It's been awhile since I've reviewed a jarred salsa. The Texas Tamale Company and Brazos Legends sent us this jar of Casa Salsa for review, and while I'm not sure who or which company is making this stuff, I enjoyed eating it and writing this review.
Texture & Appearance: – Nice medium chunky salsa with a good bit of sauce to keep things on the thick side.
Taste Straight Up: – Full mouth feel of spices, heat and flavor. There is a big hit of a "Texas Chili" flavor thanks to the cayenne pepper and cilantro. I was surprised to not see chili powder in the list.
Heat Level: – For having only jalapenos in the mix for heat, this salsa, labeled "Hot" brings the heat. Not too hot, just about right. Great for chips and dip while watching football.
Label/Graphics: – Simple graphics and label. Called "Casa Salsa", which I assume means "House Salsa", there is a Texas Tamale Company – Est. 1985 logo, with "The Mexican Deli" under it. I'm afraid Texas Tamale Company / Brazos Legends is suffering from an identity crisis. Who makes this sauce? What is the brand name? I'm confused. Going to either website will ultimately take you to the same order page.
Overall Rating (Taste on Food, Final Thoughts): – Because of the names and websites, I can't honestly ask any Chilehead "Do you like the Texas Tamale Company's Casa Salsa" OR "Do you like the Brazos Legends' Casa Salsa"? Which is it folks?
All in all, it is one of the best jarred salsas I've tried in awhile. Great taste, nice heat, good flavor and texture. A can't miss winner in my opinion. I just wish they'd pick a name and website and stick with it. And perhaps re-do the label to be a little more spicy!
It's been a while since I've had a jarred salsa, we've been making our own fresh all summer but those fresh veggies are going to be not so available soon. Might have to pick up a jar or two.
It's worth a try. So many grocery store jarred salsas are crap.
We also have a handful of other salsas lined up for review. I might use your idea to post them after the last of the fresh veggies have been harvested and eaten.
Hello, Gringold salsa is a boot strap company ran by two guys in San Antonio Texas. We are just getting started and our website will be up in a month or so. But we have three salsas we would like you to try out and give us your opinion. We have the basic Gringold salsa verde, then our rojo, and finally our east meets west. Please let me know if your interested, we would love to send you some.


text:
About 1,000 brooding octopuses found off California coast
Life, More Features
This kind of aggregation of deep-sea octopuses has been reported only once before.
Representational Image. (Photo: Pixabay)
MONTEREY - Scientists exploring the sea floor off the coast of central California for deep-water coral and sponges instead found an unprecedented sight: Hundreds of octopuses tucked between rocks with their tentacles inverted and covering groups of white eggs - a posture that is common among brooding females.
The cluster of more than 1,000 gray octopuses latching on to clean, dark rock was found last week in the Davidson Seamount, an underwater extinct volcano in the Monterey Bay National Marine Sanctuary, Chad King, a marine biologist at the sanctuary, said Tuesday.
A submersible's camera found the creatures on Oct. 23 nearly 2 miles (3 kilometers) below the surface during the first dive by the Nautilus, a vessel exploring the sanctuary and livestreaming its findings.
"This is certainly the largest cluster of brooding deep-sea octopuses that has ever been spotted," added King, who is also the chief scientist aboard the Nautilus, a research vessel with a team from the National Oceanic and Atmospheric Administration and the nonprofit Ocean Exploration Trust.
It was the first time scientists have seen a cluster of Muusoctopus robustus octopuses off California's coast, King said.
This kind of aggregation of deep-sea octopuses has been reported only once before, when earlier this year a smaller group of about 100 were spotted brooding eggs in warm ocean water off the Central American nation of Costa Rica, the nonprofit said.
The octopuses lined up in the cracks of the clean rocks near shimmering fluid seeps that looked "kind of like a heat wave off the pavement on a hot day," which suggests warm water, King said. But he added the submersible was not equipped to test the water for temperature or chemicals.
It's not clear why the creatures were gathered near fluid seeps that are associated with shrimp, anemones, snails and other animals.
"It could be because the fluid seeping out of the rock is warm or maybe they like the clean, dark rock with no sediment," King said. "There are so many questions that we have no answers for."
monterey bay national marine sanctuary
gray octopuses
davidson seamount
Octopus on ecstasy behaves the same as human, reveals secret genetic link
Study finds despite anatomical differences between octopus and human brain, there are molecular similarities in serotonin transporter gene.
Octopus skin inspires programmable camouflaging material
The pneumatically-activated material developed by researchers takes a cue from the 3D bumps, or papillae for dynamic camouflage.
Here's why zebrafish always have stripes
The process is extremely complex.

redpajama_set_name: RedPajamaCommonCrawl


text:
<?php
declare(strict_types=1);

namespace stubbles\webapp\routing\api;
/**
 * Represents a header that a resource may have.
 *
 * @since  6.1.0
 * @XmlTag(tagName='header')
 */
class Header implements \JsonSerializable
{
    /**
     * @var  string
     */
    private $name;
    /**
     * @var  string
     */
    private $description;

    /**
     * constructor
     *
     * @param  string  $name         header name
     * @param  string  $description  description of header
     */
    public function __construct(string $name, string $description)
    {
        $this->name        = $name;
        $this->description = $description;
    }

    /**
     * returns header name
     *
     * @return  string
     * @XmlAttribute(attributeName='name')
     */
    public function name(): string
    {
        return $this->name;
    }

    /**
     * returns description of header
     *
     * @return  string
     * @XmlFragment(tagName='description')
     */
    public function description(): string
    {
        return $this->description;
    }

    /**
     * returns representation suitable for encoding in JSON
     *
     * @return  array<string,string>
     * @XmlIgnore
     */
    public function jsonSerialize(): array
    {
        return ['name' => $this->name, 'description' => $this->description];
    }

}

redpajama_set_name: RedPajamaGithub


text:
Q: A ListPlot with multiple lines, fill to the axis, no thickness of the lines, and all fill in the same color I have created the ListPlot shown below (from data in a multidimensional list), but it is not exactly what I aimed for. I aimed for everything shown, except I do not want any lines to border the shading.
How do I get rid of the lines?
My current code is:
ListPlot[{Part[mat, 1, 1, 6], Part[mat, 1, 2, 6], Part[mat, 1, 3, 6], Part[mat, 1, 4, 6], Part[mat, 1, 5, 6]}, Joined -> True, Filling -> Axis, PlotStyle -> {{blue , Thickness[0]}, {blue, Thickness[0]}, {blue, Thickness[0]}, {blue, Thickness[0]}, {blue, Thickness[0]}}]

A: SeedRandom[1]
data = Accumulate @ RandomReal[10, {5, 10}];

You can use the (undocumented) sub-option "LineOpacity" -> 0 for PlotStyle:
ListLinePlot[data,
 Filling -> Axis,
 PlotStyle -> Directive[Blue, "LineOpacity" -> 0]]

ListLinePlot[data,
 Filling -> Prepend[1 -> Axis][# -> {#2} & @@@ Partition[Reverse@Range[5], 2, 1]],
 PlotStyle -> Directive["LineOpacity" -> 0],
 PlotLegends -> SwatchLegend[Automatic, Range[5]]]

Alternatively, a slight variation on Lukas's suggestion in the comments:
ListLinePlot[data,
  Filling -> Axis,
  PlotStyle -> None,
  FillingStyle -> Opacity[0.2, Blue]]

redpajama_set_name: RedPajamaStackExchange


8. No bidding, list your price clear.
free pdf delivery to your email box.
Enfamil $5 rebate for 2 22.2oz or larger tubs of Enfamil powder formula, $0.
checks ask for $5 for all.
any order needs plus shipping, except for all.
$5 共 9张。 Exp 3/31.
G家奶票3张:$3.5 过期4/1/12, $3.65 过期5/1/12, $3.65 过期6/1/12. 半价加.
on-the-go 8 fl oz bottle, exp: 3/31/12, $0.3 each.
May 31, 2012 过期 good on any Enfamil product.
转 enfamil $5 off coupon, has Kroger logo and may be only used in Kroger.
Total 11+ 0.5 shipping.go togethe.
don't need these checks anymore.
2 Enfagrow-Costco rebate forms, I'm asking for $12.5/each, plus postage.
of Enfagrow Unflavored Toddler Formula on the same receipt.
Available at select Costco Wholesale Club locations.
31/2012) and the UPC(s) from the carton(s).
* You will receive a check in the amount of either $1 or $25 via mail.
* Limit one rebate per household or address.
* Rebate valid of Costco Wholesale Club purchases only.
appear on the same receipt.
one Baozi each for one coupon. postage is also one Baozi.
2张5 off E家奶票(5/31过期),求换两张S家5 off coupon或奶票.
refill box and Enfamil D-vi-sol Vitamin Drops (exp.05/31/2012).
4. Save $7x1 on any two powder 22.2 OZ or larger (exp.06/30/2012).
Total 10+ 0.5 shipping.go together.
S家奶票$5x2,expires 6/16. S家$2off any infant formula一张,expires 5/31.
A total of 17 Similac $5 checks for sale at 40% off price, $3 each.
3 free coupons ($2, $1, $1) are included if taken together.
If sale partially, I pay for the 1st class postage.
4* enfamil check and $9.5 gerber check.
Similac $5x2. Expires on 3/24.
Similac $5x4. Expires on 5/23.
Similac $5x4. Expires on 7/25.
Gerber $2.5 x1. Expires on 6/1.
Ask for 20 baozi plus shipping.
choice of pediatricians. Limit one coupon per purchase.
我从EBAY买的,用了几张买它家的organic whole milk. Whole food加Omega3的卖4.
01/12 if you want it.
2 Enfagrow-Costco rebate forms, I'm asking for $10/each, plus postage.
送$2 off on any Enfagrow powder (manufacture's coupoun) *5=$10 by 6/30/2012.
or best offer. PM please. Thanks.


