AI can detect signs of nuclear weapons testing banned under the Comprehensive Nuclear-Test-Ban Treaty, according to research from the US Pacific Northwest National Laboratory.
A group of scientists have built a neural network to sniff out any unusual nuclear activity. Researchers from the Pacific Northwest National Laboratory (PNNL), one of the United States Department of Energy national laboratories, decided to see if they could use deep learning to sort through the different nuclear decay events to identify any suspicious behavior.
The lab, buried beneath 81 feet of concrete, rock and earth, is blocked out from energy from cosmic rays, electronics and other sources. It means that the data collected is less noisy, making it easier to pinpoint unusual activity.
The system looks for electrons emitted and scattered from radioactive particles decaying, and monitor the abundance of argon-37, a radioactive isotope of argon-39 that is created synthetically through nuclear explosions.
Argon-37 which has a half-life of 35 days, is emitted when calcium captures excess neutrons and decays by emitting an alpha particle. Emily Mace, a scientist at PNNL, said she looks for the energy, timing, duration and other features of the decay events to see if it's from nuclear testing.
Deep learning makes that process easier. Computer scientists collected 32,000 pulses and annotated their properties, teaching the system to spot any odd features that might classify a signal as 'good' or 'bad'.
"Signals can be well behaved or they can be poorly behaved," said Jesse Ward. "For the network to learn about the good signals, it needs a decent amount of bad signals for comparison." When the researchers tested their system with 50,000 pulses and asked human experts to differentiate signals, the neural network agreed with them 100 per cent of the time.
It also correctly identified 99.9 per cent of the pulses compared to 96.1 per cent from more conventional techniques.
"Deep learning is making it easier for us to filter out a small number of good events that are indicative of the activity of interest," said Craig Aalseth, a nuclear physicist and PNNL laboratory fellow.


text:
Tayo Jessica Onutor, nacida Awosusi (Karlsruhe, 8 de marzo de 1978), también conocida como TAYO, es una cantante alemana de R'n'B y soul que canta en inglés, alemán y romaní. Ella misma se describe como Afro-Sintezza. No solo es musicalmente activa sino también lo es políticamente. Hizo campaña por los derechos de los sinti y los romaníes en los grupos feministas IniRromnja y RomaniPhen, y también es políticamente activa en varias otras comunidades de color.

Trayectoria
La madre de Tayo Awosusi es la alemana Sintezza y activista de los derechos civiles Anita Awosusi, que hace campaña por los derechos de los sinti y los romaníes y es autora de investigaciones en este campo. Su padre es el músico de soul y funk Hope Awosusi. Awosusi creció en Karlsruhe, donde se graduó de la escuela secundaria en 1997. El nombre Tayo proviene del yoruba nigeriano y significa "niño de la alegría". Después de estudiar en las Universidades de Heidelberg y el Instituto Tecnológico de Karlsruhe (alemán, inglés, multimedia), Awosusi se mudó a Berlín en 2004 y desde entonces se dedica a su música.

Carrera musical
Tayo creció en una familia de músicos. A los nueve años subió por primera vez a un escenario como cantante invitada en la banda de funk y soul de su padre (The Funky Breeze Band). Más tarde se convirtió en miembro permanente de la banda y también trabajó con otros músicos internacionales.

En 2004 se trasladó a Berlín y trabajó con Rudy Stevenson ( Orquesta de Duke Ellington ) y Ferenc Snétberger, entre otros. En 2005, Awosusi grabó la canción Kelas-Let's dance junto con Snétberger para la película de anime japonesa Fullmetal Alchemist - The Movie: Conqueror of Shamballa. A finales de 2007, Awosusi fue nominado por la Fundación Africana de la Juventud - AYF en la categoría de Medios/Animadores como Mejor Artista Pop/R'n'B.

Awosusi sincronizó varias voces resp. Actuaciones vocales para series de televisión como Cold Case, Hello Holly, Todd's Amazing World y Ricky Sprocket . Awosusi está trabajando actualmente con su productor Loomis Green (guitarrista de Jan Delay) en su primer álbum.

Otras actividades
Tayo Awosusi se involucró como escritora y performer en la performance feminista Me volví decidida contra la tutela, una lectura performativa de la IniRromnja con aportes fílmicos en memoria de la cineasta Melanie Spitta.

En 2017 estrenó la película documental Phral mende - Wir über uns. Perspectivas de Sinti* y Roma* en Alemania, que se estrenó en el Museo Judío de Berlín.

Obras
 Tayo Onutor: Con una solicitud de introducción, en: Dossier: Perspectivas y análisis de los sinti y los gitanos en Alemania, ed. de la Fundación Heinrich Böll, 3. diciembre 2014
 Tayo Awosusi-Onutor: " Hemos aprendido a ser aceptados " - Pensamientos sobre #metwo, ed, por la Fundación Heinrich Böll,22. noviembre de 2018
 Tayo Awosusi-Onutor: " Kek Ducho - Una perspectiva romaní negra sobre el racismo y la cultura del recuerdo ", ed. de la Fundación Heinrich Böll, 12. junio 2020

Referencias

Enlaces externos

Nacidos en Karlsruhe
Gitanas
Música gitana
Activistas de derechos civiles
Artistas feministas
Cantantes de Alemania

redpajama_set_name: RedPajamaWikipedia


text:
1993 in Korea may refer to:
1993 in North Korea
1993 in South Korea


text:
Catch up on the latest Cushman & Wakefield news and find out what's next for our company
Zugdidi Shopping Mall
Cushman & Wakefield has been recently assigned to manage the leasing campaign of the first modern shopping center in Zugdidi.
The project is set to introduce the first contemporary, full-service shopping center in Zugdidi, combining fashion retail, hypermarket, F&B units and other stores under one roof. Negotiations are currently underway with major brand operators with established presence on the Georgian market.
Blox is the developer behind the project. Construction process will start this month and end by the end of the year. Total projected investment is 6 million USD, with developable area exceeding 15,000 square meters.
The project will provide employment opportunities on the local job market while also contributing to the economic growth of the region.
What's making the CEE real estate market tick in 2020?
Booking.com's New Office in Tbilisi
Cushman & Wakefield named world's top CRE advisor by EUROMONEY
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redpajama_set_name: RedPajamaCommonCrawl


Looking for homes close to Platt Middle School? Here you'll find all listings of homes for sale near Platt Middle School and also a quick search list to check out houses around other schools as well. Schools can be a huge factor in your home search, especially if you have a family. Our team has expert local knowledge on schools and Front Range neighborhoods in general. Let us know what you are looking for and if you would like to know anything about the areas you are looking in. We can help you find the home and lifestyle you are looking for!
Your current Search includes 55 properties. The average price of the properties currently displayed is $824,861. The median price for the real estate is $349,900.


text:
Tamara R. Crawford
tcrawford@ds-lawyers.com
Round Rock, Texas 78664
Ms. Crawford represents clients in a variety of commercial and business related matters and litigation. Her primary focus is on first party insurance litigation.
Prior to joining Doyen Sebesta & Poelma, LLLP, Ms. Crawford practiced in the Houston area and gained valuable experience by conducting depositions, successfully resolving cases through mediations, and arguing at the district and county level in contested hearings involving commercial litigation matters. Ms. Crawford successfully first chaired her first trial in the Harris County Civil District Court, gaining a favorable judgment after two days of trial.
Ms. Crawford earned her Bachelor of Arts degree in Government from The University of Texas at Austin in 2007. After graduating, Ms. Crawford spent the next few years working for the Attorney General of Texas. She then attended St. Mary's University School of Law and earned her Juris Doctorate in 2013. While at St. Mary's University, Ms. Crawford was active in Student Government and studied at the St. Mary's Institute on World Legal Problems at the University of Innsbruck in Austria.
J.D., St. Mary's University School of Law, 2013
Pro Bono Service Recipient
B.A. Government, The University of Texas at Austin, 2007
Liberal Arts Deans List
Professional & Bar Association Memberships
Texas State Bar
Houston Bar Association
Houston Young Lawyers Association
Hispanic Bar Association of Houston
Mexican American Bar Association of Houston
Tamara R. Crawford | Houston Commercial Litigation Lawyers Doyen Sebesta & Poelma, LLLP


devkitA64 release 11 is now available via pacman. Use sudo (dkp-)pacman -Syu to update or sudo (dkp-)pacman -S devkitA64 to install.
This fixes an issue with getcwd returning a path with trailing slash and removes iosupport.h from dirent.h which caused issues with some -std options.


text:
What Happens at Mile 22: Lange and Frodeno to Battle in Frankfurt
Patrick Lange running to a course record marathon in Kona last year.
Kona favorites Patrick Lange and Jan Frodeno have some history, both on this course and as a lead-up to October. What will Sunday hold for the German favorites?
by Volker Boch
It's time for the battle of the IRONMAN summer season. On Sunday, June 8, the 2018 Mainova IRONMAN European Championship, Frankfurt will welcome a thrilling duel in the men's race. All eyes will be on reigning IRONMAN world champion, German Patrick Lange, and his compatriot—the back-to-back 2015 and 2016 Kona victor, Jan Frodeno. The 17th edition of this impressive event.
Twenty-two might just be the number Patrick Lange has on his mind for this Sunday. Just a few days away from the IRONMAN European Championship, the reigning IRONMAN world champion has two reasons to focus on this number. One is that mile 22 in the marathon might be the moment for Lange to pass Frodeno. Marathoners say that there is always a reset of sorts in long-distance events—that 6 kilometers to go marks the point where marathons really start.
"I want to race the best," says Lange with his eyes on the start list for Frankfurt. This list includes numerous strong pros, and if there's something everyone should know about triathlon, Frodeno is one of them. The 36-year-old is laser-focused, remarkably talented in all three disciplines, a expert in mental strategy, and ready to go all in when it counts.
Lange, currently one of the top runners in triathlon, will be ready to put the pressure on at mile 22—just like he did in Kona. But there is another "22" he might think of when it comes down to the wire. This number has to do with Frodeno's personal Frankfurt history.
July 6, 2014 is a date to remember in Frodeno's outstanding career. It was the day he did his first IRONMAN race. Everyone in Frankfurt was talking about Frodeno as the top favorite: after winning Olympic gold six years prior in Beijing, among other impressive results, there was no doubt about his talent to reign over longer races.
Frodeno set the race on fire from the swim in Lake Waldsee, exiting the water in first. He then flatted three times on the bike—out of the game at the front, but regardless, he kept going. "This race was much more of a lesson to me than others," he says, looking back. "Never give up!" he adds—his key learning in his first IRONMAN. Once on the marathon course, the issues started again. Due to cramping legs, the triathlon star had to stop again and again. "Twenty-two times!" Frodeno recalls. He fought hard, and finished third behind experienced pros, Sebastian Kienle and Frederik van Lierde.
One year after that, Frodeno celebrated a remarkable comeback race in Frankfurt (pictured above)—three months before celebrating the first of his two consecutive Kona victories. Facing this year's European Championship, it looks like he is back to repeat, and pave the way to his third IRONMAN world title.
Despite the hype built between the two Germans, Frodeno knows that he faces more than just one strong contender. Aussie Jos Amberger is one, as is Andi Boecherer, last year's runner up. Frodeno's own training mate Nicholas Kastelein (AUS) will be racing, as well as last year's third-place Patrik Nilsson from Sweden. Other fast guys—like IRONMAN South African champ Tyler Butterfield—willl all be hungry for a podium spot as well.
A few weeks ago, Frodeno smashed it at the IRONMAN 70.3 Kraichgau. He put six minutes on Lange and nailed the fastest splits in each discipline, including the run. As two of the most gifted runners in the pro field, the last time Lange and Frodeno saw each other in a race at mile 22 of the marathon was in Kona—when Frodeno suffered a bad day and cheered for Lange out on the course to make the pass on Lionel Sanders. Lange eventually did and took the historic win.
"If there is any person who can handle a disadvantage after the bike, I should be this person," Lange said going into the Frankfurt race. He knows that he might be minutes back after the bike, just as he has been in Kona in the past two years, but knows that this race will be all about the marathon again. And he's prepared to go: "The second part of the marathon is my time," says Lange; he is hungry for number 22.
Tune in to IRONMAN Now on Facebook Watch for LIVE coverage of Sunday's big championship!


