Funeral services for Alma Black, 94, were held November 4 at the Peace United Church of Christ, Monticello with burial in Oakwood Cemetery. Rev. Robert Pike officiated at the services.
Alma K. Black of Monticello died November 1 at the Senior Home in Monticello, following an extended illness.
Alma Kathryn Arnold was born April 17, 1895 in Castle Grove Township, Jones County, the daughter of George and Trinka Marie Schoon Arnold. She received her education in the rural schools near her home. She later worked for many years at the John McDonald Hospital in the dietary department.
Alma Kathryn Arnold and John Meyer were married in 1919. The couple made their home in Monticello, John died in 1938.
Alma Arnold Meyer and Lowell Black were married June 21, 1948. The couple traveled for three years after they were married, and returned to Monticello. Lowell preceded her in death in 1955. Alma was a member of Peace United Church of Christ, Monticello. Due to declining health, she moved to the Senior Home in June 1985.
Alma is survived by a brother, Chris Arnold of Monticello, a daughter-in-law, Mabel Meyer of Monticello, five grandchildren, 11 great-grandchildren and four great-great-grandchildren. She was preceded in death by her parents, a son, two husbands, a brother and four sisters.
Casketbearers were Frank Schneiter, George Burrichter, Lewis Burrichter, L.R. Bader, Robert C. Hasler and Gene Hall.
Goettsch Funeral Home was in charge of arrangements.


Osaka Medical College is a private university in Takatsuki, Osaka, Japan. The precursor of the school was founded in 1927 as Japan's first five-year medical school, and it was chartered as a university in 1946. The incorporated educational institution Osaka Medical College consists of Osaka Medical College as a college of health sciences, the College Hospital and the Health Science Clinic as advanced treatment hospitals, and Takatsuki Junior and Senior High School as an institution for secondary education.


text:
Why do onions make you cry?
Leave a Comment / Food, Health, Human Body / By Mau / 3 minutes of reading
Ever ponder why onions cause so many people to cry? This lesson gives two explanations.
The first is a matter of survival, while the second concerns the specific chemical reactants involved in this process.
What Makes Us Cry?
What do bittersweet movies, sad songs, and onions have in common? They make us cry, of course! But why? As for movies and songs, you'll have to ask the screenwriters and songwriters, but this article will explain why onions make you cry!
It has to do with our ability to survive and different chemical substances that affect how our eyes and brains function.
Defensive Mechanisms
We can answer the question, "Why do onions make people cry?" in two different ways. One perspective is to view it from a broad perspective. Why do onions do this to us, generally speaking? What really causes the tears to start flowing?
Generally speaking, it may be that onions make people cry as a defensive mechanism. Some plants have really horrible smell, while others have very sharp thorns. All of those might serve as the plants' defense mechanisms.
They may be trying to ward off anyone or anything that may want to eat them, like herbivores. Since they are gleefully eaten everywhere in the world, onions are obviously doing a bad job when it comes to humans.
The Chemistry
Therefore, the simple explanation for why onions make people cry is a matter of survival; in other words, the whole perspective. Regarding our magnifying glass, we must consider the scientific basis of what actually makes us cry. That is to say, the small but important details behind the defensive mechanism.
Onions belong to the plant genus Allium, to which garlic (among other plants) also belong. Onions take up sulfur, a chemical element, from the earth as they grow.
Onions use sulfur to create something known as amino acid sulfoxides. These are compounds that are important precursors for the distinctive flavors and aromas of Allium vegetables.
As soon as you cut, break, or crush an onion, you break apart its cells. When cells are damaged, they produce a variety of enzymes, which are protein molecules that quicken biochemical reactions. Alliinase is the name of one of these enzymes.
It converts a specific amino acid sulfoxide, called S-1-propenyl-L-cysteine sulfoxide, into a sulfenic acid called 1-propenyl sulfenic acid. Sulfenic acid is a sulfur-based acid compound that is produce from sulfur that plants absorb from the ground.
This acid is unstable and is quickly converted by an enzyme called lachrymatory–factor synthase into syn-propanethial-S-oxide, a tear-inducing and volatile sulfur compound.
The latter chemical spreads as a vapor into the air, where it comes into contact with our eyes. A small portion of it reacts with the water present in our protective tear film to generate sulfuric acid once it comes into contact with our eyes. This acid is extremely strong.
Our eyes have incredibly sensitive receptors that can sense changes in the physical and chemical environment. The nerves associated with these receptors send signals to the brain that a chemical irritant is annoying our precious eyes!
Our lacrimal (tear) glands receive signals from the brain, which interprets them as a burning or stinging sensation and tells them to start creating tears to wipe the chemical irritants away.
Mayo Clinic Minute: Why do onions make you cry? – Mayo Clinic News Network
Why Do Onions Make You Cry? Enzymes, Treatments & More (healthline.com)
Why Chopping Onions Makes You Cry (webmd.com)
What Is the Opposite of Red?
Why Do Dogs Have Wet Noses?

redpajama_set_name: RedPajamaCommonCrawl


PHUKET: Despite some excellent chances by both sides, FC Phuket and Division One leaders Big Bang Chulalongkorn University (BBCU FC) FC played to a scoreless draw last night before a record crowd at Surakul Stadium.
The final tally barely reflected the intensity of the exciting match.
One of the best opportunities for FC Phuket came at 75 minutes, when Sarach Yooyen fired a free kick on target that was punch-saved by BBCU goalie Shaun Mark.
The ball landed on the right foot of FC Phuket captain Wiratroj Janteng, who narrowly missed the second chance.
The flamboyant BBCU goalkeeper was named "man of the match" for this and several other good saves throughout the encounter.
FC Phuket coach Surachai Jirasirichote expressed disappointment that his side failed to take away three points.
"We had several good scoring opportunities, but failed to convert. Overall the kids played better than they did in their second match and seem to be better adapted to the system, so now the key is finding the scoring touch," he told reporters.
BBCU FC coach Kiattisak Sanao-muang conceded that FC Phuket were the better team on the day and said he was satisfied a draw.
"We learned something from today's game that will help us in future matches on the road," he said.
FC Phuket's next match is a home fixture against Rangsit JW, whose home ground is Don Meuang Stadium outside Bangkok.
Rangsit JW remain winless, but have collected two points from draws.
Note: The top three teams at the end of season are promoted to the Thai Premier League. The bottom three teams are relegated to the AIS Division 2 Regional league.


We're so pleased to continue our long partnership with Birch. We reached out to them during a major rebranding process for our international network of 100+ centres and their technology provided everything we asked for. Our franchise partners now have access to a wealth of marketing materials and are able to personalise them very easily, plus they save money and everything's stays on-brand!
Birch Print's technology does a lot of the heavy-lifting, which means we get to spend more time and energy on our campaigns. We've always found Birch's quality and service to be second-to-none and we're in constant communication with the team, who are friendly and efficient – even when we pile on the pressure! We can't recommend them highly enough.
Birch explain everything well with regards to the whole printing service, therefore we understand all of the costings involved. Birch are efficient and accurate and always deliver on deadlines. Birch keep us involved with any new processes and are willing to listen and accommodate to our business requirements. The whole team is approachable at all levels and they are all on the same page as BPEC. This means that when dealing with any issues or queries, we are confident that we will be assisted in resolving those issues and will be provided with the accurate answers.
The monetary benefits have been substantial for BPEC, but it is the personal touch and excellent customer service that Birch provide, that makes them outshine any other printing provider.
Since moving to Birch for our printing and distribution of training manuals, we are now using their printing service across all business areas of BPEC.
Working with Andrea and the team from Birch Print is always delightful and rewarding. From the continuous support and bespoke promotional offerings, Birch have always helped us create very unique products that stand out and work well for what we want to achieve. We know if we need support we always get the best from Birch whether it's a last minute request or an ongoing project – the team are very helpful and knowledgeable and always keen to keep us updated with new products available. Not to mention the quality of the work produced is always as expected if not better.
After using Birch at a previous company they were the first printing firm I turned to for this crucial and complex project. I can always rely on the team to deliver high quality materials on time and to leave no stone unturned in ensuring that every element throughout the process has been well thought out and planned for. I wouldn't hesitate to recommend Birch Print to anyone looking for a fast, quality and reliable service.
As a start-up business, Birch Print has been incredibly supportive, proactive and responsive throughout the development process. This shows through in the finished stationery and packaging products. I am delighted with the quality and the range of innovative solutions proposed and delivered by the team and look forward to an exciting partnership going forward.
The introduction of the MRC has enabled us to ensure brand consistency across all of our materials. Our Instructors have a dedicated point of contact for all related queries, and we have peace of mind that their enquiries are being dealt with effectively and efficiently.
I have worked with the team at Birch for some three years now, and it is always a pleasure to work on a project with them. Nigel is extremely knowledgeable, and always goes the extra mile to ensure that the job is completed on time, to specification, and to the agreed budget. He often recommends ways of accomplishing things that we hadn't considered, and as such we always value his input on our marketing campaigns. The level of service and support we receive really is second-to-none.
Finding the right partner to support Cartridge World with the launch of our online Marketing Resource Centre was a long and drawn out process. Not only was I looking for keen pricing I wanted to be sure that the support I would receive and the quality of print would match our brand expectations. Birch Print deliver in both areas of value and quality.
Andrea and her team at Birch Print work as if they are an extension of Cartridge World, something that is rarely found in suppliers now a days and I would recommend them to anyone looking for a reliable and trustworthy partner for their print requirements.
I have known and dealt with Andrea for many years having used Birch Print for various printing requirements for National Auto Parts. Birch has consistently provided us with a high level of quality, service and support over the years which has been key to our continued use of their services.
I consider Birch Print as a partner rather than a supplier and that is largely down to the support received from Andrea and all the team at Birch Print. I have total trust in Andrea and have no hesitation in recommending her and Birch Print to anyone who requires a quality service for their print requirements.
I worked with Birch on a number of projects during my time at Vision Express. They always aim to provide the best solutions possible within budgetary constraints and will frequently offer suggestions for more effective ways to achieve an improved outcome for the client.
As Head of Marketing I selected Birch as the primary print supplier for Halifax Estate Agents based on the quality of their product, the value they offered and the holistic service offering. Over a period of 5 or 6 years they delivered print and design services that were second to none, and their pro-activity in managing a significant collateral helped drive my business forward.
Their client management was excellent and Nigel's professional approach and personal integrity are a major asset to the company. I would not hesitate in recommending Birch as an excellent supplier of quality print products.


This press platform, dedicated to journalists, provides texts, photos, videos for web and social media and B-rolls around the LVMH PRIZE 2017. It will be updated regularly with the latest press content.
These documents may only be used for editorial purposes.


text:
David 2021, Pausini mocked by Zalone: social delirium
"The Pausini beaten by Zalone? I thought it was a hoax! I swear!". The victory of Checco Zalone at the David di Donatello 2021 for the best original song 'Immigrato', which beat 'Io Sì' by Laura Pausini, sparked delirium on social networks, where the fans of the Romagna singer went wild expressing all their disappointment. "As if instead of Guardiola, Zidane, Mourinho, the 'Coach of the Year – Uefa' award went to Oronzo Canà …. Shameful !!!", one user exclaims. Read also "First reaction: shock!", Writes another, quoting the now famous sentence in English by Matteo Renzi. "But really the piece by Zalone, which nobody knows, was preferred to the award-winning Me, yes? But on what parameters did they vote?", Asks a fan. "I have no words – an Internet user urges – It ends here, I change the channel. Why is it not appreciated in Italy?". "Checco Zalone beats Pausini winner of the golden globe and nominated for an Oscar, it is already funny", observes another, referring to Pausini's reaction to the announcement of the winning song – in a video that has now become viral – in which the artist bursts into incredulous laughter. There is also someone who tries to see the positive aspect: "But maybe the memory is short. Paolo Sorrentino won the Oscar and then got fewer awards than Virzì", he writes. But there is also someone who makes fun of her and laughs: "Pausini wins golden globe, nominated for an Oscar … and is beaten by Checco Zalone … I love this country!". Zalone: "If I knew I came" "If I knew I came". With the usual irony, in connection from home with the live broadcast of Rai1, Luca Medici, aka Checco Zalone, welcomed the David to the best original song received for the song 'Immigrato', written and sung by Zalone himself for the column of the film 'Tolo Tolo'.


text:
Tiago Haron Martins Barroso (born August 31, 1993) is a Brazilian footballer who plays as an attacking midfielder for Flores da Cunha in the Campeonato Gaúcho Série B.

Career
Tiago Haron began his career in 2013 with Ferroviário , where he played for three seasons. Since February 2016, Tiago Haron plays for Ferroviário.