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Retiring To The Algarve: 5 Facts You Should Know
October 15, 2014 by Expatriate Healthcare
When it comes to overseas retirement destinations it's fair to say that Europe offers a smorgasbord of options. From the history and culture of Italy to the enviable sunshine and jaw dropping scenery of Spain there are no shortage of potential destinations to explore in your retirement.
However one often-overlooked destination is Portugal; most notably the Algarve region. Being smaller and less flashy than some of its neighbours, Portugal's subtle charms can sometimes not make the impact they deserve. However those willing to dig a little deeper will find a wealth of appeal.
Over the last few months we've seen The Algarve appear on our radar time and again, winning one award after another. As a result we thought it would be interesting to highlight five of The Algarve's greatest claims to fame. Hopefully these points will be enough to convince you that The Algarve is one of the best retirement destinations in Europe, perhaps even the world.
The Most Affordable Option in Europe for Retirees
Forbes magazine, reporting recently on the 2014 Retire Overseas Index, mentions The Algarve's reasonable cost of living in comparison to many other countries. It was described specifically as the 'most affordable option in Europe for retirees'. The report claims that one can live well in Portugal for $1,500 a month though some would argue that there are ways to enjoy this Portuguese paradise for far less.
No Tax on Pensions for New Residents
You might be surprised by just how common it is for retirement destinations to tax overseas sources of income; even when these are a retiree's pension. Portugal however has recently changed all this and now offers an enviable tax incentive scheme in order to draw retired expats from around the world.
The tax-free period is twice that of neighbour Spain, meaning that you'll naturally be far better off in Portugal than you would be in most other European countries.
World's Leading Golf Destination
The World Travel Awards are the so-called 'Oscars of the travel industry' and the results are fiercely competed as a result. Golf tourism is major business so anywhere that can claim its dominance in this area is sure to prosper.
Here Portugal was named the best golf destination in the world, largely due to an impressive combination of world-class golf resorts, combined with one of the best climates in the world, which allows for year-round golfing.
Europe's Leading Beach Destination
The same awards ceremony was also good enough to brand Portugal as Europe's leading beach resort for the 2nd year running. The cleanliness and safety of the beaches, combined with the wide diversity; from amenity-rich tourist beach destinations to wild, desolate beaches helps to ensure that there is a beach that will satisfy any visitor irrespective of taste.
Ranked Best Place in the World to Retire
Each year, the Retire Overseas Index aims to analyse the most popular retirement destinations around the world. Bearing in mind the subject of this article, it should come as no surprise that The Algarve was ranked this year's best place in the world to retire.
There were a number of reasons cited for this. Portugal successfully scored the maximum number of points for its climate, infrastructure and entertainment. The ease of integration, thanks to a large existing expat population was also praised, as was Portugal's astonishingly low crime rate.
Retirement Made Simple
Let's be honest; one of the things that holds many of back from actually retiring abroad is simply the complexity of the process. Trying to understand how to retire to a foreign country, what the rules are pertaining to property ownership, taxation and more can be difficult issues.
Now though, Algarve Senior Living are making it easier than ever before to retire to The Algarve. They offer top quality properties with views for less than the cost of living in the UK. Algarve Senior Living are even making the process of retiring to The Algarve as low-risk as possible thanks to the introduction of a new 'try-before-you-move' option.
No matter what your requirements, contact Algarve Senior Living to see how they can help to make your overseas retirement plans a reality.
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text:
Kalamita / Calamity (1982)
Despite making one of the most transgressive and unique films during the 1960s, Vera Chytilova's career never really gained much from the momentum of Daisies. Her 1966 film is rightfully beloved, as one would have a heard time erasing any of the images she crafts in her bizarre, kaleidoscopic and kinetic assault on the senses. Her political sensibility remains in Calamity, made 16 years later, but her method is a little different. On the surface, Calamity is overwhelmingly quaint and almost typical in how closely it matches the mold of an arty movie about a sad dude. But that's Chytilova's point: that the "rebellious" rhetoric of arthouse cinema is hypocritical and it has left us with the same old images. One where the plight of the young, white, heterosexual male is the center. By making a film so mundane and unremarkable, Chytilova challenges the impulse that makes us return to the well time and time again to mine pathos from a group who is least in need of representation.
Johnny takes the train home from college early. He's not exactly willing to announce the matter, but he's dropped out of school. He suggests that he' simply impatient with the college experience, but one gets the sensation that his lack of motivation lead to less than stellar performance. Whatever the case, he immediately tries to become an engine driver. While many question his ability, he seems to get through the training process and almost immediately becomes a driver. In the process, he finds himself unintentionally involved with three women, two of which he doesn't seem particularly interested in. What makes Johnny so goddamn exceptional? Nothing really, he's unmotivated and lacking in education, but he can't seem to cross the view of the opposite sex without something happening.
If there's something lacking in Chytilova's biting critique, it's that she hasn't made a film that is particularly, uh, pleasant. It has it's moments, in a technical sense, but because she is mocking a style of film that she (and I myself) believe to be dull and lifeless. As is the case, the trajectory of Johnny is somewhat, well, dull and lifeless. When the film isn't saving us from his presence with dialogue undermining his character type, there isn't much to appreciate. Sure, Chytilova loves to whip the camera all around a location, which is interesting, I guess. The stylistic flourishes are too fleeting, though, and seem like conscious attempts to make things seem exciting. It's a uphill battle, because the crux of her critique is that what is happening here is not remarkable, and the repetition of such images and narratives limits us as an audience.
The film ultimately work because it does manage to bite at times. During one potentially physical relationship, a woman, in reference to Johnny, declares that she "just loves sick idiots." Those the context of the film tells us that Johnny himself is physically ill (with a cold), out of that context its hard not to see Chytilova playfully jabbing the bellies of countless arthouse classics (Closely Watched Trains is referenced) in which women seem to gravitate towards otherwise miserable male character. Yet, she is also critical of the very fact that we're sympathetic to these men, because like Johnny, they are ultimately not that fantastic. The film ends with a train, driven by Johnny, getting stuck in an avalanche. It's not hard symbolism, but the suggestion is that like the passengers on the train, we are all stuck in the storytelling centered on the heterosexuality of white dudes. The fact that these stories are present even in a cinema that is meant to jar and clash with the conventions and molds of the mainstream is disappointing.
One can't help but get the impression that although she crafts a valid satire, that Chytilova would have liked to devoted her attention elsewhere. Following 1966, the government made it nearly impossible for her to make a film and the Soviet Union's invasion of Czechoslovakia in 1968 only worsened the situation. She was forced to pursue a safer, more easily digestible form of cinema. Simply stated, her expression was stunted by the political climate. One wonders how Chytilova's planned biopic of Božena Němcová would have turned out, but she was never given the opportunity to go ahead with that project. Even a film like Calamity was only shown in limited theaters, and the publications who were daring enough to write about it, were usually censored. Calamity as a film is remarkable because it gives us a woman, jumping through all the deliberately impossible obstacles for the sake of self-expression. The end result heavily critiques the same type of control that limited her creativity.
« Banka / Elegy of the North (1957) Underworld U.S.A. (1961) »
Date : May 14, 2014


text:
Caja de Cristal, Banco Ciudad
by Josie Watson | Oct 5, 2017 | Exhibitions Automatic City, Exhibitions WOA
Opening: 2 October 2017, 18 hs. Open until 31st of October.
Institute: Banco Ciudad, Florida 302, Buenos Aires.
Banco Ciudad, the public banck of the city of Buenos Aires is the official patreon of the exhibition "Ciudad Automática of Buenos Aires", the new icon of the city of Buenos Aires. Josie Watson presents this Art & Technology initiative with an innovative event that you will be able to experience from your cell phone in your city, live and in person.
WatsOnArt is an Argentine interactive art production company specially designed to generate intuitive, inclusive and educational artistic experiences suitable for all audiences. There will also be two interactive installations, and a tango performance by Nuria Schneller.
Created by young Argentine professionals specialized in plastic arts, technology, digital projections, programming, animation, design, theater, music and literature, WatsOnArt transmits in its products the freshness and novelty of cooperative work, a lifestyle that brings together the team members in the same vision of the future of culture, around the plastic artist Josie Watson, director of the project.
The App will be launched at the opening of the exhibition "Ciudad Automática de Buenos Aires" on October 2, 2017 in the city of Buenos Aires, at the branch of Banco Ciudad de microcentro, Florida 302. It is an interactive platform that will offer the viewer to interact with the paintings exhibited in the exhibition, with certain places in the city and with social networks.
WatsOnArt conceives technology as a generator of new forms of emotion that force art to renew itself. That is why it faces the challenge of promoting a new cultural identity that completes the content that other digital and traditional tourism and art platforms are not providing, and gives new generations the tools to generate their new cultural language.
Auction. Emerging artists of Buenos Aires. Record Sale
by Josie Watson | Oct 15, 2012 | Exhibitions Automatic City
Banco Ciudad continues its policy of supporting emerging artists
October 15, 2012 Mario Gilardoni 0 Comments
Written by Mario Gilardoni
With its "New Aires Art" label, the Bank continues to promote auctions with works by young artists who have not yet entered the commercial auction circuit. And to do so, it does not spare any of the attributes that surround its traditional operations: a first-class catalogue, advertising support and the entire sales structure at the service of this operation.
The public rewarded with the best offer a mixed technique of Josie Watson, who did not hesitate to put everything in the grill since "Avenida Corrientes – De la serie automática" obtained the first prize of the Salón Estímulo de Ramos Mejía. Made this year in a measure of 100×120 cm it was paid 8.625 pesos.
With an image that reveals the Caribbean spirit of his work, Ebenezer Leyva of Cuban origin, collected 6,400 pesos for "El vuelo del Icaro" of 136×90 cm. Third place on the price podium went to a sculpture by Federico Klarich made up of wooden assemblages that gave shape to the "Dog" of 80x110x55 cm which was paid 4,000 pesos.
The Bank's firmness in supporting this philosophy of encouraging young artists was reflected in the synthetic comment of Carlos Leiza -responsible for the sales area- "For next year we plan to hold three auctions similar to those we held in 2012".
Press Note: http://trastiendaplus.com/el-banco-ciudad-sigue-con-su-politica-de-apoyo-a-los-artista-emergentes/
"Avenida Corrientes", 100×120 cm, 2012.
First Emerging Art Auction of Buenos Aires – Banco Ciudad
by Josie Watson | Nov 10, 2011 | Uncategorized
For the first time, work by emerging artists was auctioned
It was organized at Banco Ciudad. Out of more than 70 lots, two-thirds were sold.
EMERGING ART. "Novel aesthetic proposals, without space in traditional galleries," defined the head of Art at Banco Ciudad.
Tagged as: Banco Ciudad cultura Profundo, creative, dynamic", harangue the auctioneer. Three adjectives to motivate a room that responds with laziness to the plastic works reproduced by plasma TVs and that are offered, one after the other, as "good future investments" of emerging artists that, it is assumed, will make tomorrow more than what is spent today.
The scene takes place in the Santa María de los Buenos Ayres Auditorium, at the Esmeralda branch of Banco Ciudad. For the first time, a public auction of works by emerging artists from the local plastic art scene was organized, and in this way they find their opportunity to make their way into the gallery circuit and enter the famous "market", that elusive space where silent battles for recognition are fought.
"What is emerging is not defined by the age of the artists, but by those novel aesthetic proposals, of plastics without space in the traditional galleries and that, in general, are still in full learning", defines Cristina Alonso, head of Art of the bank and one of those who made the selection of works to be finished.
The room is not crowded, but more than fifty people are watching the works and smiling with the efforts that the auctioneer Ricardo Vitaliti makes to catch buyers. At first, the works do not arouse much interest, but it is enough for one of the attendees to raise his hand in bidding, so that a small price escalation is triggered: the mechanism of the market operating in the open.
The selection of the works to be auctioned is the responsibility of the bank, but they rely on recommendations from galleries, private collectors and established artists. Two of them, Duillo Pierri and Maggie de Koenisgberg, accompany the students they recommended to the auction, who are now making their debut in the business of selling their creativity.
The aesthetic languages of the auction are multiple and range from the figurative to the geometric and expressionism. Most of the buyers are "young collectors with a mentality, with the capacity to discover what is not yet positioned in the market," says Alonso.
For the bank's press secretary, Jorge Velásquez, the experience was encouraging. "For a large auction, with more than 70 lots, two thirds have been sold, with values that doubled or tripled the base," he says. And he promises to repeat, extending the call to students from universities and art schools who want to make their debut in the market.
The link: http://www.revistaenie.clarin.com/arte/primera-subasto-obra-artistas-emergentes_0_496750533.html
"Un beso en Ciudad Automática", 120×120 cm, oil on canvas, 2010.


text:
Home > Machu Picchu Travel Information > The legend of Manco Capac and Mama Ocllo
The legend of Manco Capac and Mama Ocllo
This legend was told by one of the most important chronicles of the XVth century, Garcilaso de la Vega, El Inca, who was son of a Spanish captain and an Inca princess called Chimpu Ocllo. Because of that, Garcilaso was able to have many stories and legends from a very faithful source about the Inca beginnings and origins.
Near to the Titicaca Lake, the people lived like savages, since they don't have any kind of religious guides that bring them together. They also don't have any kind of agricultural techniques and textile abilities, and that is the reason why they were naked, living on the caves, eating wild animals and eating the fruits that are in their region.
The Sun god was concern about these people, and decides to bring them out to his sons, Manco Capac and Mama Ocllo, who were also husband and wife, to civilize these people. Manco Capac teach to the men how live in society and worship the Sun god, and Mama Ocllo teach to the women tasks like textiles and domestics labors.
Their father, the Sun god told them that search for a fertile and good land to found the empire, this land were the region of the Titicaca Lake, and gave to Manco Capac a golden stick, the one that if it sinks that's the sign of the promise land.
When they came to the Lake, the people who lived around thought that they were gods, because of his clothing and jewelry. Days go by and they can find a good place to found their city, but one day they came to a place where the golden stick sinks, and decide to found their empire.
The interpretation that follows is that on the shore of the Titicaca Lake is where the Tiawanako people lived, and they helped to the foundation of the Tahuantinsuyo, since both persons, Manco Capac and Mama Ocllo came from these lands, and because of that they were the original descendants of the Tiawanako people.
culturehistorymachu picchutravel information
Find out more Machu Picchu History
Machu Picchu History
Pachacutec, the Inca Emperor
Religious significance in Machu Picchu
The Ayar brother's legend


Economic Development Journal is the premier publication of our profession, featuring in-depth articles on important programs, projects, and trends in the United States and around the world.
Published quarterly, the Journal is free to IEDC members. Members can browse previous issues below, listed chronologically with the most recent first. Entire issues or individual articles can be downloaded as PDFs.
• Cultural Economies: What Are They and How Do We Build Them?
• Economic Development: A Misunderstood Profession and Practice?
• "What Are They Going to Call It?"
• Economic Development Financing: Developing the "Last Frontier"
• Are Economic Developers Asking Universities to Participate in the Development of Their State's Economies?
• Amenity-driven Economic Growth: Is It Worth Betting On?
• Coming Full Circle: The End of the Small Business Era?
• Do Structured International Trade Missions Improve Corporate Performance?
• BRAC-onomic Development: Pursuing the "Art-of-the-Possible"
• Does Rural Land-Use Planning and Zoning Enhance Local Economic Development?
• Small Business Vitality & Economic Development: Fact or Fiction?
• The Frontier This Time: Can the Plant and Life Sciences Move St. Louis Back to the Leading Edge?
• Can We Maintain the American Dream?: How Many Jobs Can the U.S. Afford to Lose?


text:
\section{Introduction}
In 90 years after the seminal work by Schr\"{o}dinger \cite{Sch26} analytical properties of quantum-mechanical wavefunctions describing hydrogenic-like atoms (or ions) continue to surprise. For instance, recent derivation of the Wallis formula for $\pi$ by Friedmann and Hagen \cite{Fri15} refers to the variational principle and the correspondence principle, but also employs a~counterintuitive fact that some variational approaches may truncate particular excited states of atoms with the same or better accuracy than the ground state (GS). This feature has direct analogs in many-electron hydrogenic systems and motivated the proposal of the Exact Diagonalization  -- Ab Initio method (EDABI) \cite{Ryc04}, which were recently used to discuss the influence of electron correlations on the metallization of solid hydrogen in a~rigorous manner \cite{Kad14,Kad15}. Also, as a~method putting equal footing on single- and multiparticle aspects of the correlated quantum states, EDABI seems to be a~promising candidate for the theoretical tool capable of giving a~better insight into the superconductivity mechanism in sulfur hydrides \cite{Dro15,Err15} or into the magnetic moment formation in functionalized graphenes \cite{Yaz10}.

