrams of the type
depicted in Fig.1a with only Pomerons merging vertices considered.

  Usual BK equation, describing "fan" diagrams, does not include
Pomeron's splitting vertex. Therefore, the next natural
step toward the unitarization of the amplitude is a  symmetrical consideration
of the scattering process with both vertices included.
In this case the splitting vertex
may be accounted differently in two different approximations. In the first approach
the semi-classical problem of the interest is considered and there the sum of the
diagrams depicted in Fig.1b is calculated, neglecting Pomeron loops contribution into amplitude.
The second approach is concentrated on the solution of
the full quantum problem basing on some
effective high-energy QCD inspired models, accounting
diagrams of Fig.1c type, see for example \cite{Lipatov:1995pn}.
So far several attempts of the calculations were done in both directions, on the basis of QCD-RFT
theory,  \cite{braun1,braun2,braun3} and in the framework of
CGC and dipole model, \cite{selfdual}. The NLO and one-loop contribution into the amplitude also were calculated
, see for example \cite{NLO,OneLoop}, but the full solution still seems very far from it's completeness.
In this situation it is very
natural to consider much simpler zero
transverse dimensional model, in hope that some important
properties of solutions of this model will be faced in QCD as well.

  The RFT-0 (Reggeon Field Theory
in zero transverse dimensions) model is attracted much interest during few last years,
see   \cite{0dimsol}. The approach was formulated
and studied a long time ago, even before the QCD era, see \cite{0dimc,0dimq,0dimi}.
The interest to the approach is due very attractive properties of the model.
First of all, RFT-0 is solvable at different parameters of the model.
This possibility to find full solution for the different regimes
of the theory is an important feature of the model, since we hope that we will understand
more about RFT-QCD if we will have full and semi-classical solutions for RFT-0,
see examples in \cite{BM,BondBr,Braun4}.
Therefore, the second important
fact about the RFT-0 is that in this theory we can find both
classical and quantum solutions for the amplitude that provides us with the information
about the relative contribution of the loops to the amplitude.
The possibility to consider in RFT-0
different types of the
Pomeron vertices is an another property of the RFT-0 which is very useful and which could clarify the
situation in QCD.

  In the paper we consider two  RFT-0 models, in the second one
together with usual triple Pomeron vertex we also include
the quaternary Pomeron vertex. The paper is organized as follows. In the second section we
consider the RFT-0 model with only triple Pomeron vertex.
In two subsections of the section we introduce an whole machinery of the problem for the both quantum
and classical solutions.
In the following subsections we present the results of our calculations
for the amplitude at different parameters of the model as well as the
results for the two point Green's function (''effective'' Pomeron propagator )
of the theory. In the Sec.3 we solve the same problems
for the RFT-0  theory with the quaternary Pomeron vertex included.
The next section, Sec.4, is dedicated to the possible application
of the solution to the problem of diffractive dissociation.
The Sec.6 presents a summary of the results of the calculations and
the last section, Sec.7, is the conclusion of our paper.

\section{Solution of RFT in zero transverse dimension with only triple Pomeron coupling}

 In this paper we investigate the one-dimensional problem of the interacting Pomerons
described by Lagrangian, which in the terms
of Gribov fields has the following form:
\begin{equation}\label{Lag1}
L\,=-\,\psi^{+}\,\dot{\psi}\,-\,\mu\,\psi^{+}\,\psi\,+\,i\,
\lambda\,\psi^{+}\,(\psi^{+}\,+\,\psi)\,\psi\,\,\,,
\end{equation}
where $\,\mu\,$ is the intercept of the bare Gribov field,
$\lambda$ is the triple field interactions vertex
and differentiation means differentiation on rapidity (y)
variable, which is only the variable of the problem.
Introducing the Pomeron fields, $q\,=\,i\,\psi^{+}\,$ and
$p\,=\,i\,\psi\,$ we rewrite the Lagrangian in the form
of real $\lambda\,$ coupling:
\begin{equation}\label{Lag2}
L\,=\,q\,\dot{p}\,+\,\mu\,q\,p\,-\,\lambda\,q\,(q\,+\,p)\,p\,\,.
\end{equation}
The field q and p may be understood now in the terms
of the Pomeron creation and Pomeron annihilation operators $a^{+}$
$a$:
\begin{equation}\label{Oper}
\,q\,\rightarrow\,a^{+}\,=\,q\,\,\,\,,
\,p\,\rightarrow\,a\,=\,-\,\frac{d}{d\,q}\,.
\end{equation}
The Hamiltonian of the problem in the
terms of the operators q and p has the following form
\begin{equation}\label{Ham1}
H=-\,\mu\,q\,p\,+\,\lambda\,q\,(q\,+\,p)\,p\,\,\,,
\end{equation}
being
the second order differential operator:
\begin{equation}\label{Ham2}
H=\,(\mu\,q\,-\,\lambda\,q^{2})\,\frac{d}{dq}\,+
\,\lambda\,q\,\frac{d^2}{dq^2}\,\,.
\end{equation}
The full, quantum solution of the theory
will coincide, therefore, with the solution of the quantum mechanics problem
with the Hamiltonian given by Eq.\ref{Ham2}:
\begin{equation}\label{Ham3}
H\,\Psi\,=\,\frac{d\,\Psi}{d\,y}\,\,,
\end{equation}
where the function
\begin{equation}\label{Ham33}
\Psi(y,q)\,=\,\sum_{n=0}^{n=\infty}\,\lambda_{n}\,
e^{\,-\,E_{n}\,y}\,\phi_{n}(q)\,\,
\end{equation}
is the full quantum amplitude of the process of the scattering
of two particles. Here $E_{n}$ and $\phi_{n}(q)$
denote eigenvalues and eigenfunctions of the operator
Eq.~\ref{Ham2}, and  $\lambda_{n}$ are the normalized projections
coefficients of the eigenfunctions on the value of $\,\Psi\,$ at $\,y=0\,$.

  The classical solution of the theory, i.e. the solution
with only ''tree'' diagrams without loops also may be
obtained in the given framework. It is simply the solution of equations of motion
for the Lagrangian given by Eq.~\ref{Lag2}

\begin{figure}[t]
\begin{center}
\psfig{file=diag1.eps,width=180mm}
\end{center}
\caption{\it
Examples of diagrams of the effective theory of interacting Pomerons
with triple pomeron vertices:
a) a fan diagram; b) a tree diagram defining the classical limit;
c) a diagram with quantum loops.}
\label{diag1}
\end{figure}

\subsection{The quantum solution of the model}

 An approximate solution of the Eq.\ref{Ham3} was found many years ago,
see \cite{0dimc,0dimq,0dimi}. Analytical solutions of the Eq.\ref{Ham3}
at large value of the ratio $\,\mu\,/\,\lambda\,\equiv\,\varrho\,$
for ground state and first exited state were considered at the same time as well. Nevertheless,
the spectrum of the theory and eigenfunctions for next exited
states at large value of  $\,\varrho\,$ as well as the solutions for
arbitrary values of $\,\varrho\,$ were not obtained.
Therefore, we want to  define the procedure for the
numerical calculations of
the spectrum of the theory and corresponding eigenfunctions
for the cases of the strong triple Pomeron coupling (small $\,\varrho\,$  )
and week coupling (large $\,\varrho\,$ ) as well. In the following
we will be more interesting in the case of the small triple coupling
as the case which is mostly correspond to the real perturbative QCD situation.
So, we need to solve
the second order differential equation (see more detailed
in  \cite{0dimq}):
\begin{equation}\label{Quant1}
(\varrho\,-\,q)\,\frac{d\,\Psi(y,q)}{dq}\,+
\,\frac{d^2\,\Psi(y,q)}{dq^2}\,=\frac{1}{\lambda\,q}\,
\frac{d\,\Psi(y,q)}{d\,y}\,\,,
\end{equation}
with the initial and boundary conditions on the function $\Psi(y,q)$:
\begin{eqnarray}\label{Quant2}
\,&\,&\,\Psi(y=0,q)\,=\,\sum_{n=0}^{n=\infty}\,\lambda_{n}\,
\,\phi_{n}(q)\,=\,I(q)\,\\
\,&\,&\,\Psi(y,q\,\rightarrow\,0)\,\propto\,q\,\\
\,&\,&\,\Psi(y,q\,\rightarrow\,\infty)\,\propto \,const.\,\,,
\end{eqnarray}
where form of $I(q)$
depends on the particularly considered physical problem.
Now it is constructive to make a
transformation of the eigenfunctions:
\begin{equation}\label{Quant3}
\phi_{n}(q)\,=\,e^{(q-\varrho)^2\,/4}\,f_{n}(q)\,\,.
\end{equation}
In this case the term which is proportional to the first derivative over q
is eliminated leading to hermitian Hamiltonian for the problem.
After this transformation we obtain a Shredinger type eigenvalue equation:
\begin{equation}\label{Quant4}
\frac{d^2\,f_{n}(q)}{dq^2}\,+\,\frac{f_{n}(q)}{2}\,-\,
\frac{1}{4}\,(q\,-\,\varrho)^2\,f_{n}(q)\,=\,-\,\frac{E_n}{\lambda\,q}\,
f_{n}(q)\,
\end{equation}
with the following boundary conditions on the function $f_{n}(q)$:
\begin{eqnarray}\label{Quant5}
\,&\,&\,f_n(q\,\rightarrow\,0)\,\propto\,q\,\\
\,&\,&\,f_n(q\,\rightarrow\,\infty)\,\propto \,
e^{-q^2\,/\,4\,+\,q\varrho\,/\,2}\,\,.
\end{eqnarray}
Our problem of interest is the scattering of two particles,
therefore we define initial condition for the $\Psi(y=0,q)$
in the eikonal form (see more in \cite{0dimq}) :
\begin{equation}\label{Quant6}
\Psi(y=0,q)\,=\,I(q)\,=\,I(q\,,q_{ext})\,=\,1-e^{-q\,q_{ext}}\,,
\end{equation}
here $q_{ext}$ is a value of the source
for the Pomeron field at zero rapidity. The
function $\Psi(y,q)$
results the evolution of interacting Pomeron fields for
the value of the Pomeron-external particle vertex
equal to $q_{ext}$ at zero rapidity, till the
value of the single Pomeron field equal to $q$ at rapidity y.