text:
You are here: Home1 / Analyst: These Must Happen for BTC's Uptrend
Analyst: These Must Happen for BTC's Uptrend
Benjamin Cowen, a popular crypto analyst, says he is keeping a close eye on an important indicator for the Bitcoin uptrend. But he says that even after making a strong rally, he points out that Bitcoin (BTC) still has a lot to prove.
In a new market review he posted on YouTube, Benjamin Cowen told his 782,000 YouTube subscribers that Bitcoin's "Relative Strength Index" remains at historic lows. (Relative Strength Index: It is a momentum indicator and gives information about the overbought and oversold points of the asset.)
"Very nice move from $15,000 to $21,000 by bitcoin. Note that the monthly RSI in November was lower than in June. The monthly RSI has gone all the way back to 44 to 45. What I want to do is draw a line to where it is right now. Just above this low [in 2019] and basically at the same level [in 2015.] There is still something to be proven here for Bitcoin. It should still close above these levels before we show any convincing evidence here on the monthly timeframe and Keep in mind that in 2015, the monthly RSI basically hit lows for three quarters of the year. Until September, the monthly RSI didn't really change at all. We've only been here for a few months in 2019."
Meanwhile, Cowen also mentioned that the biggest cryptocurrency needs to break above historic lows to confirm that it has turned the current resistance level into support. He also advised investors to keep their eyes on the RSI indicator.
"And what's really interesting this time is that although we're back at $21,000, we're still a little above the levels we were at in December 2018. So I think we need to keep an eye on the monthly RSI. It can really rise above the levels it was in 2015 and then in 2019. that."
January 19, 2023 /0 Comments/by ttcuser
Tags: benjamin cowen, bitcoin, btc, Relative Strength Index, Rsi
https://www.toptrendcoins.com/wp-content/uploads/2023/01/benjamin-cowen-1.jpg 534 1024 ttcuser https://www.toptrendcoins.com/wp-content/uploads/2022/10/tt-coins-logo.png ttcuser2023-01-19 14:37:052023-01-19 14:39:30Analyst: These Must Happen for BTC's Uptrend
Ripple CTO Thinks Bitcoin Should Switch To PoS
El Salvador Won't Release BTC Purchase Information
Bitcoin Mining Initiative From Oil Giant Shell?
Analyst: "This Level Like a Magnet for BTC"
Inflation Doping in Bitcoin and Cryptocurrencies
Bitcoin Price Drops Below Mining Cost
Fang: "Three Catalysts Could Move Bitcoin"
Bitcoin Move from Technology Giant Samsung
© Copyright - Top Trend Coins - Enfold Theme by Kriesi
Algorithm Sets A Price For Solana Scaramucci: "It's Bitcoin Time"


Rarely do you ever have to spend time trying to figure out the solution to a problem with the Huawei Honor 8 smartphone because most things can be fixed by simply taking a hard reset instead, a bit like utilizing a get out of jail free card.
Despite many people saying otherwise, a hard reset and a factory reset on the Huawei Honor 8 device will yield the same outcomes: the phone will be restored to its default, factory state on both occasions. In other words, the same state that it was in when it left the factory responsible for manufacturing it. This means you shouldn't expect to find any applications that you installed yourself still on the device. Any settings that you changed will be returned to how they were when you first opened the Huawei Honor 8 out of the box too. And, of course, unless you took a backup, all the data that you accumulated—pictures, documents, music, so forth—will no longer be available once the hard reset is complete.
You shouldn't worry about a hard reset causing any harm to your Huawei Honor 8 smartphone; provided that you've backed up any data that you didn't want to lose, very little else can go wrong.
1. Turn off the Huawei Honor 8 smartphone by long-pressing the Power button and then selecting the option to switch it off from the menu.
2. Boot the Huawei Honor 8 smartphone up again by pushing the Volume Up and Power buttons simultaneously.
3. Wait until you can see the Huawei logo before letting go of the two buttons.
4. Highlight the "wipe data/factory reset" by scrolling with the Volume buttons and then confirm you want to apply the factory reset by pressing the Power button.
5. When it asks you if you're sure you want to continue, select the "Yes" option.
6. Once the reset has been applied, select the "reboot system now" to get your smartphone back into its regular operating system.
That is all you need to hard reset the Huawei Honor 8 smartphone.
Applying a hard reset is probably one of the more commonly exchanged phrases for when people are referring to implementing a factory reset, but the same guide above is also good for a few other causes. The same steps would also work if you just wanted to wipe all of the data on your Huawei Honor 8 smartphone, for instance. Likewise, you could also use the same steps for bypassing the screen lock or just to restore the defaults on the Huawei Honor 8 device.


Manipal is well connected with all the major cities in Karnataka, Kerala, Goa etc. Buses run between Mangalore and Manipal frequently. Direct buses are also available from Bangalore, Goa, Hyderabad, Mumbai and other cities.


Many marketing professionals have big goals, but just don't get a voice when it matters. B2B Marketing Strategies Training: How to Set Strategy for Business Markets outlines how to get a seat at the strategy table; and earn the right to stay there.
How to set, and guide, the overall strategy for the whole business. You need it to accelerate quickly from its current position, leave its competition for dead and be strong enough to last the distance.
How to set the go-to-market strategy for your products and services. This includes deciding how much effort to give each (and which to get rid of), how to balance your portfolio and how to alter your strategy when the market changes.
How to select and evolve your product, pricing, segmentation, channel and tactics; because what's a great strategy one year can be a dud the next.
How to translate your strategy into a clear plan, and then into action; because you need to drive your killer strategy home with an execution to match.


Google Chrome for Android has been updated on Google Play. This is the first such update since Google's mobile browser exited beta at Google I/O back in June. There are no major feature additions in the new version 18.0.1025308, but there is a selection of fixes for "medium"-rated security issues, as well as stability fixes.
The new Chrome update seems to speed up the loading of image-heavy sites like your regular news website.
If you're already running Chrome, head to the "My Apps" menu in the Google Play Store to grab the new update. Note that you still need to be on Android 4.0 Ice Cream Sandwich or 4.1 Jelly Bean to install Chrome.


Please list fill in your WEEKLY expenses.
This luxury Mercedes Coupe is built with two things in mind, power and style.
The 5L V8 petrol engine makes itself known the second you turn the engine on. Needless to say this is an amazing car to drive with the capability of reaching 100 kph in around 6 seconds. There is also the option for tiptronic within the 5-speed automatic gearbox if your really looking to see what it's capable of. Packing over 300 bhp, which is 75 more than the 320 model, you really get added power that's noticeable.
Inside the cabin, this model is finished with a faultless cream leather interior that has been kept immaculate by previous owners - with no fading, yellowing or rips. The cream is complemented by a subtle wooden trim around the dash and doors. The seats in this model have been moved to a more upright position, increase driver visibility as well as avoiding the "slumped" position that was criticized in previous models. There's also the telescopic, adjustable steering wheel and added leg room in the rear seats.
In terms of features, the CLK has 10-way adjustable seats with three memory settings, ABS with emergency brake assist and electronic stability control. The 500 model comes with factory 17" 5-spoke alloys on low profile tyres (only an option on the 320).
If it's not the immaculate interior or exterior that makes you want this car, just drive it. With summer coming up this is the perfect coupe to take on the open road.
See below for the photo album and listing link.
At Drive Town, we use a range of different finance companies to help get you in your new car as soon as possible.
If you've had previous BAD CREDIT but we can see you're back on track, we may be able to offer finance where other companies wouldn't.
WORK VISA: If you're in New Zealand on a Work Visa and want finance over a period longer than your Visa, talk to us. There are situations where we may be able to help.
If you are on a LEARNER LICENCE and would like to finance a car. Guarantor/Co-Borrower is generally required.
The main benefit of our In-House Finance is that we can take each application on a case-by-case basis, giving you the best chance to get into the car of your dreams!
We can also offer the usual finance options at a competitive rate. If any of the following apply to you, get in contact today!
Lending is subject to lending criteria. Weekly rates shown on website are based on 48 month term, no deposit. Example annual fixed interest rate of 13.75%. Figure includes admin and monthly fees. Actual weekly figures may vary (higher/lower) depending on financial circumstances. Estimate only. Terms and conditions apply. Further T&C's apply. Lending subject to personal finance circumstances, may be subject to change.
Head to the Finance Page on the website to apply now!
Alternatively, come in and talk to one of the team at 522 Ellerslie-Panmure Highway, Panmure!
10 cars have just arrived from Japan, with 20 more arriving in the next two weeks. We expect these to be through compliance and appearing on the yard in the next week or so.
Below are a few photos from the port with a sneak peak at what we have coming. If you'd like to secure any vehicle before it arrives please get in contact - same goes for any first photos of cars on the port in the next few weeks.
Out of Town Special - Flights on Us!
At Drive Town, we want to ensure everyone has easy access to our vehicles and service.
Until the end of July, we are offering to fly you up to Auckland to pick up your new car. The offer is available to anyone in New Zealand, purchasing any car off our yard. All you have to do is drive it home. It's a great opportunity for you to get to know your new vehicle and become apart of the Drive Town experience.
The deal is, we'll cover the cost of your flights and pick you up from the Airport - Easy as!
Leave a reply below or send us an email if you'd like to find out more.
Offer ends 31st July 2018.


text:
Melanochromis es un género de peces de la familia Cichlidae y de la orden de los Perciformes. Es un endemismo del lago Malawi en África Oriental.