When considering a~generic second-quantized Hamiltonian with both spin and charge degrees of freedom, a~few numerical techniques can be regarded as exact ones; i.e., giving the desired correlation functions within the accuracy limited (in principle) by the machine precision only. These includes: Exact Diagonalization (ED) for relatively small systems \cite{Pre13}, Density Matrix Renormalization Group (DMRG) for low-dimensional systems \cite{DeCh08}, and Quantum Monte Carlo (QMC) for non-frustrated systems \cite{Sug86}. A~separate class is outlined by variational approaches designed to treat one-dimensional (1D) atomic chains in the insulating phase \cite{Ste11}. Even though each of these methods provides us with detailed information about the closed-system GS, it is usually a~challenging task to determine whether GS is insulating, metallic, or of a~more complex nature \cite{spnlqd}. This is primarily because standard GS signatures of the metal-insulator transition, such as a~vanishing charge gap \cite{Hir85}, are absent at any finite system size (quantified by the number of lattice sites $N$), but may appear only after the so-called finite-size scaling (with $1/N\rightarrow{}0$), a~procedure introducing systematic errors difficult to estimate in some cases.

For this reason, specially-designed GS correlation functions, not only signalling the Mott transition in the thermodynamic limit, but also showing fast convergence (with $1/N\rightarrow{}0$) for finite systems close to the metal-insulator boundary, may constitute a~valuable complement of the existing numerical techniques.

In attempt to propose such correlation functions, one need to point out the relevance of quantum fluctuations between electronic doubly-occupied sites and unoccupied sites and between singly occupied sites with spin up and spin down in lattice models, as suggested before by numerous authors \cite{Kap82,Yok90,Str99,Yok11,Zho14}. The concept quantum entanglement \cite{Ein35,Ben00}, together with entanglement measures such as the concurrence \cite{Woo98}, was adopted to quantify the above-mentioned fluctuations in various systems \cite{Ost02,Ver04,Wez05,Sch01,Zan02,SGu04,Wie08,Bog12,Gru15,Ole06,Ole12,Che15,You15,Ryc06d,Ram06,Ryc08a,Too14,Mot14,Sch13}, including basic spin models \cite{Ost02,Ver04,Wez05,Hua14}, fermionic systems \cite{Sch01,Zan02,SGu04,Wie08,Bog12,Gru15}, or systems with spin and orbital degrees of freedom \cite{Ole06,Ole12,Che15,You15}.
For an open fermionic system (namely: a~correlated quantum dot attached to the leads) we found that the concurrence, defined separately for charge and spin degrees of freedom, allowed one to distinguish between different quantum transport regimes of the system \cite{Ryc06d}. Analogs of this observation were also reported for double \cite{Ram06,Ryc08a} and triple quantum dots \cite{Too14}. For closed systems, in particular for small molecules, Mottet {\em et al.}\ \cite{Mot14} showed that the entanglement analysis provided a valuable insight into the chemical bond formation. Even noninteracting systems with complex Fermi surface topology were characterized via their entanglement spectra \cite{Sch13}. Also very recently, a~generic variational approach to correlated quantum systems, in which the output numerical precision is steered by setting maximal allowed entanglement between a~selected subsystem and the environment, was proposed \cite{Cza15}.

Sacramento {\em et al.}\ \cite{Sac13} used entanglement measures to complement the long-lasting discussion of decoupling of charge and spin degrees of freedom in 1D Hubbard model \cite{Lie68} and its relation to the charge-spin separation phenomenon in Luttinger liquids \cite{Voi93,Ber00}. In particular, it is pointed out in Ref.\ \cite{Sac13} that charge-charge fluctuations, when quantified by properly defined correlation functions, show substantially different asymptotic behavior than spin-spin fluctuations in both the metallic and the insulating phases.

\begin{figure}
\centerline{\includegraphics[width=0.7\linewidth]{chainsketch}}
\caption{ \label{fig:chain}
  A system studied numerically (schematic). Hydrogenic-like atoms, containing 1$s$-type orbitals with their radii $\alpha^{-1}$, are arranged linearly with the interatomic distance $R$. The Hamiltonian parameters, including the atomic energy $\epsilon_a$, the intraatomic Coulomb integral $U$, the hopping integrals $t_{ij}$, and interatomic Coulomb integrals $K_{ij}$, are determined for each value of  $\alpha$ and $R$ (see Sec.\ II for the details).
}
\end{figure}

Here we focus on linear chains containing up to $N=20$ equally-spaced hydrogenic-like atoms, each one containing a~single valence orbital (see Fig.\ \ref{fig:chain}). Although such chains are not directly observable due to the well-known Peierls instability, they have been intensively studied to benchmark different {\em ab initio} approaches to strongly-correlated systems \cite{Hach06}.
In this paper, two distinct physical regimes are considered: At the half filling the system shows a crossover behavior, with the increasing interatomic distance $R$, from a~{\em partly localized} quantum liquid to the Mott insulating phase \cite{Ryc04,Spa01}. At the quarter filling, the Mott transition is reconstructed with $1/N\rightarrow{}0$ \cite{Ryc04}. (The above-mentioned findings are consistent with more recent results of QMC simulations for Hubbard chains with long-range interactions \cite{Hoh12}.)
It is worth to mention that the two regimes are also significantly different from a~more fundamental point of view: in the strong-coupling (i.e., large $R$) limit the charge fluctuations are suppressed and the half-filled chain can be effectively described as 1D Heisenberg antifferomagnet, for which long-range order is absent due to the Mermin-Wagner theorem \cite{Mer66}. For the quarter-filling similar reasoning cannot be applied in the presence of long-range Coulomb interactions, and the charge-density wave phase is predicted to appear \cite{Ryc04,Eji05,Sch90,Mey09}.
In other words, we consider the two complementary model cases allowing one to test entanglement-based phase-transition indicators. We further discuss the concurrence \cite{Woo98}, defined separately for charge and spin degrees of freedom, and employ it to recognize the decoupling of charge and spin degrees of freedom accompanying a~finite-system version of the Mott transition.

The paper is organized as follows. In Sec.\ II we present the system Hamiltonian and summarize the EDABI method. In Sec.\ III the findings of Ref.\ \cite{Ryc04}, concerning the finite-size scaling of charge-energy gap, are revisited after taking larger systems under consideration. In Sec.\ IV we determine the local and pairwise entanglement, the latter for charge and spin degrees of freedom, and discuss their evolution with the interatomic distance and the system size. The conclusions are given in Sec.\ V.

\section{The exact diagonalization -- ab initio method (EDABI)}
The EDABI method, together with its application to the system of Fig.\ \ref{fig:chain}, have been presented in several works \cite{Ryc04,Kad14,Kad15,Spa01}. Here we briefly recall the main findings, to which we refer in the remaining parts of the paper.

\subsection{The Hamiltonian for a~linear chain}
The analysis starts from the second-quantized Hamiltonian, which can be written in the form
\begin{multline}
  \label{hamchain}
  \hat{\cal H}(\alpha,R)=
  \epsilon_a\sum_j\hat{n}_j
  +\sum_{i<j,\sigma}t_{ij}\left(\hat{c}_{i\sigma}^{\dagger}\hat{c}_{j\sigma}+
  \hat{c}_{j\sigma}^{\dagger}\hat{c}_{i\sigma}\right) \\
  + U\sum_i\hat{n}_{i\uparrow}\hat{n}_{i\downarrow}
  + \sum_{i<j}K_{ij}{\hat{n}_i}{\hat{n}_j} +V_{\rm ion-ion},
\end{multline}
where $\epsilon_a$ is the atomic energy (same for all $N$ sites upon applying periodic boundary conditions), $t_{ij}$ is the hopping integral between $i$-th and $j$-th site (we further set $t_{ij}\equiv{}-t$ if $i$ and $j$ are nearest neighbors, otherwise $t_{ij}\approx{}0$), $U$ is the intrasite Coulomb repulsion, $K_{ij}$ is the intersite Coulomb repulsion, and $V_{\rm ion-ion}=\sum_{i<j}e^2/R_{ij}$ (with the distance $R_{ij}\equiv{}R\min(|i-j|,N-|i-j|)$) expresses the repulsion of infinite-mass ions.

The single-particle and  interaction parameters of the Hamiltonian $\hat{\cal H}(\alpha,R)$ (\ref{hamchain}), also marked in Fig.\ \ref{fig:chain}, can be defined as follows
\begin{align}
  \langle{}w_i|T|w_i\rangle &=
  \int{}d^3r{}\,w_i^\star({\bf r})T({\bf r})w_j({\bf r}) \nonumber\\
  & \equiv{} \epsilon_a\delta_{ij}+t_{ij}(1-\delta_{ij}),
  \label{epstijdef}
\end{align}
\begin{align}
  \langle{}w_iw_j|V|w_iw_j\rangle &=
  \int{}d^3r{}d^3r'{}\,|w({\bf r})|^2V({\bf r}-{\bf r}')|w({\bf r}')|^2
  \nonumber\\
  & \equiv{} U\delta_{ij}+K_{ij}(1-\delta_{ij}),
  \label{ukijdef}
\end{align}
where $T$ is the~single particle Hamiltonian describing an electron in the medium of periodically arranged ions, and $V$ is the Coulomb repulsive interaction of two electrons.
The Wannier functions are defined via atomic (Slater-type) functions, namely
\begin{equation}
  \label{wannfun}
  w_i({\bf r})=\sum_j\beta_{ij}\psi_j({\bf r}),
\end{equation}
where the Slater 1$s$ function $\psi_i({\bf r})=(\alpha^3/\pi)^{1/2}\exp(-\alpha{}|{\bf r}\!-\!{\bf R}_i|)$, with $\alpha$ being the inverse orbital size, here taken as variational parameter, and ${\bf R}_i$ being the position of $i$-th ion. The coefficients $\beta_{ij}$ in Eq.\ (\ref{wannfun}) can be uniquely defined by imposing that (i) $\langle{}w_i|w_k\rangle=\delta_{ik}$ and that (ii) $\langle{}w_i|\psi_i\rangle$ is maximal \cite{Ryc04,Rycphd}.

The range of Coulomb interactions in the Hamiltonian  $\hat{\cal H}(\alpha,R)$ (\ref{hamchain}), quantified by parameters $K_{ij}\approx{}e^2/R_{ij}$ for $R_{ij}\gg{}R$, is {\em a~priori} limited only by the chain length $L\equiv{}NR$.
Previous studies on linear chains, both {\em ab initio} ones \cite{Ste11,Hach06} and these starting from second-quantized model Hamiltonians \cite{Hoh12,Eji05}, usually have imposed some form of charge screening reducing the range of such interactions (leaving the on-site Coulomb repulsion only in the extreme case \cite{Sch90}). This can be partly justified by possible influence of the enviroment in condensed-matter realizations such as quantum wires \cite{Mey09} or self-organised chains \cite{Blu11}. A~slightly different situation occurrs for cold atom systems, where long-range dipole-dipole interactions may be relevant \cite{Mos15}. Apart from these experiment-related premises, long-range interactions usually lead to a~noticeable slowdown of the convergence for numerical methods such as DMRG or QMC \cite{Cap00,Fan99}. This is not the case for the EDABI method, as the determination of parameters $K_{ij}$ corresponds to a~relatively small portion of the overall computation time \cite{Bib15}, and the convergence of subsequent diagonalization in the Fock space (see the next subsection) is unaffected by the fact whether or not long-range interactions are included. Also, as charge screening becomes systematically less effective when the system dimensionality is reduced \cite{Kon77}, one can expect it to be insignificant in a~hypotetical realization of chains containing $N\leqslant{}20$ atoms.