References

1993 births
Living people
Sportspeople from Fortaleza
Brazilian footballers
Association football midfielders
Ferroviário Atlético Clube (CE) players
Flores da Cunha Futebol Clube players

redpajama_set_name: RedPajamaWikipedia


The 2016 American Public Works Association – Minnesota Chapter invites you to exhibit at this year's Fall Conference, held November 17-18, 2016. For 2016, the exhibitor and sponsor booths will be in Carriage Hall for the Thursday evening reception and Friday for breaks and lunch. This will allow for an evening event on November 17 at the Earle Brown Heritage Center. This event will start with a reception from 4:15 p.m. to 6 p.m. and a Carnival from 6 p.m. to 8 p.m. This year's onsite event has a revised layout to encourage better circulation and interaction between our members, sponsors and exhibitors. The details are listed below. Please note that exhibit space is limited, and will be available on a first come, first served basis.
Participants will visit the exhibits during reception and Carnival, as well as breaks and lunch on Friday. The conference program adjourns at 4:15 p.m., and the reception is held until about 6 p.m. with the Carnival from 6 p.m. to 8 p.m. If you are interested in holding a "minute to win it" type event at your booth during the reception, please indicate this when you complete your online registration. This year's carnival will have a new prize raffle format as we look to encourage attendees to circulate to all booths. APWA MN will contact interested vendors and sponsors directly with more information.
Exhibit space will be set up in Carriage Hall. As noted above, the reception food, cash bars and Carnival will be set up within these exhibit areas.
Registration opens for all exhibitors on August 19, 2016. Priority in booth location and available space will be given to APWA MN Chapter Sponsors.
Each exhibitor is provided with two name tags. Electricity is available at your exhibit for a fee of $80. If you choose to order electricity onsite Thursday, you must pay an onsite set up fee of $130, to Earle Brown Heritage Center directly.
Take down: any time after 12:45 p.m., Friday, November 18, 2016 (all exhibits MUST be removed by 3:00 p.m. at the latest).
To request a paper copy of the registration form, contact Katherine Hagberg at cceconf3@umn.edu.
For more information, visit the APWA MN Chapter website or contact Katherine Hagberg, University of Minnesota, 612-624-4230 or ceconf3@umn.edu.


Download "THE Vienna Model. Free Exhibition & Programme of Events, Dublin April 1 st - 25 th Housing for the 21 st Century City"
7 " What is special about and makes us proud, is that you cannot tell how much someone earns simply by looking at their address" Kathrin Gaál, Executive City Councillor for Women s Issues and Housing, City of. Providing affordable housing in has been a political priority since Compared to other cities, rents and land prices in have not solely been determined by the market. 's allocation guidelines ensure that higher income earners and the middle class can access the public housing sector. - The FACTS 1 st 420,000 voted 1 ST PLACE in the international QUALITY OF LIVING RANKING by Mercer FOR 10 YEARS RUNNING 420,000 municipal and subsidised dwelling units The City of owns 220,000 dwellings and an additional 200,000 are provided by limited-profit housing associations. Two thirds of 's 1.9 million population live in public housing in each one of the 23 districts of.


text:
11 Brutally Honest Women's March Sign Ideas To Hold Up With Pride
By Charlie Beckerman
JAVIER SORIANO/AFP/Getty Images
I am a firm believer in turning pain into action, and for all of us who are watching the national travesty of Donald Trump getting sworn in as the next president of the United States on Friday, there's the potential for a lot of action. Many are planning on expressing their outrage at Saturday's Women's March on Washington, and just in case your emotional outlet happens to be through uprising arts-and-crafts, we have here 11 inspiring and brutally honest sign ideas!
A well-crafted protest sign is more than just simply writing a catchy rhyme on poster board — you want something that will stick in people's minds, make them think, and inspire them to keep fighting for what they believe in. If you can also get them to chuckle wryly while you're at it, so much the better.
To that end, here are ideas for protest signs that you can make for the Women's March, whether you're attending in Washington or at one of the sister marches taking place around the country and around the world. And whether you get wild with the puffy paint and glitter, or just have one of those giant-ass Sharpies, let your voice be heard!
You Should Express Your Anger
And There's Plenty To Be Mad About
Anger Plus Hope Is Even Better
More Importantly, You Can Draw The Battle Lines
And Let People Know Where You Stand
If You Need To, Feel Free To Get A Dig Or Two In There
The Best Digs Are Ones That Have A Political Purpose At Their Core
But Make Sure You Remember What's Really Important
So, get marching. It's time. Let's do this.


I can't find one of my forms in MY FORMS page.
i add a new form two days ago.
and i can't find in my forms list.
You cannot find your form in MY FORMS page of your account because you created it in a guest account, that means that you were not logged in when you created it.
However I have already moved it to your account marmaraweb, so kindly relog in Jotform and look for tge form G.Antep Islahiye Seçim Anketi in MY FORMS page of your accoun marmaraweb.


Saudi authorities have arrested two more women's rights activists in the last three days, according to a human rights group.
ALQST, a UK-based rights group focusing on Saudi Arabia, said Mayaa al-Zahrani was detained on Saturday for a social media post in support of Nouf Abdulaziz al-Jerawi, a fellow activist arrested on Wednesday when security forces raided her home, Al Jazeera reported.
Al-Zahrani had also posted an article written by al-Jerawi, where the latter clarified her role as volunteering to help the oppressed by putting them in contact with lawyers and human rights groups.
"Why am I considered an enemy of the state that threatens its security?" al-Jerawi wrote in a post that has been widely circulated.
"We believe the Saudi authorities are keen to suppress all activists, and all sympathy with them," the rights group said.
The arrests were also reported by Prisoners of Conscience, a rights group that documents the arrests of Saudi nationals.
"It is quite clear that the Saudi authorities want to silence and put behind bars voices that speak against their opinion," the group said on Twitter.
Other suspects were being sought, the government said at the time, while state-linked media labeled those arrested as traitors and "agents of embassies".
None of the activists have yet been officially charged, and remain incommunicado with no access to their families or lawyers.
Al-Jerawi, a journalist and TV producer, had expressed her support to the arrested women who had advocated for equal rights, the end of the male guardianship system and for the ban on women driving to be lifted.
Yahya Assiri, head of ALQST, told Al Jazeera that the activists were arrested because of their communication with international media and human rights organizations in order to "reveal the violations of the Saudi government" against its population.
"These activists are on the side of the citizen, not the authorities," Assiri said.
The activists are considered by the Saudi government as traitors, Assiri, a former air force officer, continued.
"Who is the enemy of the Saudi authorities? It is the human rights organizations and media outlets that do not repeat the propaganda of Mohammed bin Salman," he said, referring to the Saudi crown prince, also known as MBS, who has promised a series of social reforms for the kingdom, including the lifting of the driving ban on June 24.
Tyler Perry, a US specialist of human rights in the Persian Gulf, told Al Jazeera, said activists in Saudi Arabia were being targeted and arrested "solely because of their right of expression—and that is unacceptable".
"Writing reports and contacting media is a human right and not a terrorist crime," he added.


Wine is one of the oldest beverages in the world today. This is why it is considered as a special..
There used to be different kinds of businesses in the world. But, after corporate came into picture, businesses followed a..


Download Appliacation Letter for College 20 Free Usajobs Cover Letter New for Word, learn creating a very good resume, and view sample resumes. Appliacation Letter for College 20 Free Usajobs Cover Letter New placed by admin at 2018-06-28 04:18:57. Simply choose your and have started. Just scroll down and find a Microsoft Word template fitting your projects experience and sense of design. Don't fret – getting a template is perfectly acceptable. When trying to find a new job you will want to build a appliacation letter for college. After creating your resume, just buy a Appliacation Letter for College with the exact same design, fonts, and color theme to produce a professional job application. It is possible to download such resume templates free of charge and add the bullet points yourself, or that you can resume within a few minutes with .


At the core of any business is it's people, and without great people, your company's growth and revenue will stagnate. It's vital to have the right people; talent that you can trust and that will help you achieve your goals. The brutal war for new talent rages on, and expectations about company culture are continuing to evolve fast.
You need a fresh strategy for attracting employees, because Millennials, for example, who will soon comprise around 50% of the workforce by 2020, certainly require a different approach, and the companies who get a hold of this quickly are the ones who are going to have the competitive advantage and their pick of the top talent.
Understand the talent that's right for you.
What's right for someone else's team might not be right for yours, and there's no one universal standard for the perfect employee. We wouldn't be where we are today or have our clients or our awards without our hard-working Zaboura team. And they're all Zaboura material to the core. Everyone has talent, but you need to make sure you're attracting the right talent for your organization; your company might want a team of entrepreneurial outside-the-box thinkers, but another might require more deliberative, consensus-driven executives. Look at the most successful people in your organization and what they all have in common, and you'll be able to clearly outline the employee traits and abilities that you're after.
Involve your employees in why you do what you do; meaningful work means more than salaries, according to a recent survey by UNC'S Kena-Flagler Business School and the YEC. For example, when Californian paper company Ernest Packaging Solutions were struggling to attract young workers, they created a YouTube video called The Ernest Edge, focusing on the Ernest Packaging's family roots and quirky employees, which has attracted around 50 new millennial workers to the company each year since it was posted in 2011. But you don't have to be Google or a quirky family company to create a workplace that people enjoy working in. It starts at the top: you can't have a magnetic organisation without magnetic leaders and the magnetism of a company has a lot to do with who you put in charge and give responsibility to. Everyone's had a horrible boss in the past, and it's the main reason people quit their job. Managers impact your turnover and you need to equip them with the knowledge and skills to help them understand what their employees' needs are and help build a retention plan that will increase employee engagement.
They may turn up for work every morning, but are your team really engaged and motivated, and as productive as they could be? People work harder for you if you let them contribute ideas and suggestions – it gives them a sense of ownership and inclusion and they will be more engaged and committed. Take Sony, for example, annual host of the Idea Exposition, during which their scientists and engineers can display projects and ideas that they're working on. Such initiatives create a healthy and buzzing atmosphere of innovation and creativity for everyone involved, and engaged, motivated employees boost team performance, customer service, profitability and brand reputation. Even little things say a lot, like at TD Industries in Dallas, TX, one of Fortune magazine's Top 100 places to work: there's a wall within the company contains the photographs of all employees who have worked there more than five years and there are also no reserved parking spaces or other perks just for executives, because every employee is equal.
Talented people won't stick around if they see no future in their role. Millennials want to work for leaders who are going to give them room to grow and are invested in the progression of their careers – a Linkage, Inc study reported that 40% of people would consider leaving their current employer for a new job that offered the same benefits but greater challenges and better career development. Whether that's covering the costs for industry conferences or skills workshops, inviting guest speakers for in-house training sessions, or increasing employee responsibility every 12 months to give them a fresh challenge; it's up to you to pick something that will truly resonate with this generation (think digital, and flexible attendance). And whatever you do: measure. Measuring and regular evaluation is vital for continuous improvement. At Zaboura we measure attitudes, morale, turnover, employee engagement, and ensure that every employee has an individual development plan. The primary purpose of evaluation is to measure progress and determine what makes your team content and what they're dissatisfied with. Deloitte, for example, (one of the top 100 places to work) discovered during exit interviews that 70% of the talent they lost could have been re-homed within the company. Deloitte Career Connections, their resultant employee coaching program, was utilised by 2,000 employees who looked at internal job openings/self-assessments/resume-developing tools/job-seeking strategies in it's first week.
Once you've found the right talent, you want to keep hold of it, and you won't with just money and benefits. You have to keep them wanting to come back every morning. Most people like the idea of staying where they are if it makes them happy, and not uproooting themselves, so make them happy and you'll have a loyal team by your side for a long time to come. When did you last tell someone that you're really glad they're a part of your team, or gave someone a raise on the spot because of an amazing and deserving piece of work, rather than waiting for the next annual employee review? Everyone wants to feel appreciated and proud of what they accomplish, and you need to bear that in mind on a daily basis. Look at some recognition and incentive programs to get ideas flowing because they make all the difference.
To attract the workforce of the future, you need to become the workplace of the future. And part of that is understanding the importance of a proper work-life balance for today's youth, and accepting that you might need to make a few cultural shifts so that you're on the same page as your would-be new talent. More than half of those surveyed in The Hartford's 2014 Millennial Leadership Survey state the most important thing for young employees is a flexible working schedule, probably partly due to millennials being caught between home responsibilities of taking care of children or older relatives while striving for their own personal and professional goals. Even if it's simply the opportunity to work remotely one day per week, millennials are likely to respond well to such initiatives and companies who aren't aware of this aren't going to be places where such young hopefuls want to work.


