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By DifferentTeen PLATINUM, Seaford, Delaware
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DifferentTeen PLATINUM, Seaford, Delaware
"There's no such thing as true love, just spurts of insanity—falling over and over again, thinking that won't happen to me"
A man sat at the table reading the daily newspaper. His wife, across the table, was inadvertently repositioning the flowers. He thought of how she always seemed to be fixing things. "The blue flower looked absolutely at its best next to the pink one, but only if they're both behind the yellow daffodil. I won't have it any other way!" she would say to herself. Yet as he watched her arrange the very same flowers that had sat there for five years, he had never wondered why they were artificial until now.
"Why is it you don't buy real flowers for the table?" the man asked in earnest. He even touched his wife's arm in attempt to grab her attention.
She quickly stepped out of his reach to grab a ribbon from the other countertop.
"Why, real flowers fade and droop with time. Their dying colors wouldn't match the wallpaper." She wrapped the ribbon around the vase. "It's simply smarter to buy fake ones that'll always match and always look healthy and beautiful." She shook her head condescendingly and took off the ribbon again.
The husband sighed and picked up the newspaper yet again, unfolding its contents in hopes of finding something nice and exciting.
"What shall we bring to your friends the Robinsons' party tonight, dear?" he asked after a short silence.
"I'm not entirely sure, it'll come to me on the whim and I'll just whip something together." She assumed with an air, turning around and placing her hands on the sink, gazing out of the window.
"I think a nice roast would be fitting, don't you," he nodded back to her, "or maybe a baked pie?" he pushed his glasses further up the crook of his nose. She made no reply. Instead she insisted on rearranging the various items on the counter. Trying again Peter continued,
"You know, I could take the children down to the park tomorrow. I'm positive they'd love the fresh air after sitting here all night tonight with Ellen."
"Sure, sure," she answered distractedly.
"Would you care to come along?" Peter inquired.
"Oh no, I simply have too much to do."
"Yes, of course. I understand. Well, I'll be sure to take them."
He checked his watch, "I do think I'll throw myself together here, you said you were already ready didn't you? I'll go wash up. Maybe we could stop at the diner before we head over, if you'd like?"
"No no, better not. The party starts at seven, it's six now so you go on and get ready. Oh and Peter, be sure to wear the shirt I ironed and laid out for you, and don't forget to tame your hair. We'll leave when you're finished." She said, not looking at him.
"Sure sure, I'll be right down. Hey kiddo!" He rustled the hair of his oldest son who had just bustled past him and into the kitchen.
"Hi mother, I had a great day at school, do you want to hear about it?" his son sat at the table, "the teacher told us—"
Peter watched the doorway and then left to get ready.
"Dear, we should be getting a move on!" Melanie called to her husband.
Peter appeared at the top of the stairs and hurried down , greeted his wife with a kiss and a look that implied that he was ready to go. She patted down his hair.
Melanie then turned to Ellen, the babysitter, and told her to just ring the Robinsons in case of trouble; she had written the number down in the kitchen. Ellen assured her there would be no need,
"The kids will be in bed no later than nine, I'll see you when you get back Mrs. Tummel." She replied with a smile.
"Thank you so much for the favor," Melanie tilted her head slightly, "I do appreciate your help."
The couple left to continue down the sidewalk. Peter glanced over to the container his wife held, her determined, pale face illuminated by the fluorescent street lamps.
"What did you decide on dear?"
"Just a fresh salad. We had everything in the fridge so it was much easier than making a big production of everything."
"Of course." Peter hesitated.
"Well, who all will be here tonight?" Peter pushed out.
"Well the Robinsons of course, and some of our friends from childhood. It shall be interesting to see the way they've all turned out." She said, looking forward, smiling to herself. "And I suppose some of your friend might show up too." She added after a moment's pause.
"Well I'll get on well enough with your friends dear." Peter laughed to his wife.
"Hn," Melanie smiled halfheartedly back to him.
When they arrived Peter shivered as they made their way up the path. The house was nothing ornate. It was a light shade of blue, with white shutters. In the front yard some simple shrubbery lined the sidewalk. It was a breezy November night. He knocked twice on the gleaming red door.
No answer. He glanced at the window to his right, the light streaming onto the lawn. He listened to the sounds of people talking wafting through the threshold.
Melanie handed Peter the container of salad, and knocked again on the door. It immediately opened to reveal a bright faced woman who greeted her warmly and motioned to come in. She made small talk as the woman took Melanie's coat before skittering off into a side room. Peter removed his coat and held it over his arm.
They stood in a living room; a few people unfamiliar to Peter stood around and chatted amongst themselves. Melanie walked in first, making her way past them and making sure everyone had greeted her hello. After making small conversation, Melanie excused herself to the kitchen. Peter followed suit.
Baskets of fruit lay on the counter beneath a window, chicken salad, potato salad, platters of ham and turkey and chicken, bowls of dip, not to mention a variety of libations. Everyone had a plate in hand. Melanie smiled and greeted those who received her as she made her way to the counter, grabbed a plate and began piling up the home goods.
"You simply must tell me where Mrs. Robinson is?" Melanie implored, beaming, to a man behind the counter.
"I believe she should be back down soon, she went upstairs to check on the children." He returned brightly.
Everyone was quite gregarious with one another, but a tension filled the room. The couples all throughout were standing by each other. Melanie noticed Peter was standing at the outside counter.
"Peter aren't you going to make a plate?" She walked over to her husband and kissed him on the cheek, side-glancing to look for reactions.
"No that's quite all right I don't think I'm hungry." He pulled away, replying slowly and quietly, trailing off slightly at the end.
As Melanie gave him a puzzled look, Peter turned to the man behind the counter and opened his mouth to begin conversation, but the man had turned to a woman at the table behind him.
So he sat down instead. He placed the salad bowl that was still in his hands on the counter. He watched as Melanie poured herself a glass of wine and made her way to the chattering table, plate in hand. She nearly bounced over to the group of people, a hearty smile on her face, body jittery with excitement.
Melanie submerged herself in conversation. She sat at the table with two other couples. One dark haired woman laughed and shared,
"George here never brings himself out of the garage; I have to practically drag him out every night for dinner."
The man she sat next to, presumably George grinned and shrugged.
The other woman, light haired with fair skin, laughed as well, and continued,
"Chris hasn't changed since junior high, I swear on it. Memory span of a nut. I ask him to do anything and in five minute he's already forgotten it!" She shook her head, laughing.
A slight pause arrived in the conversation. Melanie smiled.
Peter sat at the counter; he grew impatient with the goose bumps on his arms. He stood up, Melanie watching all the while from the corner on her eye. He grabbed a plate from his right, opened the container of salad, and spooned a heap onto the plate. He went down the line of homemade foods, the thought of each made him want to throw up.
"Do you know?" Peter began aloud, "I'm really not hungry at all." he laughed as he made his way around the island to the white trash can on the other side, promptly dumping his plate in the contents of the black bag.
The group simply stared at him.
Melanie was horrified
Peter laughed and left the party without so much as a good-bye.
© Heather W., No., Brunswick, NJ
A fiction piece for my creative writing class, this was fun to write! Enjoy!
Confessions of an Invisible
By AmberLorraine SILVER
Olathe, Kansas
AmberLorraine SILVER, Olathe, Kansas
"If the sky were to suddenly open up, there would be no law, there would be no rule. There would only be you and your memories." - Donnie Darko
The Dove or the Sword
By angrybob13 SILVER
Fayetteville, Arkansas
angrybob13 SILVER, Fayetteville, Arkansas
The Shoes
By Chris Pumford
By Kennedy Springer
Delta, UT
She's Another Girl Gone Wrong
By Lauren Kibby


the discussions surrounding the concept of globalization.
We've examined the main international components of the world economy.
foreign direct investment and financial markets.
exactly are the implications of this for society?".
we're faced with one big problem.
accurately doesn't mean that we can't say anything at all.
So let's start with a baseline number.
the best measure we have for the output of goods and services in a society.
and here we confront our first problem.
In calculating GDP, the only part of a transaction that counted was the value added.
so there is a degree of double counting in this number.
and the value of FDI 2012 was $1.45 trillion.
So that basically deals with the real economy in its global dimension.
Now we turn our attention to the financial sector.
Let's start with equities, stocks and shares.
We have a figure for end year total valuation of $52.5 trillion.
$50 trillion, but foreign ownership is not likely to be large.
and that would give us an international trading figure of $5 trillion.
could take a trading volume of about 20 trillion in a given year.
but we noted that only a small down payment ever change hands.
So how small is small?
lower for some transaction higher for others.
real assets with real values.
Real assets with real values and real values that can and do change.
say it doesn't really matter if a dollar is a Euro.
Now these numbers are mind bogglingly huge.
So let's try and put it in an easier context.
Most common way to do this is to relate it all to GDP.
not only do say that foreign trade is equivalent to 25 percent of world GDP.
Foreign direct investment then be equivalent to 2 percent of world GDP.
Now a rough estimate for the International Trade in equities would be 7 percent of GDP.
derivative trade would still be 78 percent of GDP.
which by itself are whole 25 times bigger than GDP.
Did I just say that 25 times bigger?
And again the numbers are just becoming meaningless.
It's money we're talking about.
So let's start all over again.
and on top of that we start layering our globalization components in $100 bills.
OK, let's run quickly through the numbers again.
Let's start with foreign direct investment, $1.45 trillion.
$100 bills, the pile would reach 1,584 kilometers into space.
will have passed the international space station below us.
Add foreign trade, $18.2 trillion.
This would reach almost 20,000 kilometers into space.
and we've reached 93,000 kilometers into space.
By now we're about one-third of our way to the moon.
We haven't even had foreign exchange transactions yet, $1,855 trillion.
I hope by now the pile of notes stretches 2 million kilometers into space.
probably get to the moon and back twice over and still have a small fortune left.
If we simply carried on, we'd be on our way to Mars.
we'd be at Mars already and be on our way to the next planet.
involving infinitely larger sums of transactions.
But where do we find these numbers in the GDP statistics?
added to the stock of wealth generated then by an economy.
and profits generated by the financial services industry.
this can account for up to 10 percent of GDP.
mortgage, and banking services that we consume domestically.
a relatively small sector in relatively rich countries.
entailing massive financial risks that affect us all.
with a reading from John Maynard Keynes and I'd like to end with one.
This time I want to read from Susan Strange "Casino Capitalism."
Susan and I spent a bit of time together at the European University Institute.
journalism into the economics game and actually writes well.
system is rapidly coming to resemble nothing as much as a vast casino.
involve sums of money so large they cannot be imagined.
At night, the games go on on the other side of the world.
rooms are full of chain-smoking young men all playing these games.
Their eyes are fixed on computer screens flickering with changing prices.
electronic machines or with computer algorithms today.
odd numbers or even numbers."
and the typical bankers that older people remember, like me.
and for the careful guardianship of their customers' money.
the international financial system to make it so much like a gambling hall."
she said I'm writing in 1986 and how it came about are not clear.
but one big thing has happened.
that with the most recent financial crisis and underlying this is a loss of trust.


Restaurant Kanunnik is the place to come for breakfast, lunch, high tea, dinner or drinks. Both the kitchen team and the serving staff are not just professional, but also display an air of the best possible hospitality. They like to put together a wine package that matches your menu choice perfectly. For our little guests, we serve a wide range of children's menus. Restaurant Kanunnik is open daily and features a large terrace with outdoor bar 'Il Palazzo'. After dinner, you are welcome to have a drink in our cosy indoor bar 'de Trappist'.
Your restaurant visit is perfect for combining with a visit to the Schoorl dunes. These dunes are known as the highest, most beautiful and widest dunes in the Netherlands. The restaurant is the perfect place to start or end a hike or cycle ride through this beautiful nature reserve.
Restaurant Kanunnik owes its name to the canon (kanunnik) and painter Jan van Scorel, born in Schoorl in 1495. Chef de cuisine Sander van Doorn mans the stove and he, together with his team, has compiled a menu of dishes that only use fresh produce. Sander also uses regional and seasonal produce as far as possible, because he believes responsible cooking is important and because he aims to do business in a sustainable manner. Which is not just a noble goal, but is also to sample what he is serving his customers.