\subsection{Single-particle basis optimization}
Next, each Slater function is approximated as follows
\begin{equation}
  \psi_i({\bf r}) \approx{} \alpha^{3/2}
  \sum_{q=1}^p{}B_q\left(\frac{2\Gamma_q^2}{\pi}\right)^{3/4}
  e^{-\alpha^2\Gamma_q^2|{\bf r}-{\bf R}_i|^2},
\end{equation}
where $p$ is the number of Gaussian functions truncating $\psi_j$, and $\{B_q,\Gamma_q\}$, $q=1,\dots,p$, are adjustable parameters chosen to minimize the atomic energy for $\alpha=1$ and a~given value of $p$. Here we set $p=3$, for which the deviation from the exact energy for $1s$ function is lower that $1\%$ \cite{Rycappa}. Subsequently, the parameters $\epsilon_a$, $t_{ij}$, $U$, and $K_{ij}$ are calculated from Eqs.\ (\ref{epstijdef}), (\ref{ukijdef}) as functions of $\alpha$ and $R$. The Hamiltonian $\hat{\cal H}(\alpha,R)$ (\ref{hamchain}) is diagonalized numerically in the Fock space, using the Lanczos algorithm, and the orbital size is optimized to find $\alpha=\alpha_{\rm min}$ corresponding to the minimal GS energy $E_G(N)$ for each $R$. Defining the efective atomic energy as
\begin{equation}
  \label{eaeffdef}
  \epsilon_a^{\rm eff}=\epsilon_a
  +\frac{1}{N}\bigg(\sum_{i<j}K_{ij}+V_{\rm ion-ion}\bigg),
\end{equation}
one finds that $\alpha_{\rm min}$ and other parameters converges rapidly with $N$. This observation allows us to speed up computations by using the values obtained for smaller systems ($N\geqslant{}10$) to perform extrapolation with $1/N\rightarrow{}0$. (The results are listed in Table~\ref{tab:param}.)

\begin{table*}
\caption{ \label{tab:param}
  Microscopic parameters (specified in eV -- unless stated otherwise) of the Hamiltonian  $\hat{\cal H}(\alpha,R)$ (\ref{hamchain}) with $\alpha=\alpha_{\rm min}$ minimizing the ground-state (GS) energy $E_G$. The Bohr radius is $a_0=0.529\,$\AA. The effective atomic level $\epsilon_a^{\rm eff}$ is defined by Eq.\ (\ref{eaeffdef}); the remaining symbols are $t\equiv{}-t_{ij}$ for $j=i\pm{}1$ (mod $N$), $K_m\equiv{}K_{|i-j|}$ (with $m=1,2,3$), and  the correlated hopping integral $V\equiv\langle{}w_iw_i|V|w_iw_{i\pm{}1}\rangle$, quantifying the largest neglected term in  $\hat{\cal H}(\alpha,R)$. The numerical extrapolation with $1/N\rightarrow{}0$ is performed for all parameters.
}
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
\hline\hline
$\ R/a_0\ $  &  $\ \alpha_{\rm min}a_0\ $  &  $\ \epsilon_a^{\rm eff}\ $  &  $\ t\ $  &  $\ U\ $  &  $\ K_1\ $  &  $\ K_2\ $  &  $\ K_3\ $  & $\ V\ $  &  $\ E_G/N\ $  \\ \hline
1.5  &  1.363  &  $\ \ \ $1.36$\ $  &  $\ $11.31$\ $  &  $\ $27.95$\ $  &  $\ $15.85$\ $  &  $\ $9.08$\ $  &  $\ $6.08$\ $  & $\ $-0.597$\ $ & $\ $-10.19$\ $  \\
2.0  &  1.220  &  $\ $-7.48  &  $\ \ $6.02$\ $  &  23.58  &  12.40  &  6.82  &  4.54  & -0.324 &  -12.65  \\
2.5  &  1.122  &  10.85  &  $\ \ $3.60$\ $  &  20.83  &  10.20  &  5.46  &  3.63  & -0.203 &  -13.32  \\
3.0  &  1.062  &  12.27  &  $\ \ $2.32$\ $  &  19.14  &  $\ $8.69  &  4.54  &  3.02  & -0.150 &  -13.48  \\
3.5  &  1.031  &  12.90  &  $\ \ $1.57$\ $  &  18.16  &  $\ $7.58  &  3.89  &  2.60  & -0.128 &  -13.50  \\
4.0  &  1.013  &  13.20  &  $\ \ $1.08$\ $  &  17.57  &  $\ $6.71  &  3.40  &  2.27  & -0.119 &  -13.50  \\
5.0  &  1.004  &  13.43  &  $\ \ $0.51$\ $  &  17.12  &  $\ $5.43  &  2.72  &  1.81  & -0.096 &  -13.50  \\
6.0  &  1.001  &  13.48  &  $\ \ $0.23$\ $  &  16.99  &  $\ $4.53  &  2.27  &  1.51  & -0.058 &  -13.50  \\
7.0  &  1.000  &  13.49  &  $\ \ $0.10$\ $  &  16.97  &  $\ $3.89  &  1.99  &  1.29  & -0.027 &  -13.50  \\
\hline\hline
\end{tabular}
\end{table*}

\section{Finite-size scaling for the charge-energy gap}

\begin{figure}
\centerline{\includegraphics[width=0.9\linewidth]{delc-half}}
\caption{ \label{fig:delc-half}
  Charge-energy gap as a~function of the interatomic distance $R/a_0$
  (with the Bohr radius $a_0=0.529\,$\AA) for chains of $N=10-16$ atoms
  at the {\em half filling} ($N_{\rm el}=N$).
  Datapoints were shifted vertically by $0.05\,$Ry for $N=16$ (diamonds),
  $0.10\,$Ry for $N=14$ (solid circles), $0.15\,$Ry for $N=12$ (open circles),
  or $0.20\,$Ry for $N=10$ (stars).
  The results of the finite-size scaling with $1/N\rightarrow{}0$,
  obtained via the Richardson extrapolation of the second order \cite{Bur10},
  are also shown (triangles).  Lines are guides for the eye only.
}
\end{figure}

\begin{figure}
\centerline{\includegraphics[width=0.9\linewidth]{delc-quart-mk16}}
\caption{ \label{fig:delc-quart}
  Same as Fig.\ \ref{fig:delc-half} but for chains of $N=8-20$ atoms
  at the {\em quarter filling} ($N_{\rm el}=N/2$). No artificial datashifts
  were applied this time. Horizontal dashed line marks $\Delta{}E_C=0$,
  solid purple line represents the best-fitted function given by
  Eqs.\ (\ref{empdelc}) and (\ref{empdpar}),
  indicating the Mott metal-insulator transition in the
  $1/N\rightarrow{}0$ limit, occurring for $R=R_c\approx{}4.2\,a_0$.
}
\end{figure}

A~standard numerical approach \cite{Hir85,Hoh12}, allowing one to determine whether a~correlated system described by the Hamiltonian such as given by Eq.\ (\ref{hamchain}) is metallic or insulating in the ground state, involves calculating the charge-energy gap according to
\begin{equation}
  \label{delcdef}
  \Delta{}E_C^{N_{\rm el}}(N)=
  E_G^{N_{\rm el}+1}(N)+E_G^{N_{\rm el}-1}(N)-2E_G^{N_{\rm el}}(N),
\end{equation}
where $N_{\rm el}$ is the number of electrons. Next, the limit of $1/N\rightarrow{}0$ (with $N_{\rm el}/N=\,$const) is to be taken numerically. The limiting value of $\Delta{}E_C\rightarrow{}0$ indicates the metallic phase, whereas a~nonzero value indicates the insulating phase \cite{spingapfoo}. In the numerical examples presented here and in Sec.\ IV we consider two different physical situations: the {\em half filling} $N_{\rm el}=N$, with $N\leqslant{}16$, and the {\em quarter filling} $N_{\rm el}=N/2$, with $N\leqslant{}20$; we further restrict ourselves to even values of $N$ and $N_{\rm el}$. Due to the total spin conservation it is sufficient, for each pair $(N,N_{\rm el})$, to look for the ground-state energy $E_G^{N_{\rm el}}(N)$ in Eq.\ (\ref{delcdef}) in the subspace characterized by the total $z$-th component of spin $S_z=0$. Analogously, the values $E_G^{N_{\rm el}\pm{}1}(N)$  can be found by choosing $S_z=1/2$. In particular, the largest considered subspace dimension corresponds to $N=20$, $N_{\rm el}\!+\!1=11$, and is equal to $600\,935\,040$. Moreover, we impose the periodic ($t_{i,j+N}=t_{ij}$) or antiperiodic ($t_{i,j+N}=-t_{ij}$) boundary conditions to minimize $E_G^{N_{\rm el}}(N)$ for $N_{\rm el}=4k+2$ or $N_{\rm el}=4k$, respectively, with $k$-integer (see Table~\ref{tab:rstara}).

\begin{table}
\caption{ \label{tab:rstara}
  Boundary conditions BC (with $+/-$ marking the periodic/antiperiodic BC)
  and dimension of the largest subblock of the Hilbert space (with the total
  $z$-th component of spin $S_z=0$) for chains studied in the paper at
  the quarter filling (i.e., $N_{\rm el}=N/2$).  The last column specifies
  the values of the interatomic distance $R_\star$ at which the spin
  concurrence vanishes, with the interpolation errorbars for the last digit
  given in parenthesis. (See Sec.\ IV for details.)
}
\begin{tabular}{c|c|c|r|c}
\hline\hline
  $\,\ N\,\ $ & $\ N_{\rm el}\ $ & $\ $BC$\ $
  & $\ $dim$\,{H}(S_z^{\rm tot}\!\!=\!0)\ $ & $R_\star/a_0$ \\
\hline
 $\ $8  & $\ $4  &  $-$ & 784$\ $ & $\,\ $6.53(3)$\,\ $ \\
    12  & $\ $6  &  $+$ & 48,400$\ $ &   5.22(2)  \\
    16  & $\ $8  &  $-$ & 3,312,400$\ $ &   4.38(1)  \\
    20  &    10  &  $+$ & $\ $240,374,016$\ $ &   4.51(1)  \\
\hline\hline
\end{tabular}
\end{table}

The numerical results, corresponding to the Hamiltonian  $\hat{\cal H}(\alpha,R)$ (\ref{hamchain}) with the microscopic parameters listed in Table~\ref{tab:param}, are presented in Figs.\ \ref{fig:delc-half} and \ref{fig:delc-quart}. We notice that the datasets for $N\leqslant{}14$ at the half filling as well as for $N\leqslant{}16$ at the quarter filling were presented in Ref.\ \cite{Ryc04}, were we also used the Richardson extrapolation of the second order \cite{Bur10} to perform finite-size scaling with $1/N\rightarrow{}0$ for each value of $R$.  After adding the new datasets for $N=16$, $N_{\rm el}=N$ and $N=20$, $N_{\rm el}=N/2$ (both depicted with open diamonds) the earlier conclusions regarding the system ground state are further supported: Namely, the extrapolation with $1/N\rightarrow{}0$ (open triangles)  at $N_{\rm el}=N$ leads to $\Delta{}E_C>0$ for any accessible value of $R/a_0$ indicating the insulating phase (see Fig.\ \ref{fig:delc-half}). At $N_{\rm el}=N/2$, the finite-size scaling results can be rationalized with the empirical function
\begin{equation}
  \label{empdelc}
  \Delta{}E_C^{N/2}(1/N\!\rightarrow{}\!0)\approx
  \begin{cases}
    0 & \ \text{if } R<R_c, \\
    \Delta_0\left[(R\!-\!R_c)/a_0\right]^{\gamma}
    & \ \text{if } R\geqslant{}R_c, \\
  \end{cases}
\end{equation}
with the best-fitted parameters (numbers in parentheses are standard deviations for the last digit)
\begin{gather}
  R_{\rm c}= 4.18(2)\,a_0, \ \ \ \ \Delta_0=0.158(3)\,\text{Ry},\nonumber \\
  \gamma= 0.57(3),  \label{empdpar}
\end{gather}
depicted with solid purple line in Fig.\ \ref{fig:delc-quart}.
The gap opening at $R=R_{\rm c}$ indicates the Mott transition.

For a~sake of conciseness, our discussion of finite-size scaling estimates of the Mott transition is limited to the charge-energy gap. A~more detailed analysis, presented in Refs.\ \cite{Ryc04,Spa01,Rycphd} and involving calculations of the electron momentum distribution, the Drude weight, as well as the so-called modern theory of polarization \cite{Res02}, justify the transition appearance in the $N_{\rm el}=N/2$ case, which coincides with the results for related parametrized model studies \cite{Hoh12,Eji05,Sch90,Mos15}. In the $N_{\rm el}=N$ case, the situation is of a~slightly more complex nature: Apart from a~nonzero gap for any $R$, following from the finite-size scaling, several GS and dynamical characteristics (in particular -- the Drude weight, see Ref.\ \cite{Ryc04}) exhibit, for any finite $N$, a~crossover behavior between a~partly localized quantum liquid, appearing for small $R$, and a~fully-reconstructed Mott insulator, typically appearing for
$R/a_0\gtrsim{}4$. These findings coincides with more recent variational Monte-Carlo studies of hydrogenic chains (see Ref.\ \cite{Ste11}), and can be attributed to the increasing (with growing $R$) role of electron correlations in either the exact or variational GS wavefunction.