 The last ingredients of the theory are
projection coefficients $\,\lambda_n\,$ of the eigenfunctions
of the solution on the initial state I(q).
Having in mind, that our eigenfunctions are
orthogonal on interval of  q from $0$ to $\,\infty\,$
with weight function $\,F_{W}(q,\varrho)\,$ :
\begin{equation}\label{Quant7}
\int_{0}^{\infty}\,f_{n}(q)\,f_{m}(q)\,F_{W}(q,\varrho)\,dq\,=\,\delta_{n\,m}\,\,,
\end{equation}
we obtain for $\,\lambda_n\,$:
\begin{equation}\label{Quant8}
\lambda_n\,(q_{ext})=\,\frac{\int_{0}^{\infty}\,f_{n}(q)\,I(q\,,q_{ext})\,
F_{W}(q,\varrho)\,dq\,
}{\int_{0}^{\infty}\,f_{n}^{2}(q)\,F_{W}(q,\varrho)\,dq\,}\,\,,
\end{equation}
where the weight function $\,F_{W}(q,\varrho)\,$ has the following form
(see \cite{0dimq} and also \cite{Heun}):
\begin{equation}\label{Quant9}
F_{W}(q,\varrho)\,=\frac{\,e^{-(q\,-\,\varrho)^2\,/\,2}}{\,q\,}\,.
\end{equation}
The numerical solution of Eq.\ref{Quant4}
with the boundary conditions given by Eq.\ref{Quant5}
is the solution for the eigenvalues and eigenfunctions of the
second order differential equation with two boundary value conditions.
These solutions for eigenfunctions and eigenvalues may be found
for each values of $\mu$ and $\lambda$, i.e. for the different
values of parameter $\,\varrho\,$.
The obtained values of eigenfunctions $E_n$ and eigenfunctions $f_n(q)$
allow to complete the calculation of the full quantum scattering amplitude and , therefore, solve our problem.
So, considering
the scattering of two particles , where the vertex of interaction of the first
particle with Pomeron is equal $q_1$ and the vertex of interaction of
the second particle with Pomeron is equal $q_2$, we
define the full ,
quantum solution for the scattering amplitude at given rapidity y as
\begin{equation}\label{Quant13}
\Psi(y,q=q_2)\,=\,\sum_{n=0}^{n=\infty}\,\lambda_{n}(q_1)\,
e^{\,-\,E_{n}\,y}\,e^{(q_2-\varrho)^2\,/4}\,f_{n}(q_2)\,\,,
\end{equation}
with the $\,\lambda_{n}(q_1)$ given by Eq.~\ref{Quant8}
for the known values of eigenfunctions, see Fig.1c.
The  solution for the amplitude, Eq.~\ref{Quant13},
does not depend from which value of Pomeron field,
$q_1$ or $q_2$, the evolution begins.
If we consider the evolution of the Pomeron from $q_2$ to $q_1$ we will obtain
the same answer for the amplitude, as for the case of
evolution from $q_1$ to $q_2$.

\subsection{The classical solution of the model}

 In order to see the role
of the Pomeron loops in the scattering amplitude
it is important to calculate the classical solution of our problem,
given by the ''net'' diagrams of Fig.1b.
The comparison between two solutions will show a relative weight
of the loops in the scattering amplitude and, therefore, an applicability of
the classical solution at given values of the theory parameters.
The required classical solution we obtain solving
equations of motion for the Lagrangian given by Eq.\ref{Lag2} with the sources of
Pomeron fields at zero rapidity and final rapidity of
the process Y:
\begin{equation}\label{Class1}
L\,=\,\frac{1}{2}\,q\,\dot{p}\,-\frac{1}{2}\,\dot{q}\,p\,+
\,\mu\,q\,p\,-\,\lambda\,q\,(q\,+\,p)\,p\,+\,
\,q(y)\,p_0(y)\,+\,q_0(y)\,p(y)\,\,,
\end{equation}
where $\,p_0(y)\,\,q_0(y)\,$ are the sources of the Pomeron fields.
The equations of the motion for the fields p and q
are the following:
\begin{eqnarray}\label{Class2}
\,&\,&\,\dot{q}\,=\,\mu\,q\,-\,\lambda\,q^2\,-\,2\,\lambda\,q\,p\,\\
\,&\,&\,\dot{p}\,=\,-\,\mu\,p\,+\,\lambda\,p^2\,+\,2\,\lambda\,q\,p\,\\
\,&\,&\,q_0(y)\,=\,q_1\,\delta(y)\,\\
\,&\,&\,p_0(y)\,=\,q_2\,\delta(y-Y)\,\,.
\end{eqnarray}
This system of equations is another example of two value boundary problem
for the system of first order differential equations, and may be also solved
for different values of parameters
$\mu,\,\,\lambda,\,\,q_1,\,\,q_2\,$. The solutions of the system
Eq.\ref{Class2} at given value of final rapidity Y and given values of
parameters $\mu,\,\,\lambda,\,\,q_1\,\,q_2\,$  we will
denote by $\{q_{c}(y),p_{c}(y)\}$.
With the  solution  $\{q_{c}(y),p_{c}(y)\}$ the
''net'', classical  amplitude , represented by
diagrams Fig.1b, is defined by standard way as:
\begin{equation}\label{Class3}
\Psi_{c}(Y,\,q_{c},\,p_{c})\,=\,
1\,-\,e^{-S(Y,\,q_{c},\,p_{c})}\,\,,
\end{equation}
where
\begin{equation}\label{Class4}
S(Y,q_{c},p_{c})=\int_{0}^{Y} L(Y,q_{c}(y),p_{c}(y)) dy =
\frac{1}{2}(q_1 p_{c}(0)+q_{c}(Y)q_2)+
\frac{\lambda}{2}\int_{0}^{Y} (q^2_{c}(y)p_{c}(y)+
q_{c}(y)\,p^2_{c}(y)) dy\,.
\end{equation}
The amplitude Eq.\ref{Class3} describes
the eikonalized interactions of ''net'' diagrams
and for symmetric boundary conditions, $q_1=q_2$,
the amplitude is invariant under the duality transformations:
\begin{equation}\label{Class5}
p\,\leftrightarrow\,q\,\,\,\,and\,\,\,\,y\leftrightarrow\,Y-y\,.
\end{equation}

 The interesting feature of the classical solution of the system
defined by the
Lagrangian Eq.\ref{Class1} is that starting from some critical rapidity
$Y_c$ the solution of equation of motion is not unique
for $q_1=q_2\,<\,\varrho$.
From the  rapidity $Y_c$ there are at least three
classical trajectories  $\{q^{i}_{c}(y),p^{i}_{c}(y)\}$ which locally minimize the action and
with increasing of the rapidities the number of the trajectories is
growing. Each from them provides the local minimum of the action and the
amplitude of the theory, therefore,
must be rewritten in the following form:
\begin{equation}\label{Class6}
\Psi_{c}(Y)\,=\,1\,-\,
\sum_{i}\,\Delta_{i}\,\exp\{-S(Y,\,q^{i}_{c}(y)\,,p^{i}_{c}(y))\}\,,
\end{equation}
where $\,\Delta_{i}\,$ is the quantum weight of the corresponding
classical trajectories. In the following consideration
these weights we  take  equal 1 or -1, depending on the type
of trajectory, see more detailed consideration of this problem in \cite{0dimi}.

  In our calculations of the classical amplitude for the symmetrical case
$q_1=q_2$,
we always will use three solutions of the equations of motion.
The first one, $\{q^{1}_{c}(y),p^{1}_{c}(y)\}$,  is the  symmetrical in the sence of Eq.\ref{Class5},
which trajectory will be similar to the trajectory number 1 in the
Fig.\ref{Traj}.
Another two dominant solutions, which we use in the calculations and
which arise from some rapidity  $Y_c$ ,
are the solutions for the trajectories similar to the trajectories
2 and 3 in the Fig.\ref{Traj}.
Separately each of them is not symmetrical in the sense of
the symmetry transformation
given by the Eq.\ref{Class5}, but instead, under
the duality transformation of rapidity given by the Eq.\ref{Class5}, there is
a pair symmetry of these two solutions
\begin{equation}
q^{2}_{c}(y)\,=\,p^{3}_{c}(Y-y)\,\,\,\,and\,\,\,\,p^{2}_{c}(y)\,=\,q^{3}_{c}(Y-y).
\end{equation}
It is interesting to note, that asymptotic behavior of these
two classical solutions is well known, each of them may be described by
''fan'' diagram amplitude of Fig.1a, see  \cite{0dimq},
and being the dominant contributions these solutions lead to the
''fan'' dominance effect, see ~\cite{BM}.