Remember those lovely lambswool pillows we shared with you a bit ago? Hildegard was so pleased with the pillows, she decided she wanted to share some of Great Grandma's memories with the youngest members of the family. Using the leftover lambswool from the coat (for the ears and tail) and a sweater of Hildegard's from Germany, we created these lovable Puppy and Kitty Stuffies.
Since these Stuffies combine clothing from the recipients' Grandma and Great Grandma, these cute little guys represent a piece of family history.
What I love about these Creations is the story behind them. When gifts like these are given, the stories are passed along with them.
That just gave me an idea!!! Maybe we should include note cards as part of our gift packaging, with the intention that the giver will write down the little bit of history that goes with the gift. What do you think???
Our name, Once Upon a Time Creation, originated because EVERYTHING we create has a back story. How does every good story begin? Once Upon a Time….
Maybe it's time to encourage and facilitate the sharing of your story in written form. I would love to know what you think. Should we start including note cards as part of our gift packaging?
This entry was posted in Custom Keepsakes, Dolls and Stuffed Animals, Throw Pillow Fun and tagged memorial gifts, once upon a time creation, personalized gifts, remembrance, remembrance pillows on February 25, 2014 by onceuponatimecreation.
Hildegard saved her mother's lambswool coat for years. It was the one possession she brought back from Germany when her mother passed. The coat was beautiful and meaningful, but not able to be worn.
This entry was posted in Custom Keepsakes, Throw Pillow Fun and tagged keepsakes, memorial gifts, once upon a time creation, personalized gifts, remembrance pillows, throw pillow on February 11, 2014 by onceuponatimecreation.
These Memory Blankets were created in remembrance of Carey's Grandmom. Three blankets were made from Grandmom's clothes, one for Carey and her Mom and two for her aunts. Each one is unique and specially designed for it's recipient.
Thank you for those kind words. Our hope is that your blankets bring you peace and comfort.
This entry was posted in Custom Keepsakes, Memory Blankets and tagged memorial gifts, memory blanket, once upon a time creation, remembrance on October 1, 2013 by onceuponatimecreation.


'The Five' co-host Dana Perino discusses polling numbers and her new book.
The Daily Briefing With Dana Perino 4/19/19 | THE MUELLER REPORT | Fox News Today April 19, 2019.
Kayaking for a great cause on the Cuyahoga River.
'The Daily Briefing' anchor Dana Perino on President Trump's handling of the NYC terror attack and whether he was right to criticize the Diversity Visa Lottery.


Networks are a mathematical abstraction of the interactions between a set of entities, with extensive applications in social science, epidemiology, bioinformatics, and cybersecurity, among others. There are many fundamental problems when analyzing network data, such as anomaly detection, dense subgraph mining, motif finding, information diffusion, and epidemic spread. A common underlying task in all these problems is finding an "interesting subgraph"; that is, finding a part of the graph---usually small relative to the whole---that optimizes a score function and has some property of interest, such as connectivity or a minimum density. Finding subgraphs that satisfy common constraints of interest, such as the ones above, is computationally hard in general, and state-of-the-art algorithms for many problems in network analysis are heuristic in nature. These methods are fast and usually easy to implement. However, they come with no theoretical guarantees on the quality of the solution, which makes it difficult to assess how the discovered subgraphs compare to an optimal solution, which in turn affects the data mining task at hand. For instance, in anomaly detection, solutions with low anomaly score lead to sub-optimal detection power. On the other end of the spectrum, there have been significant advances on approximation algorithms for these challenging graph problems in the theoretical computer science community. However, these algorithms tend to be slow, difficult to implement, and they do not scale to the large datasets that are common nowadays. The goal of this dissertation is developing scalable algorithms with theoretical guarantees for various network analysis problems, where the underlying task is to find subgraphs with constraints. We find interesting subgraphs with guarantees by adapting techniques from parameterized complexity, convex optimization, and submodularity optimization. These techniques are well-known in the algorithm design literature, but they lead to slow and impractical algorithms. One unifying theme in the problems that we study is that our methods are scalable without sacrificing the theoretical guarantees of these algorithm design techniques. We accomplish this combination of scalability and rigorous bounds by exploiting properties of the problems we are trying to optimize, decomposing or compressing the input graph to a manageable size, and parallelization. We consider problems on network analysis for both static and dynamic network models. And we illustrate the power of our methods in applications, such as public health, sensor data analysis, and event detection using social media data.


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Dalkeith girl designs escape room for charity
Thursday December 17th 2020
Bryony Primrose from Dalkeith designed the Secret Spy Service
A play-at-home escape room game that was designed by a Dalkeith schoolgirl will soon be on sale raising money for Edinburgh Children's Hospital charity.
Twelve year old Bryony Primrose won the 12-14s category of a national competition to design an escape room and now her design has now been turned into a home escape room kit in time for Christmas. The escape room game is called Secret Spy Service and players must assist Commander Bryony on a top secret mission to capture the Cryptic Gang. The game combines a range of puzzles including maths, spelling, constructing and deciphering and is recommended for ages 12 and up.
Bryony said:
"My dad saw the competition online and thought I would like to enter as I have completed a couple of escape rooms. Although I didn't think I would win, I decided to give it a try anyway. I was very surprised when I got the email saying that I won, but I'm very happy that I did win because it was a lot of fun to design and all the money goes to a really good cause."
Bryony's chosen charity is Edinburgh Children's Hospital.
"My chosen charity is one close to the hearts of my family. We lost my brother 7 years ago and the Edinburgh Children's Hospital were very helpful to us while trying to deal with the loss. We have done a couple of things in the past to raise money for them so it was quite an easy decision."
The design an escape room competition was organised by Cheltenham based social enterprise, Create on the Square, which aims to "encourage creative enterprise in the local community", as a positive response to the Covid-19 pandemic.
"When the lockdown was called in April, all of our projects were immediately shut down, so we decided to run a competition to give children something to do during lockdown, and also to raise money for charities in this difficult year. We were very pleased with the response from across the country and hope the games will raise plenty of money for charity now" said Pete George, Director, Create on the Square CIC
The competition asked children, in two age categories, 12-14 and under-12s, to come up with a design for an escape room on two a4 sheets of paper. The competition judges especially liked the idea behind the story, the thoroughness of the puzzle suggestions, and the thought that was put into making it appropriate for the age group in Bryony's design and unanimously awarded her the winner in the 12-14s category.
Bryony's design was then handed over to the escape room designers at Escape Rooms Cheltenham, who have given their time freely to create and test the game.
"It has been great to bring Bryony's ideas to life, and be able to include elements that are personal to her in the game. We're really excited to launch this new game, especially for such a great cause."
The escape room game is on sale on Escape Rooms Cheltenham website with 100% of the sale price of £10 going to Bryony's chosen charity, Edinburgh Children's Hospital.

redpajama_set_name: RedPajamaCommonCrawl


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48 months profit free finance on buying apartment in Shaden Al Hail project
Business Monday 06/December/2021 17:14 PM
Muscat: Alizz Islamic Bank has strengthened its strategic partnerships with real estate development companies in the Sultanate by signing a new agreement with Shaden Development to provide easy and flexible financing facilities to customers who wish to buy an apartment in Shaden Al Hail Residential Complex, which is one of the most prestigious residential complexes in Muscat.
This agreement comes within a variety of housing financing options offered by Alizz Islamic Bank represented by the Retail Banking Department, all of which comply with the provisions and principles of Islamic Sharia.
The signing ceremony of the agreement was held at the head office of Alizz Islamic Bank in CBD, Ruwi, coinciding with the Sultanate's celebrations of the glorious National Day.
The memorandum was signed by Mohammed Al Ghassani, AGM – Chief Consumer Banking Officer on behalf of Alizz Islamic Bank, and Engineer Abdul Rahman Ba Omar, Board Member of Shaden Development Company on behalf of Shaden Development Company. The signing ceremony was attended by the CEO of Alizz Islamic Bank, Suleiman Al-Harthy, and a number of senior management members.
According to this agreement, Alizz Islamic Bank will provide financing facilities to customers who wish to purchase an apartment in Shaden Al Hail project for eligible customers with exemption from profits and transaction fees for a period of up to 48 months or 4 years.
Shaden Development Company will bear the transaction fees and profits on behalf of customers during this period, as well as the Takaful insurance fees, with the possibility of extending according to the agreement that is also concluded between the customer and the developer. The validity of this offer is from November 18 to December 31.
Alizz Islamic Bank will provide housing finance under the Diminishing Musharakah concept with a financing rate of up to 80 per cent of the value of the housing unit for eligible customers in accordance with the bank's policy, as well as the Debt Burden Ratio (DBR) and the approval of the valuation report for each financed unit from appraisers approved by the bank in accordance with the current procedures; and at a rate of 90 per cent of the value of the housing unit for customers who take home financing for the first time in accordance with the applicable internal bank policies and procedures.
In addition, Alizz Islamic Bank offers financing facilities at a reduced profit rate with a very fast completion of banking transactions, as all services and products offered by Alizz Islamic Bank have been approved by the Fatwa and Sharia Supervisory Board in accordance with the principles of the Islamic Sharia.
Under the principle of "Diminishing Musharakah", Alizz Islamic Bank will enter into a Diminishing Musharaka agreement with the customer, whereby each party contributes to the Musharaka assets a certain percentage of jointly owning the property. The parts of the bank and the customer that are joint in the partnership assets are divided into equal shares when the construction stages are completed, and the bank begins to transfer the ownership of its share in the real estate by selling its shares to the customer gradually, and the customer buys the bank's shares in the real estate from time to time by paying the sale price in scheduled instalments.
Accordingly, upon maturity of the participation, the ownership of the bank in the property must be transferred exclusively to the customer, and accordingly, the customer is considered the sole owner of the property.
As for customers who wish to take advantage of this exceptional offer and do not hold accounts with Alizz Islamic Bank, all they have to do is contact the customer service centre or visit the nearest branch of Alizz Islamic Bank, as the housing finance requires transferring the salary to the bank and enjoying a varied array of financing services provided by Alizz Islamic Bank, whether at the level of housing or personal finance, all of which are in accordance with the provisions and principles of Islamic Sharia.
It is worth mentioning that Alizz Islamic Bank has signed during the last period a variety of agreements with various real estate developers to provide a wide range of housing finance options for customers. Alizz Islamic Bank intended to stay close to its customers and listen to them and then take their desires and opinions and translate them into reality to realize every dream they deserve.
UK to allocate additional $130mn in humanitarian aid to Afghanistan: Reports


This is a list of your rights as a patient. This also gives the procedure for filing a complaint, if necessary.
You have the right to obtain from your physician complete, current information concerning your diagnosis, treatment and prognosis in terms you can reasonably expect to understand.
You have the right to receive from your physician information necessary to give informed consent prior to the start of any procedure and/or treatment.
You have the right to have an advance directive (such as a living will, health care proxy, or durable power of attorney for health care) concerning treatment or designating a surrogate decision maker with the expectation that the hospital will honor the intent of that directive to the extent permitted by law and hospital policy.
You have the right to refuse treatment to the extent permitted by law and to be informed of the medical consequences of your action.
You have the right to every consideration of your privacy concerning your own medical care program.
You have the right to expect that all communications and records pertaining to your care should be treated as confidential.
You have the right to expect that within its capacity, the hospital must make a reasonable response to your request for services. When medically permissible, you may be transferred to another facility only after you have received complete information and explanation concerning the needs for and alternatives to such a transfer.
You have the right to obtain, and get assistance to obtain, a second opinion concerning your care.
You have the right to obtain information as to any relationship of your hospital to other health care and educational institutions insofar as your care is concerned. You have the right to purchase your records.
You have the right to be advised if the hospital proposes to engage in, or perform, human experimentation affecting your care or treatment. You have the right to refuse to participate in such research projects.
You have the right to expect reasonable continuity of care.
You have the right to know what hospital rules and regulations apply to your conduct as a patient.
You have the right to be free of restraints.
You have the right to be free from seclusion.
If you feel your rights have been violated in any way, or if you have a complaint concerning your care, ask to see a member of the administrative staff. Your complaint will then be investigated and any necessary action will be taken accordingly. You will receive a response within 72 hours.
You also have the right to file a complaint with licensing agencies concerning your care or your bill.
A. For complaints regarding your care: Kansas Advocacy and Protective Services 1-800-432-8276.
You have the right to adequate pain control with the appropriate medication or treatment.
You have the right to exercise these rights and you will not be discriminated against because you have exercised these rights.


MANILA, Jan. 24 (PIA) -- President Benigno S. Aquino will lead on Tuesday afternoon, January 24, the ceremonial turnover of command of the Philippine Coast Guard at the Coast Guard Headquarters in Manila.
Vice Commandant for Operations Rear Admiral Edmund Tan will now head over 5,000 uniformed personnel and civilian employees of the Philippine Coast Guard replacing outgoing Coast Guard chief Admiral Ramon Liwag.
Admiral Liwag will retire from the military service today, coinciding with his 56th birthday. He is also a member of the Philippine Military Academy 'Matapat' Class of 1979.
Meanwhile, the newly-designated Coast Guard chief is a native of Dumaguete City and a member of the Philippine Military Academy 'Mapitagan' Class of 1980.


text:
\section{Introduction}
It has long been recognized that standard quantum field theory suffers from a plague of singularities and infinities. The author would even go so far as to say that in its usual form, a quantum field does not exist! By this we mean that it does not have a mathematically well-defined definition. The problems are compounded by employing these nonexistent quantum fields in rather complicated ways to define interaction Hamiltonians and scattering operators. Perturbation methods are then used to compute numbers that frequently agree with experiment. However, to obtain these numbers, renormalization techniques and infinity cancellations have to be employed
\cite{ps95,vel94}. Attempts have been made to develop a rigorous mathematically well-define quantum field theory \cite{sw64}. These attempts have been successful for free quantum fields, but they have not adequately included interacting fields with nontrivial scattering.