It is worth to mention here that some surprisingly efficient variational methods, such as presented in Ref.\ \cite{Ste11}, become significantly less efficient when treating open-shell configurations in attempt to calculate $\Delta{}E_C$ from Eq.\ (\ref{delcdef}), or generally when utilized away from the half filling ($N_{\rm el}\neq{}N$). Also for this reason, GS correlation functions signalling a~finite-system version of the Mott transition for $N$ small enough to be treated with some more flexible numerical techniques, are desired.

\section{Reduced density matrix and quantum entanglement}
In this section we derive explicit representations of the reduced density operator $\hat{\rho}_A=\mbox{Tr}_A|\Psi\rangle\!\langle\Psi|$ (where Tr$_A$ stands for tracing over all degrees of freedom except from these characterizing a~selected subsystem $A$) relevant when discussing the local entanglement, the pairwise entanglement for charge degrees of freedom, and the pairwise entanglement for spin degrees of freedom. The derivations remain valid for a~generic spin--$1/2$ fermionic system at any pure state $|\Psi\rangle$, which can be represented assuming four basis states for each site $j$, namely:
\begin{equation}
\label{basenjud}
\left\{\,|n_{j\uparrow}n_{j\downarrow}\rangle\,\right\}\equiv\left\{\,
|0\rangle_j,  |\!\uparrow\rangle_j, |\!\downarrow\rangle_j,  |\!\uparrow\downarrow\rangle_j\,\right\};
\end{equation}
although the numerical examples, considered in search for entanglement-based estimates of the Mott transition, all correspond to the ground state of a~linear chain described by the Hamiltonian $\hat{\cal H}(\alpha,R)$ (\ref{hamchain}) with the microscopic parameters taken from Table~\ref{tab:param}.

\subsection{General considerations}
Let us consider a~general example at first: the quantum system, which can be divided into two distinct subsystems $A$ and $B$. A~pure state $|\Psi\rangle$ can be represented as follows
\begin{equation}
\label{psialpbet}
  |\Psi\rangle=\sum_{\alpha\beta}\Psi_{\alpha\beta}|\alpha\rangle\otimes|\beta\rangle,
\end{equation}
where $\Psi_{\alpha\beta}$ denotes a~complex probability amplitude corresponding to a~basis state of the full system $|\alpha\rangle\otimes|\beta\rangle$, while $\{|\alpha\rangle\}$ and $\{|\beta\rangle\}$ are complete basis sets for the subsystems $A$ and $B$ (respectively). The reduced density operator $\hat{\rho}_A$
is defined as:
\begin{align}
  \hat{\rho}_A&=
  \mbox{Tr}_A|\Psi\rangle\!\langle\Psi|\equiv
  \sum_\beta\langle\beta|\Psi\rangle\!\langle\Psi|\beta\rangle \nonumber \\
  &=
  \sum_{\beta,\alpha,\alpha'}\Psi_{\alpha\beta}{\Psi}_{\alpha'\beta}^{\star}
  |\alpha\rangle\!\langle\alpha'|,
\end{align}
where the last equality follows from Eq.\ (\ref{psialpbet}) and $\Psi_{\alpha\beta}^{\star}$ denotes the complex conjugate of $\Psi_{\alpha\beta}$. Subsequently, the matrix elements of $\hat{\rho}_A$ are given by
\begin{equation}
\label{rhoma}
  \rho_{\alpha,\alpha'}\equiv\langle{\alpha}|\hat{\rho}_A|{\alpha'}\rangle=
  \sum_\beta\Psi_{\alpha\beta}{\Psi}_{\alpha',\beta}^{\star}.
\end{equation}

We define now projection operators $P_\alpha$ and $P_\beta$, associated with
the basis states $|\alpha\rangle$ and $|\beta\rangle$, via
\begin{equation}
\label{palpdef}
  P_{\alpha}|\alpha'\rangle=\delta_{\alpha\alpha'}|\alpha\rangle, \ \ \ \
  P_{\beta}|\beta'\rangle=\delta_{\beta\beta'}|\beta\rangle.
\end{equation}
Useful properties, directly following from Eq.\ (\ref{palpdef}) are
\begin{gather}
P_\alpha^2=P_\alpha, \ \ \ \  P_\beta^2=P_\beta, \label{palpsqu} \\
  \sum_\alpha P_\alpha = \sum_\beta P_\beta = 1, \label{palpsum}
\end{gather}
where Eq.\ (\ref{palpsum}) also employs the completeness of the basis sets $\{|\alpha\rangle\}$ and $\{|\beta\rangle\}$.
Next, we define the transfer operator
\begin{equation}
\label{tradef}
T_{\alpha\alpha'}|{\alpha'}\rangle=|{\alpha}\rangle, \ \ \ \
\langle{\alpha'}|T_{\alpha\alpha'}^{\dagger}=\langle{\alpha}|,
\end{equation}
which is unitary, i.e.\
\begin{equation}
\label{trainv}
T_{\alpha\alpha'}^{\dagger}=(T_{\alpha\alpha'})^{-1}\equiv
T_{\alpha'\alpha}.
\end{equation}
Explicit forms of  $P_\alpha$, $P_\beta$, and $T_{\alpha\alpha'}$, corresponding to the particular splittings of the system into $A$ and $B$ subsystems, are to be specified later in terms of the operators $\{\hat{c}_{i\sigma},\hat{c}_{i\sigma}^{\dagger}\}$ for lattice spin--$1/2$ fermions. Here we notice that as the indices $(\alpha,\alpha')$ refer the subsystem $A$, whereas the index $\beta$ refers to the subsystem $B$, we have
\begin{equation}
\label{ptcomm}
\left[P_\alpha,P_\beta\right]=0, \ \ \ \
\left[P_\beta,T_{\alpha\alpha'}\right]=0,
\end{equation}
for all  $(\alpha,\alpha')$ and $\beta$.

With the help of operators $P_{\alpha}$ and $P_{\beta}$ (\ref{palpdef}) one can write down, for $|\Psi\rangle$ represented according to Eq.\ (\ref{psialpbet}),
\begin{align}
  P_\alpha P_\beta|\Psi\rangle &= \Psi_{\alpha\beta}\,
  |\alpha\rangle\otimes|\beta\rangle, \\
  \langle\Psi|P_{\alpha'}P_\beta &= \langle\alpha'|\otimes\langle\beta|
  \,{\Psi}_{\alpha'\beta}^{\star}.
\end{align}
In turn, the reduced density matrix, as given by the rightmost equality in Eq.\ (\ref{rhoma}), can be rewritten as
\begin{align}
  \rho_{\alpha\alpha'}
  &=\sum_{\beta}\langle\Psi|P_{\alpha'}P_{\beta}
  T_{\alpha'\alpha}P_\alpha P_\beta|\Psi\rangle \nonumber\\
  &=
  \langle\Psi|P_{\alpha'}T_{\alpha'\alpha}P_\alpha|\Psi\rangle,
\end{align}
with the last equality following from Eqs.\ (\ref{palpsqu}), (\ref{palpsum}), and (\ref{ptcomm}). Finally, using Eq.\ (\ref{trainv}), we find
\begin{equation}
\label{rhostar}
  \left(\rho_{\alpha\alpha'}\right)^{\star} =
  \langle{\Psi}|P_{\alpha}T_{\alpha\alpha'}P_{\alpha'}|{\Psi}\rangle.
\end{equation}
Remarkably, the reduced density matrix elements in Eq.\ (\ref{rhostar}) are expressed by pure-state expectation values (correlation functions) of the operators acting only on the subsystem $A$. This feature is further explored in the remaining parts of this section.

\subsection{Entanglement entropy}
The local entanglement \cite{Zan02} exhibits quantum correlations between the local state of a~selected $j$-th site (subsystem $A$) and the rest of the system ($B$). As the basis set $\{|\alpha\rangle\}$ is simply given by Eq.\ (\ref{basenjud}), one can write down the corresponding projection operators
\begin{multline}
\label{palploc}
  \left(P_\alpha\right) =
  \left[\, (1-n_\uparrow)(1-n_\downarrow),\ n_\uparrow (1-n_\downarrow),\ \right. \\
  \left.(1-n_\uparrow)n_\downarrow,\ n_\uparrow n_\downarrow \,\right]^T,
\end{multline}
and the transfer operators
\begin{equation}
\label{traloc}
  \left(T_{\alpha\alpha'}\right)=\left(
  \begin{array}{cccc}
  1 & c_\uparrow & c_\downarrow & c_\downarrow c_\uparrow \\
  c_\uparrow^\dagger & 1 & c_\uparrow^\dagger c_\downarrow & -c_\downarrow \\
  c_\downarrow^\dagger & c_\downarrow^\dagger c_\uparrow & 1 & c_\uparrow \\
  c_\uparrow^\dagger c_\downarrow^\dagger & -c_\downarrow^\dagger & c_\uparrow^\dagger & 1 \\
  \end{array}\right),
\end{equation}
where we have omitted the site index $j$ for brevity.
For the system with the total spin and charge conservation, such as our linear chain [see Eq.\ (\ref{hamchain})], we have
\begin{gather}
\langle{c_\sigma}\rangle=
\langle{c_\sigma^\dagger}\rangle=0,
\ \ \ \
\langle{ c_\sigma c_{\bar{\sigma}} }\rangle=
\langle{c_\sigma^\dagger c_{\bar{\sigma}}^\dagger}\rangle=0,
\nonumber \\
\text{and}
\ \ \ \
\langle{c_\sigma^\dagger c_{\bar{\sigma}}}\rangle=
\langle{c_{\bar{\sigma}}^\dagger c_\sigma}\rangle=0, \label{cavezero}
\end{gather}
(with $\overline{\sigma}$ being the spin index opposite to $\sigma$) providing that the averaging takes place over an eigenstate $|{\Psi}\rangle$ of the system Hamiltonian (not necessarily the ground state). Substituting the expressions for $P_\alpha$ (\ref{palploc}) and $T_{\alpha\alpha'}$ (\ref{traloc}) into Eq.\ (\ref{rhostar}) and taking Eqs.\ (\ref{cavezero}) into account we get, after a straightforward algebra,
\begin{equation}
  \label{rholoc}
  \left(\rho_{\alpha\alpha'}\right)=
  \mbox{diag}\left( u_+,w_1,w_2,u_- \right)
\end{equation}
with
\begin{gather}
  u_+ = \left<{(1-n_{j\uparrow})(1-n_{j\downarrow})}\right>,\ \ \ \
  w_1 = \left<{n_{j\uparrow}(1-n_{j\downarrow})}\right>,
\nonumber \\
  w_2 = \left<{(1-n_{j\uparrow})n_{j\downarrow}}\right>,\ \ \ \
  u_- = \left<{n_{j\uparrow}n_{j\downarrow}}\right>. \label{locelem}
\end{gather}
Eq.\ (\ref{rholoc}) corresponds to the reduced density operator of the well-known form
\begin{multline}
\label{rholocop}
  \hat{\rho}_A =
  \sum_{\alpha\alpha'}\rho_{\alpha\alpha'}|\alpha\rangle\!\langle\alpha'|= \\
  u_+|0\rangle\!\langle{}0| +
  w_1|\!\uparrow\rangle\!\langle{}\!\uparrow\!| +
  w_2|\!\downarrow\rangle\!\langle{}\!\downarrow\!| +
  u_-|\!\uparrow\downarrow\rangle\!\langle{}\!\uparrow\downarrow\!|,
\end{multline}
where we have used the basis given by Eq.\ (\ref{basenjud}).