  The  final solution of our problem in the classical approximation
we can write with the help of these three particular solutions as
\begin{equation}\label{Class7}
\Psi_{c}(Y)=1+exp\{-S(Y,\,q^{1}_{c}(y)\,,p^{1}_{c}(y))\}-
exp\{-S(Y,\,q^{2}_{c}(y)\,,p^{2}_{c}(y))\}-
exp\{-S(Y,\,q^{3}_{c}(y)\,,p^{3}_{c}(y))\}\,,
\end{equation}
which is symmetrical in respect to the duality transformation Eq.\ref{Class5}.
In our plots for the symmetrical interactions, $q_1=q_2$,
the classical solution will be always presented
by amplitude of Eq.\ref{Class7}, which contains three parts beginning from
critical rapidity  $Y_c$, and only one part, symmetrical solution, at
rapidities smaller than  $Y_c$.  In the asymmetrical case the symmetry is
broken initially, and, therefore, only one ''fan'' configuration survives
in the classical solution.

\begin{figure}[t]
\begin{center}
\psfig{file=solution7.eps,width=100mm}
\end{center}
\caption{\it Classical solutions of the RFT-0 with
only triple Pomeron vertex:
 the $\{ q,p\}$ trajectories obtained for $Y=5\,>\,Y_c\,$, $q=p=0.7$,
$\rho\,=\,5$.}
\label{Traj}
\end{figure}

 It is important to underline, that
this picture for classical solution holds for each from the considered models,
for RFT-0 with triple Pomeron vertex only and for RFT-0 model with both
triple and quaternary vertexes.

\subsection{Parameters of the model}

 First of all, we define the range and value of parameters
for which the calculations were done in both cases, quantum and classical.
We calculated quantum and classical
amplitudes,
$\,\Psi(Y,\,q)\,$ and
$\,\Psi_{c}(Y,\,q_{c},\,p_{c})\,$, for the following value of
$\varrho\,=\,\mu\,/\,\lambda\,$:
\begin{eqnarray}\label{Num1}
\,&\,&\,\varrho\,=\,5\,\,\,\,(\mu\,=\,0.2\,,\,\lambda\,=\,0.04\,).
\end{eqnarray}
The reason of the main attention on this value of $\varrho\,$ is very simple.
The coupling constant in RFT-0 may be defined as $\,\alpha_s\,=\,1\,/\,\varrho$  and
at $\varrho\,=\,5\,$ we have, therefore, $\,\alpha_s\,=\,0.2\,$ .
This  value for the coupling constant is reasonable also in QCD that provides the possibility of some analogies between two theories.
The value of the external sources we take as follow
\begin{itemize}\label{Num2}
\item symmetrical case:
\begin{equation}
\,q_1\,=\,q_2\,=0.1\,-\,1\,\,;
\end{equation}
\item non-symmetrical case:
\begin{equation}
\,q_1\,=\,0.2\,;\,\,\,q_2\,=0.3\,-\,1\,.
\end{equation}
\end{itemize}
As an example of the case of the strong coupling , i.e. large value of the triple vertex,
we also will present the quantum solution for the symmetrical case of
interaction at
\begin{eqnarray}\label{Num11}
\,&\,&\,\varrho\,=\,1\,\,\,\,(\mu\,=\,0.1\,,\,\lambda\,=\,0.1\,).
\end{eqnarray}

 Now it is instructive to consider the spectrum of the RFT-0 theory
at different $\varrho$. The Table\ref{EFUN} presents the found eigenvalues of the
theory, for $\,\varrho\,=\,5\,$ as well as for $\,\varrho\,=\,1\,$ and  for
$\,\varrho\,=\,3\,$.
\begin{table}[t]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\,\\
$\varrho\,$ &  $E_0$ & $E_1$ & $E_2$ & $E_3$ &
 $E_4$ & $E_5$ & $E_6$ & $E_7$ & $E_8$ &
 $E_9$ \\
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, & \\
\hline
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\,\\
$\varrho\,=\,1\,$ & 0.0546 & 0.292 & 0.609 & 0.997 &
 1.447 & 1.95 & 2.503 & 3.103 & 3.745 & 4.427  \\
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\,\\
\hline
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\, \\
$\varrho\,=\,3\,$ & 0.00217 & 0.142 & 0.297 &
0.507 & 0.759 & 1.047 & 1.37 & 1.724 & 2.108 & 2.579\\
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\,\\
\hline
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\, \\
$\varrho\,=\,5\,$ &  $ 1.36\cdot\,10^{-6}$ &
 0.144 & 0.178 & 0.276 &
 0.386 & 0.519 & 0.671 & 0.841 & 1.03 & 1.228 \\
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\,\\
\hline
\end{tabular}
\caption{\it The eigenvalues of Eq.\ref{Quant4} for different values
of parameter $\varrho$.}
\label{EFUN}
\end{center}
\end{table}
The asymptotic behavior
of the amplitude is clearly seen from the Fig.\ref{EFUN77} which represents
the behavior of the ground state as a function of  $\,\varrho\,$.
\begin{figure}[hptb]
\begin{center}
\psfig{file=eigenval1.eps,width=100mm}
\end{center}
\caption{\it\large The ground state of RFT as a function of $\varrho$.}
\label{EFUN77}
\end{figure}
Being important at small rapidity, all exited states, i.e.
states with eigenvalues $E_i,\,\,\,i=1..\infty\,$, at larger rapidities
are decreasing very rapidly to zero. Therefore,
at asymptotically large rapidity only the ground state survives,
which nonzero origin is due the tunneling effect of
the order $e^{-\varrho^2\,/\,2}$ which arises
between the Coulomb and harmonic wells of the potential
of the Eq.\ref{Quant4}, see \cite{0dimc,0dimq,0dimi}.
At large enough rapidity the  amplitude for any value of $\varrho$ is zero:
for small $\varrho$ it happens rapidly already at small rapidities, at large $\varrho$ huge rapidity needs.
As we will see further, this is a particular property of the model
with only triple Pomeron vertex, the theory with additional quaternary vertex
has precisely zero ground state.

\subsection{Numerical results: loops versus ''net'' diagrams}

  The results of the calculations
of the scattering amplitude $iT$ for the quantum and classical cases
are presented in the Fig.\ref{EFUN3}-Fig.\ref{EFUN6}
for the symmetrical values of the external sources.
In the Fig.\ref{EFUN66}-Fig.\ref{EFUN666} the results for
the non symmetrical case are represented.
These results are presented in the form of 3D plots as well, as in the form
of contour plots. The white color of the contour plot corresponds
to the maximum value of the z axis function of the 3D plot, and the  black
color , correspondingly, to the minimum value of the same function.
The amplitude is given as a function
of two variables, rapidity and value of sources.
In a symmetrical case the sources are the same, in non symmetrical case
we fix one sources, $q_1\,=\,0.2\,$, and the second variable
is the value of the second, external source $q_2\,=\,0.3\,-\,1\,$.
We also present the absolute ratio of the difference
between quantum amplitude and classical amplitude to the quantum amplitude,
$|\Psi(Y,\,q)\,-\,\Psi_{c}(Y,\,q_{c},\,p_{c})|\,/\,\Psi(Y,\,q)$,
in order to illustrate the relative contribution of the loops in the full solution
comparatively to the classical solution.
The rapidity variable Y of the plots is scaled, i.e.
in the plots the variable  $\,\mu\,y\,$ was used
instead usual rapidity $y$.
Therefore, in order
to obtain the value of the amplitude at some physical rapidity
it needs to divide the Y variable of the plot on
intercept $\mu$ : $Y_{phys}\,=\,Y_{plot}\,/\,\mu\,$. For any
value of intercept $\mu$ we have the plot of the amplitude
for rapidities $Y_{phys}\,=\,Y_{plot}\,/\,\mu\,$
at triple Pomeron vertex $\lambda$ equal to
$\lambda\,=\,\mu\,/\,\varrho\,\,$.
We see, that for the maximum value
of the $Y_{plot}\,=\,8\,$ at intercept, for example,
$\,\mu\,=\,0.3\,$ the normal rapidity is large: $\,Y_{phys}\,\sim\,27\,$
and covers the reasonable rapidity range which may be interesting in the high energy physics.

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=QS3VSym.eps,width=115mm} &
\psfig{file=QS3VSymContour_1.eps ,width=60mm}\\
 &  \\
\fig{EFUN3}-a & \fig{EFUN3}-b \\
 &  \\
\end{tabular}
\caption{\it The
quantum amplitude  $\Psi(Y,\,q)\,$ of the 3P vertex model
in the form of 3D and contour plots
at $\varrho\,=\,5\,$  as a functions of scaled
rapidity Y and symmetrical values of external sources $q_1=q_2$. }
\label{EFUN3}
\end{figure}

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=SemS3VSym_1.eps,width=115mm} &
\psfig{file=SemS3VSymContour_1.eps,width=60mm}\\
&  \\
\fig{EFUN4}-a & \fig{EFUN4}-b \\
 &  \\
\end{tabular}
\caption{\it The
classical amplitude  $\,\Psi_{c}(Y,\,q_{c},\,p_{c})\,$
of the 3P vertex model in the form of 3D and contour plots at
$\varrho\,=\,5\,$ and presented as a functions of scaled
rapidity Y and symmetrical values of external sources $q_1=q_2$.}
\label{EFUN4}
\end{figure}

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=Ratio3V_1.eps,width=115mm} &
\psfig{file=Ratio3V_1Contour.eps ,width=60mm}\\
 &  \\
\fig{EFUN5}-a & \fig{EFUN5}-b \\
 &  \\
\end{tabular}
\caption{\it
The ratio $|\Psi(Y,\,q)\,-\,\Psi_{c}(Y,\,q_{c},\,p_{c})|\,/\,\Psi(Y,\,q)$
in the 3P vertex model
is represented by 3D plot and contour plot. The plots are presented as a functions
of scaled rapidity Y at $\varrho\,=\,5\,$ and at $q_1=q_2=0.2-1$.}
\label{EFUN5}
\end{figure}