Especies
 Melanochromis auratus (Boulenger, 1897)
 Melanochromis baliodigma Bowers & Stauffer, 1997
 Melanochromis benetos Bowers & Stauffer, 1997
 Melanochromis brevis Trewavas, 1935
 Melanochromis chipokae D. S. Johnson, 1975
 Melanochromis cyaneorhabdos Bowers & Stauffer, 1997
 Melanochromis dialeptos Bowers & Stauffer, 1997
 Melanochromis elastodema Bowers & Stauffer, 1997
 Melanochromis heterochromis Bowers & Stauffer, 1993
 Melanochromis interruptus D. S. Johnson, 1975
 Melanochromis joanjohnsonae (D. S. Johnson, 1974)
 Melanochromis johannii (Eccles, 1973)
 Melanochromis lepidiadaptes Bowers & Stauffer, 1997
 Melanochromis loriae D. S. Johnson, 1975
 Melanochromis melanopterus Trewavas, 1935
 Melanochromis parallelus W. E. Burgess & Axelrod, 1976
 Melanochromis perileucos Bowers & Stauffer, 1997
 Melanochromis perspicax (Trewavas, 1935)
 Melanochromis robustus D. S. Johnson, 1985
 Melanochromis simulans Eccles, 1973
 Melanochromis vermivorus Trewavas, 1935 (Purple Mbuna)
 Melanochromis xanthodigma Bowers & Stauffer, 1997

Referencias


The recipes in this category are listed as gluten-free adaptable instead of gluten-free because most of them have one or two ingredients that are gluten-based but could easily be changed to a gluten-free option (pasta, tortillas, some Asian sauces, etc.). Please read the recipes carefully to pinpoint the ingredients that need to be swapped out for a gluten-free alternative! In most cases the change should be fairly obvious giving you a great gluten-free recipe with only a few slight variations.


London_Waterloo_Interior_Rush_Hour_2,_London,_UK_-_DiliffBah hambug? Not this year according to reports that place Britons among the world's most generous shoppers, spending nearly £700 each on festive items. It's good tidings for retailers who are gearing up for another busy Christmas.
According to Route, every two weeks 1,009,646 commuters are influenced to buy products that they've seen advertised on AdGates. Last year, Argos used AdGates to promote its latest deals and offers to commuters with great success. In fact, commuters who have visited Argos in the past 3 months have generated 29,683,520 impacts every two weeks!
Argos isn't the only retailer which produces high impact and reach figures. The table below showcases the numbers of commuters who engage with AdGates and have visited popular retailers in the last 3 months.
With Christmas just around the corner, it won't be long before footfall increases in train stations as Britons take advantage of the extra hours of shopping. Call Mark on 020 7233 9777 to find out how AdGate advertising can give your client a Christmas to remember!


Here's the flyer for the Fambridge River Festival on Sunday 24 May. SSBR will be there. We shall have our Archive open for the day and will also have a tent by the riverside with lots of information about Thames sailing barges.


text:
\section{Introduction}
In September 2011 we encountered a remarkable and unbelievable
paper by the OPERA collaboration \cite{OPERA} that {\bf the speed
of neutrino exceeds that of light in vacuum}. They measured a collective
speed of mu--neutrino flying from CERN to Gran Sasso Laboratory (
see for example the Fig. 5 in \cite{OPERA}).
However, this result conflicts with special relativity in the most
basic sense.

After the paper appeared in the arXiv researchers in the world
published or are preparing many papers on this topic.
Some of them agree with the result, while others don't agree
according to each author's conviction of  the special relativity.
See the hep-ph in the arXiv.

We cannot conclude whether this experiment (detection) is correct
or not at the present time. Of course, it must be checked by other
experiment teams.

Some researchers protested against the OPERA detection strongly.
One of main reasons is due to the Kamiokande detection in 1987
\cite{Kamiokande}.
It detected lights and neutrinos coming from the Supernova SN1987A
at almost the same time. If the speed of neutrino is faster than
that of light in vacuum, it must have detected neutrinos several years
earlier.

By the way, Mecozzi and Bellini  in \cite{MB}\footnote{K.F gave a small
contribution to this paper, see Acknowledgments of the paper.}
gave a smart interpretation of the result. They suggested that
the result is due to {\bf the superluminal group velocity} of neutrinos
arising from superposition (namely, the neutrino mixing) in
Quantum Mechanics \cite{PD}.
The neutrino mixing which is well--known in particle physics is just
a quantum mechanical phenomenon.

However, coherence in Quantum Mechanics is affected by
environments and it could be destroyed in short time.
A long--distance flight from SN1987A might have destroyed coherence
of neutrinos, and as a result the superluminal group velocity was lost.
The paper \cite{MB} offers an interpretation that the OPERA result
does not conflict with Kamiokande detection.

In this note we provide mathematical reinforcements to the paper \cite{MB}
in terms of {\bf decoherence} and would like to offer a
``super--smart" interpretation to the OPERA result \cite{KF}.

\section{Superluminal Group Velocity of Neutrinos}
In this section we review the paper \cite{MB} in detail (because it is
a bit unclear from the mathematical point of view).

First, we prepare some notation for convenience. Since
we treat a two level system in the following the target space is
${\bf C}^{2}=\mbox{Vect}_{{\bf C}}(|\phi_{1}\rangle, |\phi_{2}\rangle)$
with bases
\[
|\phi_{1}\rangle=
\left(
\begin{array}{c}
1 \\
0
\end{array}
\right),\quad
|\phi_{2}\rangle=
\left(
\begin{array}{c}
0 \\
1
\end{array}
\right).
\]
Then Pauli matrices $\{\sigma_{x},\ \sigma_{y},\ \sigma_{z}\}$
with the identity $1_{2}$
\[
\sigma_{x}=
\left(
\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}
\right),\quad
\sigma_{y}=
\left(
\begin{array}{cc}
0 & -i \\
i  & 0
\end{array}
\right),\quad
\sigma_{z}=
\left(
\begin{array}{cc}
1 & 0  \\
0 & -1
\end{array}
\right),\quad
1_{2}=
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)
\]
act on the space.

The three generations of leptons are
\[
\left(
\begin{array}{c}
e         \\
\nu_{e}
\end{array}
\right),\quad
\left(
\begin{array}{c}
\mu        \\
\nu_{\mu}
\end{array}
\right),\quad
\left(
\begin{array}{c}
\tau  \\
\nu_{\tau}
\end{array}
\right).
\]
However, each neutrino is not single but slightly mixed like
\[
\nu_{\mu}^{\prime}=\cos\Theta\ \nu_{\mu}-\sin\Theta\ \nu_{\tau},\quad
\nu_{\tau}^{\prime}=\sin\Theta\ \nu_{\mu}+\cos\Theta\ \nu_{\tau}.
\]
This $\Theta$ is called the mixing angle in vacuum, which is small enough.
Therefore real generations are for example
\[
\left(
\begin{array}{c}
\mu                    \\
\nu_{\mu}^{\prime}
\end{array}
\right),\quad
\left(
\begin{array}{c}
\tau                     \\
\nu_{\tau}^{\prime}
\end{array}
\right).
\]
Note that the mixing matrix
\begin{equation}
\label{eq:mixing matrix}
R(\Theta)=
\left(
\begin{array}{cc}
 \cos\Theta & \sin\Theta \\
-\sin\Theta & \cos\Theta
\end{array}
\right)
\end{equation}
is called the Pontecorvo-Maki-Nakagawa-Sakata matrix
(\cite{Wiki}, \cite{BP}, \cite{MNS}).

Let us start with a model \cite{PM}, \cite{KP}.
Since we treat two neutrinos $\{\nu_{\mu},\ \nu_{\tau}\}$
in the paper, the Dirac equation for the two neutrinos can be
reduced to a Schr{\"o}dinger form written in terms of a two
component vector of positive energy probability amplitude
{\bf in the ultra-relativistic limit}.

Then the two neutrino system can be mapped to a two--level
quantum system with distinct energy eigenvalues along with
{\bf the assumption of equal fixed momenta} \cite{PM}.

We set that $m_{i}\ (i=1,2)$ are the neutrino masses and $E_{i}$ are
the energies given by the approximation
\[
E_{i}
=\sqrt{(cp)^{2}+(m_{i}c^{2})^{2}}
\approx cp+\frac{m_{i}^{2}c^{3}}{2p}
=cp+\frac{m_{i}^{2}c^{4}}{2pc}
\]
($p_{1}=p_{2}=p$ from the assumption) and $\Theta$ is the mixing angle.
The Hamiltonian becomes
\begin{equation}
\label{eq:hamiltonian-1}
H=H(p)
=
\left(cp+\frac{\epsilon_0}{2}\right)1_{2}+
\frac{\epsilon}{2}\{\sin(2\Theta)\sigma_{x}-
\cos(2\Theta)\sigma_{z}\}
\end{equation}
where
\[
\epsilon_0=\frac{(m_{1}^{2}+m_{2}^{2})c^{4}}{2pc},\quad
\epsilon=\frac{(m_{2}^{2}-m_{1}^{2})c^{4}}{2pc},\quad
E_{0}=cp+\frac{\epsilon_0}{2}.
\]
Here, note that $E_{0}-\frac{\epsilon}{2}=E_{1}$ and
$E_{0}+\frac{\epsilon}{2}=E_{2}$.

Let us rewrite $H$ in (\ref{eq:hamiltonian-1}) in a familiar
form by making use of Pauli matrices above
\begin{equation}
\label{eq:hamiltonian-2}
H=
\left(
\begin{array}{cc}
cp+\frac{\epsilon_0}{2}-\frac{\epsilon}{2}\cos(2\Theta) & \frac{\epsilon}{2}\sin(2\Theta) \\
\frac{\epsilon}{2}\sin(2\Theta) & cp+\frac{\epsilon_0}{2}+\frac{\epsilon}{2}\cos(2\Theta)
\end{array}
\right).
\end{equation}
For this it is easy to obtain the eigenvalues
\begin{equation}
\label{eq:eigenvalues}
E_{\pm}=cp+\frac{\epsilon_0}{2}\pm \frac{\epsilon}{2}
\equiv E_{0}\pm \frac{\epsilon}{2}
\end{equation}
and the corresponding eigenvectors
\begin{eqnarray*}
|\phi_{-}\rangle
&=&
\left(
\begin{array}{c}
\cos\Theta \\
-\sin\Theta
\end{array}
\right)
=
\cos\Theta|\phi_{1}\rangle-\sin\Theta|\phi_{2}\rangle,  \nonumber \\
|\phi_{+}\rangle
&=&
\left(
\begin{array}{c}
\sin\Theta  \\
\cos\Theta
\end{array}
\right)
=
\sin\Theta|\phi_{1}\rangle+\cos\Theta|\phi_{2}\rangle.
\end{eqnarray*}

If we define a unitary matrix
\[
\left(|\phi_{-}\rangle, |\phi_{+}\rangle \right)
=\left(
\begin{array}{cc}
 \cos\Theta & \sin\Theta \\
-\sin\Theta & \cos\Theta
\end{array}
\right)
=R(\Theta)
\]
then we can make $H$ diagonal like
\begin{equation}
\label{eq:diagonal form}
H
=
R(\Theta)
\left(
\begin{array}{cc}
E_{0}-\frac{\epsilon}{2} & 0  \\
0 & E_{0}+\frac{\epsilon}{2}
\end{array}
\right)
{R(\Theta)}^{\dagger}
=
R(\Theta)
\left(
\begin{array}{cc}
E_{1} & 0  \\
0 & E_{2}
\end{array}
\right)
{R(\Theta)}^{T}
\end{equation}
or in a spectral decomposition form
\begin{equation}
\label{eq:spectral-decomposition}
H=
\left(E_{0}-\frac{\epsilon}{2}\right)|{\phi_{-}}\rangle \langle{\phi_{-}}|+
\left(E_{0}+\frac{\epsilon}{2}\right)|{\phi_{+}}\rangle \langle{\phi_{+}}|.
\end{equation}
These forms are used in the next section.