A~quantitative measure of the entanglement between the state of $j$-th site and that of the remaining $N-1$ sites is given by the von Neumann entropy
\begin{equation}
\label{evdef}
  E_v = -u_+\log_2u_+ - w_1\log_2w_1 - w_2\log_2w_2 -u_-\log_2u_-.
\end{equation}
If the translational invariance of the system is imposed, one can define the particle density $n$ and the average number of double occupancies $d$, following
\begin{equation}
  \langle{}n_{j\uparrow}\rangle=\langle{}n_{j\uparrow}\rangle=\frac{N_{\rm el}}{2N}
  \equiv{}\frac{n}{2},
  \ \ \ \ \ \
  \langle{}n_{j\uparrow}n_{j\uparrow}\rangle\equiv{}d,
\end{equation}
where we have further assumed that the averaging takes place over an eigenstate characterized by the total $z$-th component of spin $S_z=0$. Subsequently, Eq.\ (\ref{locelem}) can be rewritten as
\begin{equation}
\label{loceltinv}
  u_+=1-n+d, \ \ \ \ w_1=w_2=\frac{n}{2}-d, \ \ \ \
  u_-=d.
\end{equation}

In particular, for the ground state of the Hamiltonian $\hat{\cal H}(\alpha,R)$ (\ref{hamchain}), for which the effective interaction between electrons can roughly be estimated by $\sim{}U-K_1$ and is always repulsive, one can show that
\begin{equation}
\label{dbounds}
  \frac{|n-1|+n-1}{2}\leqslant{}d\leqslant{}\frac{n^2}{4} \ \ \ \
  \text{for}\ \ \ \ 0\leqslant{}n\leqslant{}2,
\end{equation}
where the lower bound for $d$ corresponds to the so-called {\em strong-correlations limit} approached for $R\gg{}a_0$ (as the bandwidth $W\equiv{}4t\ll{}U-K_1$, see Table~\ref{tab:param}), whereas the upper bound for $d$ corresponds to the {\em free-electrons limit} approached for $R\lesssim{}2a_0$ ($W\gg{}U-K_1$). As the charge fluctuations are given by $\mbox{Var}\{n_j\}=\langle{}n_j^2\rangle-\langle{}n_j\rangle^2\equiv{}n-n^2+2d$, the above-mentioned limits coincide (respectively) with the minimal and the maximal fluctuations for a~given $n$. Moreover, the bounds for $d$ in Eq.\ (\ref{dbounds}) can be mapped, via Eq.\ (\ref{loceltinv}), onto
\begin{equation}
\label{evbounds}
E_v^{\rm corr}(n)\leqslant{}E_v\leqslant{}E_v^{\rm free}(n),
\end{equation}
with
\begin{multline}
  E_v^{\rm corr}(n)= -|1-n|\log_2|1-n| \\
  -\left(1-|1-n|\right)\log_2\left(\frac{1-|1-n|}{2}\right) \label{evboundc}
\end{multline}
for the strong-correlations limit, and
\begin{equation}
  E_v^{\rm free}(n)=
  -n\log_2\frac{n}{2}-(2-n)\log_2\left(1-\frac{n}{2}\right) \label{evboundf}
\end{equation}
for the free-electrons limit.
For instance, Eqs.\ (\ref{evbounds}), (\ref{evboundc}), and (\ref{evboundf}) lead to
\begin{equation}
  \label{evbounds:half}
  1\leqslant{}E_v\leqslant{}2\ \ \
  \text{for}\ \ \ N_{\rm el}=N,
\end{equation}
and
\begin{equation}
  \label{evbounds:quart}
  \frac{3}{2}\leqslant{}E_v\leqslant{}4-\frac{3}{2}\log_2{}3
  \approx{}1.623\ \ \
  \text{for}\ \ \  N_{\rm el}=\frac{N}{2}.
\end{equation}
Remarkably, the strong-correlations limit corresponds to the minimal $E_v$, whereas the free-electrons limit corresponds to the maximal $E_v$ (for any $0\leqslant{}n\leqslant{}2$). This is because when the free-electrons limit is approached, ground-state wavefunction can be truncated by a~single Slater determinant composed of fully delocalized single-particle functions (Bloch functions), resulting in the maximal entanglement between $j$-th site and the remaining $N-1$ sites \cite{Ryc06d,Ste11}. Repulsive interactions between electrons lead to the correlation-induced suppression of $\mbox{Var}\{n_j\}$, and to the decreasing $E_v$ with growing the interaction-to-bandwidth ratio.

\begin{figure}
\centerline{\includegraphics[width=0.9\linewidth]{evon-half}}
\caption{ \label{fig:evon-half}
  Local entanglement $E_v$ [see Eqs.\ (\ref{evdef})--(\ref{loceltinv})]
  as a~function of the interatomic distance $R/a_0$ for $N=10-16$
  and $N_{\rm el}=N$. Datapoints for $N=16$ are shown unmodified, the others
  were shifted vertically by $0.1$ ($N=14$), $0.2$ ($N=12$),
  or $0.3$ ($N=10$). Horizontal dashed line marks $E_v^{\rm free}(n)$
  given by Eq.\ (\ref{evboundf}); other lines are guides for the eye only.
}
\end{figure}

\begin{figure}
\centerline{\includegraphics[width=0.9\linewidth]{evon-quart}}
\caption{ \label{fig:evon-quart}
  Same as Fig.\ \ref{fig:evon-half}, but for $N=8-20$ and $N_{\rm el}=N/2$.
  The vertical shifts are: $0$ ($N=20$), $0.01$ ($N=16$), $0.02$ ($N=12$),
  and $0.03$ ($N=8$).
}
\end{figure}

General findings, presented briefly in the above, are now illustrated with the numerical examples for  $N_{\rm el}=N$ and $N_{\rm el}=N/2$ (see  Figs.\ \ref{fig:evon-half} and \ref{fig:evon-quart}). In both cases, the values of $E_v$ are close to the upper bounds given by Eqs.\ (\ref{evbounds:half}) and (\ref{evbounds:quart}) for the smallest considered value of $R/a_0=1.5$, and systematically decrease with growing $R$,  gradually approaching the lower bounds in Eqs.\ (\ref{evbounds:half}) and (\ref{evbounds:quart}). Also, a remarkably fast convergence of $E_v$ with growing $N$ is observed for any $R$, making it necessary to apply vertical shifts to the datasets in Figs.\  \ref{fig:evon-half} and \ref{fig:evon-quart}.
We further notice that $E_v\approx{}3/2$ starting from $R\gtrsim{}5\,a_0$ for $N_{\rm el}=N/2$, whereas for $N_{\rm el}=N$ we still have $E_v>1$ in this range. However, the smooth evolution of $E_v$ with $R$ is observed for both $N_{\rm el}=N$ and $N_{\rm el}=N/2$, making it difficult to consider the local entanglement as an estimate of the Mott transition. Pairwise entanglement is discussed next as the other candidate.

\subsection{The fermionic concurrence}
We consider now the subsystem $A$ consists of two spatially-separate quantum bits ({\em qubits}), one of which is associated with $i$-th lattice site and the other with $j$-th site. Each individual qubit can be realized employing charge or spin degrees of freedom.

For instance, if {\em charge qubits} are under consideration, one can choose the basis set for $A$ in a~fixed-spin sector $\sigma=\,\uparrow$, namely
\begin{align}
  \left\{\,|\alpha\rangle\,\right\}^{c} =& \left\{\,
    {|0\rangle}_i\otimes{|0\rangle}_j,
    {|0\rangle}_i\otimes{|\!\uparrow\rangle}_j, \right. \nonumber \\
    & \ \ \left.
    {|\!\uparrow\rangle}_i\otimes{|0\rangle}_j,
    {|\!\uparrow\rangle}_i\otimes{|\!\uparrow\rangle}_j
  \,\right\},
\end{align}
whereas for {\em spin qubits} we have
\begin{align}
  \left\{\,|\alpha\rangle\,\right\}^{s} =& \left\{\,
    {|\!\uparrow\rangle}_i\otimes{|\!\uparrow\rangle}_j,
    {|\!\uparrow\rangle}_i\otimes{|\!\downarrow\rangle}_j, \right. \nonumber \\
    & \ \ \left.
    {|\!\downarrow\rangle}_i\otimes{|\!\uparrow\rangle}_j,
    {|\!\downarrow\rangle}_i\otimes{|\!\downarrow\rangle}_j
  \,\right\}.
\end{align}
The corresponding projection operators are
\begin{multline}
  \left(P_\alpha^{\,c\,}\right)=
  \left[\,(1-n_{i\uparrow})(1-n_{j\uparrow}),\ (1-n_{i\uparrow})n_{j\uparrow},\ \right. \\
  \left. n_{i\uparrow}(1-n_{j\uparrow}),\ n_{i\uparrow}n_{j\uparrow}\,\right]^T,
\end{multline}
and
\begin{multline}
  \left(P_\alpha^{\,s\,}\right)=
  \left[\,
    n_{i\uparrow}(1-n_{i\downarrow})n_{j\uparrow}(1-n_{j\downarrow}),\
  \right. \\
    n_{i\uparrow}(1-n_{i\downarrow})(1-n_{j\uparrow})n_{j\downarrow},\ \\
    (1-n_{i\uparrow})n_{i\downarrow}n_{j\uparrow}(1-n_{j\downarrow}),\ \\
  \left.
    (1-n_{i\uparrow})n_{i\downarrow}(1-n_{j\uparrow})n_{j\downarrow}\,
  \right]^T,
\end{multline}
with the upper indices ($c,s$) referring to charge and spin qubits (respectively). Subsequently, the transfer operators are given by
\begin{equation}
  \left(\,T_{\alpha\alpha'}^{\,c}\right)=
  \left(
    \begin{array}{cccc}
      1 & c_{j\uparrow} & c_{i\uparrow} & c_{j\uparrow}c_{i\uparrow} \\
      c_{j\uparrow}^\dagger & 1 & c_{j\uparrow}^{\dagger}c_{i\uparrow} & c_{i\uparrow} \\
      c_{i\uparrow}^\dagger & c_{i\uparrow}^{\dagger}c_{j\uparrow} & 1 & -c_{j\uparrow} \\
      c_{i\uparrow}^{\dagger}c_{j\uparrow}^{\dagger} & c_{i\uparrow}^{\dagger} &
      -c_{j\uparrow}^{\dagger} & 1 \\
    \end{array}
  \right),
\end{equation}
and
\begin{equation}
  \left(\,T_{\alpha\alpha'}^{\,s}\right)=
  \left(
    \begin{array}{cccc}
      1 & S_j^+ & S_i^+ & S_j^+{}S_i^+ \\
      S_j^- & 1 & S_j^-{}S_i^+ & S_i^+ \\
      S_i^- & S_i^-{}S_j^+ & 1 & S_j^+ \\
      S_i^-{}S_j^- & S_i^- & S_j^- & 1  \\
    \end{array}
  \right),
\end{equation}
with the spin operators $S_i^+=c_{i\uparrow}^\dagger{}c_{i\downarrow}$,  $S_i^-=c_{i\downarrow}^\dagger{}c_{i\uparrow}$.
Substituting the above expressions into Eq.\ (\ref{rhostar}) we get
\begin{equation}
  \label{rhoalpx}
  (\,\rho_{\alpha\alpha'}^X)=
  \left(\begin{array}{cccc}
    u_+^X & 0       & 0     & 0 \\
    0     & w_1^X   & z^X   & 0 \\
    0     & (z^X)^\star & w_2^X & 0 \\
    0     & 0       & 0     & u_-^X \\
  \end{array}\right)
  \ \ \ \ \text{for}\ X=c,s,
\end{equation}
where the total spin and charge conservation is imposed. The nonzero matrix elements in Eq.\ (\ref{rhoalpx}) are given by
\begin{gather}
  u_+^c = \langle{(1-n_{i\uparrow})(1-n_{j\uparrow})}\rangle, \ \ \ \
  w_1^c = \langle{n_{i\uparrow}(1-n_{j\uparrow})}\rangle,
  \nonumber \\
  z^c = \langle{c^{\dagger}_{j\uparrow}c_{i\uparrow}}\rangle, \ \ \ \
  w_2^c = \langle{(1-n_{i\uparrow})n_{j\uparrow}}\rangle,
  \nonumber \\
  u_-^c = \langle{n_{i\uparrow}n_{j\uparrow}}\rangle,  \label{uwzchar}
\end{gather}
for charge qubits, or by
\begin{align}
  u_+^s &= \langle{
    n_{i\uparrow}(1-n_{i\downarrow})n_{j\uparrow}(1-n_{j\downarrow})
  }\rangle, \nonumber \\
  w_1^s &= \langle{
    n_{i\uparrow}(1-n_{i\downarrow})(1-n_{j\uparrow})n_{j\downarrow}
  }\rangle, \nonumber \\
  z^s &= \langle{S_j^+S_i^-}\rangle =
  \langle{
    c^{\dagger}_{j\uparrow}c_{j\downarrow}c^{\dagger}_{j\downarrow}c_{j\uparrow}
  }\rangle, \nonumber \\
  w_2^s &= \langle{
    (1-n_{i\uparrow})n_{i\downarrow}n_{j\uparrow}(1-n_{j\downarrow})
  }\rangle,
  \nonumber \\
  u_-^s &= \langle{
    (1-n_{i\uparrow})n_{i\downarrow}(1-n_{j\uparrow})n_{j\downarrow}
  }\rangle,  \label{uwzspin}
\end{align}
for spin qubits.