Where is the sheet, btw? I may have never recorded the link lost the link to it.
Might be good to stick this in the OP as well. Or one of the character thread posts. Either works.
@Muttonhawk I admit I didn't actually look at the levels when writing up the latest turn post - just Might and Free Points - so if your character levelled this turn that may not be reflected in the post even if it is in the stats file.
Don't think this escaped my notice.
Now, with this new found might, I shall await for the battle to begin and for the badassery to to truly come to fruition!
So, I was just thinking about this, given my recent comments regarding MP inflation. Would Might invested by a higher level god be more powerful than Might expended by a lower level god? A higher level god can do more things without Might, so when they invest Might it suggests that they're doing something that's powerful relative to their level. Thus, a 10 MP weapon created by a Level 1 god would not be equal to a 10 MP weapon created by a Level 10 god - if that makes sense.
I think this is hinted at (or maybe stated explicitly, can't remember) in the OP. But it brings up issues - how would this difference measured? Even if it makes sense, is it workable at all?
@Kho That feels like an incentive for players to powerplay and spend might on levels and holy sites instead of creation.
@Kho On top of Capy's valid concern, that would also require tagging creations with the level of the god at the time. That kind of extra record keeping would probably be confusing and difficult.
I've always assumed that the higher level gods can do more powerful things in a fuzzy and ill-defined way. However, I don't see a big advantage in adding a mechanical deal to it. The only time I would prefer to see raw levels being taken into account is in a tie-breaker between clashing actions by equally invested gods.
That's fair enough. I thought it could be a way of getting around the whole 'They made a 10 Might weapon. Well, I'm gonna make a 20 Might weapon.' 'Oh, damn, 20 Might weapon! I'm gonna make a 30 Might weapon!' as it would make the system less straightforward than pure Might or pure levels. But I agree that it would overcomplicate things if we try to define that all in some way. Fuzzy and ill-defined is best in this case.
Yeah, it's a fair attempt, but if the thing to counteract the might inflation is something that can be bought with might, it's not doing much to help.
Honestly, I think we're doing most of this by narrative feel anyway.
Weapon inflation is a self-solving problem. I kinda suffer a similar effect of it with the Divas, like for example, making Chronicle (8 might) instead of adding that might to the Holy Site I was making on that turn has a clear effect on my might expendability. Since I had to numbercrush a bit so I can guarantee that I can get the Mind domain ASAP, the effect of that 4 might per turn was clear to me when I made her, and it isn't nice number-wise, even if absolutely necessary for a better story.
With that in mind I can only imagine what a 10 to 20 might weapon could do in the long term to a player, especially since it provides less narrative possibilities than an avatar. Anything like a 15 or 20 might weapon not only it hinders the future might of the player, but it also isolates the player from the ongoing narrative.
And anyway, the people with the strongest weapons are also people whose characters are also in relative isolation (Logos and Niciel). I don't count Zephyrion's staff as the same as the other two because it was mostly a larger investment on Vetros, maybe Niciel's staff could have a similar effect.
Edit: The whole decision of making a 10 might weapon rests on the balance between the other options. For me for example, it would go like (Make 10 might weapon) vs (Invest in new lands, new species, new concepts) vs (Invest in might per turn generation). For the last turns I have mostly gone with the middle option, I will go a bit with the last one now because I need level 14 to have the Mind Domain, still, I numbercrushed hard to guarantee I can still keep adding a load of new things every turn instead of just banking on might.
Basically, do you think if Lugi was still around, Slough would have even spent 10 might on a weapon instead of bringing in new creatures/biomes? The way to stop a possible weapons race is by making that middle option more valuable in face of the others, and surely powering up might value per level isn't the way to do it, on the contrary, I few like it makes one less likely to spend on there, since in 4 levels or so one can make it all for free.
In light of Fate's Might Pot now having nearly 40 MP invested in it (18 of which were mine ) I felt that it was necessary to discuss what should be done with it.
We GMs have decided that as it was put forth as more of a mechanism to prevent the hoarding of MP rather than as a punishment for such, players that lose MP to the pot can determine what is done with it.
In my case, I'm thinking something along the lines of Zephyrion's power escaping from his two holy sites as something like the Winds of Change from Warhammer. (In fact, I might just outright rip off Warhammer here). Basically these tides of magic would ebb and flow around Galbar and Zeph's planet Caelum, and beings could draw upon them as sources of power for other magic or harness the winds themselves as pure change magic. Post pending; that's just an example of one thing that you could do. Alternatives might be using your spillover MP to create a new biome or some sort of monster.
As in for the question of MP that is just abandoned to the pot, that's gonna remain at GM discretion. On that topic, I would invite discussion and suggestions concerning what might be a good use of the pool in Fate's Might Pot. Even if I were to use all 18 of Zeph's spillover for something (and I might leave some in), that's more than 20 Might left over that we could do things with.
@Cyclone Give it all to Lifprasil.
IM GONNA KILL ME SOME KNIGHTS!
@Kho I made a mistake and counted double the might for ull when i was editing his stuff, it should be 22 not 31, as well as 7 FP, not 8. Edit that in pls nd soz for the mix-up.
You know what I just realised? Mammon's tunnels are still around. Those dwarves are likely to run into a forgotten beast one of these days >:D. The name works both in the Dwarf Fortress context and in our universe!
Not sure if this makes things any better, though.
I can't wait for the 12 foot NGINX PUDDING FORGOTTEN BEAST, with a terrifying 6 pairs of wings, scaly, pudding skin, breath of fire, and toxic water blood.


text:
Stephens set to face former side
Dale Stephens. Picture by Angela Brinkhurst
BRIGHTON midfielder Dale Stephens feels it will be a special occasion when he runs out against the side who gave him his Championship chance on Saturday.
Albion, who moved back to the summit after a 2-1 home success over Birmingham City last time out, welcome fourth-from-bottom Charlton to the Amex.
The Addicks are a club close to Stephens' heart, after he made more than 80 appearances for the London side.
That opportunity also saw him taste Championship football and the now Albion midfielder looks back on his time with Charlton fondly.
He said: "I enjoyed my two and a half years with them.
"A lot has changed since I came to Brighton but I have got lots of fond memories from my time there.
"They gave me the opportunity to have first-team Championship level and I'm grateful for that chance.
"I also felt I progressed a great deal as a footballer in my time there, so I have good memories of Charlton.
"But I came to Brighton because of their ambition and we have shown this year we are capable of mixing it with the top teams and that was a big factor in me coming here."
Albion remain unbeaten after 18 games and last week's victory over Birmingham saw them climb back to the top.
For some, Brighton's superb form this season will have come as a surprise but Stephens admits himself and the side have great belief.
He added: "We had a meeting pre-season and we spoke about what we wanted to achieve this year and it's going to plan.
"It's very tight still and there is a long way to go. All we can do is focus on ourselves and take it game by game."


text:
Defense bill funds IRS security
By Judi Hasson
The Internal Revenue Service is getting an extra $16 million to secure its information systems, money tucked into the fiscal 2002 Defense appropriations bill in the wake of the Sept. 11 terrorist attacks.
Most of the funds — about $13.5 million — will be used for a backup computer recovery system that will be "designed and constructed in close coordination" with the IRS Business Systems Modernization program. But the IRS declined to comment on exactly how the money would be spent.
"Prior to Sept. 11, we had a good, workable system in place. Evaluating the situation post-Sept. 11, we determined that some improvements should be made, so money was requested," an IRS spokesman said. "Because of security issues, we cannot disclose what the changes will be."
John Reece, the IRS chief information officer, told Federal Computer Week in October that the tax agency was seeking more money for security in the wake of the attacks on the World Trade Center and the Pentagon. Like many other agencies, he said, "Our priorities have changed."
The Business Systems Modernization is an ongoing project to convert the IRS' antiquated tax records into a database system as well as provide a wide range of electronic services for customers. The fiscal 2002 budget includes $391.6 million for the modernization program through 2004.


Browse listings of Member singles here at Utah Match that are associated with Filipino. Talking to others who have similar interests is a pefect way to find things to do on a first date. Signup for a 100% Free Profile to Find your perfect match!


PICO RIVERA – A brush fire broke out in the riverbed and charred up to an acre Friday night.
The fire was reported at 5:23 p.m. in the riverbed near Fairway and Bradgate Drives.
Robert Diaz, a supervising dispatcher with the Los Angeles County Fire Department, said no one was injured in the fire which was knocked down at 6:33 p.m.
Diaz said what started the fire hasn't been determined.
This entry was posted in fire, Pico Rivera by Brian Day. Bookmark the permalink.


Greeting Card Franchise opportunity supplying to Retailers on a sale or return basis. Work from Home.
Supplier of Cross-Stitch Kits reflecting the Culture of the British Isles.
Supplier of Collectible English Handmade Ceramic Teapots.
Retail and Wholesale Supplier of Watercolour Greetings Cards, Prints, Frames, Coasters, Placemats, etc.
Trade and Retail Supplier of 4mm Scale Model Narrow Boat Kits, Lock Drawings and Canal related Greetings Cards.


text:
الرئيسية/عالمية عالمية/Thiago Alcantara can avoid Manchester United frustration for Liverpool
عالمية عالمية
Thiago Alcantara can avoid Manchester United frustration for Liverpool
Facebook Twitter Google+ LinkedIn WhatsApp مشاركة عبر البريد
Thiago Alcantara knows all about playing at Anfield. The Spanish international was part of the Bayern Munich side who earned a 0-0 draw in a Champions League tie in February 2019.
He is also aware of how difficult it can be to play there as a visiting player.
Thiago only completed fewer passes in two matches where he played the full 90 minutes that season (per FBRef) and he also recorded his joint-fewest progressive passes of the 2018/19 campaign that night.
But something he knows nothing about is how it feels to run out at Anfield as a Liverpool player.
His five appearances for the Reds have all been on the road, so the visit of Manchester United on Sunday should be the 29-year-old's first opportunity to represent his current team on home turf.
That the season is already four months old when Thiago's home debut comes around is one shame, the fact the stands will be empty perhaps an even greater one.
Jurgen Klopp would give anything to have his players backed by a raucous and rocking Anfield for the tussle with the current league leaders.
Even so, the Liverpool manager will be delighted to have the former Bayern man available for the match. The visitors will look to frustrate the Reds, just as their other recent opponents have, and try to hit them on the break.
As United have scored a quarter of the counter attacking goals by all away teams in the Premier League this season (per WhoScored), they undoubtedly have the ability to hurt Liverpool in this way.
For the home side to prosper they will need to find a way through a stubborn United backline.
Enter Thiago. He has shown throughout his career that he is a master of successfully advancing the ball towards the opposition goal.
Among men with at least as many minutes as he has played this season, Liverpool's number six is second in Europe's big five leagues for progressive passes (which are "completed passes that move the ball towards the opponent's goal at least 10 yards from its furthest point in the last six passes, or any completed pass into the penalty area") per 90 minutes.
To be so high in the rankings having only played away matches is no small matter, but it's nothing new for Thiago. He completed the most progressive passes of any player in the Champions League last season, so there's no reason he can't add to his tally for Liverpool on Sunday.
Get all the latest Liverpool injury and team news as Klopp faces selection dilemma, plus breaking news and analysis of what's next for the Reds.
You'll also get the latest transfer talk and analysis straight to your inbox every day with our FREE email newsletter.
Sign up here – it only takes a few seconds!
The former Bayern man is also supreme at feeding the ball to the players who then create goal scoring opportunities.
Alcantara created 85 chances during his final three seasons in the Bundesliga, but had a total of 193 shot creating actions (per FBRef). The more traditional statistic of key passes isn't able to fully capture the incisive impact which Thiago has.
We've seen glimpses of that during his time in England. The most notable example was his injury time pass at Goodison Park which put Sadio Mane free to set up Jordan Henderson for a potential winning goal, only for an offside margin measured in pixels to rule the effort out.
But Thiago has proven to be adept at getting the ball to Liverpool's key attacking players on a consistent basis in 2020/21, and will need to do likewise against United.
In the Premier League, he has averaged the most passes per 90 minutes to Andy Robertson of any Liverpool player, and the second most to Trent Alexander-Arnold among non-defenders.
He's also joint-top, alongside Alexander-Arnold, for frequency of getting the ball to Mohamed Salah, while only Robertson and James Milner have passed to Mane more often.
Thiago is the only man who is in the top three for passing to each of the quartet of Liverpool's full-backs and their main wide forwards this season.
He can pull the strings from midfield to ensure United's packed defence is moved from side-to-side and hopefully stretched wide open.
Alcantara avoid frustration Liverpool Manchester Thiago United
Facebook Twitter Google+ LinkedIn WhatsApp مشاركة عبر البريد طباعة
Facebook Twitter Google+ LinkedIn WhatsApp Telegram Viber


Marie-Madeleine was a Fille à Marier , arriving in New France by 1659.
She married Pierre DIZY dit MONTPLAISIR 13 July 1659 in Trois-Rivières, Canada, New France . The couple had (at least) 5 children. Pierre DIZY dit MONTPLAISIR was born 21 June 1634 in Rouen, France. He died 12 June 1698 in Champlain, Québec, Canada (Notre-Dame-de-la-Visitation).
Marie-Madeleine DROUILLARD died 17 February 1719 in Champlain, Canada, New France .
Is Marie-Madeleine DROUILLARD YOUR Ancestor? Tell us more.
If you'd like to be contacted by others who are related to Marie-Madeleine DROUILLARD, leave a message here!