Next, we label the Hamiltonian by the momentum $p$. Namely,
\begin{equation}
\label{eq:full-hamiltonian}
{\bf H}={\bf H}(p)=H(p)\otimes |p\rangle\langle p|.
\end{equation}
This is in a certain sense the graph of a function.

If we define eigenvectors of ${\bf H}$ in (\ref{eq:full-hamiltonian}) as
simultaneous ones of both flavor and momentum
\[
H\otimes {1}|{\phi_{\pm},p}\rangle=E_{\pm}|{\phi_{\pm},p}\rangle, \quad
{1}\otimes \hat{p}|{\phi_{\pm},p}\rangle=p|{\phi_{\pm},p}\rangle,
\]
then we have
\begin{eqnarray}
\label{eq:eigenvectors}
|\phi_{-},p\rangle
&=&
\left(
\begin{array}{c}
\cos\Theta \\
-\sin\Theta
\end{array}
\right)\otimes |p\rangle
=
\cos\Theta|\phi_{1},p\rangle-\sin\Theta|\phi_{2},p\rangle,  \nonumber \\
|\phi_{+},p\rangle
&=&
\left(
\begin{array}{c}
\sin\Theta  \\
\cos\Theta
\end{array}
\right)\otimes |p\rangle
=
\sin\Theta|\phi_{1},p\rangle+\cos\Theta|\phi_{2},p\rangle.
\end{eqnarray}
Therefore, the spectral decomposition of ${\bf H}$ is given by
\begin{equation}
\label{eq:spectral decomposition}
{\bf H}=
\left(E_{0}-\frac{\epsilon}{2}\right)|{\phi_{-},p}\rangle \langle{\phi_{-},p}|+
\left(E_{0}+\frac{\epsilon}{2}\right)|{\phi_{+},p}\rangle \langle{\phi_{+},p}|.
\end{equation}

\vspace{5mm}
Next, we consider a time--evolution of the system.

\vspace{10mm}
\input{time-evolution.fig}

\vspace{10mm}
The state $|\psi_{t}\rangle$ at time $t$ is given by
\begin{equation}
\label{eq:}
|\psi_{t}\rangle=\int dp^{\prime}e^{-it{\bf H}(p^{\prime})/\hbar}\ |\psi_{0}\rangle
\end{equation}
with the initial state $|\psi_{0}\rangle$, and straightforward calculation
with (\ref{eq:spectral decomposition}) gives
\begin{equation}
\label{eq:evolution-1}
|\psi_{t}\rangle
=
\int dp^{\prime}e^{-itE_{0}/\hbar}
\left\{
e^{-it\frac{\epsilon}{2\hbar}}\langle{\phi_{+},p^{\prime}}|\psi_{0}\rangle
|{\phi_{+},p^{\prime}}\rangle
+
e^{it\frac{\epsilon}{2\hbar}}\langle{\phi_{-},p^{\prime}}|\psi_{0}\rangle
|{\phi_{-},p^{\prime}}\rangle
\right\}
\end{equation}
where $E_{0}=E_{0}(p^{\prime}),\ \epsilon=\epsilon(p^{\prime}),\
\epsilon_{0}=\epsilon_{0}(p^{\prime})$ for simplicity.

From now on we assume some conditions :

\noindent
(a)\ \ we start with one neutrino flavor, namely,
$\langle{\phi_{2},p}|\psi_{0}\rangle=0$,

\noindent
(b)\ \ the initial amplitude of the neutrino waveform is
$\langle{\phi_{1},p}|\psi_{0}\rangle=\langle{p}|\psi_{0}\rangle$.

\noindent
These assumptions seem to be natural.

Then it is easy to see that (\ref{eq:evolution-1}) can be rewritten as
\begin{eqnarray}
\label{eq:evolution-2}
&&|\psi_{t}\rangle =
\int dp^{\prime} e^{-itE_{0}/\hbar}\langle{p^{\prime}}|\psi_{0}\rangle \times
\nonumber \\
&&
\left[
e^{-it\frac{\epsilon}{2\hbar}}
\left\{
\sin^{2}\Theta |{\phi_{1},p^{\prime}}\rangle +
\sin\Theta \cos\Theta |{\phi_{2},p^{\prime}}\rangle
\right\}
+
e^{it\frac{\epsilon}{2\hbar}}
\left\{
\cos^{2}\Theta |{\phi_{1},p^{\prime}}\rangle -
\sin\Theta \cos\Theta |{\phi_{2},p^{\prime}}\rangle
\right\}
\right]  \nonumber \\
&{}&
\end{eqnarray}
by use of (\ref{eq:eigenvectors}).

Now let us start detection of neutrino :

\noindent
(a)\ \ first, we perform a flavor measurement (for example,
flavor 1),

\noindent
(b)\ \ immediately after (a), we perform a position measurement.

When flavor 1 is detected the collapsed state becomes
\begin{equation}
\label{eq:collapsed state}
|\psi_{t}\rangle \longrightarrow
|\psi_{t}^{\prime}\rangle
=\frac{1}{\sqrt{{\cal D}}}
\int dp^{\prime}|{\phi_{1},p^{\prime}}\rangle \langle{\phi_{1},p^{\prime}}|\psi_{t}\rangle
=\frac{1}{\sqrt{{\cal D}}}{\cal P}|\psi_{t}\rangle
\end{equation}
where ${\cal D}$ is the normalization factor given by
\begin{equation}
\label{eq:normalization factor}
{\cal D}
=\int dp^{\prime}
\langle{\psi_{t}}|{\phi_{1},p^{\prime}}\rangle \langle{\phi_{1},p^{\prime}}|\psi_{t}\rangle
=\langle\psi_{t}|{\cal P}|\psi_{t}\rangle
\end{equation}
($\langle{\psi_{t}^{\prime}}|\psi_{t}^{\prime}\rangle=1$) and
\[
{\cal P} \equiv
\int dp^{\prime}|{\phi_{1},p^{\prime}}\rangle \langle{\phi_{1},p^{\prime}}|
\]
is the projection operator to the flavor 1 state
(: ${\cal P}^{2}={\cal P},\ {\cal P}^{\dagger}={\cal P}$).

It is in general difficult to perform a position measurement immediately
after flavor 1 is detected, so we average the positions of neutrinos.
We believe that this replacement is not so bad.

The expectation value of position measurement on the collapsed state
$|\psi_{t}^{\prime}\rangle$ is given by
\[
\langle{x_{t}}\rangle=
\int dp^{\prime}\langle{\psi_{t}^{\prime}}|p^{\prime}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle p^{\prime}|\psi_{t}^{\prime}\rangle.
\]
This is a kind of definition. Note that
\[
i\hbar \frac{\partial}{\partial p}
\]
is a position operator, because
\[
i\hbar \frac{\partial}{\partial p} \langle{p}|{x}\rangle
=
i\hbar \frac{\partial}{\partial p} e^{-ixp/\hbar}
=xe^{-ixp/\hbar}=x\langle{p}|{x}\rangle.
\]

We must evaluate the expectation value of position
$\langle{x_{t}}\rangle$. From (\ref{eq:collapsed state})
and noting the formula
\begin{equation}
\label{eq:delta function}
\langle{\phi_{1},p^{\prime}}|{\phi_{1},p^{\prime\prime}}\rangle=
\delta(p^{\prime}-p^{\prime\prime})
\end{equation}
we have and set
\begin{equation}
\label{eq:position}
\langle{x_{t}}\rangle=\frac{1}{{\cal D}}
\int dp^{\prime}\langle{\psi_{t}}|{\phi_{1},p^{\prime}}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{\phi_{1},p^{\prime}}|{\psi_{t}}\rangle
\equiv \frac{{\cal N}}{{\cal D}}
\end{equation}
for simplicity.

First, let us calculate $\langle{\phi_{1},p^{\prime}}|{\psi_{t}}\rangle$.
From (\ref{eq:evolution-2}) and the formula (\ref{eq:delta function})
it is easy to see
\begin{equation}
\label{eq:relation}
\langle{\phi_{1},p^{\prime}}|{\psi_{t}}\rangle=
\langle{p^{\prime}}|\psi_{0}\rangle F(p^{\prime})
\end{equation}
where
\begin{equation}
\label{eq:F formula}
F(p^{\prime})=e^{-itE_{0}/\hbar}
\left(
\sin^{2}\Theta e^{-it\epsilon/2\hbar}+\cos^{2}\Theta e^{it\epsilon/2\hbar}
\right).
\end{equation}
Then the normalization factor ${\cal D}$ in (\ref{eq:normalization factor})
becomes
\[
{\cal D}=
\int dp^{\prime} |\langle{p^{\prime}}|\psi_{0}\rangle |^{2} |F(p^{\prime})|^{2}.
\]

Now, we make another assumption. The initial distribution of the neutrino
momentum $|\langle{p^{\prime}}|\psi_{0}\rangle |^{2}$ is narrow with respect
to $F(p^{\prime})$, and centered on $p^{\prime}=p$ ($p$ is fixed). Namely,
\begin{equation}
\label{eq:initial distribution}
|\langle{p^{\prime}}|\psi_{0}\rangle |^{2} \approx \delta(p^{\prime}-p).
\end{equation}
We believe this one natural. Then
\begin{equation}
\label{eq:value of D}
{\cal D}\approx |F(p)|^{2}.
\end{equation}

For later convenience we calculate (\ref{eq:value of D}).
Here is an elementary formula

\vspace{3mm}\noindent
{\bf Formula}\ \ for $\alpha, \ \beta \in {\bf R}$
\[
|\alpha e^{-i\theta}+\beta e^{i\theta}|^{2}
=
(\alpha + \beta)^{2}-4\alpha \beta \sin^{2}\theta.
\]
This gives
\begin{eqnarray}
\label{eq:|F(p)|^{2}}
|F(p)|^{2}
&=&
|\sin^{2}\Theta e^{-it\epsilon/2\hbar}+\cos^{2}\Theta e^{it\epsilon/2\hbar}|^{2}
\nonumber \\
&=&\left(\sin^{2}\Theta +\cos^{2}\Theta\right)^{2}-
4\sin^{2}\Theta\cos^{2}\Theta \sin^{2}\left(\frac{t\epsilon}{2\hbar}\right)
\nonumber \\
&=&1-\left(2\sin\Theta\cos\Theta\right)^{2}
\sin^{2}\left(\frac{t\epsilon}{2\hbar}\right)    \nonumber \\
&=&1-\sin^{2}(2\Theta)\sin^{2}\left(\frac{t\epsilon}{2\hbar}\right)
\end{eqnarray}
by use of  (\ref{eq:F formula}) (note that $\epsilon=\epsilon(p)$).