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=Ratio3V_2.eps,width=90mm} &
\psfig{file=Ratio3V_3.eps,width=90mm} \\
 &  \\
\fig{EFUN6}-a & \fig{EFUN6}-b \\
 &  \\
\end{tabular}
\caption{\it
The same ratio $|\Psi(Y,\,q)\,-\,\Psi_{c}(Y,\,q_{c},\,p_{c})|\,/\,\Psi(Y,\,q)$
in the 3P vertex model
is represented by 3D plots, now
at  the values of the sources $q_1=q_2=0.1-0.4$ in the Fig.7-a
and at  the values of the sources $q_1=q_2=0.4-1$ in the Fig.7-b
correspondingly. The plots are presented as a functions
of scaled rapidity Y at $\varrho\,=\,5\,$ . }
\label{EFUN6}
\end{figure}

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=QS3VAntisym_1.eps,width=115mm} &
\psfig{file=QS3VAntisymContour_1.eps,width=60mm}\\
 &  \\
\fig{EFUN66}-a & \fig{EFUN66}-b \\
 &  \\
\end{tabular}
\caption{\it
The quantum amplitude  $\Psi(Y,\,q)\,$ of the 3P vertex model
for the non symmetrical case
is represented by 3D plot and contour plot. The plots are presented as a functions
of scaled rapidity Y at $\varrho\,=\,5\,$ and at fixed $q_1=0.2$ with  $q_2=0.3-1$.}
\label{EFUN66}
\end{figure}

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=Ratio3VAntiSym_1.eps,width=115mm} &
\psfig{file=Ratio3VAntiSymContour_1.eps,width=60mm} \\
 &  \\
\fig{EFUN666}-a & \fig{EFUN666}-b \\
 &  \\
\end{tabular}
\caption{\it
The ratio $|\Psi(Y,\,q)\,-\,\Psi_{c}(Y,\,q_{c},\,p_{c})|\,/\,\Psi(Y,\,q)$
in the 3P vertex model
is represented by 3D plot and contour plot
at  the non symmetrical values of the
sources:  $q_1=0.2\,$ is fixed and  $q_2=0.3-1$ is a variable.
The plots are presented as a functions
of scaled rapidity Y at $\varrho\,=\,5\,$ . }
\label{EFUN666}
\end{figure}

  We also present a quantum amplitude  calculated
for the small  value of $\varrho\,=\,1$.
This value of $\varrho\,$ means large value of triple Pomeron vertex
and relatively large value of the ground state energy, see table
\ref{EFUN}.
From the plot Fig\ref{EFUN7} is clearly seen, that at large enough
rapidities the amplitude approaches
zero as it must be for such value of $\varrho\,$.

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=QS3VSymRo1.eps,width=120mm} &
\psfig{file=QS3VSymContourRo1.eps ,width=55mm}\\
 &  \\
\fig{EFUN7}-a & \fig{EFUN7}-b \\
 &  \\
\end{tabular}
\caption{\it The
quantum amplitude  $\Psi(Y,\,q)\,$ of the 3P vertex model
in the form of 3D and contour plots
at $\varrho\,=\,1\,$  as a functions of scaled
rapidity Y and symmetrical values of external sources $q_1=q_2$.}
\label{EFUN7}
\end{figure}

\subsection{''Effective'' Pomeron propagator}

 With the knowledge of the spectrum of RFT, the calculation
of the  ''effective'' Pomeron propagator $P_{eff}(y,q)$ is easy task.
First of all, we change the initial condition for our equations,
instead Eq.\ref{Quant6} now we have

\begin{equation}\label{Prop1}
\Psi(y=0,q)\,=\,I(q)\,=\,I(q\,,q_{ext})\,=\,q\,q_{ext}\,.
\end{equation}
The scattering amplitude $\Psi(y,q)$ now describes the Green's function
of transition from one Pomeron created from
source $q$ at rapidity zero
to any number N  Pomerons interacting with
sources $q_{ext}$ at final rapidity.

\begin{figure}[hptb]
\begin{center}
\begin{tabular}{ c c}
\psfig{file=green.eps,width=80mm} &
\psfig{file=greenro.eps,width=80mm}\\
 &  \\
\fig{EFUN8}-a & \fig{EFUN8}-b \\
 &  \\
\end{tabular}
\end{center}
\caption{\it The  $P_{amp}(y,q)$ and $P_{eff}(y,q)$
functions of the 3P model
at $\varrho\,=\,5\,$  and
$q_1\,=\,q_{2}\,=1/ \varrho\,=0.2\,$ as a functions of scaled
rapidity Y .}
\label{EFUN8}
\end{figure}
Therefore, in order to obtain requested "effective" Pomeron propagator
with only one Pomeron interacting with one source at final rapidity,
we also need to take the derivative of  $\Psi(y,q)$
over Pomeron field $q$ at $\,q\,=\,0\,$
and multiply obtained function on Pomeron field $q$ :
\begin{equation}\label{Prop2}
P_{eff}(y,q)\,=\,\left(\frac{d\,\Psi(y,q)}{d\,q}\right)_{q=0}\,\cdot\,q\,\,.
\end{equation}
The $P_{eff}(y,q)$ function is simply the second term of the Tailor expansion
of the $\Psi(y,q)$ around $q=0$.
Dividing obtained propagator $P_{eff}(y,q)$ on the sources $q\,q_{ext}\,$,
we obtain the Green's function of the theory, which does not depend on the values of the sources:
\begin{equation}\label{Prop22}
P_{amp}(y,q)\,=\,\frac{P_{eff}(y,q)}{q\,q_{ext}\,}.
\end{equation}
$P_{amp}(y,q)$ does not depend on the sources of the Pomerons fields
and in order to obtain requested propagator with given sources
we need simply to multiply $P_{amp}(y,q)$ on the values of these sources
$q\,q_{ext}\,$. Fig\ref{EFUN8} presents a plot of the  $P_{amp}(y,q)$ function
and plot for the $P_{eff}(y,q)$ function at $q\,=\,q_{ext}\,=1/ \varrho\,=0.2\,$.
The interesting question,
which we can address in this calculations, it is the question about the importance of $P_{eff}(y,q)$ in  RFT-0.
Namely, let's define with the help of $P_{eff}(y,q)$  the eikonalized ''effective''
Pomeron amplitude
\begin{equation}
\Psi_{eff}^{eik}(y,q)\,=\,1-e^{-P_{eff}(y,q)}\,
\label{EikProp}
\end{equation}
which account loops in the "effective" propagator
and does not account interactions between different propagators
and between propagators and external sources.
This amplitude could clarify the precision of this approximation in comparison with the full solution.
So, the required ratio is presented in the Fig\ref{EFUN10} .

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=EikRatio.eps,width=115mm} &
\psfig{file=ContEikRatio.eps,width=60mm}\\
 &  \\
\fig{EFUN10}-a & \fig{EFUN10}-b \\
 &  \\
\end{tabular}
\caption{\it The ratio
$|\Psi(Y,\,q)\,-\,\Psi_{eff}^{eik}(y,q)\,|\,/\,\Psi(Y,\,q)$
in the 3P model
in the form of 3D and contour plotshe 3P model
at $\varrho\,=\,5\,$  as a functions of scaled
rapidity Y and symmetrical values of external sources $q_1=q_2=0.1-0.2$.}
\label{EFUN10}
\end{figure}

\section{Solution of RFT in zero transverse dimension with triple and
quaternary Pomeron vertices}

 In this section we discuss a
RFT-0 model given by the following Lagrangian:
\begin{equation}\label{FV1}
L\,=\,q\,\dot{p}\,+\,\mu\,q\,p\,-\,\lambda\,q\,(q\,+\,p)\,p\,+
\,\lambda^{'}\,q^2\,p^2\,,
\end{equation}
where new quaternary Pomeron vertex $\,\lambda^{'}\,$ is introduced in
comparison to the Eq.\ref{Lag2}.
In the terms of the $q,\,p\,$ operators, see  Eq.\ref{Oper},
the Hamiltonian of the problem has the form
\begin{equation}\label{FV2}
H=\,(\mu\,q\,-\,\lambda\,q^{2})\,\frac{d}{dq}\,+
\,(\lambda\,q\,-\,\lambda^{'}\,q^{2})\,\frac{d^2}{dq^2}\,\,,
\end{equation}
and the full, quantum solution of the theory
will coincide with the solution of the quantum mechanical problem
determined by the new Hamiltonian  Eq.\ref{FV2}:
\begin{equation}\label{FV3}
H\,\Psi\,=\,\frac{d\,\Psi}{d\,y}\,\,,
\end{equation}
where the function
\begin{equation}
\Psi(y,q)\,=\,\sum_{n=0}^{n=\infty}\,\lambda_{n}\,
e^{\,-\,E_{n}\,y}\,\phi_{n}(q)\,\,
\end{equation}
is the full quantum amplitude
of two scattering particles. As before, $E_{n}$ and $\phi_{n}(q)$
denote eigenvalues and eigenfunctions of the Hamiltonian given
by the
Eq.\ref{FV2}, coefficients $\lambda_{n}$ are the normalized projections
coefficients of the eigenfunctions on the value of $\,\Psi\,$ at $\,y=0\,$.