It seems to us that the best way to overcome these difficulties is to postulate that spacetime is discrete \cite{gud68,hei30,rus54}. Although such a granular structure for spacetime has not been experimentally observed, with increasingly accurate instruments it may become evident in the future, perhaps in an indirect manner. The granular structure may manifest itself in terms of elementary lengths and times at Planck scales of about $10^{-33}$cm. and $10^{-43}$sec., respectively. The simplest such framework would be a rigid cubic 4-dimensional lattice. The author has previously studied a tetrahedral lattice that appears to have certain advantages \cite{gud16} but for simplicity and ease of computation, we shall only consider the cubic lattice ${\mathcal S}$ here.

It follows from this description that the speed of light is the only nonzero instantaneous speed for a particle. We observe slower speeds because we actually measure average speeds of objects. The dual space $\widehat{\sscript}$ corresponds to a 4-dimensional cubic lattice of energy-momentum. This description implies that there is a discrete set of possible particle masses, which we then determine.

We next define discrete scalar quantum fields on ${\mathcal S}$. We show that these fields exist mathematically and derive some of their properties. We mention one can also study discrete vector quantum fields that involve spin, but  for simplicity we only consider the spin zero case here. Various fields are combined to form interaction Hamiltonians. We postulate that the scattering operator satisfies a ``second quantization'' Schr\"odinger equation and derive its form. Besides the definition of these operators, our main assumption is conservation of energy-momentum for scattering processes. Due to this conservation law, we show that a scattering process usually possesses only a finite number of possible outgoing states. These can then be summed over to find scattering amplitudes and probabilities.

The article concludes with various examples of perturbation approximations. These include simplified versions of electron-electron and electron-proton scattering as well as simple decay processes. We also define scattering cross-sections, decay rates and lifetimes within this formalism. We compute these quantities in simple cases. Finally, some speculations are made about the theory's ability to predict the existence of dark energy and dark matter.

\section{Discrete Spacetime and Energy-Momentum}
Our basic assumption is that spacetime is discrete and has the structure of a 4-dimensional cubic lattice ${\mathcal S}$. We regard ${\mathcal S}$ as a framework or scaffolding in which the vertices of the lattice (or network) represent tiny cells of Planck scale that may or may not be occupied by a particle. The edges between vertices represent space directions in which particles can propagate at a given time. Let
\begin{equation*}
{\mathbb Z} =\brac{0,\pm 1,\pm 2,\ldots}
\end{equation*}
be the integers and ${\mathbb Z} ^+=\brac{0,1,2,\ldots}$ be the nonnegative integers. We have that ${\mathcal S} ={\mathbb Z} ^+\times{\mathbb Z} ^3$ where ${\mathbb Z} ^+$ represents discrete time and ${\mathbb Z} ^3$ represents discrete 3-space. If $x=(x_0,x_1,x_2,x_3)\in{\mathcal S}$, we write $x=(x_0,\bfx )$ where $x_0\in{\mathbb Z} ^+$ is time and
$\bfx\in{\mathbb Z} ^3$ is a 3-space point. We have that ${\mathcal S}$ is a module in the sense that ${\mathcal S}$ is closed under addition and multiplication by elements of
${\mathbb Z} ^+$. The vectors
\begin{align*}
d&=(1,\bfzero )=(1,0,0,0),\enspace e=(0,\bfe )=(0,1,0,0)\\
f&=(0,\bff )=(0,0,1,0),\enspace g=(0,\bfg )=(0,0,0,1)
\end{align*}
form a basis for ${\mathcal S}$ and every $x\in{\mathcal S}$ has the unique form
\begin{equation*}
x=nd+me+pf+qg
\end{equation*}
$n\in{\mathbb Z} ^+$, $m,p,q\in{\mathbb Z}$.

We equip ${\mathcal S}$ with the Minkowski distance
\begin{equation*}
\doubleab{x}_4^2=x_0^2-\doubleab{\bfx}_3^2=x_0^2-x_1^2-x_2^2-x_3^2
\end{equation*}
As usual $x,y\in{\mathcal S}$ are \textit{time-like separated} if $\doubleab{x-y}_4^2\ge 0$ and \textit{space-like separated} if $\doubleab{x-y}_4<0$. For $x\in{\mathcal S}$, the set
\begin{equation*}
{\mathcal C} ^+(x)=\brac{y\in{\mathcal S}\colon y_0\ge x_0,\doubleab{y-x}_4^2\ge 0}
\end{equation*}
is the \textit{future light cone at} $x$. Of course, we are using units in which the speed of light is 1. With our interpretation of the structure of ${\mathcal S}$, it seems natural that a particle at $\bfx$ can either stay at $\bfx$ or move in one of the six space directions to a point $\bfy$ in one time unit, where
$\bfy =\bfx\pm\bfe, \bfx\pm\bff ,\bfx\pm\bfg$. In this way, the only nonzero instantaneous speed of a particle is the speed of light~1. This gives a primitive reason why the speed of light is the upper limit for all signal speeds. The reason that we observe slower speeds is that we usually measure average speeds and not instantaneous ones. If a particle moves from $\bfx$ to $\bfy$, its \textit{average speed} is $\overline{v} =\doubleab{\bfy -\bfx}_3/t$ where $t$ is the elapsed proper time; that is, the time measured along the particle trajectory. For example, suppose a particle initially at $\bfx$, stays at $\bfx$ for one time unit and then moves to $\bfy =\bfx +\bfe$ in the next time unit. Then the proper time $t=2$ and the average speed is $\overline{v}=\doubleab{\bfe}_3/2=1/2$. As another example, suppose a particle initially at $\bfzero$ moves to $\bfe$ at the first time unit and then to $\bfe +\bff$ at the second time unit. The proper time is again $t=2$ and $\overline{v}=\doubleab{\bfe +\bff}_3/2=1/\sqrt{2}$. We conclude that photons must travel in straight lines along coordinate axes.

The dual of ${\mathcal S}$ is denoted by $\widehat{\sscript}$. We regard $\widehat{\sscript}$ as having the identical structure as ${\mathcal S}$ and that $\widehat{\sscript}$ again has basis $d,e,f,g$.
The only difference is that we denote elements of $\widehat{\sscript}$ by
\begin{equation*}
p=(p_0,\bfp )=(p_0,p_1,p_2,p_3)
\end{equation*}
and interpret $p$ as the energy-momentum vector for a particle. In fact, we sometimes even call $p\in\widehat{\sscript}$ a particle. Moreover, we only consider the forward light cone
\begin{equation*}
{\mathcal C} ^+(0)=\brac{p\in\widehat{\sscript}\colon\doubleab{p}_4\ge 0}
\end{equation*}
in $\widehat{\sscript}$. For a particle $p\in\widehat{\sscript}$, we call $p_0\ge 0$ the \textit{total energy}, $\doubleab{\bfp}_3\ge 0$ the \textit{kinetic energy}
and $m=\doubleab{p}_4\ge 0$ the \textit{mass} of $p$. The integers $p_1,p_2,p_3$ are \textit{momentum components}. Since
\begin{equation}
\label{eq21}
m^2=\doubleab{p}_4^2=p_0^2-\doubleab{\bfp}_3^2
\end{equation}
we conclude that Einstein's energy formula $p_0=\sqrt{m^2+\doubleab{\bfp}_3^2}$ holds.

It is clear that any nonnegative integer can be a total energy. However, the square of a kinetic energy must be the sum of three squares
$\doubleab{\bfp}_3^2=p_1^2+p_2^2+p_3^2$. A number theory theorem \cite{vde87} says that $n\in{\mathbb Z} ^+$ is a sum of three squares if and only if $n$ does not have the form $4^k(8s+7)$, $k,s\in{\mathbb Z} ^+$. We conclude that $7,15,23,28,\ldots$ are eliminated. Thus, $\doubleab{\bfp}_3=\sqrt{n}$, $n\in{\mathbb Z} ^+$ is a kinetic energy if and only if $n\ne 7,15,23,28,\ldots\,$. It follows that every nonnegative number $\sqrt{n}$, $n\in{\mathbb Z} ^+$ is a possible mass and vice versa, although it may not appear for certain total energy $p_0\ge\sqrt{n}$. Therefore, this theory predicts that there are a countable number of admissible particle masses. In contrast to the speeds considered previously, we also have the concept of a \textit{geometric speed} for $p\in\widehat{\sscript}$ given by
\begin{equation*}
v=\frac{\doubleab{\bfp}_3}{p_0}=\frac{\sqrt{p_0^2-m^2}}{p_0}=\sqrt{1-\frac{m^2}{p_0^2}}
\end{equation*}
whenever $p_0\ne 0$. Of course, $0\le v\le 1$ and $v=1$ if and only if $m=0$. Moreover, $v=0$ if and only if $\doubleab{\bfp}=0$.

Table~1 lists kinetic energy values, the corresponding momentum vectors and the number of such vectors. Notice, as previously observed, that $\sqrt{7}$, $\sqrt{15}$, $\sqrt{23}$ are missing.

We can apply Table~1 to find the possible particle masses $m$ corresponding to total energy $p_0$ as well as the number of particle kinetic energies. In this work, we assume that $p_0\ne 0$.
\newpage