By inserting the equation (\ref{eq:relation}) into ${\cal N}$ in
(\ref{eq:position}) we have and set
\begin{eqnarray*}
{\cal N}
&=&
\int dp^{\prime} \langle{\psi_{t}}|{\phi_{1},p^{\prime}}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{\phi_{1},p^{\prime}}|{\psi_{t}}\rangle  \\
&=&
\int dp^{\prime} \bar{F}(p^{\prime})\langle{\psi_{0}|p^{\prime}}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\{\langle{p^{\prime}}|\psi_{0}\rangle F(p^{\prime})\}  \\
&=&
\int dp^{\prime} |F(p^{\prime})|^{2}\langle{\psi_{0}|p^{\prime}}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{p^{\prime}}|\psi_{0}\rangle
+
\int dp^{\prime} |\langle{p^{\prime}}|\psi_{0}\rangle |^{2}
\bar{F}(p^{\prime})
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
F(p^{\prime})  \\
&\equiv& {\cal N}_{1}+{\cal N}_{2}
\end{eqnarray*}
for simplicity. Next, let us calculate ${\cal N}_{1}$ and ${\cal N}_{2}$
separately.

From
\[
{\cal N}_{1}=\mbox{Re}{\cal N}_{1}+i\mbox{Im}{\cal N}_{1}
\]
we have
\begin{eqnarray}
\mbox{Im}{\cal N}_{1}
&=&
\frac{1}{2i}\left({\cal N}_{1}-\bar{{\cal N}}_{1}\right)  \nonumber \\
&=&
\frac{\hbar}{2}
\int dp^{\prime} |F(p^{\prime})|^{2}
\left\{
\langle{\psi_{0}|p^{\prime}}\rangle
\frac{\partial}{\partial p^{\prime}}
\langle{p^{\prime}}|\psi_{0}\rangle
+
\langle{p^{\prime}}|\psi_{0}\rangle
\frac{\partial}{\partial p^{\prime}}
\langle{\psi_{0}|p^{\prime}}\rangle
\right\}  \nonumber \\
&=&
\frac{\hbar}{2}
\int dp^{\prime} |F(p^{\prime})|^{2}
\frac{\partial}{\partial p^{\prime}}
\left\{
\langle{p^{\prime}}|\psi_{0}\rangle \langle{\psi_{0}|p^{\prime}}\rangle
\right\}  \nonumber \\
&=&
\frac{\hbar}{2}
\int dp^{\prime} |F(p^{\prime})|^{2}
\frac{\partial}{\partial p^{\prime}} |\langle{p^{\prime}}|\psi_{0}\rangle|^{2}
\nonumber \\
&{}& (\mbox{integration by parts})  \nonumber \\
&=&
-\frac{\hbar}{2}
\int dp^{\prime}
|\langle{p^{\prime}}|\psi_{0}\rangle|^{2}
\frac{\partial}{\partial p^{\prime}} |F(p^{\prime})|^{2} \nonumber \\
&\approx &
-\frac{\hbar}{2}
\int dp^{\prime} \delta(p^{\prime}-p)
\frac{\partial}{\partial p^{\prime}} |F(p^{\prime})|^{2}
=-\frac{\hbar}{2}\frac{\partial}{\partial p} |F(p)|^{2}
\end{eqnarray}
by use of the assumption in (\ref{eq:initial distribution}). Similarly,
we have
\begin{eqnarray*}
\mbox{Re}{\cal N}_{1}
&=&
\frac{1}{2}\left({\cal N}_{1}+\bar{{\cal N}}_{1}\right)  \\
&=&
\frac{1}{2}
\int dp^{\prime} |F(p^{\prime})|^{2}
\left\{
\langle{\psi_{0}|p^{\prime}}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{p^{\prime}}|\psi_{0}\rangle
-
\langle{p^{\prime}}|\psi_{0}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{\psi_{0}|p^{\prime}}\rangle
\right\}.
\end{eqnarray*}
The range of integration is narrow enough because of
the assumption in (\ref{eq:initial distribution}), so we
approximate the integration like

\begin{eqnarray}
\mbox{Re}{\cal N}_{1}
&\approx&
|F(p)|^{2}\ \frac{1}{2}
\int dp^{\prime}
\left\{
\langle{\psi_{0}|p^{\prime}}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{p^{\prime}}|\psi_{0}\rangle
-
\langle{p^{\prime}}|\psi_{0}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{\psi_{0}|p^{\prime}}\rangle
\right\} \nonumber \\
&{}& (\mbox{integration by parts})  \nonumber \\
&=&
|F(p)|^{2}
\int dp^{\prime}
\left\{
\langle{\psi_{0}|p^{\prime}}\rangle
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
\langle{p^{\prime}}|\psi_{0}\rangle
\right\}  \nonumber \\
&=&|F(p)|^{2}\ \langle{x_{0}}\rangle.
\end{eqnarray}
Therefore, we obtain the approximate value
\begin{equation}
\label{eq:N_{1}}
{\cal N}_{1}=\mbox{Re}{\cal N}_{1}+i\mbox{Im}{\cal N}_{1}
\approx
|F(p)|^{2}\langle{x_{0}}\rangle -\frac{i\hbar}{2}\frac{\partial}{\partial p} |F(p)|^{2}.
\end{equation}

For ${\cal N}_{2}$ we have
\begin{eqnarray}
\label{eq:N_{2}}
{\cal N}_{2}
&=&
\int dp^{\prime} |\langle{p^{\prime}}|\psi_{0}\rangle |^{2}
\bar{F}(p^{\prime})
\left(i\hbar \frac{\partial}{\partial p^{\prime}}\right)
F(p^{\prime})  \nonumber \\
&\approx &
\bar{F}(p)\left(i\hbar \frac{\partial}{\partial p}\right)F(p)
=
i\hbar \bar{F}(p)\frac{\partial}{\partial p}F(p)
\end{eqnarray}
by use of the assumption in (\ref{eq:initial distribution}).

Then (\ref{eq:N_{1}}) and (\ref{eq:N_{2}}) give
\begin{eqnarray}
\label{eq:N}
{\cal N}={\cal N}_{1}+{\cal N}_{2}
&=&
|F(p)|^{2}\langle{x_{0}}\rangle -\frac{i\hbar}{2}\frac{\partial}{\partial p} |F(p)|^{2}+
i\hbar \bar{F}(p)\frac{\partial}{\partial p}F(p)  \nonumber \\
&=&
|F(p)|^{2}\langle{x_{0}}\rangle+\frac{i\hbar}{2}
\left\{
\bar{F}(p)\frac{\partial}{\partial p}F(p)-\frac{\partial}{\partial p}\bar{F}(p) F(p)
\right\}
\end{eqnarray}
and (\ref{eq:position}) and (\ref{eq:value of D}) give (the approximate value)
\begin{equation}
\label{eq:value of position}
\langle{x_{t}}\rangle=\frac{{\cal N}}{{\cal D}}=\frac{{\cal N}}{|F(p)|^{2}}=
\langle{x_{0}}\rangle +\frac{i\hbar}{2}
\frac{
\bar{F}(p)\frac{\partial}{\partial p}F(p)-\frac{\partial}{\partial p}\bar{F}(p) F(p)
       }{|F(p)|^{2}}.
\end{equation}

Next, let us calculate the right hand side of (\ref{eq:value of position})
by use of (\ref{eq:F formula}) :
\[
F(p)=e^{-itE_{0}/\hbar}
\left(
\sin^{2}\Theta e^{-it\epsilon/2\hbar}+\cos^{2}\Theta e^{it\epsilon/2\hbar}
\right).
\]
Noting $E_{0}=E_{0}(p),\ \epsilon=\epsilon(p)$ we have
\[
\frac{\partial}{\partial p}F(p)
=
-i\frac{t}{\hbar}\frac{\partial E_{0}}{\partial p}F(p)
-i\frac{t}{2\hbar}\frac{\partial \epsilon}{\partial p} e^{-itE_{0}/\hbar}
\left(
\sin^{2}\Theta e^{-it\epsilon/2\hbar}-\cos^{2}\Theta e^{it\epsilon/2\hbar}
\right)
\]
and
\begin{eqnarray*}
\bar{F}(p)\frac{\partial}{\partial p}F(p)
&=&
-i\frac{t}{\hbar}\frac{\partial E_{0}}{\partial p}|F(p)|^{2} \\
&{}&
-i\frac{t}{2\hbar}\frac{\partial \epsilon}{\partial p}
\left(
\sin^{2}\Theta e^{-it\epsilon/2\hbar}-\cos^{2}\Theta e^{it\epsilon/2\hbar}
\right)
\left(
\sin^{2}\Theta e^{it\epsilon/2\hbar}+\cos^{2}\Theta e^{-it\epsilon/2\hbar}
\right) \\
&=&
-i\frac{t}{\hbar}\frac{\partial E_{0}}{\partial p}|F(p)|^{2}
-i\frac{t}{2\hbar}\frac{\partial \epsilon}{\partial p}
\left(\sin^{4}\Theta -\cos^{4}\Theta + ** \right)  \\
&=&
-i\frac{t}{\hbar}\frac{\partial E_{0}}{\partial p}|F(p)|^{2}
-i\frac{t}{2\hbar}\frac{\partial \epsilon}{\partial p}
\left(\sin^{2}\Theta -\cos^{2}\Theta + ** \right)
\end{eqnarray*}
where $**$ is the terms which will be neglected at the final stage.
This gives
\begin{eqnarray*}
\frac{i\hbar}{2}
\left(
\bar{F}(p)\frac{\partial}{\partial p}F(p)-\mbox{c.c.}
\right)
&=&
t\frac{\partial E_{0}}{\partial p}|F(p)|^{2}+
\frac{t}{2}
\frac{\partial \epsilon}{\partial p}\left(\sin^{2}\Theta -\cos^{2}\Theta\right)  \\
&=&
t\frac{\partial E_{0}}{\partial p}|F(p)|^{2}-
\frac{t}{2}
\frac{\partial \epsilon}{\partial p}\cos(2\Theta)
\end{eqnarray*}
and $\langle{x_{t}}\rangle$ in (\ref{eq:value of position}) is given by
\begin{equation}
\label{eq:}
\langle{x_{t}}\rangle=
\langle{x_{0}}\rangle +t\frac{\partial E_{0}}{\partial p} -
\frac{t}{2}\frac{\cos(2\Theta)}{|F(p)|^{2}}
\frac{\partial \epsilon}{\partial p}.
\end{equation}

\vspace{5mm}\noindent
{\bf Definition}\ \ The group velocity $v_{g}$ is given by
\[
\langle{x_{t}}\rangle - \langle{x_{0}}\rangle =v_{g}t
\Longleftrightarrow v_{g}=
\frac{\langle{x_{t}}\rangle - \langle{x_{0}}\rangle}{t}.
\]
Therefore it becomes
\begin{equation}
\label{eq:group velocity}
v_{g}=\frac{\partial E_{0}}{\partial p}-\frac{1}{2}
\frac{\cos(2\Theta)}{|F(p)|^{2}}\frac{\partial \epsilon}{\partial p}
\end{equation}
from the result above.