  The
value of the quaternary vertex $\,\lambda^{'}\,$ is important for our following calculations.
If the vertex is very small,
$\,\lambda^{'}\,<\,\lambda\,/\,\varrho\,$ then all results of the Section 1
will stay the same with only very small corrections to the amplitude, it is was known many years
ago, see  \cite{0dimq}. Our calculations support this conclusion.
Therefore, for our calculations more interesting to take another
value of the vertex, so called ''magic'' value of $\,\lambda^{'}\,$, which is given by the following expression
\begin{equation}\label{FV33}
\lambda^{'}\,=\,\frac{\lambda}{\varrho}\,.
\end{equation}
The important property of the theory with this ''magic'' value of
$\,\lambda^{'}\,$ is that in this theory , unlike previous cases,
the ground state is precisely zero, as we will see later.
Another important feature of the Hamiltonian Eq.\ref{FV2}
with the ''magic'' value of the vertex
is that this model has precise correspondence
with the s-channel reaction-diffusion models, see \cite{BMMSX}, and with conformal
approach for the QCD Pomeron, see \cite{Korch,BondLang}.

   As in the previous model,  in this section we will also consider only the
calculations with the
value of the parameter  $\varrho\,=\,5$.
Nevertheless, the qualitative understanding of the behavior of the amplitude
at smaller values of $\varrho$ also will be clear.

\subsection{The quantum solution of the second model}

 The quantum solution
of the Hamiltonian Eq.\ref{FV2} is a solution of
the following second order differential equation
\begin{equation}\label{QFV1}
\,\lambda^{'}\,q\,(\,\lambda\,/\,\lambda^{'}\,-\,q\,)\,
\,\frac{d^2\,\Psi(y,q)}{dq^2}\,+\,\,\lambda\,q\,(
\varrho\,-\,q)\,\frac{d\,\Psi(y,q)}{dq}\,=\,\frac{d\,\Psi(y,q)}{d\,y}\,\,
\end{equation}
with the initial and boundary conditions on the function $\Psi(y,q)$:
\begin{eqnarray}\label{QFV2}
\,&\,&\,\Psi(y=0,q)\,=\,\sum_{n=0}^{n=\infty}\,\lambda_{n}\,
\,\phi_{n}(q)\,=\,I(q)\,\\
\,&\,&\,\Psi(y,q\,\rightarrow\,0)\,\propto\,q\,\\
\,&\,&\,\Psi(y,q\,\rightarrow\,\varrho)\,\propto \,const.\,\,,
\end{eqnarray}
where as before, the form of the  function $I(q)$ depends on the particular physical problem.
Very important property of Eq.\ref{QFV1} is the existing of the ground state
$\phi_{0}(q)$ with the zero energy $E_0\,=\,0\,$.
In this case the solution for Eq.\ref{QFV1} is trivial:
\begin{equation}\label{QFV3}
\phi_{0}(q)\,=\,1\,-\,e^{-\varrho\,q}\,.
\end{equation}
This ground state is not orthogonal to the other
eigenfunctions of the Hamiltonian, $\phi_{i},\,\,i=\,1\,..\,\infty\,$,
which are orthogonal to each other:
\begin{equation}\label{QFV4}
\int_{0}^{\varrho}\,\phi_{i}(q)\,\phi_{j}(q)\,F_{W}(q,\varrho)\,dq\,=
\,Const\,\delta_{i\,j}\,
\end{equation}
with the weight function
\begin{equation}
F_{W}(q,\varrho)\,=\,\frac{e^{\varrho\,q}}{q\,(\varrho\,-\,q)}\,.
\end{equation}
where
\begin{eqnarray}\label{QFV55}
\,&\,&\,\phi_{i}(q=0)\,=\,0,\,\,\,\,i=\,1\,..\,\infty\,;\\
\,&\,&\,\phi_{i}(q=\varrho)\,=\,0,\,\,\,\,i=\,1\,..\,\infty\,.
\end{eqnarray}
Using these properties of th eigenfunctions we obtain for the
$\,\lambda_0$ projection coefficient:
\begin{equation}\label{QFV5}
\lambda_0\,=\,\frac{I(\varrho)}{\phi_{0}(\varrho)}\,=\,\frac{I(\varrho\,,q_{ext})}
{\,1\,-\,e^{-\varrho^2}\,}
\end{equation}
All other projection coefficients, $\lambda_{i},\,\,i=\,1\,..\,\infty\,$,
may be calculated using the following expression:
\begin{equation}\label{QFV6}
\lambda_{i}(q_{ext})\,=\,\frac{\int_{0}^{\varrho}\,\phi_{i}(q)\,I(q\,,q_{ext})\,
F_{W}(q,\varrho)\,dq\,
}{\int_{0}^{\varrho}\,\phi_{i}^{2}(q)\,F_{W}(q,\varrho)\,dq\,}\,-\,
\lambda_0\,\frac{\int_{0}^{\varrho}\,\phi_{i}(q)\,\phi_{0}(q)\,
F_{W}(q,\varrho)\,dq\,
}{\int_{0}^{\varrho}\,\phi_{i}^{2}(q)\,F_{W}(q,\varrho)\,dq\,}\,.
\end{equation}
For the functions $\phi_{i},\,\,i=\,1\,..\,\infty\,$ the equation Eq.\ref{QFV1}
has the form
\begin{equation}\label{QFV7}
\,\frac{d^2\,\phi_i(q)}{dq^2}\,+\,\varrho\,
\frac{d\,\phi_i(q)}{dq}\,=\,-\,\frac{E_i}{\lambda^{'}\,q\,(\varrho-q)}\,\phi_i(q)\,,
\end{equation}
and with the help of the transformation
\begin{equation}\label{QFV8}
\phi_i(q)\,=\,f_i(q)\,e^{-\varrho\,q\,/\,2}
\end{equation}
this equation gets the following standard Shredinger form:
\begin{equation}\label{QFV9}
\,\frac{d^2\,f_i(q)}{dq^2}\,-\,\frac{\varrho^2}{4}\,f_i(q)\,=\,-
\,\frac{E_i}{\lambda^{'}\,q\,(\varrho-q)}\,f_i(q)\,
\end{equation}
for the $\,f_i(q)\,$ functions.
The function $f_i(q)$ are defined on the edges of interval as
\begin{eqnarray}\label{QFV10}
\,&\,&\,f_{i}(q=0)\,=\,0,\,\,\,\,i=\,1\,..\,\infty\,;\\
\,&\,&\,f_{i}(q=\varrho)\,=\,0,\,\,\,\,i=\,1\,..\,\infty\,,
\end{eqnarray}
and they are orthogonal each to other on the interval from $\,0\,$
to $\,\varrho\,$ with the new weight function $F_{W}^{f}(q,\varrho)$
\begin{equation}
F_{W}^{f}(q,\varrho)\,=\,\frac{1}{q\,(\varrho\,-\,q)}\,.
\end{equation}
Our problem we also can solve using hermitian
Hamiltonian defined by Eq.\ref{QFV9} with the projection coefficients
\begin{equation}\label{QFV66}
\lambda_{i}(q_{ext})\,=\,\frac{\int_{0}^{\varrho}\,f_{i}(q)\,I(q\,,q_{ext})\,
\,e^{\varrho\,q\,/\,2}\,F_{W}^{f}(q,\varrho)\,dq\,
}{\int_{0}^{\varrho}\,f_{i}^{2}(q)\,F_{W}^{f}(q,\varrho)\,dq\,}\,-\,
\lambda_0\,\frac{\int_{0}^{\varrho}\,f_{i}(q)\,\phi_{0}(q)\,
\,e^{\varrho\,q\,/\,2}\,F_{W}^{f}(q,\varrho)\,dq\,
}{\int_{0}^{\varrho}\,f_{i}^{2}(q)\,F_{W}^{f}(q,\varrho)\,dq\,}\,.
\end{equation}

 As was done in the previous section, solving the two value boundary problem
for differential equation Eq.\ref{QFV7} or Eq.\ref{QFV9},
we obtain the spectrum and eigenfunctions of the model and find
an amplitude for the different values of the sources $q_1,\,q_2\,$
and parameter $\varrho$:
\begin{equation}\label{QFV11}
\Psi(y,q=q_2)\,=\,\frac{I(\varrho\,,q_{ext})\,(1\,-\,e^{-\varrho\,q}\,)\,}
{\,1\,-\,e^{-\varrho^2}\,}\,+\,
\,\sum_{n=1}^{n=\infty}\,\lambda_{n}(q_1)\,
e^{-E_{n}\,y}\,\phi_{n}(q_2)\,\,.
\end{equation}