\vglue -5pc
{\parindent=-3pc
\begin{tabular}{c|c|c}
Kinentic&&\hfill\hfill\\
Energy&Momentum Vectors&Number\hfill\hfill\\
\hline
1&$\pm\bfe ,\ \pm\bff ,\ \pm\bfg$\hfill\hfill&6\\\hline
$\sqrt{2}$&$\pm\bfe\pm\bff ,\ \pm\bfe\pm\bfg ,\ \pm\bff\pm\bfg$\hfill\hfill&12\\\hline
$\sqrt{3}$&$\pm\bfe\pm\bff\pm\bfg$\hfill\hfill&8\\\hline
2&$\pm 2\bfe ,\ \pm 2\bff ,\ \pm 2\bfg$\hfill\hfill&6\\\hline
$\sqrt{5}$&$\pm 2\bfe\pm\bff ,\ \pm 2\bfe\pm\bfg ,\ \pm 2\bff\pm\bfe ,\ \pm 2\bff\pm\bfg ,\ \pm 2\bfg\pm\bfe ,\ \pm 2\bfg\pm\bff$\hfill\hfill&24\\\hline
$\sqrt{6}$&$\pm 2\bfe\pm\bff\pm\bfg ,\ \pm2\bff\pm\bfe\pm\bfg ,\ \pm 2\bfg\pm\bfe\pm\bff$\hfill\hfill&24\\\hline
$\sqrt{8}$&$\pm 2\bfe\pm 2\bff ,\ \pm 2\bfe\pm 2\bfg ,\ \pm 2\bff\pm 2\bfg$\hfill\hfill&12\\\hline
3&$\pm 3\bfe ,\ \pm 3\bff ,\ \pm 3\bfg ,\ \pm 2\bfe\pm 2\bff\pm\bfg ,\ \pm 2\bfe\pm 2\bfg\pm\bff ,\ \pm 2\bff\pm 2\bfg\pm\bfe$\hfill\hfill&30\\\hline
$\sqrt{10}$&$\pm 3\bfe\pm\bff ,\ \pm 3\bfe\pm\bfg ,\ \pm 3\bff\pm\bfe ,\ \pm 3\bff\pm\bfg ,\ \pm 3\bfg\pm\bfe ,\ \pm 3\bfg\pm\bff$\hfill\hfill&24\\\hline
$\sqrt{11}$&$\pm 3\bfe\pm\bff\pm\bfg ,\ \pm 3\bff\pm\bfe\pm\bfg ,\ \pm 3\bfg\pm\bfe\pm\bff$\hfill\hfill&24\\\hline
$\sqrt{12}$&$\pm 2\bfe\pm 2\bff\pm 2\bfg$\hfill\hfill&8\\\hline
$\sqrt{13}$&$\pm 3\bfe\pm 2\bff ,\ \pm 3\bfe\pm 2\bfg ,\ \pm 3\bff\pm 2\bfe ,\ \pm 3\bff\pm 2\bfg ,\ \pm 3\bfg\pm 2\bfe ,\ \pm 3\bfg\pm 2\bff$\hfill\hfill&24\\\hline
$\sqrt{14}$&$\pm 3\bfe\pm 2\bff\pm\bfg ,\ \pm 3\bfe\pm 2\bfg\pm\bff ,\ \pm 3\bff\pm 2\bfe\pm\bfg ,\ \pm 3\bff\pm 2\bfg\pm\bfe$,\hfill\hfill&\\
&$\pm 3\bfg\pm 2\bfe\pm\bff ,\ \pm 3\bfg\pm 2\bff\pm\bfe$\hfill\hfill&48\\\hline
4&$\pm 4\bfe ,\ \pm 4\bff ,\ \pm 4\bfg$\hfill\hfill&6\\\hline
$\sqrt{17}$&$\pm 4\bfe\pm\bff ,\ \pm 4\bfe\pm\bfg ,\ \pm 4\bff\pm\bfe ,\ \pm 4\bff\pm\bfg ,\ \pm 4\bfg\pm\bfe ,\ \pm 4\bfg\pm\bff$,\hfill\hfill&\\
&$\pm 3\bfe\pm 2\bff\pm 2\bfg ,\ \pm 3\bff\pm 2\bfe\pm 2\bfg ,\ \pm 3\bfg\pm 2\bfe\pm 2\bff$\hfill\hfill&48\\\hline
$\sqrt{18}$&$\pm 4\bfe\pm\bff\pm\bfg ,\ \pm 4\bff\pm\bfe\pm\bfg ,\ \pm 4\bfg\pm\bfe\pm\bff ,\ \pm 3\bfe\pm 3\bff ,\ \pm 3\bfe\pm 3\bfg ,\ \pm 3\bff\pm 3\bfg$\hfill\hfill&36\\\hline
$\sqrt{19}$&$\pm 3\bfe\pm 3\bff\pm\bfg ,\ \pm 3\bfe\pm 3\bfg\pm\bff ,\ \pm 3\bff\pm 3\bfg\pm\bfe$\hfill\hfill&24\\\hline
$\sqrt{20}$&$\pm 4\bfe\pm 2\bff ,\ \pm 4\bfe\pm 2\bfg ,\ \pm 4\bff\pm 2\bfe ,\pm 4\bff\pm 2\bfg ,\ \pm 4\bfg\pm 2\bfe ,\ \pm 4\bfg\pm 2\bff$\hfill\hfill&24\\\hline
$\sqrt{21}$&$\pm 4\bfe\pm 2\bff\pm\bfg ,\ \pm 4\bfe\pm 2\bfg\pm\bff ,\ \pm 4\bff\pm 2\bfe\pm\bfg ,\ \pm 4\bff\pm 2\bfg\pm\bfe$,\hfill\hfill&\\
&$\pm 4\bfg\pm 2\bfe\pm\bff ,\ \pm 4\bfg\pm 2\bff\pm\bfe$\hfill\hfill&48\\\hline
$\sqrt{22}$&$\pm 3\bfe\pm 3\bff\pm 2\bfg ,\ \pm 3\bfe\pm 3\bfg\pm 2\bff ,\ \pm 3\bff\pm 3\bfg\pm 2\bfe$\hfill\hfill&24\\\hline
$\sqrt{24}$&$\pm 4\bfe\pm 2\bff\pm 2\bfg ,\ \pm 4\bff\pm 2\bfe\pm 2\bfg ,\ \pm 4\bfg\pm 2\bff\pm 2\bfe$\hfill\hfill&24\\\hline
5&$\pm 5\bfe ,\ \pm 5\bff ,\ \pm 5\bfg ,\ \pm 4\bfe\pm 3\bff ,\ \pm 4\bfe\pm 3\bfg ,\ \pm 4\bff\pm 3\bfe$,\hfill\hfill&\\
&$\pm 4\bff\pm 3\bfg ,\ \pm 4\bfg\pm3\bfe ,\ \pm 4\bfg\pm 3\bff$\hfill\hfill&30\\\hline
$\vdots$&\hfill\hfill&\\\hline
6&$\pm 6\bfe ,\ \pm 6\bff ,\ \pm 6\bfg ,\ \pm 4\bfe\pm 4\bff\pm 2\bfg ,\ \pm 4\bfe\pm 4\bfg\pm 2\bff ,\ \pm 4\bff\pm 4\bfg\pm 2\bfe$\hfill\hfill&30\\
\hline\noalign{\medskip}
\multicolumn{3}{c}{\textbf{Table 1}}\\
\end{tabular}
\parindent=18pt}
\newpage

\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
Total Energy&\multicolumn{2}{|c|}{1}&\multicolumn{5}{|c|}{2}&\multicolumn{9}{|c|}{3}\\
\hline
mass${}^2$&1&0&4&3&2&1&0&9&8&7&6&5&4&3&1&0\\
\hline
number&1&6&1&6&12&8&6&1&6&12&8&6&24&24&12&30\\
\hline
\end{tabular}
\vskip 2pc

\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
Total Energy&\multicolumn{15}{|c|}{4}\\
\hline
mass${}^2$&16&15&14&13&12&11&10&8&7&6&5&4&3&2&0\\
\hline
number&1&6&12&8&6&24&24&12&30&24&24&8&24&48&6\\
\hline
\end{tabular}
\vskip 2pc

{\parindent=-5pc
\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}
Total Energy&\multicolumn{23}{|c|}{5}\\
\hline
mass${}^2$&25&24&23&22&21&20&19&17&16&15&14&13&12&11&9&8&7&6&5&4&3&1&0\\
\hline
number&1&6&12&8&6&24&24&12&30&24&24&8&24&48&6&48&36&24&24&48&24&24&30\\
\hline\noalign{\medskip}
\multicolumn{24}{c}{\textbf{Table 2}}\\
\end{tabular}
\parindent=18pt}
\bigskip

We classify the mass zero particles (photons) into two types, the one dimensional photons (e.g. $\bfe , 2\bff ,\ldots$) that we call \textit{light photons}
(pun intended) and the 2-dimensional and 3-dimensional photons (e.g. $4\bfe +3\bff , 2\bfe +2\bff+\bfg ,\ldots$) that we call \textit{dark photons}. Both of these types have geometric speed~1. However, the light photons propagate along coordinate axes so we can consider their average speed to be 1 while the dark photons do not move directly along coordinate axes so we can consider their average speed to be less than 1. We speculate that these latter particles correspond to dark energy. In a similar way the positive mass particles are either 1, 2 or 3-dimensional and we speculate that the first two types correspond to matter and the last type corresponds to dark matter.

\section{Free Quantum Fields}
In this section, we study free discrete quantum fields. Since we are considering scalar fields, a particle is essentially determined by its mass. We assume that we are describing a physical system that contains particles of a finite number of various types. For illustrative purposes, suppose there are two types of particles under consideration which we call $p$-particles and $q$-particles with masses $m$ and $M$, respectively, $m\ne M$. The sets
\begin{align*}
\Gamma _m&=\brac{p\in\widehat{\sscript}\colon\doubleab{p}_4=m}\\
\Gamma _M&=\brac{q\in\widehat{\sscript}\colon\doubleab{q}_4=M}
\end{align*}
are called the mass \textit{hyperboloids} for the particles. To describe these particles quantum mechanically, we construct a complex Hilbert space $K$. Technically speaking,
$K$ is a symmetric Fock space, but the details are not important here. All we need to know is that $K$ exists and has a very descriptive orthonormal basis of the form
\begin{equation*}
\ket{p_1p_2\ldots p_nq_1q_2\ldots q_s}
\end{equation*}
that represents the quantum state in which there are $n$ $p$-particles and $s$ $q$-particles where $p_i,q_j\in\widehat{\sscript}$, $i=1,\ldots ,n$, $j=1,\ldots ,s$. The order of the $p$'s and $q$'s is immaterial and different states are mutually orthogonal unit vectors. For example, $\ket{p_1p_2}=\ket{p_2p_1}$ and $\elbows{p_1p_2\mid p_1q_1q_2}=0$. The
\textit{vacuum state} in which there are no particles present is the unit vector $\ket{0}$. The one-particle states of type $p$ are $\ket{p}$, $p\in\Gamma _m$, the two-particle states of the type $p$ are $\ket{p_1p_2}$, etc.

For $p\in\Gamma _m$, we define the \textit{annihilation operator} $a(p)$ on $K$ by $a(p)\ket{0}=0$ where 0 is the zero vector and
\begin{equation*}
a(p)\ket{pp\ldots pp_1\ldots p_nq_1\ldots q_s}=\sqrt{n}\,\ket{pp\ldots pp_1\ldots p_nq_1\ldots q_s}
\end{equation*}
where $n$ $p$'s appear in the vector on the left side and $n-1$ $p$'s appear in the vector on the right side. We interpret $a(p)$ as the operator that annihilates a particle with energy-momentum $p$. For example, $a(p)\ket{p}=\ket{0}$, $a(p)\ket{ppq}=\sqrt{2}\,\ket{pq}$ and
\begin{equation*}
a(q)\ket{p_1p_2qqq}=\sqrt{3}\,\ket{p_1p_2qq}
\end{equation*}
The adjoint $a(p)^*$ of $a(p)$ is the operator on $K$ defined by
\begin{equation*}
a(p)^*\ket{pp\ldots pp_1\ldots p_nq_1\ldots q_s}=\sqrt{n+1}\,\ket{pp\ldots pp_1\ldots p_nq_1\ldots q_s}
\end{equation*}
where $n$ $p$'s appear in the vector on the left side and $n+1$ $p$'s appear in the vector on the right side. For example $a(p)^*\ket{0}=\ket{p}$ and
$a(p)^*\ket{p}=\sqrt{2}\,\ket{pp}$. We interpret $a(p)^*$ as the operator that creates a particle with energy-momentum $p$. To show that $a(p)^*$ is indeed the adjoint of $a(p)$, suppose that
\begin{equation*}
\ket{\alpha}=\ket{pp\ldots pq_1\ldots q_s}
\end{equation*}
has $n$ $p$'s. Then the inner product of $a(p)^*\ket{\alpha}$ with any other basis vector $\ket{\beta}$ is zero unless $\beta$ has the same form as $\ket{\alpha}$ except now there are $n+1$ $p$'s. In this case we have
\begin{equation*}
\bra{\beta}a(p)^*\ket{\alpha}=\sqrt{n+1}\,\elbows{\beta\mid\beta}=\bra{\alpha}a(p)\ket{\beta}
\end{equation*}
which is the defining relationship for the adjoint.

The operators $a(p)$ and $a(p)^*$ have the characteristic property that their commutator $\sqbrac{a(p),a(p)^*}=I$. We illustrate this for two cases which should convince the reader that this holds. If $\ket{p_1\ldots p_nq_1\ldots q_s}$ contains no $p$'s, then
\begin{align*}
a(p)a(p)^*\ket{p_1\ldots p_nq_1\ldots q_s}&=a(p)\ket{pp_1\ldots p_nq_1\ldots q_s}\\
   &=\ket{p_1\ldots p_nq_1\ldots q_s}
\end{align*}
and $a(p)^*a(p)\ket{p_1\ldots p_nq_1\ldots q_s}=0$. Next, if $\ket{\alpha}$ and $\ket{\beta}$ are the vectors in the previous paragraph, then
\begin{equation*}
a(p)a(p)^*\ket{\alpha}=\sqrt{n+1}\,a(p)\ket{\beta}=(n+1)\ket{\alpha}
\end{equation*}
Since $a(p)^*a(p)\ket{\alpha}=n\ket{\alpha}$ we have that $\sqbrac{a(p),a(p)^*}\ket{\alpha}=\ket{\alpha}$.

One reason that $a(p)$ and $a(p)^*$ are important is that they can be employed to describe physically relevant operators. For example the \textit{free total energy operator} for $p$-particles is defined by
\begin{equation*}
P_0=\sum _{p\in\Gamma _m}p_0a(p)^*a(p)
\end{equation*}
The eigenvectors of $P_0$ have the form $\ket{\alpha}=\ket{p_1p_2\ldots p_n}$, $p_j\in\Gamma _m$, $j=1,\ldots ,n$, with eigenvalues the total energy
\begin{equation*}
E_\alpha =(p_1)_0+\cdots +(p_n)_0
\end{equation*}
In a similar way, we define the \textit{free momentum operators}
\begin{equation*}
P_j=\sum _{p\in\Gamma _m}p_ja(p)^*a(p)
\end{equation*}
for $j=1,2,3$. We also define the \textit{number operator}
\begin{equation*}
N=\sum _{p\in\Gamma _m}a(p)^*a(p)
\end{equation*}
Then $N\ket{\alpha}=n\ket{\alpha}$ where $n$ is the number of $p$-particles in the basis vector $\ket{\alpha}$. Of course, all these operators are self-adjoint.