More often than not, when the University of North Carolina gets talked about in December, people assume it's all about college basketball. Very rarely is football in the mix, especially considering the Tar Heels haven't won a conference title in nearly 40 years. Everything is different this year, though.
During halftime of one of the marquee hoops games of the weekend, when UNC upset Gonzaga, the Dean E. Smith Center was a full-blown celebration of both sports. That's because head coach Mack Brown has returned to the sidelines and has hired a dynamite staff to bring winning football back to Chapel Hill.
The first introduction and reception from the fans is exactly what you would expect for the winningest active coach in college football and 2018 College Football Hall of Fame inductee.
From 1988-1997, Brown guided the Tar Heels to a 69-46-1 record before leading the Texas Longhorns to two Big 12 titles and one national championship in Austin. He was hired at UNC to replace Larry Fedora and has immediately brought a buzz back to the football program.
Brown certainly has some work to do to bring UNC back to ACC and national prominence, of course, but it's definitely time for fans to get as excited as he is and hop aboard the train before it takes off next season.


text:
package com.ivan.dourachok.engine.memory;

import static com.ivan.dourachok.engine.utils.TestUtils.apple;
import static com.ivan.dourachok.engine.utils.TestUtils.givenApple;
import static com.ivan.dourachok.engine.utils.TestUtils.givenBasket;
import static org.assertj.core.api.Assertions.assertThat;

import java.io.IOException;

import org.junit.Test;

import com.google.common.collect.Lists;
import com.ivan.dourachok.TestEngine;
import com.ivan.dourachok.common.AbstractDourachokTest;
import com.ivan.dourachok.engine.compiler.exception.DourachokCompileException;
import com.ivan.dourachok.engine.parser.exception.DourachokParseException;
import com.ivan.dourachok.engine.runtime.exception.DourachokRuntimeException;
import com.ivan.dourachok.external.AnyService;

/**
 * @author Christophe Dame
 */
public class ExternalRulesTest extends AbstractDourachokTest {

	public TestEngine givenEngine() throws DourachokRuntimeException, DourachokCompileException, DourachokParseException, IOException {
		TestEngine testEngine = givenEngine("rules/06_externalRules.drc");
		testEngine.addExternal("anyService", new AnyService());
		return testEngine;
	}

	@Test
	public void crossReferenceFromExternalTest() throws DourachokRuntimeException, DourachokCompileException, DourachokParseException, IOException {
		TestEngine testEngine = givenEngine();
		testEngine.run(Lists.newArrayList((Object)givenApple(false)));
		assertThat(testEngine.wasFired("cross reference from external")).isTrue();
		assertThat(testEngine.wasFired("should never match from external")).isFalse();

		assertThat(testEngine.getModifiedObjects().size()).isGreaterThan(0);
	}

	@Test
	public void loadFromExternal() throws DourachokRuntimeException, DourachokCompileException, DourachokParseException, IOException {
		TestEngine testEngine = givenEngine();
		testEngine.run(Lists.newArrayList((Object)givenBasket(apple), (Object)givenBasket(apple)));
		assertThat(testEngine.countFired("load from external")).isEqualTo(1);//only one load by cycle
		assertThat(testEngine.getModifiedObjects().size()).isGreaterThan(1);
	}

}

redpajama_set_name: RedPajamaGithub


text:
Harmanjot Khabra handed two-match ban, fined for violent conduct
ISL Media
Khel Now
The right-back will miss Kerala Blasters' last two league matches.
The All India Football Federation (AIFF) Disciplinary Committee sentenced Kerala Blasters FC's Harmanjot Singh Khabra to a two-match suspension and a fine of INR 1.5 Lacs.
Khabra was found guilty of 'hitting an opponent when not challenging for the ball' after being charged with 'violent conduct' in an Indian Super League (ISL) match against Hyderabad FC last month.
Upon investigating video footage of the incident, the AIFF body show caused the player. In his written response, Khabra regretted "making an error in sporting judgment and acting in a manner which is not in line with the laws of the game and it was never his intention to hurt Sahil Tavora."
The Committee called upon for an additional hearing and passed their verdict on Tuesday, March 1. Harmanjot Khabra will remain suspended for Kerala Blasters FC's last two ISL league-stage matches this season against Mumbai City FC (March 2) and FC Goa (March 6).
The 33-year-old will be eligible for selection next should Kerala Blasters FC reach the Semi-finals.
All transfers completed so far in January Window in ISL 2022-23
The incident occurred during the Kerala Blasters vs Hyderabad FC match at the Athletic Stadium on February 23 (Wednesday). Khabra, frustrated by the scoreline, had an off-the-ball incident with Sahil Tanvora in the final moments of the game. The right-back did play Kerala Blasters' next match against Chennaiyin FC, which they won. He will, however, miss the Blasters' final two league matches against Mumbai City FC and FC Goa.
With Ivan Vukomanovic's men chasing a playoff spot, Harmanjot Khabra's absence will be a big miss for them. Sandeep Singh is likely to man the right flank in his absence.
Kerala Blasters in ISL 2021-22
The Blasters will face defending champions Mumbai City FC in a virtual quarter-final clash for a place in final four later today. A defeat could end their hopes of making the playoffs. Kerala are currently fifth with 30 points from 18 matches, a point behind Mumbai who have played the same number of games.
For more football updates, follow Khel Now on Twitter, Instagram and join our community on Telegram.
by Sathvik Bharadwaj
Ivan Vukomanovic shares what he looks for while scouting a striker
by Tarkesh Jha
Ratings: Dimitrios' brace against NorthEast United guides Kerala Blasters to their fifth consecutive home win
by Keshava Verma
AIFFHarmanjot Singh KhabraHyderabad FCISL-Indian Super LeagueKerala Blasters FCKhel NowMumbai City Football Club


text:
Michigan State Tennis Sweeps Doubleheader
The Michigan State men's tennis team defeated both Ball State and Eastern Illinois, 7-0 on Saturday.
"It went great, we didn't lose a set, and we had a good vibe," senior Aaron Pfister said.
The Spartans began the doubleheader by facing off against head coach Gene Orlando's alma mater and former coach Bill Richards and the Ball State Cardinals.
"It was bittersweet," coach Orlando said.
Pfister and freshman Brett Forman set the tone with an 8-2 win in doubles competition vs. the Cardinals.
The teams of John Patrick Mullane and Drew Lied and Harry Jadun and Gijs Linders also won their doubles competitions by a score of 8-6.
The most exciting game of the early session came courtesy of Gijs Linders who won 7-5, 7-6.
"There was ups and downs," Linders said. "I'm glad the team stood behind me and loved the energy we got from the home crowd."
Michigan State kept the momentum going against Eastern Illinois.
The Spartans made easy work of the Panthers, winning all the doubles competitions by at least three sets.
Again, the singles competition provided the most exciting finish.
Redshirt senior Will Davis grinded out a victory over Kevin Bauman 6-4, 6-4.
Michigan State improved to 4-3 overall and will travel to Lexington, Ky. next Sunday to play No. 13 Kentucky.
"There will definitely be some added motivation," Davis said on playing against his hometown university. "I'm looking forward to seeing some old hitting friends."
Pfister said the key to pulling an upset at Kentucky will be taking care of their bodies over the weekend and having a good practice on Monday.
Troy Jefferson is a multimedia journalist for Impact Sports.
Photo: Michigan State University Men's Tennis


On Dec. 1, 1974, Officer Murphy was shot after stopping a vehicle at West Main Street and U.S. Highway 281 in Alice, Texas. Several minutes after the initial stop, another officer found Murphy suffering from gunshot wounds. Murphy was transported to a local hospital, where he later died.
Submit a tip online: https://www.tipsubmit.com/WebTipsCSI.aspx?L=E&AgencyID=650.
Created in January 2016, the Fallen Hero Reward program is designed to help generate new investigative leads in unsolved cases involving a Texas law enforcement officer or first responder who was murdered or suffered an incapacitating injury while in the line of duty or while acting in an official capacity. The program is funded by the Governor's Criminal Justice Division and administered by DPS.
Agencies that may have an unresolved or cold case that meets the criteria for the Fallen Hero Reward program are encouraged to email Texas Crime Stoppers at txcrimestoppers@gov.texas.gov.


text:
Football Recruiting Irish Stew Men's Basketball Women's Basketball More Sports
Purring Notre Dame offense wipes out Pittsburgh Panthers
Tribune Correspondent
SOUTH BEND — In a blink, Pitt was shoved into a pit.
The Panthers never had a prayer of clawing their way out, either.
Keyed by a pair of 21-0 runs during an overwhelming first-half performance, No. 2 Notre Dame crushed Pittsburgh, 100-44, in each team's Atlantic Coast Conference women's basketball opener Thursday night at Purcell Pavilion.
With each starter's damage coming during the first three quarters, Marina Mabrey scored 20 points and dealt seven assists, Arike Ogunbowale netted 18 points and Jackie Young tallied 16 as the Irish improved to 13-1.
The Panthers dropped to 9-6 in front of a crowd announced at 8,566.
Young scored her sixth point of the game and 1,000th of her career just 3:13 into the evening, becoming the 40th player in program history to reach a grand.
The junior received a warm ovation from the crowd and from her teammates when her milestone was announced during the next timeout
Notre Dame, beginning its sixth season in the ACC, improved to 41-0 in the league all-time at home.
Jessica Shepard compiled an odd first-half line, going 0-of-7 from the field, but pulling in 10 rebounds and dishing five assists to blend with 4-of-4 at the line. She finished the night at a game-high 11 boards and six points.
Brianna Turner added all of her nine points in the first half and closed with three blocked shots.
Ogunbowale poured in 17 points and Mabrey 14 over the opening two periods. Notre Dame went on 21-0 sprees in both the first and second quarters on its way to taking a 60-14 lead by halftime.
The purring Irish offensive machine scored on 11 of its first 12 possessions while grabbing a 23-2 lead with the first 21-0 spurt, that one spanning six minutes.
ND's second 21-0 run occurred over a stretch of merely five minutes to push the spread to 53-12 at four minutes left in the opening half.
During that one, the Irish scored their points in threesomes five consecutive times — a pair of triples by Mabrey, one by Jordan Nixon, and Turner and Ogunbowale each converting free throws after baskets.
The Irish played mostly zone while building their commanding advantage, mixing in occasional half-court traps.
By halftime, ND already had 15 assists and seven steals, three from Mabrey.
Pittsburgh committed 14 turnovers before the break to five for the hosts, who also had a 31-17 rebounding advantage during the opening 20 minutes.
The Irish were whistled for just two fouls before the intermission to the Panthers' 13.
The 14 points that Notre Dame allowed was its second-fewest given up during a half in an ACC game, behind only the 11 allowed during the opening half of an 80-41 win over Virginia Tech in January 2016.
Pitt made just 6-of-31 field goals in the first half, including 2-of-11 on 3s, to go with 0-of-1 at the line. The Irish were 21-of-39 from the field by halftime, including 5-of-9 outside the arc, to go with 13-of-17 at the stripe.
In the third quarter, ND added two more double-digit runs, each of them 10-0, the second giving the Irish their largest lead of the night, 81-19, at 2:35 to go in the period.
Freshman Nixon and sophomore Mikayla Vaughn each scored nine points off the bench for Notre Dame.
Jasmine Whitney led the Panthers, who had won seven of their last eight going in, with nine points.
The Irish play their first ACC road game of the season Sunday when they visit Georgia Tech (11-3) for a 3 p.m. tip. That's followed by next Thursday's 7 p.m. home showdown against No. 3 Louisville on ESPN.
Five, grand
Young reaching 1,000 points gives Notre Dame five 1,000-point career scorers on its active roster for the second straight season.
Based on information shared through Stats Inc., on Dec. 20, the ND women are believed to be the only NCAA Division I program to feature five 1,000-point career scorers two straight seasons — men or women — in at least 15 years.
Last season, Ogunbowale, Mabrey, Shepard, Lili Thompson and Kathryn Westbeld all wound up 1,000-point scorers (transfers Shepard and Thompson were, in fact, over 1,000 before the season began), and they were all active together at some point during the season for the Irish.
This season, Young joins Ogunbowale, Mabrey, Shepard and Brianna Turner in forming the 1,000-point fivesome. Turner sat out all of last season while recovering from a knee injury.
The Dec. 20 compilation by Stats Inc., included players who began their careers in 1999-2000 or later.
At the time, the only women's programs to have five 1,000-point scorers active in the same season all came within these last three seasons. They included Florida State and Louisville in 2016-17, Connecticut and Notre Dame last season, and Marquette this season.
On the men's side, the lone teams that had five 1,000-point scorers active in the same season were Providence and national champion Florida in 2006-07, George Washington and North Florida in 2015-16, and Nevada already this season.
Contact Us Support Local Business Advertise Your Business Advertising Terms and Conditions Help Center Licensing & Reprints Subscriber Guide My Account Give Feedback
Subscribe Today Newsletters Facebook YouTube
© 2023 www.ndinsider.com. All rights reserved.


text:
\section{\label{subsec: sm magnetization operator}Derivation of the magnetization and current operators}