We use now the {\em concurrence} ${\cal C}$, as a~quantitative measure of quantum entanglement in the two-qubit subsystem $A$. The closed-form expression was derived by Wootters \cite{Woo98} and reads
\begin{equation}
\label{conc:def}
  \mathcal{C} = \max\left\{0,\sqrt{\lambda_1}-\sqrt{\lambda_2}-
  \sqrt{\lambda_3}-\sqrt{\lambda_4}\right\},
\end{equation}
where $\lambda_1\geqslant\lambda_2\geqslant\lambda_3\geqslant\lambda_4$ are eigenvalues  of the matrix product
\begin{equation}
  \label{matprod}
  \rho\cdot (\sigma_i^y\otimes\sigma_j^y)\rho^* (\sigma_i^y\otimes\sigma_j^y)
\end{equation}
with $\rho=(\,\rho_{\alpha\alpha'}^X)$ given by Eqs.\ (\ref{rhoalpx}), (\ref{uwzchar}), and (\ref{uwzspin}), and $\sigma_i^y$ ($\sigma_j^y$) being the second Pauli matrix acting on the qubit associated with $i$-th ($j$-th) lattice site. It was also shown in Ref.\ \cite{Woo98} that the entanglement of formation for a~pair of qubits was uniquely determined by ${\cal C}$, namely
\begin{equation}
  E_f=-\xi\log_2\xi-(1-\xi)\log_2(1-\xi),
\end{equation}
where $\xi=(1+\sqrt{1-\mathcal{C}^2})/2$. In fact, ${\cal C}$ can be interpreted as a~distance between a~given quantum state and the nearest separable state \cite{Woo01}. For these reasons, ${\cal C}$ can be used to quantify the entanglement of two qubits instead of $E_f$, and is hereinafter called {\em pairwise entanglement}.

The eigenvalues of the matrix defined by Eq.\ (\ref{matprod}) can be written as
\begin{align}
  \tilde{\lambda}_1 &= \tilde{\lambda}_4 = u_+^X u_-^X, \nonumber \\
  \tilde{\lambda}_{2,3} &= \left(\sqrt{w_1^Xw_2^X}\pm|{z^X}|\right)^2,
\end{align}
where $\tilde{\lambda}_1,\dots,\tilde{\lambda}_4$ are yet unsorted. After some straightforward steps, Eq.\ (\ref{conc:def}) leads to
\begin{equation}
  \label{conc:valx}
  {\cal C}=2\max\left\{0,|z^X|-\sqrt{u_+^X u_-^X}\right\},
\end{equation}
where the relevant correlation functions are given by Eq.\ (\ref{uwzchar}) for $X=c$, or by Eq.\ (\ref{uwzspin}) for $X=s$. The correlation functions $w_1^X$ and $w_2^X$ are absent in Eq.\ (\ref{conc:valx}) as they contribute to $\sqrt{\lambda_1}-\sqrt{\lambda_2}-\sqrt{\lambda_3}-\sqrt{\lambda_4}$ if and only if  $\sqrt{\lambda_1}-\sqrt{\lambda_2}-\sqrt{\lambda_3}-\sqrt{\lambda_4}<0$.

Generalizations of ${\cal C}$ for multifermionic states, appearing when larger subsystem $A$ is under consideration, are also possible \cite{Eck02}. Such generalizations are, however, beyond the scope of this paper. We focus here on the pairwise entanglement, mainly because the form of Eq.\ (\ref{conc:valx}) makes it possible to be determined with no significant computational costs once standard (spin or charge) correlation functions are calculated.

\begin{figure}
\centerline{\includegraphics[width=0.9\linewidth]{conc-half-mk16}}
\caption{ \label{fig:conc-half}
  Nearest-neighbor
  pairwise entanglement calculated from Eq.\ (\ref{conc:valx}) for charge
  (top panel) and spin (bottom panel) degrees of freedom;
  $N=N_{\rm el}=10-16$.
  The vertical shifts, applied to the datasets in both panels, are:
  $0$ ($N=16$), $0.01$ ($N=14$), $0.02$ ($N=12$), and  $0.03$ ($N=10$).
  Lines are guides for the eye only.
}
\end{figure}

\begin{figure}
\centerline{\includegraphics[width=0.9\linewidth]{conc-quart-mk16}}
\caption{ \label{fig:conc-quart}
  Same as Fig.\ \ref{fig:conc-half}, but for $N=8-20$ and $N_{\rm el}=N/2$.
  The vertical shifts, applied to the datasets in the top panel only,
  are: $0$ ($N=20$), $0.01$ ($N=16$), $0.02$ ($N=12$), and  $0.03$ ($N=8$).
}
\end{figure}

Our numerical results for the pairwise entanglement are presented in Figs.\ \ref{fig:conc-half} and \ref{fig:conc-quart}, constituting the central findings of this paper. The presentation is limited to the cases when sites $i$ and $j$ are the nearest neighbors in a~linear chain (i.e., $j=i\pm{}1\,$mod$\,N$), as we have found that ${\cal C}=0$ for more distant neighbors \cite{cdistfoo}. Also, correlation functions in Eq.\ (\ref{conc:valx}) are averaged over the reference site number $i$, to reduce the finite-precision effects.

For $N_{\rm el}=N$, we have ${\cal C}=0$ for the charge degrees of freedom (so-called {\em charge concurrence}) provided that $R/a_0\gtrsim{}4$ (see top panel in Fig.\ \ref{fig:conc-half}).
In such a~range, the separability of a~quantum state in the position representation, appearing if a~bipartite subsystem ($A$) is selected in a~fixed-spin sector, justifies the above-used notion of a~fully-reconstructed Mott insulator for a~finite $N$ (see  Sec.\ III and Refs.\ \cite{Ryc04,Ste11}).
For smaller $R$, stronger charge fluctuations manifest itself via nonzero charge concurrence, providing an insight into the nature of a~partly-localized quantum liquid. Remarkably, spin concurrence (see bottom panel in Fig.\ \ref{fig:conc-half}) is positive for any considered $N$ and $R/a_0$, reaching the maximum in the crossover range of $4\lesssim{}R/a_0\lesssim{}5$. For this reason, spin concurrence still can be considered as a~prospective signature of the Mott transition.

For $N_{\rm el}=N/2$ the evolution of the pairwise entanglement with $R/a_0$ (see Fig.\ \ref{fig:conc-quart}) is significantly different than for the $N_{\rm el}=N$ case. For each $N$, we have a~nonzero charge concurrence for any considered $R/a_0$ (top panel in  Fig.\ \ref{fig:conc-quart}), whereas the spin concurrence (bottom panel) indicates a~finite-system version of the Mott transition \cite{lanczfoo}. In brief, for spin degrees of freedom ${\cal C}=0$ for $R>R_\star(N)$, where the nodal value $R_\star(N)$ corresponds to $|z^s|-\sqrt{u_+^s{}u_-^s}=0$,  and is determined numerically via least-squares fitting of a~line to the actual datapoints for a~given $N$ (see the last column in Table \ref{tab:rstara}). For larger $N$, the values of $R_\star(N)$ are close to $R_c\approx{}4.2\,a_0$, determined in Sec.\ III via the finite-size scaling for $\Delta{}E_C$.

Additionally, we observe that the nearest-neighbor pairwise entanglement may indicate the decoupling of charge and spin degrees of freedom in correlated systems. This happens for $R/a_0\gtrsim{}4$ in the $N_{\rm el}=N$ case (as well as for $R>R_\star(N)$ in the $N_{\rm el}=N/2$ case), where  nonzero spin (charge) concurrence is accompanied by vanishing charge (spin) concurrence. In the above-mentioned ranges, quantum states of a~bipartite subsystem ($A$) are entangled when spin, but separable when charge degrees of freedom are under consideration ($N_{\rm el}=N$), or vice versa ($N_{\rm el}=N/2$). Such a~decoupling should be distinguished from charge-spin separation predicted to appear in 1D systems showing the Luttinger liquid phase \cite{Sac13,Lie68,Voi93,Ber00}. Our numerical examples also show there is  no one-to-one relation between the decoupling and the Mott transition, as the format can be identified either for the system showing the crossover behavior only or undergoing the Mott transition (in the large-$N$ limit).
Finally, the pairwise-entanglement analysis presented above allows one to identify noticeably different behavior of charge and spin correlation functions without referring to their asymptotic behavior, as previously proposed in the literature (see the first paper in Ref.\ \cite{Ste11} and Refs.\ \cite{Sac13,Gal16}).

\section{Conclusions}
In this study we have demonstrated that pairwise entanglement, quantified by the fermionic concurrence determined separately for charge and spin degrees of freedom, can serve as a~convenient indicator for a~finite-system version of the Mott transition. In particular, standard finite-size scaling estimates of the Mott transition for linear chains of hydrogenic-like atoms are revisited utilizing the exact diagonalization -- ab initio metod (EDABI) at the half and the quarter electronic filling. In the latter case, we find that not merely the charge gap indicates the transition for $1/N\rightarrow{}0$ (with $N$ being the number of atoms) at the interatomic distance $R_c\approx{}4.2\,a_0$ (where $a_0$ denotes the Bohr radius), but also the spin concurrence vanishes for a~finite $N$ at $R_\star(N)$ taking the values relatively close to $R_c$. Charge concurrence remains nonzero in both the metallic and the insulating phases, signaling a~decoupling between charge and spin degrees of freedom.

At the half filling the Mott transition is not observed. Instead, the crossover from a~partly localized quantum liquid to a~fully-reconstructed Mott insulator occurs. The charge concurrence vanishes at the crossover point, where the spin concurrence shows a~broad peak; the latter remains nonzero in the entire parameter range.

It is worth to stress here that calculations of the fermionic concurrence, employed in this paper, generate essentially no extra computational costs, as the concurrence is determined solely via pairwise ground-state correlation functions for charge or spin degrees of freedom, without referring to the dynamical properties such as the optical or {\em dc} conductivity.
The analytic relations between entanglement and the correlation functions [see Eq.\ (\ref{conc:valx})] are also derived in this paper.

A~striking feature is that the analysis holds true regardless how the exact (or approximated) ground state is obtained, making it possible to employ the entanglement-based signatures of the Mott transition when discussing a~generic correlated-electron system.
What is more, Eq.\ (\ref{conc:valx}) also applies when correlation functions on its right-hand side are determined using more powerful numerical techniques, such as DMRG or QMC.
For these reasons,
we believe the approach we propose may shed new light on various open problems in the field, such as the long-standing metallization of solid hydrogen \cite{Kad14}, or the recently-raised phase diagram of the repulsive Hubbard model on a~honeycomb lattice \cite{spnlqd}.

\section*{Acknowledgements}
I thank Andrzej Biborski, Andrzej K\k{a}dzielawa, and Prof.\ J\'ozef Spa{\l}ek for many discussions,  and Yichen Huang for pointing out the role of asymptotic properties of the reduced density matrix.
The work was supported by the National Science Centre of Poland (NCN)
via Programme SONATA BIS, Grant No.\ 2014/14/E/ST3/00256.
Computations were performed using the PL-Grid infrastructure.