The most important operators in this work are the \textit{free quantum fields}
\begin{equation}
\label{eq31}
\phi (x)=\sum _{p\in\Gamma _m}\frac{1}{p_0}\sqbrac{a(p)e^{i\pi px/2}+a(p)^*e^{-i\pi px/2}}
\end{equation}
where $x\in{\mathcal S}$ and $px$ is the indefinite inner product
\begin{equation*}
px=p_0x_0-p_1x_1-p_2x_2-p_3x_3
\end{equation*}
The self-adjoint operator $\phi (x)$ is an observable representing a quantum field for a particle of mass $m$ at the spacetime point $x$. In standard quantum field theory, the summation in \eqref{eq31} is replaced by an integral over the mass hyperboloid and it is questionable whether such an integral exists. (It is not even clear whether the underlying Fock space $K$ exists.) However, in the discrete case, the summation in \eqref{eq31} does exist. Although this holds in general, we shall illustrate it for a simple, but important case. Denoting the mass hyperboloid for light photons by $\Gamma '_0$ we have that
\begin{equation*}
\phi (0)\ket{0}=\sum _{p\in\Gamma '_0}\frac{1}{p_0}\,\ket{p}
\end{equation*}
Now $\sigma (0)\ket{0}\in K$ with
\begin{equation*}
\doubleab{\phi (0)\ket{0}}^2=6\sum _{n=1}^\infty\frac{1}{n^2}=\pi ^2
\end{equation*}
Hence, $\doubleab{\phi (0)\ket{0}}=\pi$.

Let $\phi (x)$ be a general free quantum field given by \eqref{eq31}. The next result gives an expression for the commutator $\sqbrac{\phi (x),\phi (y)}$.
\begin{thm}
\label{thm31}
The commutator
\begin{equation*}
\sqbrac{\phi (x),\phi (y)}=2i\sum _{p\in\Gamma _m}\frac{1}{p_0^2}\,\sin\sqbrac{\pi p_0(x_0-y_0)/2}\cos\sqbrac{\bfp\mathrel{\mathlarger\cdot} (\bfx -\bfy )/2}I
\end{equation*}
\end{thm}
\begin{proof}
We have that
\begin{align*}
\phi (x)\phi (y)&=\sum _{p\in\Gamma _m}\frac{1}{p_0}\,\sqbrac{a(p)e^{i\pi px/2}+a(p)^*e^{-i\pi px/2}}\\
  &\quad\mathrel{\mathlarger\cdot}\sum _{q\in\Gamma _m}\frac{1}{q_0}\sqbrac{a(q)e^{i\pi qy/2}+a(q)^*e^{-i\pi qy/2}}\\
  &=\sum _{p,q\in\Gamma _m}\frac{1}{p_0q_0}\left[a(p)a(q)e^{i\pi (px+qy)/2}+a(p)a(q)^*e^{i\pi (px-qy)/2}\right.\\
  &\quad\left. +a(p)^*a(q)e^{i\pi(qy-px)/2}+a(p)^*a(q)^*e^{-i\pi (px+qy)/2}\right]
\end{align*}
Similarly,
\begin{align*}
\phi (y)\phi (x)&=\sum _{p,q\in\Gamma _m}\frac{1}{p_0q_0}\,\left[a(q)a(p)e^{i\pi (px+qy)/2}+a(q)a(p)^*e^{i\pi (qy-px)/2}\right.\\
  &\quad\left. +a(q)a(p)^*e^{i\pi (px-qy)/2}+a(q)^*a(p)^*e^{-i\pi (px+qy)/2}\right]
\end{align*}
Since $\sqbrac{a(p),a(q)}=\sqbrac{a(p)^*,a(q)^*}=0$ and $\sqbrac{a(p),a(q)^*}=\delta _{q,p}I$, we have that
\begin{align*}
\sqbrac{\phi (x),\phi (y)}&=\sum _{p\in\Gamma _m}\frac{1}{p_0^2}\,\brac{\sqbrac{a(p),a(p)^*}e^{i\pi p(x-y)/2}-\sqbrac{a(p),a(p)^*}e^{-i\pi p(x-y)/2}}I\\
  &=2i\sum _{p\in\Gamma _m}\frac{1}{p_0^2}\,\sin\sqbrac{\pi p(x-y)/2}I\\
  &=2i\sum _{p\in\Gamma _m}\frac{1}{p_0^2}\,\sin\sqbrac{\pi p_0(x_0-y_0)/2-\pi\bfp\mathrel{\mathlarger\cdot} (\bfx -\bfy )/2}I\\
  &=2i\sum _{p\in\Gamma _m}\frac{1}{p_0^2}\,\left[\sin\pi p_0(x_0-y_0)/2\cos\pi\bfp\cdot (\bfx -\bfy )/2\right.\\
  &\hskip 6pc\left. -\cos\pi p_0(x_0-y_0)/2\sin\pi\bfp\mathrel{\mathlarger\cdot} (\bfx -\bfy )/2\right]I
\end{align*}
If  $(p_0,\bfp)\in\Gamma _m$, then $(p_0,-\bfp)\in\Gamma _m$ and the result follows because $\sin$ is an odd function.
\end{proof}

The next result is called the \textit{equal-time commutation relation}.
\begin{cor}
\label{cor32}
If $x_0=y_0$, then $\sqbrac{\phi (x),\phi (y)}=0$.
\end{cor}

\textit{Quantum locality} says that if $\doubleab{x-y}_4^2<0$, then $\sqbrac{\phi (x),\phi (y)}=0$. That is, if $x$ and $y$ are space-like separated, then a field measurement at $x$ cannot affect a field measurement at $y$. If $x_0=y_0$ and $x\ne y$ then $\doubleab{x-y}_4^2<0$ and by Corollary~\ref{cor32}, quantum locality holds. However, in contrast to standard quantum field theory \cite{vel94}, quantum locality does not hold, in general. We show this with the following counterexample. Suppose we have light photons and let $x=(1,2\bfe )$, $y=0$ so that $\doubleab{x-y}_4^2=-3<0$. Applying Theorem~\ref{thm31} we have that
\begin{align*}
\sqbrac{\phi (x),\phi (y)}&=2i\sum _{p\in\Gamma '_0}\frac{1}{p_0^2}\,\sin (\pi p_0/2)\cos\pi\bfp\mathrel{\mathlarger\cdot}\bfe I\\
  &=4i\sum _{n=1}^\infty\frac{1}{n^2}\,\sin (\pi n/2)\cos\pi nI=4i\sum _{n=1}^\infty\frac{(-1)^n}{n^2}\,\sin\pi n/2I\\
  &=-4i\sum _{n=1}^\infty\,\frac{1}{(2n-1)^2}\,\sin (2n-1)\pi /2I=4i\sum _{n=1}^\infty\frac{(-1)^n}{(2n-1)^2}\,I\ne 0
\end{align*}

\section{Interacting Quantum Fields}
The important part of quantum fields occurs when we have interactions because then we obtain nontrivial scattering. This section considers interaction Hamiltonians and scattering operators. These are relevant because most of modern theoretical and experimental physics involves some kind of scattering. We shall eventually construct some examples of interaction Hamiltonians, but for now let $H(x_0)$ be self-adjoint operators on $K$ that describe an interaction, where $x_0=0,1,2,\ldots$, represents time. The corresponding \textit{scattering operators} $S(x_0)$ satisfy a ``second quantization'' equation
\begin{equation}
\label{eq41}
\nabla _{x_0}S(x_0)=iH(x_0)S(x_0)
\end{equation}
Of course, \eqref{eq41} is a generalization of Schr\"odinger's equation and for this work $\nabla _{x_0}$ is the difference operator
\begin{equation*}
\nabla _{x_0}S(x_0)=S(x_0+1)-S(x_0)
\end{equation*}
Starting with the initial condition $S(0)=I$, we obtain from \eqref{eq41} that
\begin{equation*}
S(1)=I+iH(0)
\end{equation*}
Continuing, we conclude that
\begin{align}
\label{eq42}
S(2)&=S(1)+iH(1)S(1)=\sqbrac{I+iH(1)}S(1)=\sqbrac{I+iH(1)}\sqbrac{I+iH(0)}\notag\\
S(3)&=S(2)+iH(2)S(2)=\sqbrac{I+iH(2)}S(2)=\sqbrac{I+iH(2)}\sqbrac{I+iH(1)}\sqbrac{I+iH(0)}\notag\\
\vdots&\notag\\
S(n)&=\sqbrac{I+iH(n-1)}\sqbrac{I+iH(n-2)}\cdots\sqbrac{I+iH(1)}\sqbrac{I+iH(0)}
\end{align}

In general, the operators $H(j)$ do not commute so the order in \eqref{eq42} must be retained. This is called a \textit{time ordered product} of operators. Most of the complication in quantum field theory results from trying to solve \eqref{eq41}. In fact, it appears that solving \eqref{eq41} exactly is intractable in general and all we can accomplish is to solve \eqref{eq42} approximately using ``perturbation'' techniques. For $n\ne 0$, $S(n)$ is not unitary, in general. However, presumably the \textit{limiting scattering operator}
$S=\lim\limits _{n\infty}S(n)$ should be unitary in order to preserve probability. Of course, this depends on the interaction Hamiltonian $H(n)$. The author does not know reasonable conditions on $H(n)$ that ensure the unitarity of $S$.

If we multiply \eqref{eq42} out, we obtain the useful form
\begin{align}
\label{eq43}
S(n)&=I+i\sum _{j=0}^{n-1}H(j)+i^2\sum _{j_2<j_1}^{n-1}H(j_1)H(j_2)+i^3\sum _{j_3<j_2<j_1}^{n-1}H(j_1)H(j_2)H(j_3)\notag\\
&\qquad +\cdots +i_nH(n-1)H(n-2)\cdots H(0)
\end{align}
Equation~\eqref{eq43} gives a quantum inclusion-exclusion principle with the interaction Hamiltonian again time ordered. The interaction Hamiltonian $H(x_0)$ is usually constructed from an \textit{interaction Hamiltonian density} consisting of self-adjoint operators on $K$ denoted by ${\mathcal H} (x)$, $x\in{\mathcal S}$. Letting
\begin{equation*}
V(x_0)=\ab{\brac{\bfx\in{\mathbb Z} ^3\colon\doubleab{\bfx}_3\le x_0}}
\end{equation*}
be the cardinality of the set in brackets (called the \textit{space volume} at $x_0$) we define
\begin{equation}
\label{eq44}
H(x_0)=\frac{1}{V(x_0)}\sum\brac{{\mathcal H} (x_0,\bfx )\colon\doubleab{\bfx}_3\le x_0}
\end{equation}
The first few terms of \eqref{eq44} are: $H(0)={\mathcal H} (0)$
\begin{align*}
H(1)&=\tfrac{1}{7}\sqbrac{{\mathcal H} (1,\bfzero )+{\mathcal H} (1,\bfe )+{\mathcal H} (1,-\bfe )+{\mathcal H} (1,\bff )+{\mathcal H} (1,-\bff )+H(1,\bfg )+H(1,-\bfg )}\\
H(2)&=\tfrac{1}{33}\sqbrac{{\mathcal H} (2,\bfzero )+{\mathcal H} (2,\bfe )+\cdots +{\mathcal H} (2,2\bfg )+{\mathcal H} (2,-2\bfg )}
\end{align*}

We now illustrate this theory with a simplified example that contains the essential elements that can be generalized to more complicated and realistic situations. We consider the scattering of two electrons with mass $m$ and initial energy-momentum $p,q\in\widehat{\sscript}$, $p\ne q$. We assume that the electrons interact by exchanging light photons and arrive with final energy-momentum $p',q'\in\widehat{\sscript}$. As usual in this article we are neglecting spin so we are really describing spin-zero particles like neutral pions. We say that the \textit{input state} is $\ket{p,q}\in K$ and the \textit{output state} is $\ket{p'q'}\in K$. Let $\phi (x)$, $\sigma (x)$ be the quantum fields for the electrons and light photons, respectively
\begin{align*}
\phi (x)&=\sum _{p\in\Gamma _m}\frac{1}{p_0}\sqbrac{a(p)e^{i\pi px/2}+a(p)^*e^{-i\pi px/2}}\\
\sigma (x)&=\sum _{k\in\Gamma '_0}\frac{1}{k_0}\sqbrac{a(k)e^{i\pi kx/2}+a(k)^*e^{-i\pi kx/2}}
\end{align*}

The interaction Hamiltonian density ${\mathcal H} (x)$ is frequently constructed by interacting the fields $\phi$ and $\sigma$. A linear combination of $\phi$ and $\sigma$ will not produce any scattering so we take a simple nonlinear combination of the form \cite{vel94}
\begin{equation}
\label{eq45}
{\mathcal H} (x)=g\phi (x)^2\sigma (x)
\end{equation}
where $g$ is called the \textit{coupling constant}. Ideally, we can now find the scattering operator $S$ and compute the \textit{scattering amplitude} $\bra{p'q'}S\ket{pq}$. The probability of the interaction becomes
\begin{equation*}
P\paren{\ket{pq}\to\ket{p'q'}}=\ab{\bra{p'q'}S\ket{pq}}^2
\end{equation*}
Unfortunately, we cannot find $S$ exactly so we must be content with computing the lower level perturbation terms, $\bra{p'q'}S(n)\ket{pq}$, $n=0,1,2,\ldots\,$. (We shall stop at $n=2$. The furthest anyone usually can go is $n=5$.) As we shall see, these terms contain a considerable amount of symmetry which could imply conservation of energy-momentum. However, we have not proved this so we shall postulate the conservation law
\begin{equation}
\label{eq46}
p'+q'=p+q
\end{equation}
As one would expect, \eqref{eq46} gives some simplifications.