Remembering
\[
E_{0}=cp+\frac{\epsilon_0}{2},\quad
\epsilon_0=\frac{(m_{1}^{2}+m_{2}^{2})c^{4}}{2pc},\quad
\epsilon=\frac{(m_{2}^{2}-m_{1}^{2})c^{4}}{2pc}
\]
from (\ref{eq:hamiltonian-1}) simple calculation gives
\[
\frac{\partial E_{0}}{\partial p}=c-\frac{\epsilon_{0}}{2p},\quad
\frac{\partial \epsilon}{\partial p}=-\frac{\epsilon}{p}
\]
and by inserting the above into (\ref{eq:group velocity}) we have
\begin{equation}
v_{g}=c-\frac{\epsilon_{0}}{2p}+{\cal S}\frac{\epsilon}{2p}
\end{equation}
where ${\cal S}$ is given by
\begin{equation}
\label{eq:S}
{\cal S}=\frac{\cos(2\Theta)}{|F(p)|^{2}}=
\frac{\cos(2\Theta)}
{1-\sin^{2}(2\Theta)\sin^{2}\left(\frac{t\epsilon}{2\hbar}\right)}
\end{equation}
from (\ref{eq:|F(p)|^{2}}).

As a result we obtain

\vspace{5mm}\noindent
{\bf Theorem} (Mecozzi and Bellini)\ \ The group velocity $v_{g}$
is given by
\begin{equation}
\label{eq:theorem}
v_{g}=c-\frac{\epsilon_{0}}{2p}+
\frac{\cos(2\Theta)}
{1-\sin^{2}(2\Theta)\sin^{2}\left(\frac{t\epsilon}{2\hbar}\right)}\frac{\epsilon}{2p}.
\end{equation}

\vspace{5mm}
Note that the term ${\cal S}$ was obtained from a quantum effect (the neutrino
mixing) and this plays a definite role in {\bf Superluminal Group Velocity}.

Let us analyze the theorem. For the purpose we set
\[
\alpha\equiv \sin^{2}\left(\frac{t\epsilon}{2\hbar}\right)
\]
to look for some condition satisfying ${\cal S}>1$. Namely,
\begin{eqnarray*}
&&{\cal S}=\frac{\cos(2\Theta)}{1-\alpha\sin^{2}(2\Theta)}>1 \\
\Longleftrightarrow
&&\cos(2\Theta)>1-\alpha\sin^{2}(2\Theta)=1-\alpha(1-\cos^{2}(2\Theta)) \\
\Longleftrightarrow
&&(1-\cos^{2}(2\Theta))\alpha>1-\cos(2\Theta)\quad (\mbox{excepting} \
1=\cos(2\Theta))  \\
\Longleftrightarrow
&&\alpha>\frac{1}{1+\cos(2\Theta)}>\frac{1}{2}.
\end{eqnarray*}
Therefore, we obtain
\begin{equation}
\label{eq:}
\alpha=\sin^{2}\left(\frac{t\epsilon}{2\hbar}\right)>\frac{1}{2}
\ \Longrightarrow\ {\cal S}>1.
\end{equation}

Here, we assume $m_{2}\gg m_{1}$. Then $\epsilon \approx \epsilon_{0}$
from (\ref{eq:hamiltonian-1}) and ${\cal S}>1$ gives
\[
v_{g}\approx c+({\cal S}-1)\frac{\epsilon_{0}}{2p}>c.
\]
As a result

\vspace{5mm}\noindent
{\bf Corollary 1}\ \ Under the conditions $\alpha>\frac{1}{2}$ and
$m_{2}\gg m_{1}$ we have the superluminal group velocity
\begin{equation}
\label{eq:superlunimal}
v_{g}>c.
\end{equation}

\vspace{3mm}\noindent
{\bf Note}\ \
In general, the higher the generation, the heavier corresponding mass.
Therefore, the assumption $m_{2}\gg m_{1}$ is not unnatural.

\vspace{3mm}
If $\Theta=0$ (no neutrino mixing) then ${\cal S}=1$ from
(\ref{eq:S}) and we have
\[
v_{g}
\approx c-\frac{\epsilon_{0}}{2p}+\frac{\epsilon_{0}}{2p}
=c
\]
under $m_{2}\gg m_{1}$. As a result

\vspace{5mm}\noindent
{\bf Corollary 2}\ \ If $\Theta=0$ we have the usual velocity
\begin{equation}
\label{eq:usual}
v_{g}=c
\end{equation}
under $\alpha>\frac{1}{2}$ and $m_{2}\gg m_{1}$.

\vspace{5mm}
We cannot help admitting a mechanism which accelerates
neutrinos arising from the neutrino oscillation.
How do we interpret the result ?  What is the relation to
special relativity ? In the last part of the paper \cite{MB}
they write :

\vspace{3mm}\noindent
``Of course, this does not mean that the speed of a possible
signal transmitted with a neutrino wave-packet exceeds
the speed of light, it is just a property that comes from
the wave-packet deformation caused by the interference
of the two possible quantum paths that a neutrino may
follow before reaching the detector".

\vspace{3mm}
Unfortunately, the author cannot understand what they meant
and therefore present the following

\vspace{3mm}\noindent
{\bf Problem}\ \ Give a mathematical expression to their claim.

\section{Decoherence of the Neutrino Oscillation}
In this section, for the model in the previous section
we take {\bf decoherence} into consideration in order to
make it more realistic. Then we can build a bridge between
Corollary 1 and Corollary 2.
For a general introduction to this topic see for example
\cite{BandP}.

First of all we present the following

\vspace{3mm}\noindent
{\bf Problem}\ \ Is there no problem to apply theory of
decoherence to neutrinos in a long--distance flight ?

\vspace{3mm}\noindent
Although this problem is very subtle, let us proceed to the
discussion of decoherence.

Since the two neutrino system can be mapped to the
two level system we prepare some notation from
Quantum Optics. For
\[
\sigma_{+}\equiv \frac{1}{2}(\sigma_{x}+i\sigma_{y})=
\left(
\begin{array}{cc}
0 & 1 \\
0 & 0
\end{array}
\right),\quad
\sigma_{-}\equiv \frac{1}{2}(\sigma_{x}-i\sigma_{y})=
\left(
\begin{array}{cc}
0 & 0 \\
1 & 0
\end{array}
\right)
\]
it is easy to see
\[
\sigma_{+}\sigma_{-}=
\left(
\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array}
\right),\quad
\sigma_{-}\sigma_{+}=
\left(
\begin{array}{cc}
0 & 0 \\
0 & 1
\end{array}
\right).
\]

Let us remember
\begin{eqnarray}
\label{eq:diagonal form again}
H
&=&
R(\Theta)
\left(
\begin{array}{cc}
E_{1} & 0  \\
0 & E_{2}
\end{array}
\right)
{R(\Theta)}^{T} \nonumber \\
&=&
\left(
\begin{array}{cc}
E_{1}\cos^{2}\Theta+E_{2}\sin^{2}\Theta & (E_{2}-E_{1})\sin\Theta \cos\Theta \\
(E_{2}-E_{1})\sin\Theta \cos\Theta & E_{1}\sin^{2}\Theta+E_{2}\cos^{2}\Theta
\end{array}
\right)
\end{eqnarray}
from (\ref{eq:diagonal form}) and set
\begin{equation}
\label{eq:diagonal matrix}
H_{0}=
\left(
\begin{array}{cc}
E_{1} & 0  \\
0 & E_{2}
\end{array}
\right).
\end{equation}
Note that $H$ and $H_{0}$ are symmetric matrices ($H=H^{T}$,\
$H_{0}=H_{0}^{T}$).

To treat decoherence in a correct manner we must change
models based on from a pure state to a density matrix.
The general definition of density matrix $\rho$ is given by both
$\rho^{\dagger}=\rho$ and $\tr{\rho}=1$, so we can write
$\rho=\rho(t)$ as
\begin{equation}
\label{eq:density matrix}
\rho=
\left(
\begin{array}{cc}
a         & b  \\
\bar{b} & d
\end{array}
\right)
\quad (a=\bar{a},\ d=\bar{d},\ a+d=1).
\end{equation}

The general form of master equation is well--known to be
\begin{equation}
\label{eq:master equation}
\frac{d}{dt}\rho=-i[H, \rho]+D\rho \quad (\Leftarrow \hbar=1\
\mbox{for simplicity})
\end{equation}
where
\[
D\rho=
\mu\left(\sigma_{-}\rho\sigma_{+}-\frac{1}{2}\sigma_{+}\sigma_{-}\rho-\frac{1}{2}\rho\sigma_{+}\sigma_{-}\right)
+
\nu\left(\sigma_{+}\rho\sigma_{-}-\frac{1}{2}\sigma_{-}\sigma_{+}\rho-\frac{1}{2}\rho\sigma_{-}\sigma_{+}\right)
\]
and $\mu>\nu>0$ \footnote{we don't know how to determine
the precise values of $\mu$ and $\nu$ in our system}.
We must solve the equation.

If we write $H$ in (\ref{eq:diagonal form again}) as
\begin{equation}
\label{eq:diagonal form again 2}
H=
\left(
\begin{array}{cc}
h & k \\
k & l
\end{array}
\right)
\quad (h,\ k,\ l\ \in {\bf R})
\end{equation}
for simplicity, then the master equation above can be rewritten as
\begin{equation}
\label{eq:master equation 2}
\frac{d}{dt}
\left(
\begin{array}{c}
a \\
b \\
\bar{b} \\
d
\end{array}
\right)
=
\left(
\begin{array}{cccc}
-\mu & ik & -ik & \nu                        \\
ik & i(l-h)-\frac{\mu+\nu}{2} & 0 & -ik   \\
-ik & 0 & -i(l-h)-\frac{\mu+\nu}{2} & ik  \\
\mu & -ik & ik & -\nu
\end{array}
\right)
\left(
\begin{array}{c}
a \\
b \\
\bar{b} \\
d
\end{array}
\right).
\end{equation}
We leave the derivation to readers.