In this Section we derive the explicit form of the magnetization operator for electrons hopping in a single ``molecule'' in a second-quantized fashion.
The position operator $\hat{\bm r}_k$, associated with the $k$-th  molecule, can be written in terms of electron creation and annihilation operators, $\hat{c}^\dagger_{j,k}$ and $\hat{c}_{j,k}$ for an electron roaming on a polygon with $n_{\rm s}$ sides:
\begin{equation}
\hat{\bm r}_k=\sum^{n_{\rm s}-1}_{j=0} {\bm r}_j\hat{c}^\dagger_{j,k} \hat{c}_{j,k}~.
\end{equation}
Here, ${\bm r}_j= d_r [\cos(\gamma_j),\sin(\gamma_j)]^{\rm T}$ is the position of the $j$-th site measured from the center of the molecule, $d_r$ is the distance of the site from the center and $\gamma_j=2\pi j/n_{\rm s}$ is the angle subtended between the position vector of the first site and the one of the $j$-th site.
The magnetic moment of a single molecule is defined in terms of the position operator as \cite{Jackson}
\begin{equation}
\label{eq:Mk}
\hat{\bm M}_k
=\frac{1}{2}   \hat{\bm r}_k \times \Big(-\frac{e}{c} \dot{\hat{\bm r}}_k\Big)~.
\end{equation}
We use the Heisenberg equation of motion to find an expression for the velocity operator $\dot{\hat{\bm r}}_k$,
\begin{equation}
\dot{\hat{\bm r}}_k= i [\hat{\cal H},\hat{\bm r}_k]~.
\end{equation}
Replacing this result in the definition of the magnetic moment (Eq.~\eqref{eq:Mk}) we get
\begin{equation}
\label{eq:Mk1}
 \hat{\bm M}_k=\frac{-e}{c}\frac{i }{2}    \hat{\bm r}_k \times  [\hat{\cal H},\hat{\bm r}_k]~.
\end{equation}
We now consider the magnetic moment along the ${\bm e}_z$ direction,  $\hat{M}_{z,k}$, and we use the explicit form of the total Hamiltonian $\hat{\cal H}$ in Eq.~(\ref{eq:Mk1}). We find
\begin{align}\label{eq:Sm_BMz_k-th_molecule}
    \hat{M}_{z,k} &=\frac{\lambda }{B_0 \sqrt{N}} \times \nonumber \\ \sum_{j = 0}^{n_{\rm s} - 1}
   & \times \left[ i t_{j,j+1} e^{-i \frac{\lambda}{\sqrt{N}} (\hat{a}+\hat{a}^\dagger)} \hat{c}_{j,k}^\dagger \hat{c}_{j+1,k} + {\rm H.c.} \right]~.
\end{align}
In the presence of $N$ identical molecules, the total magnetization operator is expressed as
\begin{equation}
\hat{M}_z=\sum^N_{k=1}\hat{M}_{z,k}~.
\end{equation}
In the mean-field approach described in the main text,  we can trace out the photonic degrees of freedom by projecting the total magnetization onto a coherent state $\ket{\phi_{\alpha}}$, i.e.~onto a state such that $\hat{a}\ket{\phi_{\alpha}}=\alpha\sqrt{N}\ket{\phi_{\alpha}}$.
In the thermodynamic limit ($N \to \infty $), this mean-field procedure reduces to the following formal replacement $\hat{a}\to \alpha\sqrt{N}$. In this limit,
the total magnetization operator becomes
\begin{equation}\label{eq:SM_B0Mz0}
    \hat{M}_z(\alpha)= \sum^N_{k=1}\hat{M}_{z,k}(\alpha)~,
\end{equation}
where
\begin{equation}\label{eq:SM_B0Mz1}
    \hat{M}_{z,k}(\alpha)=\frac{\lambda}{B_0  \sqrt{N} } \sum^{n_{\rm s}-1}_{j=0 }
    \left( i t_{j,j+1} e^{-i2 \lambda \alpha} \hat{c}^\dagger_{j,k} \hat{c}_{j+1,k} + {\rm H.c.} \right)~.
\end{equation}
To map this operator into a more familiar object with physical dimensions of a magnetic moment, it is necessary to introduce an effective mass $m_{\rm eff}$ for our electrons hopping through the lattice sites: $m_{\rm eff} = 1/(2 |t| A_{\rm p})$, where $A_{\rm p}$ is the area of a plaquette and $t$ is the hopping parameter that we have introduced in the main text. For instance, for a square plaquette of side $d$ ($n_{\rm s}=4$), the effective mass is $m_{\rm eff}=1/(2|t| d^2)$. Introducing the effective mass into the definition of the magnetization operator we find

\begin{align}\label{eq:SM_B0Mz2}
\hat{M}_z(\alpha)=-\frac{e}{2 m_{\rm eff} c}\sum_{k, j}\left( i \frac{t_{j,j+1}}{n_s |t|} e^{-i 2 \lambda \alpha} \hat{c}_{j,k}^\dagger \hat{c}_{j+1,k} + {\rm H.c.} \right)~,
\end{align}
where the prefactor has the natural form of a magnetic moment $e/(2m_{\rm eff} c)$.

The total current operator $\hat{J}$ can be derived by starting from a discretized form of the continuity equation for the density operator. The local density $\hat{n}_{j,k}\equiv\hat{c}^\dagger_{j,k} \hat{c}_{j,k}$ obeys the Heisenberg equation of motion
\begin{equation}\label{eq:Heis}
\dot{\hat{n}}_{j,k}= i [\hat{\cal H},{\hat{n}}_{j,k}]~.
\end{equation}
By this comparing this equation to the following discretized continuity equation,
\begin{equation}\label{eq:Cont1}
\dot{\hat{n}}_{j,k}= -\frac{1}{d}\big(\hat{J}_{j,k}-\hat{J}_{j-1,k}\big)~,
\end{equation}
where $\hat{J}_{j,k}$ is the current flowing from site $j$ to site $j+1$ and $d$ is the lattice spacing, we obtain the following expression for the local current,
\begin{equation}\label{eq:Cont2}
{\hat{J}}_{j,k}= d\big(it_{j,j+1} e^{-i \lambda (\hat{a}+\hat{a}^\dagger)/\sqrt{N}}\hat{c}^\dagger_{j+1,k} \hat{c}_{j,k}-{\rm H.c}\big)~.
\end{equation}
The total current $\hat{J}$ is the sum of all local terms, i.e.~$\hat{J}=\sum_{k,j} {\hat{J}}_{j,k}$, and is given by
\begin{equation}\label{eq:Curr1}
{\hat{J}}= d\sum_{k,j}\big(it_{j,j+1}e^{-i \lambda (\hat{a}+\hat{a}^\dagger)/\sqrt{N}}\hat{c}^\dagger_{j+1,k} \hat{c}_{j,k}+{\rm H.c}\big)~.
\end{equation}
As usual, in the mean-field approximation, we replace $\hat{a}\to \alpha\sqrt{N}$ obtaining a current operator acting only on the matter degrees of freedom:
\begin{equation}\label{eq:Curr2}
{\hat{J}}(\alpha)= d \sum_{k,j}\big( it_{j,j+1}e^{-i2 \lambda\alpha }\hat{c}^\dagger_{j+1,k} \hat{c}_{j,k}+{\rm H.c}\big)~.
\end{equation}
As expected, the current operator  ${\hat{J}}(\alpha)$ and the magnetization operator $\hat{M}_z(\alpha)$ in Eq.~\eqref{eq:SM_B0Mz2} are directly proportional to each other. Specifically, $\hat{M}_z(\alpha)=-({e}/{c})  {\hat{J}}(\alpha) [{A_{\rm p}}/({n_s d})]$.

\section{\label{subsec: sm external field}The effect of an external classic field}

As we have seen in the main text, the orbital magnetic response changes sign when $\tau>0$, i.e.~when one of hopping parameters in the single-molecule Hamiltonian (\ref{eq:single_molecule_Hamiltonian_no_cavity}) has a different sign with respect to the others (we remind the reader that $t<0$). We now explain how one can achieve this frustrated condition on the hoppings by considering the action of an external classical magnetic field.
Starting from the mean-field Hamiltonian [cfr.~Eq.~\eqref{eq:mean-field Hamiltonian per plaquette} in the main text],
\begin{align}
\frac{\hat{\mathcal{H}}_{\rm MF}(\alpha)}{N} &=\frac{\mel{\phi_{\alpha}}{\mathcal{\hat{H}}}{\phi_{\alpha}}}{N} = \omega_{\rm c} \alpha^2 +\frac{1}{N} \sum^N_{k=1} \hat{h}_{{\rm e},k}(\alpha)~,\nonumber\\
\hat{h}_{{\rm e},k}(\alpha)&=
\sum_{j = 0}^{n_{\rm s} -1} \left( e^{-i 2  \lambda \alpha  } t_{j, j+1} \hat{c}^\dagger_{j,k} \hat{c}_{j+1,k}+ {\rm H.c.} \right)~, \nonumber
\end{align}
and considering the special case in which all the hopping parameters are equal to $t < 0$ but for $t_{0,1} = e^{-i \Theta} t$, we see that the matter Hamiltonian reduces to
\begin{eqnarray*}
\hat{h}_{{\rm e},k} (\alpha)&=& \sum_{j = 1}^{n_{\rm s} -1} \left( t e^{-i 2  \lambda \alpha  }  \hat{c}^\dagger_{j,k} \hat{c}_{j+1,k} + {\rm H.c.} \right)\\
&+& \left( te^{-i \Theta}  e^{-i 2  \lambda \alpha  }  \hat{c}^\dagger_{0,k} \hat{c}_{1,k} + {\rm H.c.} \right)~.
\end{eqnarray*}
By applying the unitary transformation $\hat{c}_{0,k}\to e^{-i \Theta}  \hat{c}_{0,k}$ and $ \hat{c}_{j,k}\to e^{-i \Theta (j / n_{\rm s})}  \hat{c}_{j,k}$ for $j\neq 0$,  we get
\begin{align}
\hat{h}_{{\rm e},k}(\alpha) &=  \sum_{j = 0}^{n_{\rm s} -1} \left( t e^{-i \Theta/ n_{\rm s}}  e^{-i 2  \lambda \alpha  }  \hat{c}^\dagger_{j,k} \hat{c}_{j+1,k}+ {\rm H.c.} \right)~.
\end{align}
This suggests that a classical magnetic field $\bm{B}_{\rm cl}=B_{\rm cl} \bm{e}_z$ with $B_{\rm cl} = - c \Theta / (e A_{\rm p})$ can be used to the change the sign of one of the hopping parameters, thereby paving the way for orbital paramagnetism in our toy model and the occurrence of photon condensation.

\section{\label{subsec: condensation criteria}Condensation criteria}
In this Section we derive a general criterion to study the onset of photon condensation. Here, we will consider the energy functional $E[\alpha,\psi]\equiv\langle\psi| \hat{\mathcal{H}}_{\rm MF}(\alpha)|\psi\rangle/N$ as a function of the photonic order parameter $\alpha$ and an arbitrary trial wave-function $|\psi\rangle$.
By analyzing the function ${\cal E}(\alpha)={\rm min}_{\psi}E[\alpha,\psi]=\langle\psi_{0}(\alpha)| \hat{\mathcal{H}}_{\rm MF}(\alpha)|\psi_{0}(\alpha)\rangle/N$, the appearance of a finite order parameter $\tilde{\alpha}$ corresponds to the instability of the normal ground state, ${\cal E}(\alpha)\leq{\cal E}(0)$. Since we are interested only in the onset of the phase transition, in this derivation we retain only terms up to second order in $\alpha$.

First of all, we remind the reader that  $\bar{\epsilon}_n(0)$ and $\ket{\psi_n(0)}$ are the eigenvalues and eigenvectors of the operator $\hat{\mathcal{H}}_0\equiv{\hat{\mathcal{H}}_{\rm e}}/{N}$ (see Eq.~\eqref{eq: Hamiltonian Mp and Md expansion} in the main text).

Now we expand the energy functional $E[\alpha,\psi]$ up to second order in $\alpha$. Hence, we can consistently approximate the diamagnetic magnetization with the one in absence of light, $\mel{\psi}{\hat{\mathcal{M}}_{\rm d} (0)}{\psi}\approx  \mel{\psi_0(0)}{\hat{\mathcal{M}}_{\rm d} (0)}{\psi_0(0)}= \mathcal{M}^{0,0}_{\rm d}(0)$. Up to the second order in $\alpha$ the energy functional reads
\begin{equation}
    \label{eq: Energy functional}
  E[\alpha,\psi] =\Omega\alpha^2 + \mel{\psi}{\hat{\mathcal{H}}_0}{\psi} + \mel{\psi}{\hat{\mathcal{M}}_{\rm p} (0)}{\psi} \alpha~,
\end{equation}
where $\Omega\equiv \omega_{\rm c} +  \mathcal{M}^{0,0}_{\rm d}(0) / 2  $. By minimizing with respect to $\psi$ we get ${\cal E}(\alpha)\equiv{\rm min}_{\psi}E[\alpha,\psi]$.
Hence, we need to calculate
\begin{equation}
    \label{eq: Energy functional 2}
 {\cal E}(\alpha) =\Omega\alpha^2+{\rm min}_{\psi}\{\mel{\psi}{(\hat{\mathcal{H}}_0 + \hat{\mathcal{M}}_{\rm p} (0) \alpha)}{\psi} \}~,
\end{equation}
Using second-order perturbation theory we obtain
\begin{eqnarray}
\label{eq:HAlt2}
&~&{\rm min}_{\psi}\{\mel{\psi}{(\hat{\mathcal{H}}_0 + \hat{\mathcal{M}}_{\rm p} (0) \alpha)}{\psi} \}=\\ &=& \bar{\epsilon}_0(0)+\frac{\chi_{\rm p}(0)}{2}\alpha^2+ { \mathcal{M}^{0,0}_{\rm p}(0)}\alpha~,\nonumber
\end{eqnarray}
where
\begin{eqnarray}
\label{eq: M_p0 definition}
{ \mathcal{M}^{0,0}_{\rm p}(0)}&\equiv& \mel{\psi_0}{\hat{\mathcal{M}}_{\rm p} (0)}{\psi_0}~,\\
\label{eq: chi_p0 definition}
{\chi_{\rm p}} (0) &\equiv& -2 \sum_{n\neq 0} \frac{\abs{\mel{\psi_n}{\hat{\mathcal{M}}_{\rm p} (0)}{\psi_0}}^2}{\bar{\epsilon}_n(0) - \bar{\epsilon}_0(0)} \leq 0~.
\end{eqnarray}