Applying \eqref{eq43} with $n=2$ we have that
\begin{align}
\label{eq47}
\bra{p'q'}S(2)\ket{pq}&=\elbows{p'q'\mid pq}+i\bra{p'q'}H(0)\ket{pq}\notag\\
  &\qquad +i\bra{p'q'}H(1)\ket{pq}-\bra{p'q'}H(1)H(0)\ket{pq}
\end{align}

As is usually done, we shall assume that $\ket{p'q'}\ne\ket{pq}$ \cite{ps95,vel94}. We have three reasons for this assumption. One is that $\ket{pq}\to\ket{pq}$ is not of interest because there is no scattering in this case. Another is that this case is unlikely so it would have small probability. Finally, as we shall later show, this case is very hard to compute. Because of this assumption, the first term in \eqref{eq47} vanishes. The second and third terms contain one $H(n)$ and hence only one $\sigma$-field. This applied to a state
$\ket{pq}$ that does not have a $\sigma$-particle (light photon) gives 0 for the $a(k)$ part or a state containing a $\sigma$-particle for the $a(k)^*$ part. The inner product of such a state with the state $\ket{p'q'}$ containing no $\sigma$-particle is 0. Similarly, any product of an odd number of $H(n)$ gives zero between states without $\sigma$-particles.

We conclude that \eqref{eq47} reduces to
\begin{equation}
\label{eq48}
\bra{p'q'}S(2)\ket{pq}=-\bra{p'q'}H(1)H(0)\ket{pq}
\end{equation}
If we continue to the next order of perturbation, we obtain
 \begin{align}
\label{eq49}
\bra{p'q'}S(3)\ket{pq}&=-\bra{p'q'}H(1)H(0)\ket{pq}-\bra{p'q'}H(2)H(0)\ket{p}\notag\\
 &\qquad -\bra{p'q'}H(2)H(1)\ket{pq}
\end{align}
As shown earlier, the next term in \eqref{eq43} does not appear in \eqref{eq49}. With some more work we could compute \eqref{eq49} instead of \eqref{eq48}, but to save space we shall only consider \eqref{eq48}. The operator in \eqref{eq48} has the form
\begin{align}
\label{eq410}
H(1)H(0)&=\frac{g^2}{7}\sum _{x,y}\phi (x)^2\sigma (x)\phi (y)^2\sigma (y)\notag\\
  &=\frac{g^2}{7}\sum _{x,y}\phi (x)^2\phi (y)^2\sigma (x)\sigma (y)
\end{align}
In order to get a nonzero inner product in \eqref{eq48}, we must have that $H(1)H(0)\ket{pq}=\alpha\ket{pq}$ for some $\alpha\in{\mathbb C}$, $\alpha\ne 0$, which we write
$\ket{pq}\to\ket{p'q'}$. As far as $\sigma (x)\sigma (y)$ is concerned, we only have the possibility $a(k)a(k)^*$, $k\in\Gamma '_0$ which corresponds to an exchange of photons. For $\phi (x)^2\phi (y)^2$ we have six possibilities. These can be described by six Feynman diagrams, but we shall employ a symbolic notation.

We could first annihilate $p$ and $q$ using $\phi (y)^2$ and then create $p'$ and $q'$ using $\phi (x)^2$. Another possibility is to first annihilate $p$ and create $p'$ using
$\phi (y)^2$ and then annihilate $q$ and create $q'$ using $\phi (x)^2$. The six cases can be described symbolically as follows, with the first two cases given above.
{\obeylines
(Case 1)\enspace$\ket{pq}\to\ket{0}\to\ket{p'q'}$
(Case 2)\enspace$\ket{pq}\to\ket{p'q}\to\ket{p'q'}$
(Case 3)\enspace$\ket{pq}\to\ket{q'q}\to\ket{p'q'}$
(Case 4)\enspace$\ket{pq}\to\ket{pp'}\to\ket{p'q'}$
(Case 5)\enspace$\ket{pq}\to\ket{pq'}\to\ket{p'q'}$
(Case 6)\enspace$\ket{pq}\to\ket{pqp'q'}\to\ket{p'q'}$
}\bigskip

Note that although we have conservation of energy-momentum \eqref{eq46} for the input and output states of $\ket{pq}\to\ket{p'q'}$, we do not conserve energy-momentum at intermediate times in the interaction. For example, in Case~1 we do not have $p+q=0$. During the interaction, energy-momentum is annihilated and created by the quantum fields.

We now explain specifically why we assume that $\ket{p'q'}\ne\ket{pq}$. In the situation in which the input and output states are both $\ket{pq}$ we have cases like
\begin{align*}
\ket{pq}&\to\ket{p'q}\to\ket{pq}\\
\ket{pq}&\to\ket{pq'}\to\ket{pq}\\
\ket{pq}&\to\ket{pqp'q'}\to\ket{pq}
\end{align*}
But now $p',q'$ are arbitrary elements of $\Gamma _m$ so we would have to sum over these infinite number of elements which is quite difficult.

The next lemma shows that for a fixed input state $\ket{pq}$, there are only a finite number of output states $\ket{p'q'}$ satisfying $\ket{pq}\to\ket{p'q'}$. In this way we can sum probabilities to get various alternatives.

\begin{lem}
\label{lem41}
There exist only finitely many $\ket{p'q'}$ such that $\ket{pq}\to\ket{p'q'}$.
\end{lem}
\begin{proof}
If $\ket{pq}\to\ket{p'q'}$, then by \eqref{eq46} $p'+q'=p+q$. Hence,
\begin{equation*}
p'_0,q'_0\le p'_0+q'_0=p_0+q_0
\end{equation*}
We conclude that there are only a finite number of possible values for $p'_0,q'_0$. Since for $p'\in\Gamma _m$ we have that $\doubleab{\bfp '}_3^2=(p'_0)^2-m^2$, there are only a finite number of possible $\bfp '$ and hence only a finite number of possible $p'$.
\end{proof}

Notice that Lemma~\ref{lem41} is general and does not depend on a particular mass $m$. If $\ket{p'q'}\ne\ket{pq}$, then we have seen that there are only six possible cases for each of the finite number of scattering situations. We thus have only a finite number of terms to compute.

\section{Scattering Examples}
For our first example, we consider the simplified electron scattering process discussed in Section~4. The interaction Hamiltonian density is given by \eqref{eq45} and we want to compute the first order perturbation scattering amplitude \eqref{eq48}. Writing \eqref{eq410} in detail, we have that
\begin{align*}
H(1)H(0)&=\frac{1}{7}\,\sqbrac{{\mathcal H} (1,\bfzero )+{\mathcal H} (1,\bfe )+\cdots +{\mathcal H} (1,-\bfg )}{\mathcal H} (0)\\
  &=\frac{g^2}{7}\left[\phi (1,\bfzero )^2\sigma (1,\bfzero )+\phi (1,\bfe )^2\sigma (1,\bfe )\right.\\
  &\hskip 3pc\left. +\cdots +\phi (1,-\bfg )^2\sigma (1,-\bfg )\right]\phi (0)^2\sigma (0)\\
  &=\frac{g^2}{7}\left[\phi (1,\bfzero )^2\phi (0)^2\sigma (1,\bfzero )\sigma (0)+\phi (1,\bfe )^2\phi (0)^2\sigma (1,\bfe )\sigma (0)\right.\\
  &\hskip 3pc\left. +\cdots +\phi (1,-\bfg )^2\phi (0)^2\sigma (1,-\bfg)\sigma (0)\right]
\end{align*}
Taking the photon field $\sigma$ first, we have that
\begin{align*}
\sigma (1,\bfzero )\sigma (0)&=\sum _{k\in\Gamma '_0}\frac{1}{k_0}\,\sqbrac{a(k)e^{i\pi k_0/2}+a(k)^*e^{-i\pi k_0/2}}\\
  &\hskip 3pc\mathrel{\mathlarger\cdot}\sum _{k'\in\Gamma '_0}\frac{1}{k'_0}\sqbrac{a(k')+a(k')^*}
\end{align*}
By our argument in Section~4, $\sigma (1,\bfzero )\sigma (0)$ contributes the coefficient
\begin{equation*}
c_1=6\sum _{n=1}^\infty\frac{1}{n^2}\,e^{i\pi n/2}
\end{equation*}
The next term is
\begin{align*}
\sigma (1,\bfe )\sigma (0)&=\sum _{k\in\Gamma '_0}\frac{1}{k_0}\,\sqbrac{a(k)e^{i\pi (k_0-k_1)/2}+a(k)^*e^{-i\pi (k_0-k_1)/2}}\\
  &\hskip3pc\mathrel{\mathlarger\cdot}\sum _{k'\in\Gamma '_0}\frac{1}{k'_0}\,\sqbrac{a(k')+a(k')^*}
\end{align*}
Since $k_1=\pm k_0$, this term contributes the coefficient
\begin{align*}
c_2&=\sum _{k\in\Gamma '_0}\frac{1}{k_0^2}\,e^{i\pi (k_0-k_1)/2}=\sum _{n=1}^\infty\frac{1}{n_2}\,(1+e^{i\pi n}+4e^{i\pi n/2})\\
  &=\frac{\pi ^2}{12}\,+4\sum _{n=1}^\infty\frac{1}{n^2}\,e^{i\pi n/2}
\end{align*}
Since the other terms contribute this same coefficient, the nonzero part of $H(1)H(0)$ in \eqref{eq48} becomes
\begin{align}
\label{eq51}
H(1,0)&=\frac{g^2}{7}\left[c_1\phi (1,\bfzero )^2\phi (0)^2+c_2\phi (1,\bfe )^2\phi (0)^2\right.\notag\\
&\hskip 3pc\left. +\cdots +c_2\phi (1,-\bfg )^2\phi (0)^2\right]
\end{align}

We now consider the terms of $H(1,0)$ in \eqref{eq51}. We have that
\begin{align}
\label{eq52}
\phi (1,\bfzero )^2\phi (0)&=\brac{\sum _{p\in\Gamma _m}\frac{1}{p_0}\,\sqbrac{a(p)e^{i\pi p_0/2}+a(p)^*e^{-i\pi p_0/2}}}^2\notag\\
  &\hskip 3pc\mathrel{\mathlarger\cdot}\brac{\sum _{p\in\Gamma _m}\frac{1}{p_0}\,\sqbrac{a(p)+a(p)^*}}
\end{align}
For the six cases, this term contributes the following coefficients where\newline $\beta = 1/p_0q_0p'_0q'_0$.\bigskip

{\obeylines
(Case 1)\enspace$\beta e^{-i\pi (p'_0+q'_0)/2}$
(Case 2)\enspace$\beta e^{i\pi (q_0-q'_0)/2}$
(Case 3)\enspace$\beta e^{i\pi (q_0-p'_0)/2}$
(Case 4)\enspace$\beta e^{i\pi (p_0-q'_0)/2}$
(Case 5)\enspace$\beta e^{i\pi (p_0-p'_0)/2}$
(Case 6)\enspace$\beta e^{i\pi (p_0+q_0)/2}$
}\bigskip

\noindent Adding these six coefficients and applying conservation of energy-momentum gives the following contribution of \eqref{eq52}
\begin{equation*}
d_1=2\beta\sqbrac{\cos\frac{\pi}{2}\,(p_0+q_0)+\cos\frac{\pi}{2}\,(q_0-q'_0)+\cos\frac{\pi}{2}\,(q_0-p'_0)}
\end{equation*}

The next term in \eqref{eq51} is
\begin{align*}
\phi (1,\bfe )^2\phi (0)^2&=\brac{\sum _{p\in\Gamma _m}\frac{1}{p_0}\,\sqbrac{a(p)e^{i\pi (p_0-p_1)/2}+a(p)^*e^{-i\pi (p_0-p_1)/2}}}^2\notag\\
  &\hskip 3pc\mathrel{\mathlarger\cdot}\brac{\sum _{p\in\Gamma _m}\frac{1}{p_0}\,\sqbrac{a(p)+a(p)^*}}^2
\end{align*}
Again, we have the six cases:\bigskip

{\obeylines
(Case 1)\enspace$\beta e^{-i\pi (p'_0+q'_0-p'_1-q'_1)/2}$
(Case 2)\enspace$\beta e^{i\pi (q_0-q'_0-q_1+q'_1)/2}$
(Case 3)\enspace$\beta e^{i\pi (q_0-p'_0-q_1+p'_1)/2}$
(Case 4)\enspace$\beta e^{i\pi (p_0-q'_0-p_1+q'_1)/2}$
(Case 5)\enspace$\beta e^{i\pi (p_0-p'_0-p_1+p'_1)/2}$
(Case 6)\enspace$\beta e^{i\pi (p_0+q_0-p_1-q_1)/2}$
}\bigskip