\subsection{The classical solution for the second model}

 Comparison between the contribution to the amplitude of diagrams with loops and without loops
may be performed again if we additionally will calculate classical solutions
for the Lagrangian Eq.\ref{FV1}:
\begin{equation}\label{CFV1}
L\,=\,\frac{1}{2}\,q\,\dot{p}\,-\frac{1}{2}\,\dot{q}\,p\,+
\,\mu\,q\,p\,-\,\lambda\,q\,(q\,+\,p)\,p\,+\,
\,\lambda^{'}\,q^2\,p^2\,+\,
\,q(y)\,p_0(y)\,+\,q_0(y)\,p(y)\,\,.
\end{equation}
The equation of motion for the fields p and q
in this case are the following:
\begin{eqnarray}\label{CVF2}
\,&\,&\,\dot{q}\,=\,\mu\,q\,-\,\lambda\,q^2\,-\,2\,\lambda\,q\,p\,+\,2\,
\lambda^{'}\,q\,p^2\,\\
\,&\,&\,\dot{p}\,=\,-\,\mu\,p\,+\,\lambda\,p^2\,+\,2\,\lambda\,q\,p\,-\,2\,
\lambda^{'}\,q^2\,p\,\\
\,&\,&\,q_0(y)\,=\,q_1\,\delta(y)\,\\
\,&\,&\,p_0(y)\,=\,q_2\,\delta(y-Y)\,\,.
\end{eqnarray}
The solutions of these equations, of course, are different
from the solutions of the equations of motion for the Lagrangian
Eq.\ref{Class1}, but the representation of the amplitude $\Psi_{c}(Y)$
in the terms of three classical solutions $\{q^{i}_{c},p^{i}_{c}\}$ of the system
Eq.\ref{CVF2} is the same as in the Subsection 2.2 ( see also
Fig.\ref{Traj4}) :
\begin{equation}\label{CVF3}
\Psi_{c}(Y)=1+exp\{-S(Y,\,q^{1}_{c}(y)\,,p^{1}_{c}(y))\}-
exp\{-S(Y,\,q^{2}_{c}(y)\,,p^{2}_{c}(y))\}-
exp\{-S(Y,\,q^{3}_{c}(y)\,,p^{3}_{c}(y))\}\,.
\end{equation}
Therefore, here
we will not repeat all steps of the  Subsection 2.2.
The picture for the classical solution derived for the
model with only triple Pomeron vertex is the same for the model
with both triple and quaternary vertices. The only difference between two models
is the value of critical rapidity $Y_c$ from which the additional
solutions arising, breaking the target-projectile symmetry of the initial
symmetrical solution, but for our consideration it is not important.

\begin{figure}[t]
\begin{center}
\psfig{file=solution8.eps,width=100mm}
\end{center}
\caption{\it Classical solutions of the RFT-0 with
triple and quaternary Pomeron vertexes:
 the $\{ q,p\}$ trajectories obtained for $Y=5\,>\,Y_c\,$, $q=p=0.7$,
$\rho\,=\,5$.}
\label{Traj4}
\end{figure}

\subsection{Parameters and asymptotic behavior of the amplitude in the second model}

  As in  the previous case, for the second model we will consider
quantum amplitude
$\,\Psi(Y,\,q)\,$  for the following value of
$\varrho\,=\,\mu\,/\,\lambda\,$:
\begin{eqnarray}\label{CVF4}
\,&\,&\,\varrho\,=\,5\,\,\,\,(\mu\,=\,0.2\,,\,\lambda\,=\,0.04\,);
\end{eqnarray}
and for the following values of the external sources (only symmetrical case ):
\begin{equation}\label{CVF5}
\,q_1\,=\,q_2\,=\,0.1\,-\,1\,.
\end{equation}
The ground state of this model has precisely zero eigenvalue, as we
showed before,
see  Eq.\ref{QFV3} and for other eigenvalues of the model see Table.\ref{QFV6}.
\begin{table}[t]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\,\\
$\varrho\,$ &  $E_0$ & $E_1$ & $E_2$ & $E_3$ &
 $E_4$ & $E_5$ & $E_6$ & $E_7$ & $E_8$ &
 $E_9$ \\
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, & \\
\hline
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\, \\
$\varrho\,=\,5\,$ & 0 & 0.182 & 0.189 & 0.307 &
 0.338 & 0.42 & 0.513 & 0.626 & 0.756 & 0.905 \\
\, & \,& \,& \, &\, &\, &\, &\, &\, &\, &\,\\
\hline
\end{tabular}
\caption{\it The eigenvalues of Eq.\ref{QFV7} for the value
of the parameter $\varrho$ equal 5.}
\label{CVF6}
\end{center}
\end{table}
Indeed,
now, at asymptotically
large rapidity, the amplitude is fully defined by the ground state:
\begin{equation}\label{CVF7}
\Psi_{asymp.}(q)\,=\,\lambda_0\,\phi_{0}(q)\,=
\,\frac{I(\varrho\,,q_{ext})\,(\,1\,-\,e^{-\varrho\,q}\,)}
{\,1\,-\,e^{-\varrho^2}\,}\,.
\end{equation}
In the case of the eikonal type initial conditions $I(q\,,q_{ext})$
the quantum amplitude $\Psi(y,q)\,$ at $q\,=\,\varrho$
has the following form
\begin{equation}\label{CVF8}
I(\varrho\,,q_{ext})\,=\,1\,-\,e^{-\varrho\,q_{ext}}\,,
\end{equation}
and, therefore, we obtain the following asymptotic behavior of our quantum amplitude :
\begin{equation}\label{CVF9}
\Psi_{asymp.}(q)\,=
\,\frac{(1\,-\,e^{-\varrho\,q_{ext}})\,(1\,-\,e^{-\varrho\,q})}
{\,1\,-\,e^{-\varrho^2}\,}\,.
\end{equation}
For the ''effective'' Pomeron propagator, $P_{eff}(y,q)$
the initial condition is different:
\begin{equation}\label{CVF10}
I(\varrho\,,q_{ext})\,=\,\varrho\,q_{ext}\,.
\end{equation}
Using the definition of the propagator given by Eq.\ref{Prop2} at
asymptotically large rapidity we obtain:
\begin{equation}\label{CVF11}
P_{eff}^{asymp.}(q)\,=\,
\,\frac{(\varrho\,q_{ext})\,(\varrho\,q)}
{\,1\,-\,e^{-\varrho^2}\,}\,.
\end{equation}
As we see, for each particular values of sources $q_1,q_2$ and parameter $\varrho$
neither quantum amplitude nor full Pomeron propagator are not decreasing
to zero at asymptotically large rapidity, approaching instead some constant
values defined by  Eq.\ref{CVF9}  and Eq.\ref{CVF11}  correspondingly.

\subsection{Numerical results for the  the second model}

  The results of our calculations
of the scattering amplitude $iT$ for the quantum case
are presented in the same from as for the first model.
Due the similarities between these two models, in this section
we present only quantum solution for the amplitude
for the symmetrical case of interactions, see
Fig.\ref{NFV1}. We also present
the ratio between the difference of the
triple Pomeron vertex model quantum amplitude and
quantum amplitude of the second model to the the
quantum amplitude of the first model :
$|\,\Psi^{3P}\,-\,\Psi^{4P}\,|\,/\,\Psi^{3P}$ ,
see Fig.\ref{NFV2}.

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=QS4VSym.eps ,width=115mm} &
\psfig{file=QS4VSymContour_1.eps ,width=60mm}\\
 &  \\
\fig{NFV1}-a & \fig{NFV1}-b \\
 &  \\
\end{tabular}
\caption{\it
The
quantum amplitude  $\Psi(Y,\,q)\,$ of the second model
in the form of 3D and contour plots
at $\varrho\,=\,5\,$  as a functions of scaled
rapidity Y and symmetrical values of external sources $q_1=q_2$.}
\label{NFV1}
\end{figure}

\begin{figure}[hptb]
\begin{center}
\psfig{file=QSRatio3V_4V.eps,width=170mm}
\end{center}
\caption{\it
The ratio of the
quantum amplitudes
$|\,\Psi^{3P}\,-\,\Psi^{4P}\,|\,/\,\Psi^{3P}$
in the second model
in the form of 3D  plot
at $\varrho\,=\,5\,$  as a functions of scaled
rapidity Y and symmetrical values of external sources $q_1=q_2$.}
\label{NFV2}
\end{figure}

\subsection{''Effective'' Pomeron propagator in the second model}

 The ''effective''  Pomeron propagator we define as
in the previous model
\begin{equation}\label{PFV1}
P_{eff}(y,q)\,=\,\left(\frac{d\,\Psi(y,q)}{d\,q}\right)_{q=0}\,\cdot\,q\,\,,
\end{equation}
with the initial condition given by
\begin{equation}\label{PFV2}
\Psi(y=0,q)\,=\,I(q)\,=\,I(q\,,q_{ext})\,=\,q\,q_{ext}\,.
\end{equation}
The asymptotic behavior of the $\,P_{eff}(y,q)\,$ at asymptotically large rapidities
is defined by expression Eq.\ref{CVF11}:
\begin{equation}
P_{eff}^{asymp.}(q)\,=\,\,\frac{(\varrho\,q_{ext})\,(\varrho\,q)}
{\,1\,-\,e^{-\varrho^2}\,}\,.
\end{equation}
For the values $q_{ext}\,=\,q\,=\,1\,/\,\varrho\,$, we obtain that at large rapidity
this function reaches unitarity limit see plot Fig.\ref{NFV3} and more detailed derivation
in ~\cite{BMMSX}.
The plot of the Green's function of the theory
\begin{equation}\label{Prop222}
P_{amp}(y,q)\,=\,\frac{P_{eff}(y,q)}{q\,q_{ext}\,},
\end{equation}
is presented in Fig.\ref{NFV3}.