At this point there are two different cases to distinguish: {\it i)} the normal ground state pararamagnetic moment is zero,  ${ \mathcal{M}^{0,0}_{\rm p}(0)}=0$ or {\it ii)} the normal phase has a finite paramagnetic character ${ \mathcal{M}^{0,0}_{\rm p}(0)}\neq0$.
In the case {\it i)}, Eq.~\eqref{eq:HAlt2} reads
\begin{equation}
 \label{eq:HAlt3A}
 {\cal E}(\alpha)= \bar{\epsilon}_0(0)+\big(\Omega +\frac{\chi_{\rm p}(0)}{2}\big)\alpha^2~.
\end{equation}

It is possible to define the magnetic susceptibility $\chi_{\rm M}$ as the concavity of the molecular energy $\bar{\epsilon}_0(\alpha)$

\begin{equation}
 \label{eq:HAltA}
     \chi_{\rm M} \equiv -\frac{1}{2} \partial^2_{\alpha}  \bar{\epsilon}_0(\alpha) \Big|_{{\alpha}=0}~.
\end{equation}
In order to calculate $ \chi_{\rm M}$ it is useful to notice that $ \bar{\epsilon}_0(\alpha)= {\cal E}(\alpha)-\omega_{\rm c}\alpha^2~$.
By comparing Eq.~\eqref{eq:HAlt3A} with Eq.~\eqref{eq:HAltA}, we enstablish the a relation between the magnetic susceptibility $\chi_{\rm M}$ and the following microscopic quantities
\begin{eqnarray}
\label{eq: chi_M definition}
  - \chi_{\rm M}+\omega_{\rm c} &=&\big(\Omega +\frac{\chi_{\rm p}(0)}{2}\big)~,
\end{eqnarray}
or equivalently $- \chi_{\rm M}={ \mathcal{M}^{0,0}_{\rm d}(0)}/2+\chi_{\rm p}(0)/{2}$.
Hence, the criterion for the occurrence of photon condensation ${\cal E}(\alpha)\leq{\cal E}(0)$ is given by
\begin{equation}
\label{eq: chi_M criteria for superradiance to occur}
   \chi_{\rm M}\geq  \omega_{\rm c}~.
\end{equation}

Now, we focus on the case {\it ii)}, where ${ \mathcal{M}^{0,0}_{\rm p}(0)}\neq 0$.
Eq.\eqref{eq: Energy functional 2} can be expressed at the linear order in $\alpha$ as
\begin{equation}
    \label{eq:H9}
  {\cal E}(\alpha) =  \bar{\epsilon}_0(0)+ { \mathcal{M}^{0,0}_{\rm p}(0)}\alpha~.
\end{equation}
In this case, it is always possible to find a $\bar{\alpha}$ such that ${\cal E}(\bar{\alpha})\leq{\cal E}(0)$ by choosing $\bar{\alpha}$ to have a different sign with respect to
${ \mathcal{M}^{0,0}_{\rm p}(0)}$. In this case, the system is always unstable and displays photon condensation.
$\mathcal{M}^{0,0}_{\rm p}(0)$ can be different from zero if an additional classic field is introduced. Fig.~\ref{fig: 3d_fig}(b) reproduces both conditions \eqref{eq: chi_M criteria for superradiance to occur} and \eqref{eq:H9} in the case of an ensemble of squared rings, varying one hopping parameter and the external classic field.

\section{\label{subsec: sm Bosonization and polaritons study}Bosonization and polaritons}
In order to study the polaritonic properties of our coupled light-matter system, we start from Eq.~\eqref{eq: initial hamiltonian}. First of all, as stated in the main text, we shift the cavity photon operators, $\hat a =\alpha \sqrt{N}+\delta \hat a $, where $\delta \hat a$ describes zero-average fluctuations around the mean-field solution $\alpha\sqrt{N}$. Hence, in the thermodynamic limit, and expanding the shifted Hamiltonian up to second order in the fluctuations $\delta \hat a$, we find the following Hamiltonian:
\begin{eqnarray}
    \hat{\mathcal{H}} &=& \omega_{\rm c} \left[ \delta\hat a ^\dagger \delta\hat a + \alpha \sqrt{N} \left( \delta\hat a + \delta\hat a ^\dagger \right) + \alpha^2 N \right] \\
    &+& \sum_{k = 1}^N \hat{ h}_{{\rm e},k} (\alpha)
    + \frac{\sqrt{N}}{2 } \hat{\mathcal{M}}_{\rm p} (\alpha) \left(\delta \hat a +\delta \hat a ^\dagger \right) \nonumber\\ &+& \frac{1}{8 } \hat{\mathcal{M}}_{\rm d} (\alpha) \left(\delta \hat a + \delta\hat a ^\dagger \right)^2~. \nonumber
\end{eqnarray}

It is now useful to introduce fermionic operators, $\hat{\varphi}_{l, k}$ and $\hat{\varphi}_{l, k}^\dagger$, that  destroy and create an electron in the
energy eigenstate $\ket{\varphi_{l}(\alpha) }_k$ of the Hamiltonian $\hat{h}_{{\rm e},k}(\alpha)$, defined in Eq.~\eqref{eq:Hmatk} of the main-text, i.e.~$\hat{h}_{{\rm e},k}(\alpha) \ket{\varphi_{l}(\alpha) }_k = \epsilon_l (\alpha) \ket{\varphi_{l}(\alpha) }_k$.
Notice that $\hat{\varphi}_{l, k}$ depends on the photonic mean-field $\alpha$ and we dropped that explicit dependence for the ease of readability.
Here, the eigenvalues $\epsilon_l (\alpha)$ are $k$-independent, since all molecules are identical.
Expressing $\hat{\mathcal{H}}$ in terms of the operators $\hat{\varphi}_{l, k}$, we obtain
\begin{eqnarray}\label{eq:sm_Hdeltaa}
\hat{\mathcal{H}} &=& \omega_{\rm c} \left[\delta \hat a ^\dagger\delta \hat a + \alpha \sqrt{N} \left(\delta \hat a + \delta\hat a ^\dagger \right) + \alpha^2 N \right] \\
&+& \sum_{l = 0}^{n_{\rm s} - 1} \epsilon_l(\alpha) \sum_{k = 1}^N \hat{\varphi}_{l, k}^\dagger\hat{\varphi}_{l, k} \nonumber \\
&+& \frac{1}{2 \sqrt{N}} \left( \delta\hat a + \delta\hat a ^\dagger \right) \sum_{l, m = 0}^{n_{\rm s} - 1}\mathcal{M}_{\rm p}^{l, m} (\alpha) \sum_{k = 1}^{N} \hat{\varphi}_{l, k}^\dagger\hat{\varphi}_{m, k} \nonumber \\
&+& \frac{1}{8N } \left(\delta \hat a + \delta\hat a ^\dagger \right)^2 \sum_{l, m = 0}^{n_{\rm s} - 1}\mathcal{M}_{\rm d}^{l, m} (\alpha)  \sum_{k = 1}^{N} \hat{\varphi}_{l, k}^\dagger\hat{\varphi}_{m, k}~,\nonumber
\end{eqnarray}
where
\begin{align}
\mathcal{M}_{\rm p}^{l, m} (\alpha) &= \sum^N_{k=1} \tensor[_{k}]{\langle}{} {\varphi_{l}(\alpha)}|{\hat{\cal M}_{\rm p}(\alpha)}\ket{\varphi_{m}(\alpha)}_k\label{eq:sm_Mp}\nonumber\\
&=- \frac{2 B_0}{\sqrt{N}}  \sum^N_{k=1} \tensor[_{k}]{\langle}{}{\varphi_{l}(\alpha)}|{\hat{M}_{z,k}(\alpha)}\ket{\varphi_{m}(\alpha)}_k
\end{align}
and
\begin{align}
\mathcal{M}_{\rm d}^{l, m} (\alpha) &= \sum^N_{k=1} \tensor[_{k}]{\langle}{} {\varphi_{l}(\alpha)}|{\hat{\cal M}_{\rm d}(\alpha)}\ket{\varphi_{m}(\alpha)}_k\label{eq:sm_Md}\nonumber\\
&=- \frac{2 B_0}{\sqrt{N}}  \sum^N_{k=1} \tensor[_{k}]{\langle}{} {\varphi_{l}(\alpha)}|{\partial_\alpha \hat{M}_{z,k}(\alpha)}\ket{\varphi_{m}(\alpha)}_k~,
\end{align}
In writing Eq.~\eqref{eq:sm_Hdeltaa} we used that the matrix elements $\tensor[_{k}]{\langle}{}{\varphi_{l}(\alpha)}|{\hat{M}_{z,k}(\alpha)}\ket{\varphi_{m}(\alpha)}_k$ and $ \tensor[_{k}]{\langle}{}{\varphi_{l}(\alpha)}|{\partial_\alpha \hat{M}_{z,k}(\alpha)}\ket{\varphi_{m}(\alpha)}_k$ are independent of the $k$ index.
Making use of the collective notation
\begin{equation}\label{eq:Sigma_operators}
\hat{\Sigma}_{l, m} \equiv \sum_k \hat{\varphi}_{l, k}^\dagger\hat{\varphi}_{m, k}~,
\end{equation}
 the Hamiltonian takes a more compact form
\begin{eqnarray}
\label{eq: alpha-shifted second order hamiltonian in the energy basis (collective)}
    \hat{\mathcal{H}} &=& \omega_{\rm c} \left[ \delta \hat a ^\dagger \delta \hat a + \alpha \sqrt{N} \left( \delta\hat a + \delta \hat a ^\dagger \right) + \alpha^2 N \right]   \\
    &+&  \sum_{l = 0}^{n_{\rm s} - 1} \epsilon_l (\alpha) \hat{\Sigma}_{l, l}+\frac{1}{2 \sqrt{N}} \left( \hat a + \hat a ^\dagger \right) \sum_{l, m = 0}^{n_{\rm s} -1} \left[ \mathcal{M}_{\rm p}^{l, m} ( \alpha ) \hat{\Sigma}_{l, m} \right] \nonumber \\
    &+& \frac{1}{8N} \left( \hat a + \hat a ^\dagger \right)^2 \sum_{l, m = 0}^{n_{\rm s} -1} \left[ \mathcal{M}_{\rm d}^{l, m} ( \alpha ) \hat{\Sigma}_{l, m} \right]~.\nonumber
\end{eqnarray}

By following Ref.~\cite{kurucz_pra_2010}, we focus on the symmetric Hilbert subspace, which is spanned  by the occupation number states defined as
\begin{eqnarray}\label{eq:occstate}
&&\ket{n_0,m_1,\cdots,p_{n_{\rm s}-1}}=\frac{1}{\sqrt{n!m!\cdots p!}}
\sum_{\rm perm}
\ket{\varphi_0(\alpha)}_1\cdots  \nonumber\\
&\times&\ket{\varphi_0(\alpha)}_n \ket{\varphi_1(\alpha)}_{n+1}\cdots \ket{\varphi_1(\alpha)}_{n+m}\dots
\nonumber\\
&\times&\ket{\varphi_{n_{\rm s}-1}(\alpha)}_{N-p+1}\cdots \ket{\varphi_{n_{\rm s}-1}(\alpha)}_{N}~,
\end{eqnarray}
which means that $n$ molecules have a single electron that occupies the state $\ket{\varphi_0(\alpha)}$,
$m$ molecules have a single electron that occupies the state $\ket{\varphi_1(\alpha)}$, $\ldots$, and
$p$ molecules have a single electron that occupies the state   $\ket{\varphi_{n_{\rm s}-1}(\alpha)}$, such that $N=n+m+\ldots+p$.
By applying an occupation number state on collective operators defined in Eq.~\eqref{eq:Sigma_operators}, we find the following properties
\begin{eqnarray}
\hat{\Sigma}_{l,l'}\ket{n_l,m_{l'},\ldots}&=&\sqrt{(n+1)m} \nonumber\\
&\times&\ket{(n+1)_l,m_{l'}-1,\ldots}, \label{eq:Sigma_prop1} \\
\hat{\Sigma}_{l,l}\ket{n_l,m_{l'},\ldots}&=&n\ket{n_l,m_{l'},\ldots}, \label{eq:Sigma_prop2}
\end{eqnarray}
where $l \neq l'$. Moreover, the collective operators fulfill the commutator identity $[\hat{\Sigma}_{l,l'}, \hat{\Sigma}_{m,m'}]=\delta_{l',m}\hat{\Sigma}_{l,m'}-
\delta_{m',l}\hat{\Sigma}_{m,l'}$.
Here, we introduce $n_{\rm s}-1$ couple of bosonic creation and annihilation operators
 $\hat{b}_l^\dagger$ and $\hat{b}_l$ such that
\begin{eqnarray}
\hat{b}_l \ket{n_0,m_l,\cdots}&=&\sqrt{m}
 \ket{(n+1)_0,(m-1)_l,\cdots}~, \label{eq:b_prop1}\\
 \hat{b}_l^\dagger \hat{b}_l \ket{n_0,m_l,\cdots}&=&m
\ket{n_0,m_l,\cdots}~,  \label{eq:b_prop2}
\end{eqnarray}
where $l=1,\ldots,n_{\rm s}-1$, and the mean-field groundstate $\ket{\psi_0(\alpha)}=\ket{N_0,0,\cdots,0}=\prod^N_{k=1} \ket{\varphi_0(\alpha)}_k $,  that corresponds to all molecules with a single electron in the state $\ket{\varphi_0(\alpha)} $,  acts as the vacuum state for any $\hat{b}_l$.
Creation operators $\hat{b}^\dagger_l$ applied on the vacuum state $\ket{\psi_0(\alpha)}$ describe the collective matter excitations, which represents the bright modes.
By comparing Eqs.~\eqref{eq:Sigma_prop1}-\eqref{eq:Sigma_prop2} with Eqs.~\eqref{eq:b_prop1}-\eqref{eq:b_prop2}, we write the collective operators
 in terms of the bosonic fields accordingly to a multilevel Holstein-Primakoff transformation~\cite{kurucz_pra_2010},
\begin{eqnarray}
\hat{\Sigma}_{0,0}&=&N-\sum_{l>0} \hat{b}_l^\dagger\hat{b}_l~,\\
\hat{\Sigma}_{l,0}&=&\hat{b}_l^\dagger \sqrt{N-\sum_{l^\prime>0} \hat{b}_{l^\prime}^\dagger\hat{b}_{l^\prime} }~,\\
\hat{\Sigma}_{l,l^\prime}&=&\hat{b}_l^\dagger\hat{b}_{l^\prime}~,
\end{eqnarray}
where $l, l^\prime>0$.
In the proximity of the mean-field matter ground state $\ket{\psi_0(\alpha)}$, namely for a small number of collective matter excitations, one can approximate
\begin{eqnarray}
\hat{\Sigma}_{l,0}&\approx &\sqrt{N}\hat{b}_l^\dagger~.
\end{eqnarray}
By using the matter collective bosonic fields  $\hat{b}_l^\dagger$ and $\hat{b}_l$ to rewrite the Hamiltonian in Eq.~\eqref{eq: alpha-shifted second order hamiltonian in the energy basis (collective)}, and we get rid off all terms beyond the second order in the (light and matter) bosonic fields
\begin{eqnarray}\label{eq:Hquadratic}
    \hat{\mathcal{H}} &\simeq& \omega_{\rm c} \delta\hat a ^\dagger \delta\hat a + \omega_{\rm c} \sqrt{N} \left[ \alpha + \frac{\mathcal{M}_{\rm p}^{0,0} ( \alpha )}{2 \omega_{\rm c}} \right] \left(\delta \hat a + \delta\hat a ^\dagger \right)  \nonumber \\
    &+&N \omega_{\rm c} \alpha^2 + N \epsilon_0 (\alpha) + \sum_{l = 1}^{n_{\rm s} - 1} \left[ \epsilon_l (\alpha) - \epsilon_0 (\alpha) \right] \hat{b}_l^\dagger \hat{b}_l \nonumber \\
    &+& \frac{1}{2} \left( \delta \hat a + \delta \hat a ^\dagger \right) \sum_{l = 1}^{n_{\rm s} - 1} \left[ \mathcal{M}_{\rm p}^{l,0} ( \alpha ) \hat{b}_l^\dagger + \mathcal{M}_{\rm p}^{0,l} ( \alpha ) \hat{b}_l \right] \nonumber\\
    &+& \frac{1}{8} \left( \delta  \hat a + \delta  \hat a ^\dagger \right)^2 \mathcal{M}_{\rm d}^{0, 0} ( \alpha )~,
\end{eqnarray}