\noindent The sum of these coefficients gives:
\begin{align*}
d_2&=2\beta\left[\cos\frac{\pi}{2}\,(p_0+q_0-p_1-q_1)+\cos\frac{\pi}{2}\,(q_0-q'_0-q_1+q'_1)\right.\\
  &\hskip 3pc\left. +\cos\frac{\pi}{2}\,(q_0-p'_0-q_1+p'_1)\right]
\end{align*}

The next term in \eqref{eq51} has the form $\phi (1,-\bfe )^2\phi (0)^2$ and the sum of the corresponding coefficients $d_3$ will be the same as $d_2$ except the $p_1,q_1,p'_1,q'_1$ terms will be the negatives of those in $d_3$. We then have
\begin{align*}
d_2+d_3&=4\beta\left[\cos\frac{\pi}{2}\,(p_0+q_0)\cos\frac{\pi}{2}\,(p_1+q_1)\right.\\
  &\left. +\cos\frac{\pi}{2}\,(q_0-q'_0)\cos\frac{\pi}{2}\,(-q_1+q'_1)+\cos\frac{\pi}{2}\,(q_0-p'_0)\cos\frac{\pi}{2}\,(-q_1+p'_1)\right]\end{align*}
The other terms will be similar, so adding all these terms gives
\begin{align*}
&\bra{p'q'}S(2)\ket{pq}=\frac{-g^2}{7}\,\sqbrac{c_1d_1+c_2(d_2+d_3+d_4+d_5+d_6)}\\
&=\frac{-2g^2}{7}\,\beta\left\{\cos\frac{\pi}{2}\,(p_0+q_0)\sqbrac{c_1+2c_2\paren{\cos\frac{\pi}{2}\,(p_1+q_1)+\cos\frac{\pi}{2}\,(p_2+q_2)+\cos\frac{\pi}{2}\,(p_3+q_3)}}\right.\\
\noalign{\medskip}
&\quad +\cos\frac{\pi}{2}\,(q_0-q'_0)\sqbrac{c_1+2c_2\paren{\cos\frac{\pi}{2}\,(q'_1-q_1)+\cos\frac{\pi}{2}\,(q'_2-q_2)+\cos\frac{\pi}{2}\,(q'_3-q_3)}}\\
\noalign{\medskip}
&\left.\quad +\cos\frac{\pi}{2}\,(q_0-p'_0)\sqbrac{c_1+2c_2\paren{\cos\frac{\pi}{2}\,(p'_1-q_1)+\cos\frac{\pi}{2}\,(p'_2-q_2)+\cos\frac{\pi}{2}\,(p'_3-q_3)}}\right\}\\
\end{align*}

We now consider a specific example. Let $p=(2,1,1,1)$, $q=(3,-2,-2,0)$, $p'=(2,-1,1,-1)$ and $q'=(3,0,-2,2)$. In this case, the mass $m=1$ and we have conservation of energy-momentum because
\begin{align*}
p+q&=p'+q'=(5,-1,-1,1)\\
\intertext{Moreover,}
\beta&=\frac{1}{p_0q_0p'_0q'_0}=\frac{1}{36}
\end{align*}
Applying our previous formula, the amplitude becomes
\begin{equation*}
\bra{p'q'}S(2)\ket{pq}=\frac{-g^2}{126}\,\sqbrac{c_1+2c_2(-1+1-1)}=\frac{g^2}{126}\,(2c_2-c_1)
\end{equation*}
To the first order perturbation we have that
\begin{equation*}
P\paren{\ket{pq}\to\ket{p'q'}}=\ab{\bra{p'q'}S(2)\ket{pq}}^2=\frac{g^4}{(126)^2}\,(2c_2-c_1)^2
\end{equation*}
Since $g$ is unknown, this does not tell us anything. However, we can use this calculation to compare probabilities. Let $p,q$ be as before and let $p''=(2,1,-1,-1)$,
$q''=(3,-2,0,2)$. We again obtain
\begin{equation*}
\bra{p''q''}S(2)\ket{pq}=\frac{g^2}{126}\,(2c_2-c_1)
\end{equation*}
which is not surprising due to the symmetry of the situation. It does however, exhibit some consistency. To illustrate a more interesting comparison, suppose the initial state is
$p=(3,2,2,0)$, $q=(3,-2,-2,0)$ and the final state is $p'=(3,0,2,2)$, $q'=(3,0,-2,-2)$. We then obtain
\begin{equation*}
\bra{p'q'}S(2)\ket{pq}=\frac{g^2}{126}\,(8c_2-c_1)
\end{equation*}
which is quite different from what we obtained before.

Our second example is a simplified version of electron-proton scattering due to the exchange of photons. As before, we have the electron and photon fields $\phi (x)$,
$\sigma (x)$, but now we include a proton field of mass $M$
\begin{equation*}
\psi (x)=\sum _{p\in\Gamma _M}\sqbrac{a(p)e^{i\pi px/2}+a(p)^*e^{-i\pi px/2}}
\end{equation*}
Let the interaction Hamiltonian density be ${\mathcal H} =g\phi ^2\sigma -g\psi ^2\sigma$ where the negative sign is because the electron and proton have opposite charge. Let
$p\in\Gamma _m$ be an initial electron, $q\in\Gamma _M$ be as initial proton and we consider the scattering process $\ket{pq}\to\ket{p'q'}$ where
$\ket{p'q'}\ne\ket{pq}$ is the final state. As before, we study the first perturbation term \eqref{eq48} where $H(1)H(0)$ now has the form
\begin{align*}
H(1)H(0)&=\frac{g^2}{7}\,\sum _{x,y}\sqbrac{\phi (x)^2\sigma (x)-\psi (x)^2\sigma (x)}\sqbrac{\phi (y)^2\sigma (y)-\psi (y)^2\sigma (y)}\\
  &=\frac{g^2}{7}\,\sum _{x,y}\sqbrac{\phi (x)^2\phi (y)^2-\phi (x)^2\psi (y)^2-\psi (x)^2\phi (y)^2+\psi (x)^2\psi (y)^2}\sigma (x)\sigma (x)
\end{align*}
To map $\ket{pq}$ to $\ket{p'q'}$, we again have the term $a(k)a(k)^*$ for $k\in\Gamma '_0$ and we obtain the same two constants $c_1$ and $c_2$. For the electron-proton terms we have only two cases:\bigskip

{\obeylines
(Case 1')\enspace$\ket{pq}\to\ket{pq'}\to\ket{p'q'}$
(Case 2')\enspace$\ket{pq}\to\ket{p'q}\to\ket{p'q'}$
}\bigskip

As in \eqref{eq51}, the nonzero part of $H(1)H(0)$ is
\begin{align*}
H(1,0)&=\frac{-g^2}{7}\,\left[c_1\phi (1,\bfzero )^2\psi (0)^2+c_1\psi (1,\bfzero )^2\phi (0)^2+c_2\phi (1,\bfe )^2\psi (0)^2\right.\\
  &\quad\left. +c_2\psi (1,\bfe )^2\phi (0)^2+\cdots +c_2\phi (1,-\bfg )^2\psi (0)^2+c_2\psi (1,-\bfg )^2\phi (0)^2\right]
\end{align*}
The first term on the right side gives
\begin{align*}
\phi (1,\bfzero )^2\psi (0)^2&=\brac{\sum _{p\in\Gamma _m}\frac{1}{p_0}\,\sqbrac{a(p)e^{i\pi p_0/2}+a(p)^*e^{-i\pi p_0/2}}}^2\\
  &\qquad\mathrel{\mathlarger\cdot}\brac{\sum _{p\in\Gamma _M}\frac{1}{q_0}\,\sqbrac{a(q)+a(q)^*}}^2
\end{align*}
This applies to Case~1' and contributes the coefficient
\begin{equation*}
\beta e^{i\pi (p_0-p'_0)/2}
\end{equation*}
In a similar way, the second term on the right side applies to Case~2' and contributes the coefficient
\begin{equation*}
\beta e^{i\pi (q_0-q'_0)/2}
\end{equation*}
The third term contributes
\begin{equation*}
\beta e^{i\pi (p_0-p'_0)/2}e^{i\pi (p'_1-p_1)/2}
\end{equation*}
while the fourth term contributes
\begin{equation*}
\beta e^{i\pi(q_0-q'_0)/2}e^{i\pi (q'_1-q_1)/2}
\end{equation*}
The other terms are similar. Adding all these terms gives
\begin{align*}
\bra{p'q'}S(2)\ket{pq}&=\frac{-g^2\beta}{7}\,\left\{e^{i\pi (p_0-p'_0)/2}\sqbrac{c_1+2c_2\sum _{j=1}^3\cos\frac{\pi}{2}\,(p_j-p'_j)}\right.\\
&\qquad\left. +e^{i\pi (q_0-q'_0)/2}\sqbrac{c_1+2c_2\sum _{j=1}^3\cos\frac{\pi}{2}(q_j-q'_j)}\right\}
\end{align*}
Applying conservation of energy-momentum, this becomes
\begin{equation}
\label{eq53}
\bra{p'q'}S(2)\ket{pq}=\frac{-2g^2\beta}{7}\,\brac{\cos\frac{\pi}{2}\,(p_0-p'_0)\sqbrac{c_1+2c_2\sum _{j=1}^3\cos\frac{\pi}{2}\,(p_j-p'_j)}}
\end{equation}\smallskip

\noindent Notice that \eqref{eq53} is independent of $q$ and $q'$. This may not be true for higher order perturbations such as $S(3)$.

By Lemma~\ref{lem41}, we know that there are only a finite number of possible final states $\ket{p^jq^j}$ for the initial state $\ket{pq}$. As usual, we assume that
$\mathop{Prob}\paren{\ket{pq}\to\ket{pq}}$ is small. The \textit{flux} is the number of particles, per unit surface area, per unit time and this is given by the geometric speed
\begin{equation*}
\frac{\doubleab{\bfp ^j}_3}{p_0^j}+\frac{\doubleab{\mathbf{q} ^j}_3}{q_0^j}
\end{equation*}
The \textit{cross section} $\sigma$ is defined to be the probability, per unit time, for unit flux summed over the final states. Since the total time, in our case, is 2, we have that
\begin{align}
\label{eq54}
\sigma&=\frac{1}{2}\,\sum _j\sqbrac{\frac{\doubleab{\bfp ^j}_3}{p_0^j}+\frac{\doubleab{\mathbf{q} ^j}_3}{q_0^j}}\mathop{Prob}\paren{\ket{pq}\to\ket{p^jq^j}}\notag\\
&=\frac{1}{2}\,\sum _j\sqbrac{\frac{\doubleab{\bfp ^j}_3}{p_0^j}+\frac{\doubleab{\mathbf{q} ^j}_3}{q_0^j}}\ab{\bra{p^jq^j}S(2)\ket{pq}}^2
\end{align}

We now restrict our attention to cases of the form $p=(p_0,0,0,p_3)$, $q=(q_0,0,0,0)$. Thus, electrons are moving along the $z$-axis and the proton is initially stationary. Furthermore, we make the approximation that $m=0$ so that $p_3=p_0$ and the electron acts like a photon. This a good approximation because the electron is much less massive than the proton (about 1 to 2000). To be specific, consider the simple example $p=(5,0,0,5)$, $q=(5,0,0,0)$. It is easy to check that conservation of energy-momentum implies that the only possible outgoing state $\ket{p'q'}\ne\ket{pq}$ is given by $p'=(3,2,2,1)$, $q'=(7,-2,-2,4)$. Substituting these values into \eqref{eq53} gives
\begin{equation*}
\bra{p'q'}S(2)\ket{pq}=\frac{2g^2\beta}{7}\,(c_1+2c_2)
\end{equation*}
Applying \eqref{eq54}, the cross section becomes
\begin{equation*}
\sigma =\frac{2}{49}\,g^4\beta ^2(c_1+2c_2)^2
\end{equation*}