Note and set
\begin{eqnarray*}
&&\left(
\begin{array}{cccc}
-\mu & ik & -ik & \nu                        \\
ik & i(l-h)-\frac{\mu+\nu}{2} & 0 & -ik   \\
-ik & 0 & -i(l-h)-\frac{\mu+\nu}{2} & ik  \\
\mu & -ik & ik & -\nu
\end{array}
\right) \\
&=&
\left(
\begin{array}{cccc}
0 & ik & -ik & 0        \\
ik & i(l-h) & 0 & -ik   \\
-ik & 0 & -i(l-h) & ik  \\
0 & -ik & ik & 0
\end{array}
\right)
+
\left(
\begin{array}{cccc}
-\mu & 0 & 0 & \nu                 \\
0 & -\frac{\mu+\nu}{2} & 0 & 0  \\
0 & 0 & -\frac{\mu+\nu}{2} & 0  \\
\mu & 0 & 0 & -\nu
\end{array}
\right)
\equiv \widehat{H}+\widehat{D}.
\end{eqnarray*}
The general solution of (\ref{eq:master equation 2})
is given by
\begin{equation}
\label{eq:general solution}
\left(
\begin{array}{c}
a(t) \\
b(t) \\
\bar{b}(t) \\
d(t)
\end{array}
\right)
=
e^{t\left(\widehat{H}+\widehat{D}\right)}
\left(
\begin{array}{c}
a(0) \\
b(0) \\
\bar{b}(0) \\
d(0)
\end{array}
\right).
\end{equation}
However, it is not easy to calculate the term
$e^{t\left(\widehat{H}+\widehat{D}\right)}$ exactly, so
we use a simple approximation
\[
e^{t\left(\widehat{H}+\widehat{D}\right)}=
e^{t\left(\widehat{D}+\widehat{H}\right)}\approx
e^{t\widehat{D}}e^{t\widehat{H}}.
\]
In general, we must use the Zassenhaus formula, see
for example \cite{CZ}, \cite{Five}. Therefore, we treat
the approximate solution
\begin{equation}
\label{eq:general approximate solution}
\left(
\begin{array}{c}
a(t) \\
b(t) \\
\bar{b}(t) \\
d(t)
\end{array}
\right)
\approx
e^{t\widehat{D}}e^{t\widehat{H}}
\left(
\begin{array}{c}
a(0) \\
b(0) \\
\bar{b}(0) \\
d(0)
\end{array}
\right).
\end{equation}

\vspace{3mm}
First, we calculate $e^{t\widehat{D}}$. For the purpose
we set
\[
K=
\left(
\begin{array}{cc}
-\mu & \nu \\
\mu & -\nu
\end{array}
\right)
\]
and calculate $e^{tK}$. The eigenvalues of $K$ are
$\{0,-(\mu+\nu)\}$ and corresponding eigenvectors (
not normalized) are
\[
0\longleftrightarrow
\left(
\begin{array}{c}
\nu \\
\mu
\end{array}
\right),\quad
-(\mu+\nu)\longleftrightarrow
\left(
\begin{array}{c}
1  \\
-1
\end{array}
\right).
\]
If we define the matrix
\[
O=
\left(
\begin{array}{cc}
\nu & 1   \\
\mu & -1
\end{array}
\right)
\Longrightarrow
O^{-1}=\frac{1}{\mu+\nu}
\left(
\begin{array}{cc}
1 & 1          \\
\mu & -\nu
\end{array}
\right)
\]
then it is easy to see
\[
K=
O
\left(
\begin{array}{cc}
0 &                 \\
  & -(\mu+\nu)
\end{array}
\right)
O^{-1}
\]
and
\[
e^{tK}=
O
\left(
\begin{array}{cc}
1 &                         \\
   & e^{-t(\mu+\nu)}
\end{array}
\right)
O^{-1}
=
\frac{1}{\mu+\nu}
\left(
\begin{array}{cc}
\nu+\mu e^{-t(\mu+\nu)} & \nu-\nu e^{-t(\mu+\nu)}  \\
\mu-\mu e^{-t(\mu+\nu)} & \mu+\nu e^{-t(\mu+\nu)}
\end{array}
\right).
\]
Therefore, we have
\begin{equation}
e^{t\widehat{D}}
=
\left(
\begin{array}{cccc}
\frac{\nu+\mu e^{-t(\mu+\nu)}}{\mu+\nu} & 0 & 0 & \frac{\nu-\nu e^{-t(\mu+\nu)}}{\mu+\nu}  \\
0 & e^{-t\frac{\mu+\nu}{2}} & 0 & 0                                                                                \\
0 & 0 & e^{-t\frac{\mu+\nu}{2}} & 0                                                                                \\
\frac{\mu-\mu e^{-t(\mu+\nu)}}{\mu+\nu} & 0 & 0 & \frac{\mu+\nu e^{-t(\mu+\nu)}}{\mu+\nu}
\end{array}
\right)
\approx
\frac{1}{\mu+\nu}
\left(
\begin{array}{cccc}
\nu & 0 & 0 & \nu   \\
0 & 0 & 0  & 0        \\
0 & 0 & 0  & 0        \\
\mu & 0 & 0 & \mu
\end{array}
\right)
\end{equation}
if $t$ is large enough ($t\gg 1/\nu$).

Next, we calculate $e^{t\widehat{H}}$. Since we need
some properties of tensor product in the following see
for example \cite{Five}. We can write the equation as
\[
\widehat{H}=-i\left(H\otimes 1_{2}-1_{2}\otimes H\right)
\quad (\Longleftarrow H=H^{T}).
\]
In fact,
\begin{eqnarray*}
\widehat{H}
&=&-i
\left\{
\left(
\begin{array}{cc}
h & k \\
k & l
\end{array}
\right)
\otimes
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)
-
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)
\otimes
\left(
\begin{array}{cc}
h & k \\
k & l
\end{array}
\right)
\right\} \\
&=&-i
\left\{
\left(
\begin{array}{cccc}
h & 0 & k & 0 \\
0 & h & 0 & k \\
k & 0 & l  & 0 \\
0 & k & 0 & l
\end{array}
\right)
-
\left(
\begin{array}{cccc}
h & k & 0 & 0 \\
k & l  & 0 & 0 \\
0 & 0 & h & k \\
0 & 0 & k & l
\end{array}
\right)
\right\}
=-i
\left(
\begin{array}{cccc}
0 & -k & k & 0       \\
-k & -(l-h) & 0 & k \\
k & 0 & l-h  & -k    \\
0 & k & 0-k& 0
\end{array}
\right).
\end{eqnarray*}

It is well--known that
\[
e^{t\widehat{H}}
=
e^{-it\left(H\otimes 1_{2}-1_{2}\otimes H\right)}
=
e^{-it H\otimes 1_{2}}e^{it 1_{2}\otimes H}
=
\left(e^{-itH}\otimes 1_{2}\right)
\left(1_{2}\otimes e^{itH}\right)
=
e^{-itH}\otimes e^{itH}.
\]
Since
\[
H=
R(\Theta)
\left(
\begin{array}{cc}
E_{1} & 0 \\
0 & E_{2}
\end{array}
\right)
R(\Theta)^{T}
\]
we have
\[
e^{-itH}=
R(\Theta)
\left(
\begin{array}{cc}
e^{-itE_{1}} & 0 \\
0 & e^{-itE_{2}}
\end{array}
\right)
R(\Theta)^{T}
\]
and
\begin{eqnarray*}
e^{t\widehat{H}}
&=&
\left\{
R(\Theta)
\left(
\begin{array}{cc}
e^{-itE_{1}} & 0 \\
0 & e^{-itE_{2}}
\end{array}
\right)
R(\Theta)^{T}
\right\}
\otimes
\left\{
R(\Theta)
\left(
\begin{array}{cc}
e^{itE_{1}} & 0 \\
0 & e^{itE_{2}}
\end{array}
\right)
R(\Theta)^{T}
\right\} \\
&=&
\left(R(\Theta)\otimes R(\Theta)\right)
\left\{
\left(
\begin{array}{cc}
e^{-itE_{1}} & 0 \\
0 & e^{-itE_{2}}
\end{array}
\right)
\otimes
\left(
\begin{array}{cc}
e^{itE_{1}} & 0 \\
0 & e^{itE_{2}}
\end{array}
\right)
\right\}
\left(R(\Theta)\otimes R(\Theta)\right)^{T} \\
&=&
\left(R(\Theta)\otimes R(\Theta)\right)
\left(
\begin{array}{cccc}
1 &  &  &                         \\
  & e^{it(E_{2}-E_{1})} &  &    \\
  &   & e^{-it(E_{2}-E_{1})} & \\
  &   &   & 1
\end{array}
\right)
\left(R(\Theta)\otimes R(\Theta)\right)^{T}.
\end{eqnarray*}

Since
\[
R(\Theta)=
\left(
\begin{array}{cc}
 \cos\Theta & \sin\Theta \\
-\sin\Theta & \cos\Theta
\end{array}
\right)
\]
we have
\[
R(\Theta)\otimes R(\Theta)
=
\left(
\begin{array}{cccc}
\cos^{2}\Theta & \cos\Theta\sin\Theta & \cos\Theta\sin\Theta & \sin^{2}\Theta    \\
-\cos\Theta\sin\Theta & \cos^{2}\Theta & -\sin^{2}\Theta & \cos\Theta\sin\Theta \\
-\cos\Theta\sin\Theta & -\sin^{2}\Theta & \cos^{2}\Theta & \cos\Theta\sin\Theta \\
\sin^{2}\Theta & -\cos\Theta\sin\Theta & -\cos\Theta\sin\Theta & \cos^{2}\Theta
\end{array}
\right)
\]
and, by setting $J=e^{it(E_{2}-E_{1})}=e^{it\epsilon}$ for simplicity,
\begin{eqnarray*}
e^{t\widehat{H}}
&=&
R(\Theta)\otimes R(\Theta)
\left(
\begin{array}{cccc}
1 &  &  &          \\
  & J &  &        \\
  &   & J^{-1} & \\
  &   &   & 1
\end{array}
\right)
\left(R(\Theta)\otimes R(\Theta)\right)^{T} \\
&=&
\left(
\begin{array}{cccc}
\alpha_{11} & \alpha_{12} & \alpha_{13} & \alpha_{14}  \\
* & * & * & *                                                     \\
* & * & * & *                                                     \\
\alpha_{41} & \alpha_{42} & \alpha_{43} & \alpha_{44}
\end{array}
\right)
\end{eqnarray*}
where
\begin{eqnarray*}
\alpha_{11}&=&\cos^{4}\Theta+(J+J^{-1})\cos^{2}\Theta\sin^{2}\Theta+\sin^{4}\Theta, \\
\alpha_{12}&=&-\cos^{3}\Theta\sin\Theta+J\cos^{3}\Theta\sin\Theta
-J^{-1}\cos\Theta\sin^{3}\Theta+\cos\Theta\sin^{3}\Theta, \\
\alpha_{13}&=&-\cos^{3}\Theta\sin\Theta-J\cos\Theta\sin^{3}\Theta
+J^{-1}\cos^{3}\Theta\sin\Theta+\cos\Theta\sin^{3}\Theta, \\
\alpha_{14}&=&\cos^{2}\Theta\sin^{2}\Theta-(J+J^{-1})\cos^{2}\Theta\sin^{2}\Theta
+\cos^{2}\Theta\sin^{2}\Theta
\end{eqnarray*}
and
\begin{eqnarray*}
\alpha_{41}&=&\cos^{2}\Theta\sin^{2}\Theta-(J+J^{-1})\cos^{2}\Theta\sin^{2}\Theta
+\cos^{2}\Theta\sin^{2}\Theta, \\
\alpha_{42}&=&-\cos\Theta\sin^{3}\Theta-J\cos^{3}\Theta\sin\Theta
+J^{-1}\cos\Theta\sin^{3}\Theta+\cos^{3}\Theta\sin\Theta, \\
\alpha_{43}&=&-\cos\Theta\sin^{3}\Theta+J\cos\Theta\sin^{3}\Theta
-J^{-1}\cos^{3}\Theta\sin\Theta+\cos^{3}\Theta\sin\Theta, \\
\alpha_{44}&=&\sin^{4}\Theta+(J+J^{-1})\cos^{2}\Theta\sin^{2}\Theta+\cos^{4}\Theta.
\end{eqnarray*}
Note that $*$'s in the matrix are elements not used in the latter.
We leave this derivation to readers.