This Hamiltonian still depends upon the mean-field displacement $\alpha$.
 In order to fix the value of $\alpha$, we select the value which nullifies the derivative of the energy functional ${\cal E}(\alpha)=\bra{\psi_0(\alpha)} \hat{\cal H}_{\rm MF} (\alpha)\ket{\psi_0(\alpha)}/N$ with respect $\alpha$, i.e.
\begin{equation}
\frac{d {\cal E}(\alpha)}{d \alpha}=2 \omega_{\rm c} \alpha+
\frac{d \bar{\epsilon}_0(\alpha)}{d \alpha}~.
\end{equation}
By using the Hellmann–Feynman theorem to compute $({d  \bar{\epsilon}_0(\alpha)}/{d \alpha})$,
\begin{eqnarray}
\frac{d\bar{\epsilon}_0(\alpha)}{d \alpha}&=&\frac{1}{N} \sum^N_{k=1} \bra{\psi_0(\alpha)} \frac{d \hat{h}_{{\rm e},k}(\alpha)}{d \alpha}\ket{\psi_0(\alpha)}~, \\
&=&\bra{\psi_0(\alpha)} \hat{\cal M}_{\rm p}(\alpha) \ket{\psi_0(\alpha)}=\mathcal{M}_{\rm p}^{0, 0} (\alpha),\nonumber
\end{eqnarray}
we can write a non-linear equation to determine $\tilde{\alpha}$:
\begin{equation}\label{eq: sm solutions for alpha}
\frac{d {\cal E}(\alpha)}{d \alpha}\bigg|_{\alpha=\tilde{\alpha}}=2\omega_{\rm c} \tilde{\alpha}+\mathcal{M}_{\rm p}^{0, 0} (\tilde{ \alpha} )=0~.
\end{equation}
By imposing $\alpha=\tilde{\alpha}$, the linear terms in the bosonic operators in Eq.~\eqref{eq:Hquadratic} vanish. We clearly see that Eq.~(\ref{eq: sm solutions for alpha}) coincides with Eq.~(\ref{eq: solutions for alpha}) in the main text.

Once we set $\alpha=\tilde{\alpha}$, the approximate Hamiltonian in Eq.~\eqref{eq:Hquadratic} becomes a {\it polaritonic} Hamiltonian $\hat{\cal H}_{\rm pol}$ the low-energy hybrid excitations (polaritons) on top of the mean-field ground-state solution,
\begin{eqnarray}
\label{eq: sm quadratic bosonic Hamiltonian}
\hat{\cal H}_{\rm pol} &=& \omega_{\rm c} \delta \hat a ^\dagger \delta \hat a
+ \sum_{l=1}^{n_{\rm s} - 1} [\epsilon_{l}(\tilde{\alpha}) -\epsilon_{0}(\tilde{\alpha})]\hat{b}_l^\dagger \hat{b}_l \nonumber\\
&+& \frac{1}{2} (\delta \hat a + \delta \hat a ^\dagger) \sum_{l=1}^{n_{\rm s} - 1} \left[ \mathcal{M}_{\rm p}^{l,0} ( \tilde{\alpha} ) \hat{b}_l^\dagger + \mathcal{M}_{\rm p}^{0,l} (\tilde{\alpha}) \hat{b}_l \right] \nonumber \\
&+& \frac{1}{8} \mathcal{M}_{\rm d}^{0, 0} ( \tilde{\alpha} ) (\delta \hat a + \delta \hat a ^\dagger)^2~.
\end{eqnarray}
Polaritons are linear combinations of light and matter operators,
\begin{equation}
\label{eq:polariton}
\hat{p}_{\nu}=X_\nu \delta \hat{a} + Y_\nu \delta \hat{a}^\dagger +
\sum^{n_{\rm s}-1}_{l=1} \big(W_{\nu,l}   \hat{b}_l + Z_{\nu,l} \hat{b}^\dagger_l\big)~.
\end{equation}

Being the polariton a proper bosonic excitation of the system, the operator $\hat{p}_{\nu}$ fulfills the equation of motion of a harmonic ladder operator
\begin{equation}
\label{eq:polaritonEOM}
[\hat{\cal H}_{\rm pol},\hat{p}_{\nu}]= -\Omega_{{\rm p},\nu}  \hat{p}_{\nu}~.
\end{equation}
Being the polariton a combination of $\hat{a},\hat{a}^\dagger,\hat{b}_l,\hat{b}_l^\dagger $, in order to calculate Eq.~\eqref{eq:polaritonEOM} we need the equations of motion for these light and matter operators.
These are given by the following commutators
\begin{eqnarray}
[\hat{\cal H}_{\rm pol},\delta \hat a]&=&-\omega_{\rm c} \delta \hat a - \frac{1}{4} \mathcal{M}_{\rm d}^{0, 0} (\tilde{\alpha})(\delta\hat a+\delta\hat a ^\dagger) \nonumber \\
&-& \frac{1}{2} \sum^{n_{\rm s}-1}_{l=1} (\mathcal{M}_{\rm p}^{l,0}(\tilde{\alpha}) \hat{b}_l^\dagger + \mathcal{M}_{\rm p}^{0,l}(\tilde{\alpha}) \hat{b}_l)~,
\end{eqnarray}
\begin{eqnarray}
[\hat{\cal H}_{\rm pol},\delta \hat a ^\dagger]&=&\omega_{\rm c}  \delta \hat a ^\dagger + \frac{1}{4} \mathcal{M}_{\rm d}^{0, 0}(\tilde{\alpha})(\delta \hat a+\delta \hat a ^\dagger)\nonumber \\
&+& \frac{1}{2}
 \sum^{n_{\rm s}-1}_{l=1} (\mathcal{M}_{\rm p}^{l,0} (\tilde{\alpha}) \hat{b}_l^\dagger + \mathcal{M}_{\rm p}^{0,l}(\tilde{\alpha}) \hat{b}_l)~,
\end{eqnarray}
\begin{equation}
[\hat{\cal H}_{\rm pol},\hat{b}_l]=-[\epsilon_{l}(\tilde{\alpha}) -\epsilon_{0}(\tilde{\alpha})] \hat{b}_l -
  \frac{1}{2} \mathcal{M}_{\rm p}^{0,l} (\tilde{\alpha})  (\delta\hat a+\delta\hat a ^\dagger)~,
\end{equation}
\begin{equation}
[\hat{\cal H}_{\rm pol},\hat{b}_l^\dagger]=\-[\epsilon_{l}(\tilde{\alpha}) -\epsilon_{0}(\tilde{\alpha})] \hat{b}_l^\dagger +
 \frac{1}{2} \mathcal{M}_{\rm p}^{l,0}(\tilde{\alpha}) (\delta\hat a+\delta\hat a ^\dagger)~,
\end{equation}
By using these commutators and Eq.~\eqref{eq:polariton}, the equation of motion expressed in Eq.~\eqref{eq:polaritonEOM} can be mapped in the following eigenvalue problem
\begin{equation}
\label{eq:eigenvalueP}
\Xi {\bm v}_{\nu}=\Omega_{\rm p,\nu}   {\bm v}_{\nu}~,
\end{equation}
where ${\bm v}_\nu=(X_\nu,Y_\nu,{\bm W}_\nu,{\bm Z}_\nu)^{\rm T}$, and
\begin{equation}
\label{eq:sm_Hopfield_Matrix}
\Xi=
\begin{pmatrix}
\omega_{\rm c} + \frac{1}{4} \mathcal{M}_{\rm d}^{0,0}(\tilde{\alpha}) & - \frac{1}{4} \mathcal{M}_{\rm d}^{0,0}(\tilde{\alpha}) &   {\bm g}(\tilde{\alpha}) & - {\bm g}^*(\tilde{\alpha})\\
\frac{1}{4} \mathcal{M}_{\rm d}^{0,0}(\tilde{\alpha}) & -\omega_{\rm c} - \frac{1}{4} \mathcal{M}_{\rm d}^{0,0}(\tilde{\alpha})&    {\bm g}(\tilde{\alpha}) & -  {\bm g}^*(\tilde{\alpha})\\
  {{\bm g}^*}^{\rm T}(\tilde{\alpha}) &-{{\bm g}^*}^{\rm T}(\tilde{\alpha}) & {\bm \Omega}(\tilde{\alpha}) & 0\\
  {\bm g}^{\rm T}(\tilde{\alpha}) &-{\bm g}^{\rm T}(\tilde{\alpha}) & 0 & - {\bm \Omega}(\tilde{\alpha})\\
\end{pmatrix}~,
\end{equation}
where ${\bm g}(\tilde{\alpha})= ({1}/{2}) [\mathcal{M}_{\rm p}^{1,0}(\tilde{\alpha}), \dots ,\mathcal{M}_{\rm p}^{n_{\rm s}-1,0}(\tilde{\alpha})]$ and $ {\bm \Omega}(\tilde{\alpha}) = {\rm diag} [\epsilon_{1}(\tilde{\alpha}) -\epsilon_{0}(\tilde{\alpha}), \dots, \epsilon_{n_{\rm s}-1}(\tilde{\alpha}) -\epsilon_{0}(\tilde{\alpha})]$.