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=green4v.eps,width=85mm} &
\psfig{file=green4vro.eps,width=85mm}\\
 &  \\
\fig{NFV3}-a & \fig{NFV3}-b \\
 &  \\
\end{tabular}
\caption{\it
The  $P_{amp}(y,q)$ and $P_{eff}(y,q)$
functions of the second model
at $\varrho\,=\,5\,$  and
$q_1\,=\,q_{2}\,=1/ \varrho\,=0.2\,$ as a functions of scaled
rapidity Y .}
\label{NFV3}
\end{figure}

\section{Diffractive dissociation process in RFT-0}

 As an example of applications of the methods developed
in the paper, we present calculations
of the diffractive dissociation process in the RFT-0 with only
triple Pomeron vertex.
The algorithm of the calculations is defined as follows.
We have the quantum solution of the theory
\begin{equation}\label{DD1}
\,\Psi_{1}(y,q)\,=\,\sum_{n=0}^{n=\infty}\,\lambda_{n}\,
e^{\,-\,E_{n}\,y}\,\phi_{n}(q)\,\,
\end{equation}
defined at all range of rapidity $Y$ at some values of the sources
$q_1$ and $q_2$. Let's now consider this solution at the rapidity interval
$Y_1\,<\,Y$.
With the help of the $\,\Psi_{1}(Y_1,q)\,$
we  define another function $\,\Psi_{2}(y,q)\,$
\begin{equation}\label{DD2}
\,\Psi_{2}(y,q)\,=\,\sum_{n=0}^{n=\infty}\,\tilde{\lambda}_{n}\,
e^{\,-\,E_{n}\,\left( y\,-\,Y_1\right)}\,\phi_{n}(q)\,
\end{equation}
on the rapidity interval $\,Y_1<\,\,y\,<\,Y\,$
with the following initial condition:
\begin{eqnarray}\label{DD3}
\,&\,&\,\Psi_{2}(y=Y_1,q)\,=
\,\sum_{n=0}^{n=\infty}\,\tilde{\lambda}_{n}\,
\,\phi_{n}(q)\,=\,
\,(\,\Psi_{1}(Y_1,q)\,)^{2}\,\,.
\end{eqnarray}
The illustration of this construction see in the
Fig.\ref{diffr}.

\begin{figure}[t]
\begin{center}
\psfig{file=diffr.eps,width=110mm}
\end{center}
\caption{\it
The single diffractive dissociation process for rapidity $Y_1$ at total rapidity
$Y$.}
\label{diffr}
\end{figure}

The only numbers that we need
to calculate now
are the coefficients $\tilde{\lambda}_{n}\,$. But their calculation
is trivial, they simply
\begin{equation}\label{DD4}
\tilde{\lambda}_n\,(Y_1)=\,\frac{\int_{0}^{\infty}\,\phi_{n}(q)\,
(\,\Psi_{1}(Y_1,q)\,)^{2}\,e^{-(q-\varrho)^2\,/4}\,
F_{W}(q,\varrho)\,dq\,
}{\int_{0}^{\infty}\,\phi_{n}^{2}(q)\,F_{W}(q,\varrho)\,
e^{-(q-\varrho)^2\,/2}\,dq\,}\,\,,
\end{equation}
where the weight function $\,F_{W}(q,\varrho)\,$
is the same as in the Eq.\ref{Quant9}.
The coefficients $\tilde{\lambda}_{n}\,$ fully determine requested
$\,\Psi_{2}(y,q)\,$ function, which represents the differential cross section of the
single diffraction process
at given and fixed rapidity interval
$$
y=Y_2=Y-Y_1
$$
Integrating this function over rapidity interval $\,Y_1<\,\,y\,<\,Y\,$
we obtain as the answer the sum of the total diffractive dissociation
cross sections on this rapidity interval and elastic cross section
\begin{equation}
\sigma_{SD}^{Tot}\,+\,\sigma_{el}\,=\,\int_{Y_1}^{Y}\,\Psi_{2}(y,q)\,dy
\end{equation}
So, the value of the
single difractive cross section integrated
over rapidity interval $Y_2$ is the following
\begin{equation}\label{DD5}
\sigma_{SD}^{Tot}=\,\int_{Y_1}^{Y}\,\Psi_{2}(y,q)\,dy\,-\left( \Psi_{1}(Y,q)\right)^{2}\,,
\end{equation}
where $Y$ is the total rapidity of the process,
$Y_1$ is the rapidity "gap" of the process,
$Y_2$ is the value of the rapidity taken by produced diffractive state and
elastic cross section of the process is defined as
$\sigma_{el}=\left( \Psi_{1}(Y,q)\right)^{2}$.
In general we do not expect that the numbers given by  RFT-0 theory
will be correct. The only interesting value in the given model, therefore,
is the ratio of the single diffractive cross section to the
total cross section.
\begin{equation}\label{DD6}
R(Y_1)\,=\,\frac{\sigma_{SD}^{Tot}}{\sigma_{tot}}
\end{equation}
where $\sigma_{tot}=2\Psi_{1}(Y,q)$.
Considering this ratio we can not prove,
of course, that the same value of the $R$
we will obtain in QCD as well, but, nevertheless,  it is interesting to see
what new about this ratio we obtain using full quantum solution for the amplitudes.
Another interesting ratio, which we can calculate in the given framework, this
is a ratio of differential single diffraction cross section to the
total cros section:
\begin{equation}\label{DD7}
R_1(Y_1)\,=\,\frac{\Psi_{2}(Y_2)}{\sigma_{tot}}
\end{equation}
In $R_1$ ratio the dependence on diffractive mass $Y_2$ is present, and it allows
to as to trace the energy dependence of the ratio of the diffractive state
to the total cross section. Therefore, the following calculations are presented.
We fix the value of diffractive state $Y_2$ equal 1-4 units of
scaled rapidity as a one variable and as a second variable in the 3D plot
we consider the value of total rapidity $Y$, which we take equal 4-8 units.
These plots we make for the two different symmetrical values of the
external sources: 0.2 and 0.7.
Obtained plots, see Fig.\ref{diffr1} for the ratio $R$ and
Fig.\ref{diffr2} for the ratio  $R_1$, show the changes of the ration accordingly to
the total rapidity changes.

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=TDifraction0_2.eps ,width=115mm} &
\psfig{file=ContTDifraction0_2.eps,width=60mm}\\
 &  \\
\fig{diffr1}-a & \fig{diffr1}-b \\
 &  \\
\psfig{file=TDifraction0_7.eps ,width=115mm} &
\psfig{file=ContTDifraction0_7.eps,width=60mm}\\
 &  \\
\fig{diffr1}-c & \fig{diffr1}-d \\
 &  \\
\end{tabular}
\caption{\it
The ratio $R$ for the case of symmetrical sources $q_1=q_2=0.2$
in the ~\fig{diffr1}-a - ~\fig{diffr1}-b
and for the symmetrical
sources $q_1=q_2=0.7$ in the
~\fig{diffr1}-c - ~\fig{diffr1}-d
correspondingly. The ratios are plotted
at $\varrho\,=\,5\,$  as a functions of scaled
rapidity Y and scaled rapidity variable $Y_2$.}
\label{diffr1}
\end{figure}

\begin{figure}[hptb]
\begin{tabular}{ c c}
\psfig{file=DDifraction0_2.eps ,width=115mm} &
\psfig{file=ContDDifraction0_2.eps,width=60mm}\\
 &  \\
\fig{diffr2}-a & \fig{diffr2}-b \\
 &  \\
\psfig{file=DDifraction0_7.eps ,width=115mm} &
\psfig{file=ContDDifraction0_7.eps,width=60mm}\\
 &  \\
\fig{diffr2}-c & \fig{diffr2}-d \\
 &  \\
\end{tabular}
\caption{\it
The ratio $R_1$ for the case of symmetrical sources $q_1=q_2=0.2$
in the ~\fig{diffr2}-a - ~\fig{diffr2}-b
and for the symmetrical
sources $q_1=q_2=0.7$ in the
~\fig{diffr2}-c - ~\fig{diffr2}-d
correspondingly. The ratios are plotted
at $\varrho\,=\,5\,$  as a functions of scaled
rapidity Y and scaled rapidity variable $Y_2$.}
\label{diffr2}
\end{figure}

\section{Discussion of results}

 The are the following main results of the paper.
First of all, we obtained the
full quantum solutions for the models and
compared it with the  classical solutions, illustrating
the relative contribution of the loops to the amplitude.
In order to find an useful approximation to the full quantum solution
we also considered an eikonalized amplitude which was built with
the help of the full two point Green's functions. All these amplitudes were calculated
for different values of external vertexes that allows to trace the applicability
of each approximation separately in different parameter's regions.
Another important result is a comparison of quantum solutions with and without quaternary Pomeron vertex
that clarify the contribution of this vertex to the process amplitude .

Plots
Fig.\ref{EFUN3}-Fig.\ref{EFUN666} represents the result of the calculations in the RFT-0 approach with
triple Pomeron vertex only.
The quantum solution of the model is presented in the plot
Fig.\ref{EFUN3}, whereas the classical solution is presented in the plot Fig.\ref{EFUN4}.
The comparison between these two solutions could be found in the
plots Fig.\ref{EFUN5}-Fig.\ref{EFUN6}.
From the Fig.\ref{EFUN3} we see, that our full solution for
small values of external sources does not reach unitarity really.
Instead of "black"
disk the "grey" one is achieved. This is may be easily explained
if we will remember that our initial conditions have a eikonal form,
whereas in the s-channel model, see \cite{BMMSX} and references therein, the propagator form of initial
conditions is always applied. In RFT-0 model the black disk limit, as it must be, achieved only at large values of the sources, that
corresponds to the nuclei interaction in usual QCD, see  \cite{braun1,braun2,BGLM}.
Other interesting results are represented in the Fig.\ref{EFUN5}-Fig.\ref{EFUN6}. Indeed, the natural question, which in fact is very
important in any practical calculations with BK equation involved, is a question about applicability of classical solutions
of the model. Namely, at which values of external sources the difference between the classical and quantum solutions
is acceptable from the point of view of the precision of the amplitude calculations. In real QCD this answer may be obtained only approximately, at least. In RFT-0 calculations the answer is clear, at large values of the external sources the classical solution is acceptable
only at large values of the sources. In the symmetrical case of interactions with the small sources involved, the relative difference between
the amplitudes is very large. The difference pretty fast decreases with the increasing of the value of the sources, but still the difference
is not negligible
even when $q=0.3\,-\,0.4$, see Fig.\ref{EFUN5}-Fig.\ref{EFUN6}. Important to underline, that one from the main reasons that provides the applicability
of the classical solution is a presence of the third classical solution arose at some
rapidity $Y_c$ ( see Section 2.2 an \eq{Class6}) in the framework.
The third solution provides the decrease of the classical amplitude
at large values of the total rapidity and without this solution the precision of the model was even worse.
The situation with the non-symmetrical case of interactions is better, see  Fig.\ref{EFUN66}-Fig.\ref{EFUN666}.
Due the initially broken symmetry of the interaction, the fan diagrams are initially dominant and provide the better
precision of the classical solution in comparison to the symmetrical case of interactions.