\noindent
Here, we list very important relations among $\{\alpha\}$
\begin{equation}
\label{eq: important relations}
\alpha_{11}+\alpha_{41}=1,\quad
\alpha_{12}+\alpha_{42}=0,\quad
\alpha_{13}+\alpha_{43}=0,\quad
\alpha_{14}+\alpha_{44}=1.
\end{equation}

Therefore, from (\ref{eq:general approximate solution}) and
(\ref{eq: important relations}) we obtain
\begin{eqnarray}
\label{eq:Fujii I}
\left(
\begin{array}{c}
a(t) \\
b(t) \\
\bar{b}(t) \\
d(t)
\end{array}
\right)
&\approx&
\frac{1}{\mu+\nu}
\left(
\begin{array}{cccc}
\nu & 0 & 0 & \nu   \\
0 & 0 & 0  & 0        \\
0 & 0 & 0  & 0        \\
\mu & 0 & 0 & \mu
\end{array}
\right)
\left(
\begin{array}{cccc}
\alpha_{11} & \alpha_{12} & \alpha_{13} & \alpha_{14}  \\
* & * & * & *                                                     \\
* & * & * & *                                                     \\
\alpha_{41} & \alpha_{42} & \alpha_{43} & \alpha_{44}
\end{array}
\right)
\left(
\begin{array}{c}
a(0) \\
b(0) \\
\bar{b}(0) \\
d(0)
\end{array}
\right) \nonumber \\
&=&
\frac{1}{\mu+\nu}
\left(
\begin{array}{cccc}
\nu & 0 & 0 & \nu   \\
0 & 0 & 0  & 0        \\
0 & 0 & 0  & 0        \\
\mu & 0 & 0 & \mu
\end{array}
\right)
\left(
\begin{array}{c}
a(0) \\
b(0) \\
\bar{b}(0) \\
d(0)
\end{array}
\right)
\end{eqnarray}
for $t\gg 1/\nu$.

\vspace{3mm}
For $H_{0}$ in (\ref{eq:diagonal matrix})
\[
H_{0}=
\left(
\begin{array}{cc}
E_{1} & 0  \\
0 & E_{2}
\end{array}
\right)
\]
we can perform the same process much easily.
The master equation is
\[
\frac{d}{dt}
\left(
\begin{array}{c}
a \\
b \\
\bar{b} \\
d
\end{array}
\right)
=
\left(
\begin{array}{cccc}
-\mu & 0 & -0 & \nu                       \\
0 & i(l-h)-\frac{\mu+\nu}{2} & 0 & 0   \\
0 & 0 & -i(l-h)-\frac{\mu+\nu}{2} & 0  \\
\mu & 0 & 0 & -\nu
\end{array}
\right)
\left(
\begin{array}{c}
a \\
b \\
\bar{b} \\
d
\end{array}
\right)
\]
and the (exact) solution is given by
\[
\left(
\begin{array}{c}
a(t) \\
b(t) \\
\bar{b}(t) \\
d(t)
\end{array}
\right)
=
\left(
\begin{array}{cccc}
\frac{\nu+\mu e^{-t(\mu+\nu)}}{\mu+\nu} & 0 & 0 & \frac{\nu-\nu e^{-t(\mu+\nu)}}{\mu+\nu}  \\
0 & e^{it(l-h)}e^{-t\frac{\mu+\nu}{2}} & 0 & 0                                                                      \\
0 & 0 & e^{-it(l-h)}e^{-t\frac{\mu+\nu}{2}} & 0                                                                     \\
\frac{\mu-\mu e^{-t(\mu+\nu)}}{\mu+\nu} & 0 & 0 & \frac{\mu+\nu e^{-t(\mu+\nu)}}{\mu+\nu}
\end{array}
\right)
\left(
\begin{array}{c}
a(0) \\
b(0) \\
\bar{b}(0) \\
d(0)
\end{array}
\right).
\]
When $t\gg 1/\nu$ we obtain
\begin{equation}
\label{eq:Fujii II}
\left(
\begin{array}{c}
a(t) \\
b(t) \\
\bar{b}(t) \\
d(t)
\end{array}
\right)
\approx
\frac{1}{\mu+\nu}
\left(
\begin{array}{cccc}
\nu & 0 & 0 & \nu   \\
0 & 0 & 0  & 0        \\
0 & 0 & 0  & 0        \\
\mu & 0 & 0 & \mu
\end{array}
\right)
\left(
\begin{array}{c}
a(0) \\
b(0) \\
\bar{b}(0) \\
d(0)
\end{array}
\right).
\end{equation}

As a result we have

\vspace{5mm}\noindent
{\bf Theorem}\ \ Two systems (master equations) whose
Hamiltonians are $H$ and $H_{0}$ have the same asymptotic
behavior (\ref{eq:Fujii I}) and (\ref{eq:Fujii II}) under
our approximation.

\vspace{5mm}
This theorem implies

\vspace{5mm}\noindent
{\bf Corollary}\ \ The mixing angle $\Theta$ will become
$0$ if $t$ is large enough.

\vspace{3mm}
From both this corollary and corollary 2 in the preceding section
we can conclude that the speed of neutrinos (after a long--distance
flight) is just that of light in vacuum.

By the way, our calculation in this section is based on a simple
approximation. This is a bit poor, so we present the following

\vspace{3mm}\noindent
{\bf Problem}\ \ Give the explicit (full) calculation.

\vspace{3mm}
As for interesting topics of decoherence (which is essential
in Quantum Physics) arising from Quantum Optics or Quantum
Computation see our papers
\cite{EFS}, \cite{FS2} and \cite{FS-1}, \cite{FS-2}.

\vspace{3mm}
At the end of this section, one comment is in order.
It seems to the author that Neutrino Physics gets along with
Quantum Optics or Quantum Computation, see for example \cite{CW}.
In order to make some (deep) relations clear further studies
will be required.

\section{Concluding Remarks}
In this paper we re-examined the paper \cite{MB} by Mecozzi and Bellini
in detail and tried to give mathematical reinforcements to it by taking
decoherence into consideration. Our conclusion is
\begin{center}
{\bf Neutrinos are latently superluminal.}
\end{center}
We would like to present the following

\vspace{3mm}\noindent
{\bf Problem}\ \ Re-check our result from a different point of view.

\vspace{3mm}
Let us write once more that our argument is based on {\bf group velocity}.
Therefore, it is a bit unsatisfactory.

Whether the OPERA experiment is correct or not is not concluded
at the present time and it must be checked by other experiment teams.
However, such a check will take time.
Therefore, it is very important for us to state

\vspace{3mm}\noindent
{\bf Problem}\ \ Make some (inside) questions clear
from a theoretical point of view.

\vspace{3mm}\noindent
Regarding papers related to this topic see for example \cite{several}.

\vspace{3mm}
The work is a great challenge to not only (young) Physicists
but also (young) Mathematicians, so we conclude the paper
by citing famous sentences by late Steve Jobs \footnote{\ Steven
Paul Jobs (1955--2011)}
\begin{center}
Stay hungry,\ \ Stay foolish
\end{center}
(see the Concluding Remarks in \cite{KF}).

\vspace{5mm}\noindent
{\it Acknowledgments}\ \
We wish to thank Shin'ichi Nojiri and Ryu Sasaki for useful
suggestions and comments.

redpajama_set_name: RedPajamaArXiv


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text:
When Is a Mosque Invited to Host a Religious Event?
The San Antonio, Mich.
mosque is hosting a religious event, and its organizers are worried about the reaction.
According to the Islamic Center of Michigan, the event will be held at the Islamic Centre of Michigan on Sept. 11, but the event is being held at a location at a different location.
The mosque has no official location, but says it is hosting the event because it believes it has a duty to provide a welcoming place for all.
The mosque was created in 1926 and is located on the outskirts of the city of San Anto, Mich., about 30 miles east of Grand Rapids.
The Islamic Center is a member of the National Council of American Islamic Relations, a national group dedicated to promoting the values of Islam and promoting interfaith dialogue.
"The organization has taken a very hard stance on Islamophobia," said the group's executive director, Mohammad Farooq.
"This is not something we do lightly, and it is not a thing we have to do lightly.
We're trying to create a welcoming space for all."
The Islamic Center's position was supported by its executive director and two members of the Grand Rapids Muslim community.
"We think it's an important thing to have a welcoming atmosphere for all faiths and all ethnicities, and we think it can be a positive thing to build bridges," said Grand Rapids Imam Mohamed al-Bashir.
"I don't think there's a place in this world that is not welcoming to people."
The mosque has been hosting interfaith events for the past decade, including a Muslim festival on Sept, 11, and a Christian festival on Oct. 13.
The event was moved to the next day because of a rain storm, but it was not rescheduled.
"There's no way we're going to be able to do this every year," said Imam Bashir.
"I know that this has been going on for a while," said San Anonio Imam Nabil Ali.
"We've had people coming from all over Michigan to come and visit us, and now they're coming to this event."
Muhammad Ali, an Iraq War veteran who lives in San Ano, Mich, said the mosque's stance is the reason he hasn't been able to attend the event.
"It's not a place where I can go and do this," said Ali.
Ali said the Islamic center is not the only one with concerns about what will happen if a large number of people attend the religious event.
"It's very important to us that it's a peaceful event," he said.
"A lot of people will be surprised, and I'm not surprised, that the event has gone forward," Ali said.
"They've been asking us, 'How long have you been there?'
And I'm like, 'Since January.'
We've had so many people come and see us."
Ali said the event was organized with the goal of providing a place for people to meet other Muslims and learn more about Islam.
"This is a way to bring people together and to learn about Islam, and also to share the religion," Ali explained.
"And so we're glad that this event has come forward, but I think we've been neglecting the bigger issues."
The group said they plan to file a lawsuit against the city, the mosque and the Grand County Board of Commissioners.
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