Finally, we briefly consider particle decay rates and lifetimes. The decay probability per unit time is the \textit{decay rate} and the inverse of the decay rate is the
\textit{lifetime}. Suppose we have $\phi$-particles and $\sigma$-particles, but now we assume that the $\sigma$-particle is more massive than two $\phi$-particles so the
$\sigma$ decays into two $\phi$'s. We again consider an interaction Hamiltonian density ${\mathcal H} =g\phi ^2\sigma$. Let $k$ the initial energy-momentum of the $\sigma$ and
$p,q$ the final energy-momentum of the two $\phi$-particles. At the lower order perturbation, we have only one case for the decay, namely $\ket{k}\to\ket{pq}$. The last term of the scattering operator
\begin{equation*}
S(2)=I+i\sqbrac{H(0)+H(1)}-H(1)H(0)
\end{equation*}
gives $\bra{pq}H(1)H(0)\ket{k}=0$ and we have
\begin{align}
\label{eq55}
\bra{pq}S(2)\ket{k}&=i\sqbrac{\bra{pq}H(0)\ket{k}+\bra{pq}H(1)\ket{k}}\notag\\
  &=ig\biggl\{\bra{pq}\phi (0)^2\sigma (0)\ket{k}+\frac{1}{7}\left[\bra{pq}\phi (1,\bfzero )^2\sigma (1,\bfzero )\ket{k}\right.\biggr.\notag\\
  &\quad\biggl.\left. +\bra{pq}\phi (1,\bfe )^2\sigma (1,\bfe )\ket{k}+\cdots +\bra{pq}\phi (1,-\bfg )^2\sigma (1,-\bfg )\ket{k}\right]\biggr\}\notag\\
  &=\frac{ig}{k_0p_0q_0}\,\biggl\{1+\frac{1}{7}\,e^{-i\pi (p_0+q_0)/2}\left[ e^{i\pi k_0/2}+e^{i\pi (k_0-k_1)/2}e^{i\pi (p_1+q_1)/2}\right.\biggr.\notag\\
  &\hskip 4pc\biggl.\left. +\cdots +e^{i\pi (k_0-k_3)/2}e^{i\pi (p_3+q_3)/2}\right]\biggr\}
\end{align}

Now suppose we are in the $\sigma$ rest system so that $k=k_0$ and $\mathbf{k} =0$. By conservation of energy-momentum, $p_0+q_0=k_0$ and $\bfp +\mathbf{q} =\bfzero$. Then
$\bfp =-\mathbf{q}$ so $q_0=p_0$ and $k_0=2p_0$. Hence, \eqref{eq55} simplifies to
\begin{equation}
\label{eq56}
\bra{pq}S(2)\ket{k}=\frac{8ig}{k_0^3}
\end{equation}
We can now use \eqref{eq56} for the first approximation to decay rates and lifetimes.

redpajama_set_name: RedPajamaArXiv


text:
Jesus Christ Superstar Zone
The largest online community for Jesus Christ Superstar
Chilina Kennedy
All rights belong to their respective authors. If you wish to use this material, please contact us.
Interview conducted by Andrew J. Simpson.
Q. When did you first hear JCS? What did you think?
A. I was young. I was listening to the [Brown Album] in my parents' basement. I loved it and there were auditions for Jesus Christ Superstar at Theatre Aquarius. I was just out of school, so 20 or 21. I didn't know the show intimately so I did it at Aquarius and then this production [at the Stratford Festival]. I've always been a huge fan of the show. I think it's a great show. It's a great story. It's great to tell it in that way. It's a gorgeous and smart score. I love it.
Q. On that note, what's your favorite song from the score?
A. Oh that's hard! I was thinking of that the other day. Every time a song in the show comes up you're like "Oh I forgot this one…" or "That's my favorite one!" I don't know. I'd have to say that it changes from day to day. Of my tunes, I'd have to say "Everything's Alright." I love "Gethsemane (I Only Want to Say)." It's tough to pick one.
Q. Your thoughts on this — 40 years on, why is JCS still performed and relevant?
A. This is a bit of a guess, but… superstardom has become more and more of a craze in our culture. The message is hitting me harder than it has before when Judas sings "Superstar" at the end, you know, it's not a fun rock and roll song. He asks some very pertinent questions. Sometimes you take somebody's teachings, and then suddenly, you turn that person into a megastar. It becomes a focus, the focus becomes less about the teachings and more about the individual who's getting this fame and height, I guess like any spiritual leader. So, it gives us a chance to examine that a little bit and to ask some of those questions again, especially in the age of the Paris Hiltons. The superstars of the world are fascinating. You tend to ask questions like "Why?" and why do we tend to have these people with that status? I think every culture needs heroes, and we all need leaders. It's interesting how far we can go. Look at the [Royal] wedding. It was a phenomenon, and you're like, "Wow!" It's crazy, our need and our desire to get obsessive about people. What is it about? What is our need to do this? I think it's a very interesting question. I also think the story itself is really powerful. Christianity is a huge religion. It affects so many people differently. So, it's interesting to tell this story in this particular way. People have very passionate opinions on it — negative and positive. It's always a source of debate, and it's a great show.
Q. Have you ever interacted with any of the original cast, be it in a production of the show or the course of your career?
A. I was actually on tour with Mamma Mia! in Boston and the JCS tour with Carl Anderson was also playing there. So, I got to hang out with Carl after the show and talk to him. That was a thrill. I grew up listening to him on the soundtrack record, and here I am on tour with a show and I'm hanging out with him in the same hotel! What a huge loss when he passed away.
Q. So, Mary Magdalene, huh? What's it like playing the only major female role in a very male-oriented show?
A. (Laughs) It is interesting, right? The great thing is Des [McAnuff] has a real original interpretation of the characters in the show, especially keeping the characters of Mary, Judas, and Jesus as a very close-knit "triad" — you know, a trio, three friends who have been very, very close, and who are starting to fall apart. So, Mary is in a lot of scenes that maybe in other productions she isn't seen in traditionally, like "The Last Supper" for example… I don't want to give too much away by telling you this. But I think that when you get a large pack of men and one woman, you get certain dynamics going on. I was on tour with a show one time and we were in Toronto and I was the only female, I think — other than the director — and maybe ten men. We were on the road, and it was bizarre how I was treated among those males, and how they jockeyed for position. It's interesting because that's what happens in the animal kingdom, and I think it happens with people whether we recognize it or not. So, we see the same thing happening in the show, there are different sorts of tensions and dominance between Mary and the other Apostles. It's very interesting. Then Jesus is there trying to teach all of his lessons about everybody being equal and treating everybody the same, that sort of thing. So, there's a lot of good dramatic tension there.
I think "I Don't Know How to Love Him" is so interesting because there are so many ways you can interpret it. But Mary has never really fallen in love before, and this is the first time, and it's with Jesus. That's a huge journey for somebody. I think Mary has a lot of intuition. She seems to understand Christ's teachings in very deep ways and perhaps in ways the other Apostles don't. Jesus says that at one point in the show: "She alone has tried to give me / What I need right here and now." There is something about Mary that is very intuitive and smart, and I think she's a really good listener. I think she knows that something is going to happen to Christ — he is going to be taken away. He says that in the play, "The end…" and "…when I'm gone…" So, how do you love a man that is going to be leaving you? That alone, in and of itself, is difficult, and then you add to it her past and loving so many men before him, you know, what is love and all those questions. So, I think she goes through a real struggle.
Q. How do you view Mary Magdalene personally, and how will that factor into your portrayal?
A. Well, I have to say that I was raised with a very feminist mother in a very open-minded home, so you know, everything was to be questioned, and be interpreted, and it's best to go and read a lot about something before you reach a final decision with that kind of attitude in mind. For me, Mary Magdalene can represent so many different things. If people believe she was a prostitute, then what a beautiful message that is, that she asked Jesus' forgiveness and he forgave her, and brought her on to be someone that, I think, he had great esteem for. What a beautiful lesson for all women who have maybe gone astray in life — that you can be forgiven and choose a new path. For me, that sends an excellent message. For those people who don't believe she was a prostitute, she's a great image for women either way. I think she's a very strong female figure from whom people can take a lot of hope. I think she's also a beautiful symbol of service. If you take a look at her actions in the play, I'm not talking about the Bible, if you look at her actions in the show, she serves. She serves food; she is a witness to what Christ is going through in his journey. So, I think she is a very humble, strong female follower. I think that a lot of people don't associate "strength" with "following," but there is great strength in being humble enough to set aside ego and all of the rest to simply follow along the path, and to do what is required of you on that journey.
Q. You mentioned doing a lot of reading before deciding something. What research did you do into the role?
A. A couple of books. The Bible, and there's another book on Mary Magdalene by a woman named Lynne Picknett. She's sort of an expert on Mary Magdalene and John the Baptist; I read a lot of her stuff. I took it with a grain of salt like you have to do with everything, but she had a lot of interesting ideas about who she was, because so much of the stuff you read about Mary is vague, and no one is ever really sure if she's the Mary that they list as the Mary of Bethany. There are lots of different theories of who she was and what she was all about. I tried to make the best choices for telling the story, for the story that we're telling — and the version that we're telling in our production. It's not as clear as if you were playing, say, Judas or Jesus. Obviously, Des has lots of ideas of how he wants her to be played, so I just took the best of all the worlds and tied together something that made sense to me.
Q. Are there any previous Mary performances (such as Yvonne Elliman, etc.) that will influence your version?
A. Oh, definitely. I think Yvonne has influenced me. We're trying to stay as close to the flavor of the original album as possible — especially the concept album. I have a record player in my dining room so I've been rocking it out with the old Brown Album. I love that style. It was written in the '70s with all that was happening at the time in rock and roll and all of those influences. I think that it's important that our production creates something new, but keeps the styles that Lloyd Webber originally wrote it in. So, I'm trying to do my best version of that, and what that was, with what Yvonne was doing.
Q. You've played roles in other Rice-Webber pieces, for example, Eva Peron in Evita, and also similar female leads like Maria in West Side Story. How do these roles compare to that of Mary Magdalene, and is there anything you will be drawing on from these past experiences in your performance?
A. Oh, interesting. I guess the subject of faith has come across for me in the last couple of years, starting with Maria. I get asked this question a lot with my relationship with religion and God because Maria was so devoted to her faith, I had to explore a lot of that in that first year here. And again, with Evita, Eva was deeply spiritual. I don't think she always agreed with the church, but I think she had a deep devotion to God. Looking at that, as well as looking at Mary Magdalene in her journey with faith… she has absolute faith in Jesus' teachings. She's confused sometimes about whether or not she sees him as God until the end, but she believes in him, and she is living proof of his work. She is sort of a channel for that; she is made better because of his teachings. I think Maria was sort of the beginning of me digging deep, and about the other, and figuring out what all that is, what it means, and what it feels like inside different characters. So, it's an interesting journey.
Q. What do you most want the audience to "get" out of your portrayal of the role?
A. Oh boy, that's a good question. It's interesting. I'm still trying to find that out, 'cause I never thought of it that way. I'm trying to play the role as truthfully as possible, and I'm still trying to discover what that is. I think that, if they take away a sense of somebody who has maybe gone astray in life, has gotten back on track, and is stronger for it…. she's a strong woman despite what her past was. I think that's the most important thing, especially for young women. We've all made mistakes in life, and I think the point of Jesus' teachings is that it's about forgiveness and moving forward, not looking back and feeling bad about what has happened in the past, and there's always hope for people who ask for forgiveness. I think it's a beautiful message. I think if people can take that away from it, then that's great.
Q. Now we get to the production itself. What's it like working with such talented people, including Des McAnuff?
A. Oh, it's the best. It's incredible. Des is the best. Some of his ideas for this show are just brilliant. And I love the way he has kept the flavor of the original but brought it into the present. There's a lot of hip-hop and those sorts of influences in the show. I think that's going to connect the story with young people, I think that's important for the story and the Festival. Working with Paul [Nolan] is incredible. He's an amazing, amazing Jesus. I think he's the best I've seen, and the work he's doing is fantastic. He's a great actor. I love working with Josh [Young], Brent Carver, and Bruce Dow. We have this huge company, and I'm just trying to get better, and you can't help but get better with people around you all the time.
Q. How big is the company?
A. Actually, I don't know. I should check that out online. (Laughs) Let's see… there must be about 25 people in the cast.
Q. When does the show open officially?
A. June 3rd at the Avon Theatre, and we run until October 29th. (The show later extended to run an additional week until November 6; at the time of the interview, neither the California nor Broadway transfer had been discussed.) We're just wrapping up tech rehearsals, and today we're doing a full dress rehearsal, so it's pretty exciting.
Q. Thank you for talking to us today, Ms. Kennedy.
A. Thank you, Andy; it was my pleasure. We'll see you in a couple of weeks!
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Trailer for the upcoming project.
If your interested for more details or want to help fund, check out more here.
It's been a busy week so far, with school, Nerfing with friends, and my birthday occurring recently. I have a review in the making, but in the meantime enjoy this trailer by Wrecko Productions.
Made by Society of Meth of Youtube.
Following Clarkson's first-ever game of Humans v. Zombies, I put a 'trailer' together for the potential re-match in 2012. This was submitted as my final project for a video production class.
The entire video from concept to completion was done in five days. The production involved over a dozen people, but was conceived, shot, edited, color-corrected, and finished by myself.
The audio track is "In the House, In a Heartbeat" composed by John Murphy. It is utilized under 'fair use' for learning/ education.
Special thanks to Zach for the use of his equipment, and Jimmy for the tips and advice.