Considering the second possible
candidate on the role of "good amplitude approximation", eikonalized function \eq{EikProp}, see Fig.\ref{EFUN10},
we see that at small values of the external sources this amplitude is indeed better than the classical solution. In the region
of large values of the sources both amplitude are equal more or less - the unitarization corrections are small
for the large sources. We also could compare the effective Pomeron propagator
with the asymptotic results obtained  in the framework of s-channel model, see \cite{BMMSX}.
Looking in the
Fig.\ref{EFUN8} and Fig.\ref{NFV3}, we see, that the value of the
full propagator, which considered as an amplitude in s-channel model, is close to one, especially
for the second considered model with quaternary vertex included, which has direct relations with
s-channel model,  see again \cite{BMMSX}.
But, in general, for an arbitrary value of a source there is no
propagator's unitarization and the way to achieve the  unitarization in this
case is the eikonalization of the propagator only.

Interesting problem, which also could be investigated in the framework of RFT-0, is a problem of the
influence of the value of the triple Pomeron vertex on the behavior of the amplitude. Changing value
of the $\rho$ parameter from the $5$ till $1$ we change the value of the triple vertex $\lambda$
from $0.04$ till $0.1$. The resulting amplitude is depicted in the Fig.\ref{EFUN7}. From this plot we see,
that already at relatively small values of rapidity this amplitude is decreasing till zero.
This fast of the decreasing of the amplitude at
large values of $\lambda$ denote a main difference between the models of RFT-0 with and without
quaternary vertex. In RFT-0 without the quaternary vertex the amplitude, as it mentioned above, approaches zero
whereas in RFT-0 with the quaternary vertex the amplitude approach constant with any value of
triple vertex and correctly adjusted value of quaternary vertex, see \eq{FV33}, \eq{CVF9} and \eq{CVF11}.
Of course, when the value of the triple vertex is small, two models do not so differ, see
Fig.\ref{NFV2}. This difference in the behavior of the amplitudes in two models may be
also analyzed in the terms of s-channel model, see \cite{BMMSX} for the details and explanations.

As a possible example of the application of the model we calculated
the values of the ratios of the single diffractive and differential single diffractive cross
sections to the total cross section in the given framework, see Eq.\ref{DD6} and Eq.\ref{DD7}.
The results of these calculations are present in the
Fig.\ref{diffr1} - Fig.\ref{diffr2}. The plot
Fig.\ref{diffr1} represents the ratio of single diffractive cross section
(integrated over rapidity of the diffractive state) to the value of the total cross
section. In two cases when the sources are equal 0.2 and 0.7, we obtain, that
for the fixed value of the diffractive state this ratio almost does not change with energy.
More precisely,  in both cases there is a tiny change in the ratio behavior
at small rapidities 4-5, and constant behavior at large rapidiies 6-8.
Tracing the relative contribution of the diffractive state at fixed rapidity
in both cases we see, that the contribution of the
large mass diffractive state, $Y_2=4$, approximately
two times large than the contribution from small , "elastic", diffractive state
$Y_2=1$,
as it must be in reality. We see, that the simple RFT-0 model correctly reproduce
main futures of the real QCD. It is also interesting to note, that
the relative contribution of the large mass diffractive state is larger for the case of
small external sources, there is $R=0.11$  for source $q=0.2$ against
$R=0.028$  for source $q=0.7$.
Concerning the calculations of the second ratio,  $R_1$, see Fig.\ref{diffr2},
we obtain that these two cases are different. At small source, $q=0.2$, the maximum of the ratio is at
small diffractive state, which has almost constant behavior with total rapidity.
At large vale of source, $q=0.7$ , the situation is opposite, the maximim contribution
comes from the large mass diffractive state. The contribution of the small diffractive state
at $q=0.7$ is zero in Fig.\ref{diffr2} at large values of Y, that means
constant value of single diffraction cross section at small values of $Y_2$
and large values of $Y$. In general, considering Fig.\ref{diffr1} - Fig.\ref{diffr2},
we can conclude, that the description of the diffractive states
in RFT in the case when we include in calculations all possible re-scattering corrections
is different from the "naive" diffraction models, where only part of corrections
is included. Therefore, there is a hope, that using the same
receipts of calculations in the QCD RFT, we will be able correctly describe diffraction
data and their energy dependence.

\section{Conclusion}

In this paper we performed calculations of the different amplitudes in the framework of the RFT-0 model.
Comparing different approximation we analyze the applicability of the different
approximation schemes and their dependence on the parameters of the model. The main conclusion of the calculations is that we could trust
to the classical approximation for the amplitude only when the value of the triple Pomeron vertex is small.
In this region the difference between models with and without quaternary vertexes is negligible. Nevertheless, the important condition for this
approximation in the case of interacting of symmetrical particles is an including of the third classical solution in the set of classical solutions, which arises at some critical value
of rapidity. Without this solution the classical approximation is not good, even at not small values of the external sources.
For the case of the small values of these vertexes the eikonalized Green's function amplitude is more precise solution then the classical one. Unfortunately,
in real QCD this amplitude also could not be calculated precisely and therefore we could not trust to the classical solutions at all when the
values of the external sources is not large enough. Does the proton is "large enough" in the case of real QCD is an open question,
which could not be answered, unfortunately, in the given framework.

When the value of the triple vertex is growing,
the picture is changing drastically.
In this case we could not trust anymore to the classical solutions of the models. Also, there is
a large difference in the behavior of the amplitude in models with and without quaternary vertex, it's presence
changes the asymptotic behavior of the amplitude.
In real QCD it could means, that the different evolution equations must be
applied in the different regions of the impact parameter space with the different values of the coupling constant.
If we assume, that the influence of the non-perturbative effects is in the change of the value of the
coupling constant only, then we must separate the contribution of the perturbative spots in impact parameter space,
which evolution is described by the BK equation, from the non-perturbative regions where different evolution equations must be applied.
In this case the overall amplitude is a sum of the amplitudes described by the different evolution equations in different regions,
with the BK equation applicable in the regions of high partons density only.
This picture will lead to the factorization of the non-perturbative effects from the perturbative ones, first of all, and also it
could explain the possible applicability of the BK equation in case of proton-proton collisions. Namely, at high enough energies
the overall contribution of "black" spots may be larger then the contribution of the "white" and "grey" ones and
we could describe even inclusive data by the BK equation formalism. The large but constant non-perturbative contributions
into the amplitude in this case may be accounted by the adjusting of the values of the external sources in interaction of interests.

Concluding we underline, that RFT-0 model is a very interesting and important ground for an initial implementation
of the different ideas which further could be applied in real QCD calculations as well.

\section*{Acknowledgments}

I am grateful to Leszek Motyka for his participation in the development
of the main themes of the paper and to M.Braun for his support and interest to the paper during
period of it's writing.

\newpage

redpajama_set_name: RedPajamaArXiv


text:
Questlove, Book News, News, Music News
Questlove Announces New Book 'Music Is History'
Kathryn Milewski April 15, 2021
The Roots' Questlove announced he will be releasing a new nonfiction book called Music Is History. Set to appear on shelves October 12, the book will look at the last half-century of music and explain the importance of one song from every year since 1971.
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A post shared by Questlove (@questlove)
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Article Image: Questlove performing at a music festival. (benhoudijk via DepositPhotos.)
About Kathryn Milewski
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Posted 26 October, 2018
Seven Announces Launch Of 7News.com.au
By B&T Magazine
The Seven Network today announced that it will launch a new online home for its news and public affairs brand 7 News.
A standalone online news service, 7News.com.au, will debut in the first quarter of 2019 across desktop and mobile, with integration into major third-party news aggregators. This major advancement will also feature premium content Seven's breakfast and morning shows, Sunrise and The Morning Show, and Sunday Night.
Seven West Media's chief digital officer, Clive Dickens, said: "Over two years ago SWM began the process of taking control of our direct to consumer digital products.
"Today we are announcing the next big step of that strategy with the launch of 7News.com.au, which will utilise leading technology to deliver best in class user experience, deep client integration and on and off network distribution.
"We are confidently predicting 7News.com.au will become Australia's number one online news brand within six months of launch."
Seven's director of news and public affairs, Craig McPherson, said: "This multi-million dollar investment will marry Seven's hundreds of world-class journalists reporting from across the globe with best in class technology.
We are delighted to finally be fully in control of our news online, and can't wait to bring 7News.com.au to our audiences."
Seven's chief revenue officer, Kurt Burnette, said: "Seven News is Australia's leading and most trusted news brand, and with a predicted daily audience of 1.5 million, this new online platform will provide advertisers with a far more effective approach to client solutions, matching short form video, display, first-party data and native content to our already leading BVOD position."
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