Eigenvalues of the Hopfield matrix $\Xi$ can be determined by the roots of the determinant ${\cal D}(\Omega_{{\rm p},\nu})$
\begin{equation}
\label{eq:det}
{\cal D}(\Omega_{{\rm p},\nu})={\rm Det}(\Omega_{{\rm p},\nu} \openone-\Xi)=0~.
\end{equation}
The determinant ${\cal D}(\Omega_{{\rm p},\nu})$ can be calculated by using the following algebraic property of the block matrices
\begin{equation*}
{\rm Det}
\begin{pmatrix}
A&B\\
C&D
\end{pmatrix}
={\rm Det}(D){\rm Det}(A-BD^{-1}C)~,
\end{equation*}
which leads to
\begin{widetext}
\begin{eqnarray}
{\cal D}(\Omega_{{\rm p},\nu})&=&
\prod^{n_{\rm s}-1}_{l=1}\{\Omega_{{\rm p},\nu}^2 - [\epsilon_{l}(\tilde{\alpha}) -\epsilon_{0}(\tilde{\alpha})]^2\}\\
&\times&{\rm Det}
\begin{pmatrix}
\omega_{\rm c} + [\frac{1}{4} \mathcal{M}_{\rm d}^{0, 0} (\tilde{\alpha})+ \frac{1}{4} \chi^{(\tilde{\alpha})}_{\rm p}(\Omega_{{\rm p},\nu})]
-\Omega_{{\rm p},\nu}&- [\frac{1}{4} \mathcal{M}_{\rm d}^{0, 0}(\tilde{\alpha}) + \frac{1}{4} \chi^{(\tilde{\alpha})}_{\rm p}(\Omega_{{\rm p},\nu})]  \\
 [\frac{1}{4} \mathcal{M}_{\rm d}^{0, 0}(\tilde{\alpha}) + \frac{1}{4} \chi^{(\tilde{\alpha})}_{\rm p}(\Omega_{{\rm p},\nu})]& - \omega_{\rm c} -  [\frac{1}{4} \mathcal{M}_{\rm d}^{0, 0} (\tilde{\alpha})+ \frac{1}{4} \chi^{(\tilde{\alpha})}_{\rm p}(\Omega_{{\rm p},\nu})]-\Omega_{{\rm p},\nu}
\end{pmatrix}~, \nonumber
\end{eqnarray}
\end{widetext}
where
\begin{eqnarray}
\label{eq: sm chi di omega}
\chi^{(\tilde{\alpha})}_{\rm p}(\omega)&=&
2 \sum^{n_{\rm s}-1}_{l=1}
\frac{\abs{\mathcal{M}_{\rm p}^{l,0}(\tilde{\alpha})}^2 [\epsilon_l(\tilde{\alpha})-\epsilon_0(\tilde{\alpha})]}{\omega^2-[\epsilon_l(\tilde{\alpha})-\epsilon_0(\tilde{\alpha})]^2}~.
\end{eqnarray}
The static limit  of $\chi^{(\tilde{\alpha})}_{\rm p}(\omega)$ generalizes the static paramagnetic susceptibility
for a finite value of $\tilde{\alpha}$, such that $\chi^{(0)}_{\rm p}(0)=\chi_{\rm p}(0)$, where $\chi_{\rm p}(0)$ has been defined in the main text as
\begin{equation}
\label{eq:chipS}
\chi_{\rm p}(0)= -2 \sum_{n \neq 0}\frac{ \abs{\mel{\psi_n(0)}{\hat{\mathcal{M}}_{\rm p} (0)}{\psi_0(0)}}^2}{\bar{\epsilon}_n(0) - \bar{\epsilon}_0(0)}~.
\end{equation}
To verify this generalization, we note that the matrix element $\mel{\psi_n(0)}{\hat{\mathcal{M}}_{\rm p} (0)}{\psi_0(0)}$ connects the mean-field many-body ground state $\ket{\psi_0(0)}$ with a mean-field many-body state $\ket{\psi_n(0)}$ with only an excited molecule, namely the $N$-tuple $\{l_{1,n},\ldots,l_{k,n},\ldots,l_{N,n}\}$ associated with the many-body excited state $\ket{\psi_n(0)}$ consists of $N-1$ zeroes and a single non-zero element $l_{k^\star,n}=l^\star_n>0$.

Thus, the energy difference between the many-body excited state and the many-body ground state can be expressed in terms of the single-particle energy difference between excited states and the ground state as $\bar{\epsilon}_n(0)-\bar{\epsilon}_0(0)=[\epsilon_{l^\star_n}(0)-\epsilon_0(0)]/N$.
Moreover, since we are dealing with $N$ identical molecules, there are $N$ different excited states (with a single excitation) associated with the gap $\bar{\epsilon}_n(0)-\bar{\epsilon}_0(0)$.
Thanks to these remarks and by means of Eq.~(\ref{eq:sm_Mp}), the static paramagnetic susceptibility (in Eq.~\eqref{eq:chipS}) can be expressed as
\begin{eqnarray}
\label{eq:chi1s}
\chi_{\rm p}(0)=-2 \sum^{n_{\rm s}-1}_{l=1}\frac{\abs{\mathcal{M}_{\rm p}^{l,0}(0)}^2}{[\epsilon_l({0})-\epsilon_0(0)]}~.
\end{eqnarray}
Comparing this expression with Eq.~\eqref{eq: sm chi di omega} evaluated at $\omega=0$, we conclude that
\begin{eqnarray}
\label{eq:chi=chi}
\chi_{\rm p}^{(0)}(0)=\chi_{\rm p}(0)~.
\end{eqnarray}

With few algebraic manipulations, the determinant can be expressed as
\begin{eqnarray}
{\cal D}(\Omega_{{\rm p},\nu}) &=&
\prod^{n_{\rm s}-1}_{l=1}\{\Omega_{{\rm p},\nu}^2 - [\epsilon_{l}(\tilde{\alpha}) -\epsilon_{0}(\tilde{\alpha})]^2\}\bigg\{\Omega_{{\rm p},\nu}^2 + \nonumber\\ &-& \omega_{\rm c}\left[ \omega_{\rm c}
+  \left( \frac{1}{2} \mathcal{M}_{\rm d}^{0, 0} (\tilde{\alpha})+ \frac{1}{2} \chi^{(\tilde{\alpha})}_{\rm p}(\Omega_{{\rm p},\nu}) \right) \right]\bigg\}~.\nonumber\\
\end{eqnarray}
Hence, polaritonic eigenenergies are given by the following non-linear equation
\begin{equation}\label{eq:eiP_pol}
\Omega_{{\rm p},\nu}^2-\omega_{\rm c}\left\{\omega_{\rm c} + \left[ \frac{1}{2} \mathcal{M}_{\rm d}^{0, 0}(\tilde{\alpha}) + \frac{1}{2} \chi^{(\tilde{\alpha})}_{\rm p}(\Omega_{{\rm p},\nu}) \right]\right\}=0~,
\end{equation}
where we remind that $\chi^{(\tilde{\alpha})}_{\rm p}(\Omega_{{\rm p},\nu})$ is a function of $\tilde{\alpha}$.
The onset of the superradiant phase transition corresponds to a softening of a polariton, i.e. a polariton with zero energy.
Eq.~\eqref{eq:eiP_pol} for $\Omega_{{\rm p},\nu}=0$ reduces to
\begin{equation}\label{eq:pol-instability}
\omega_{\rm c} +  \frac{1}{2} \mathcal{M}_{\rm d}^{0, 0}(\tilde{\alpha}) + \frac{1}{2} \chi^{(\tilde{\alpha})}_{\rm p}(0)=0~.
\end{equation}
 We note that for $\tilde{\alpha}=0$, employing Eq.~\eqref{eq:chi=chi}, the previous equation coincides with the instability criterion shown in Eq.~\eqref{eq: chi_M criteria for superradiance to occur}.

\section{\label{subsec: sm Van Vleck}Microscopic theory of Van Vleck paramagnetism}

In this Section we briefly remind the reader about the microscopic theory of Van Vleck paramagnetism.

We consider a generic many-electron Hamiltonian of the form
\begin{equation}
\hat{\cal{H}}_0=\sum_{k}\Big[\frac{\hat{\bm{p}}^2_k}{2m}+V(\hat{\bm{r}}_k)\Big] +\frac{1}{2}\sum_{k\neq k^\prime} v(\hat{\bm{r}}_k-\hat{\bm{r}}_k^\prime)~,
\end{equation}
where $m$ is the electronic mass, $\hat{\bm{p}}_k$ is the momentum of the $k-$ electron, $V(r)$ is a external potential and $v(\hat{\bm{r}}_k-\hat{\bm{r}}_k^\prime)$ is the electron-electron interaction.
The coupling to a uniform magnetic field ${\bm B}$ is made as customary via the minimal coupling substitution, i.e.~$\hat{\bm{p}}_k\to \hat{\bm{p}}_k+(e/c)\bm{A}(\bm{r}_k)$.

In the symmetric gauge, the vector potential can be expressed as follows
\begin{equation}\label{eq:symmetric gauge static B field}
\bm{A}(\bm{r})=\frac{1}{2}B\hat{\bm{e}}_z\times \bm{r},
\end{equation}
where $\bm{B}=B {\bm e}_z$ is the magnetic field. Notice that the symmetric-gauge vector potential obeys also the Coulomb condition, i.e. $\bm{\nabla}\cdot\bm{A}(\bm{r})=0$.
Carrying out the minimal coupling in the symmetric gauge, we find the following Hamiltonian of the many-electron systems coupled to a uniform external magnetic field $\bm{B}$:
\begin{equation}
\hat{\cal H}=\hat{\cal H}_0+\frac{e}{2mc} \bm{B}\cdot \hat{\bm{L}} +\frac{e^2}{8mc^2}B^2\sum_{k=1}^N |\hat{\bm{r}}_k|^2~,
\end{equation}
where $\hat{\bm{L}} =\sum_{k=1}^N(\hat{\bm{r}}_k\times \hat{\bm{p}}_k)$ is the (paramagnetic) angular momentum operator and the last term is the magnetostatic energy. Note that, accordingly to the main text, we have neglected the Zeeman coupling.

The magnetization $\hat{\bm{M}}$ is related to the angular momentum and magnetic field by the following relation:
\begin{align}
\hat{\bm{M}}&\equiv -\frac{ei}{2c}   \sum_{k=1}^N\hat{\bm r}_k  \times  [\mathcal{H}, \hat{\bm r}_k ]\nonumber\\
&= -\frac{e}{2mc}\hat{\bm{L}}- \frac{e^2}{2mc^2} \sum_{k=1}^N  \hat{\bm{r}}_k\times \bm{A}(\bm{r}_k)\nonumber\\
&=- \frac{e}{2mc}\hat{\bm{L}}- \frac{e^2B}{4mc^2}\Big( \sum_{k=1}^N  |\hat{\bm{r}}_k|^2\Big) \hat{\bm{z}}\nonumber\\
&\equiv \hat{\bm{M}}_{\rm p}+\hat{\bm{M}}_{\rm d}~,
\end{align}
where the first (second) term is the paramagnetic (diamagnetic) contribution.

In order to find the orbital magnetic susceptibility, we need to study the energy variation under the applied magnetic field. This splits in two terms:
$\Delta E=\Delta E_{\rm P}
+\Delta E_{\rm D}$. Introducing the exact eigenstates and eigenvalues of the Hamiltonian $\hat{\cal H}_0$, we find:
\begin{align}
\Delta E_{\rm D}&\equiv\frac{e^2}{8mc^2}B^2\langle \psi_0|\sum_{k=1}^N| \hat{\bm{r}}_k|^2|\psi_0\rangle\nonumber\\
&=-\frac{1}{2} \langle\psi_0|\hat{{M}}_{{\rm d},z}|\psi_0\rangle B^2~,
\end{align}
where $\hat{{M}}_{{\rm d},z}$ is the $\bm{e}_z$ component of the vector $\hat{\bm{M}}_{\rm d}$ and
\begin{align}
\Delta E_{\rm P}&\equiv\Big(\frac{e}{2mc}\Big)^2\sum_{n\neq 0}\frac{|\langle \psi_n|  \bm{B}\cdot \hat{\bm{L}} |\psi_0\rangle|^2}{E_n-E_0}\nonumber\\
&= \sum_{n\neq 0}\frac{|\langle \psi_n|  \bm{B}\cdot \hat{\bm{M}}_{\rm p} |\psi_0\rangle|^2}{E_n-E_0}~.
\end{align}
$\Delta E_{\rm B}$ is the magnetic contribution, while
by construction $\Delta  E_{\rm D}>0$ and $\Delta  E_{\rm P}<0$  i.e. they are a diamagnetic and a paramagnetic contribution, respectively. If the first excited state $n=1$ is nearly degenerate with the ground state $E_{1}-E_{0}\approx 0$, the paramagnetic contribution is dominant and the system has a paramagnetic response (known as Van Vleck paramagnetism).

The paramagnetic contribution can be recast as $\Delta E_{\rm P}=-(1/2){\chi_{\rm p}}B^2$ where $\chi_{\rm p}$ is the magnetization response function,
\begin{eqnarray}
\chi_{{\rm p}}&\equiv&-2\sum_{n\neq 0}\frac{|\langle \psi_n|  \hat{\bm{z}}\cdot \hat{\bm{M}}_{\rm p} |\psi_0\rangle|^2}{E_n-E_0},\\
&=&-2\Big(\frac{e}{2mc}\Big)^2\sum_{n\neq 0}\frac{|\langle \psi_n|  \hat{\bm{z}}\cdot \hat{\bm{L}} |\psi_0\rangle|^2}{E_n-E_0},\label{eq:ChiLL}
\end{eqnarray}

Thus, orbital paramagnetism is governed by the angular momentum-angular momentum response function. From the expression for the response function in Eq. \eqref{eq:ChiLL}, it can be seen that non-zero orbital paramagnetic response occurs only if the system does not have rotational invariance around the $\hat{\bm{z}}$ axis. Indeed, if $[\hat{\cal H}_0, \hat{\bm{z}}\cdot \hat{\bm L}]=0$, one can choose a common eigenstate basis of the energy and
 the angular momentum projection $\hat{\bm{z}}\cdot {\bm L}$ and the matrix element $\bra{\psi_0} \hat{\bm{z}}\cdot \hat{\bm L}\ket{\psi_n}$ vanishes, precluding the presence of Van Vleck paramagnetism in rotationally invariant systems, such as closed shell atoms.

\end{document}

redpajama_set_name: RedPajamaArXiv