redpajama_set_name: RedPajamaArXiv


text:
Beto O'Rourke in a pinch, overshadowed by Joe Biden's experience and Pete Buttigieg's fresher face
Filed under Politics at May 12
Todd J. Gillman, Washington Bureau Chief
Connect with Todd J. Gillman
EXETER, N.H. - Joe Biden's entry into the 2020 campaign halted the momentum of a number of rivals. Beto O'Rourke was among the casualties.
Like 20 other contenders, the Texan can't touch the former vice president's experience. But at least he could harness a thirst for youthful energy and an upbeat demeanor -- until small-town mayor Pete Buttigieg boxed him out on that flank, becoming the flavor of the month.
How long can the Texan survive in these twin shadows?
"Obama was the underdog by a longshot. It's still early," said Ross Mainville, 30, a union carpenter who came to see O'Rourke in Salem, N.H., with his wife Kirsten.
She's 28 and works in insurance and became such a fan last year that she started sending $5 or $10 a month for his Senate race.
They describe themselves as "very liberal." They love Vermont Sen. Bernie Sanders and New York Rep. Alexandria Ocasio-Cortez, a self-described socialist. They see O'Rourke as a centrist -- and like him because of that, not despite of it, because it makes him more electable.
"He hits a note with a lot of people," Kirsten Mainville said. "He bridges the gap, where I haven't seen it in the other 22 yet. If there were just three people he would be way up there."
Biden now commands support from two in five Democratic primary voters nationwide. That's double the support for the current runner-up, Sanders, who has slipped dramatically from a high of about 27%. Put another way, the top two contenders have soaked up 60% of support, leaving 20 others scrambling for crumbs.
"There's a lot of time left, a lot of people to meet, a lot of town halls to hold," O'Rourke told reporters in Salem.
After eight years as vice president, many Democrats view the 76-year-old Biden as the most electable candidate in the era of Trump, despite any qualms about his age, relative centrism, handling of the Clarence Thomas confirmation hearings and fizzled campaigns of 1988 and 2008.
A Quinnipiac University Poll released April 30 found that 56% of Democrats said Biden has the best chance of beating President Donald Trump. He was the top choice of 38%, up from 29% a month earlier, as O'Rourke's support plunged from 12% to 5%.
Biden holds double-digit leads in Iowa and New Hampshire. In South Carolina, nearly half the Democratic voters pick him over anyone else -- and Buttigieg places third in all three states.
"No one thinks Biden is the future of the party. They're just hoping that he can restore the past," said Dante Scala, a University of New Hampshire political scientist. "I'm not sure it's so much excitement" as a sense that he's "a safe spot for voters to land" as the race plays out.
"At least for now, Biden has put Beto pretty far back in the pack," he said.
O'Rourke won't be outworked, though.
Biden hits New Hampshire on Monday and Tuesday for his first foray there. He has three events scheduled -- a relatively leisurely pace. Unlike O'Rourke, Biden is not heading to living rooms to schmooze a couple dozen voters at a time.
O'Rourke defends his approach and shrugs off the suggestion that he's lost momentum.
"If we're only looking for short-term returns and a bounce in the polls next week, there's one strategy that we could use," he told reporters in New London, N.H. "If over the long term we want to build a committed movement of voters and caucus-goers, then there's another strategy. ...You don't see us doing a ton of big rallies. You don't see me doing a lot of television."
That's changing, though. He's booked Monday night on Rachel Maddow's show on MSNBC, and the next morning on ABC's The View.
"Maybe it's good to do a slow build as opposed to immediately becoming a front-runner," Scala said, "with all of the attention that goes with that, and all of the scrutiny."
Beto O'Rourke stumps at a house party in Lebanon, N.H., on May 10, 2019.
(Todd J. Gillman/Staff)
Beto O'Rourke stumps at a house party in Bedford, N.H., on May 11.
(Todd J. Gillman
Beto O'Rourke autographs a campaign sign after an outdoor town hall event in Exeter, N.H., at Town Hall Common on May 11, 2019. He wrote "Daniel! Suerte con todo! Beto." ("Good luck with everything!")
Beto O'Rourke stumps at the Town Hall Common in Exeter, N.H., on May 11, 2019.
In Salem, N.H., Beto O'Rourke stumped for president on May 9, 2019. Kirsten and Ross Mainville of Fremont, N.H., each put him in their top three in the Democratic primary, even though they view him as a moderate and describe themselves as "very liberal." For them, electability is paramount, and they're impressed at his ability to connect with voters and his sincerity at trying to bridge divides.
Stolen thunder
Biden isn't the only candidate putting a squeeze on O'Rourke.
He's also has been eclipsed by Buttigieg, mayor of Indiana's fourth-largest city. If South Bend were in Texas, it would rank 38th in population -- behind Denton, Allen, McKinney, Plano, Richardson and Garland, among others just in North Texas.
"Buttigieg stole his thunder," said Scala.
O'Rourke only sort of qualified as a fresh face on the national scene, coming off a Senate race in which he raked in a record-smashing $80 million. By contrast, Scala said, "Buttigieg was the unknown phenom, out of nowhere. I don't know if it lasts, but ... even for someone like O'Rourke, it's tough to be the fresh face forever."
O'Rourke is hardly the only rival suffering the indignity of trailing a small town mayor with no national political experience.
Unfazed by sinking poll numbers, Beto O'Rourke keeps plugging away on presidential bid
Progressives -- a key part of the Democratic base -- rank Buttigieg third, behind Sanders and Massachusetts Sen. Elizabeth Warren and ahead of Biden and everyone else. Straw polls taken by Democracy for America, a leading progressive group, showed O'Rourke in second place in December; he has since plummeted to seventh.
"Beto needs to find some way to stay relevant," said Charles Chamberlain, chairman of Democracy for America. "The biggest threat is Buttigieg."
Progressives, he said, look askance at O'Rourke and Biden for some of the same reasons -- their support for NAFTA and other trade deals. Biden also faces doubts on the left because he opposed school busing and supported a bankruptcy bill viewed as favoring credit card companies over consumers.
Both support a national $15 minimum wage and other worker protections, cheaper college, repeal of the Trump tax cuts, and expanded health coverage.
O'Rourke has long called for ending the federal prohibition on marijuana. Decriminalization is a central element of his agenda, along with a vow to expunge the records of people convicted of minor drug offenses.
He's never drawn the linkage to Biden. But this could be a line of attack later. As a senator, Biden backed a 1994 crime bill that led to mass incarcerations. He also backed harsher sentences for crack cocaine offenses, which O'Rourke and others blame for racial disparities in prison -- positions for which Biden now expresses regret.
Still unsettled
The race is anything but settled.
"This time four years ago, Mike Huckabee was the great hope of the Republican Party," noted Sam Sands, 30, a public policy researcher at the University of New Hampshire who drove across the state at a friend's urging to see O'Rourke at Dartmouth College on Friday.
He likes that O'Rourke keeps his distance from the party line. That, in his view, "is the way out of this dark, dark place" in American politics. But he's also keeping an eye on Buttigieg, because "he's got a real generational message.... I'm not super excited about Biden."
Micah Hundley, 31, a Lubbock native studying for a masters degree in globalization at Dartmouth, recalled that Howard Dean and John Edwards went through hot streaks, only to fade once voters decided they'd rather bet on electability than a message.
"I think he has legs," he said of O'Rourke. "He reminds me of George W. Bush in 2000 -- relatable, a guy you can have a beer with. He seems like a normal reasonable guy. Not all the Democrats have that. If he can survive long enough, he could have a chance."
The first televised debates are in late June, with the huge field split into two groups sharing a stage in Miami on back-to-back nights. Voters say they're hoping for clarity and even some winnowing.
O'Rourke's prep work has already begun. When he's home in El Paso, he said, "I'm reading a ton."
"I haven't done any mock debates," he said. "I'm sure we will. But no, we haven't gotten that far."
Biden is a seasoned debater with a checkered record, capable of both a sizzling zinger and a painful gaffe. O'Rourke's high-stakes debate experience is confined to two televised showdowns against Sen. Ted Cruz. He held his own but couldn't stop the senator from painting him as too liberal for Texas.
There won't be much time to make an impression.
"Somebody did some math and determined that ... each candidate's going to have between four to seven minutes to talk, total," O'Rourke said. "I hope that it is a true competition of ideas and policies... That's my plan. That's what I hope to be able to show when I'm on that stage."
Trump's take
Trump declared Biden the prohibitive front-runner on Friday, telling Politico he sees parallels to his own quick ascent in 2016. He also mocked Buttigieg, comparing him to a Mad magazine character: "Alfred E. Neuman cannot become president of the United States."
In March, he told Fox Business Network that he relished a challenge from Biden, Sanders or O'Rourke: "The press seems to have chosen Beto. And when I watch Beto, I say, 'We could dream about that.'"
With O'Rourke's prospects waning, Trump taunted him Wednesday night during a rally in Panama City Beach, Fla., recalling the rally he'd held in El Paso -- the one that gave O'Rourke the opening to stage a mass counter-rally and reclaim attention he'd lost as he weighed his options.
"I went to Texas and made a speech. And Beto -- Beto, boy has he fallen like a rock," Trump told the crowd. "What the hell happened to Beto?"
That was five days after O'Rourke came out forcefully for impeachment.
Shallow support for Biden
Anecdotally, it appears that support for Biden may be shallow.
In dozens of interviews with uncommitted New Hampshire Democrats in the last few days, few expressed enthusiasm. They respect Biden and would eagerly support him if he wins the nomination. Most, however, view him as a safe -- and perhaps temporary -- choice as the race plays out.
"He's a known quantity," said Elizabeth Lowell, 75, a retired teacher. She added, "I don't think that's what we need in 2020."
She came to hear O'Rourke on Friday at Colby-Sawyer College in New London, where he spoke to about 100 voters in a library built from two Civil War-era barns. "Joe Biden and Bernie Sanders are my generation. I like both of them but they need to step aside," she said.
Robert Rushton, 59, a software engineer from Brookline, N.H., said Biden represents "incremental change" at a moment that calls for bold leadership. But he would still support him if it comes to that.
"I wouldn't turn him down if he happened to be the nominee," he said. "He's the darling of the moment."


I really like paying with this stamp set so I thought I'd share another card I made with the dress and mannequin.
I usually try to keep away from paper piecing but this stamp just begs for it in my opinion.
I love this pearl strip and thought it was a perfect accent on this card.


Kristen Stewart has been dogging the media for quite some time now after the cheating scandal. Kristen Stewart has finally made a public appearance. The Twilight star was promoting her new film On the Road, an adaptation of Jack Kerouac's novel, at its premiere during this year's Toronto International Film Festival (TIFF).
As always Kristen Stewart was stunning!
Kristen skipped out of the film's London premiere last month over the fallout from her "indescretion" when she cheated on boyfriend of three years Robert Pattinson. Fans branded her a "Trampire" after she confessed to stories of her infidelities with her Snow White and the Huntsman director Rupert Sanders.
K-Stew wore a black lace Zuhair Murad floral dress with lots of scarlet flowers on it (but no scarlet letters) for the TIFF premiere and signed plenty of autographs for forgiving fans. She was joined by co-stars Garrett Hedlund and Kirsten Dunst.
After her initial jitters over her first public appearance after her split from Pattinson, Kristen revealed that she was given a warm welcom by fans.
Meanwhile, ex-Pattinson was in Los Angeles Thursday night for the MTV Video Music Awards to present fans with the latest trailer for The Twilight Saga: Breaking Dawn – Part 2. The film opens in November; and we're sure there will be plenty of tension on that red carpet.
SNL & Daniel Craig Parodies James Bond!


text:
La English Football League Championship 2021-22, también llamada Sky BET Championship por razones de patrocinio, es la sexta temporada bajo el nombre de English Football League y la vigésima edición de la Football League Championship desde su fundación en 2004. Un total de 24 equipos disputan la liga, incluyendo 18 equipos de la English Football League Championship 2020-21, tres relegados de la Premier League 2020-21 y tres promovidos de la English Football League One 2020-21.

Relevos

Información

Cambios de entrenadores

Localización

Condados de Inglaterra

Condados preservados de Gales

Desarrollo

Clasificación
Los dos primeros equipos de la clasificación ascienden directamente a la Premier League, los clubes ubicados del tercer al sexto puesto disputan un play-off para determinar un tercer ascenso.

Evolución

Resultados
 Los horarios corresponden al huso horario del Reino Unido (Hora de Europa Occidental): UTC+0 en horario estándar y UTC+1 en horario de verano.

Partidos

|}

Tabla de resultados cruzados

Play-offs

Cuadro de desarrollo
{{Copa
| RD1= Semifinales
| RD2= Final
| group1=
| group2=
| group3=
| group4=
| group5=

| seed-width=20
| team-width=200
| score-width=20

| RD1-date=13 y 14 de mayo de 2022 (ida)16 y 17 de mayo de 2022 (vuelta)
| RD1-seed01=3
| RD1-team01=Huddersfield Town
| RD1-score01-1= 1
| RD1-score01-2= 1
| RD1-score01-3= 2
| RD1-seed02=6
| RD1-team02= Luton Town
| RD1-score02-1= 1
| RD1-score02-2= 0
| RD1-score02-3= 1

| RD1-seed03=4
| RD1-team03= Nottingham Forest
| RD1-score03-1= 2
| RD1-score03-2= 1
| RD1-score03-3= 3
| RD1-seed04=5
| RD1-team04= Sheffield United
| RD1-score04-1= 1
| RD1-score04-2= 2
| RD1-score04-3= 3

| RD2-date=29 de mayo de 2022
| RD2-seed01=3
| RD2-team01=Huddersfield Town
| RD2-score01-1= 0
| RD2-seed02=4
| RD2-team02=Nottingham Forest
| RD2-score02-1= 1
}}

 Semifinales
Los horarios corresponden al huso horario de verano británico BST (UTC+1).

 Final
El horario corresponde al huso horario de verano británico BST (UTC+1).

 Estadísticas
 Goleadores

 Tripletes o más

 Máximos asistentes

 Disciplina
 Jugadores
 Más tarjetas amarillas: 10 Matt Crooks

 Más tarjetas rojas: 2 Gary Gardner (Birmingham City)
 Fakanty Dabo (Coventry City)
 Jefferson Lerma (Bournemouth)
 Jake Livermore (West Bromwich Albion)
 Ryan Manning (Swansea City)
 Jordan Lawrence-Gabriel (Blackpool)

 Clubes
 Más tarjetas amarillas: 62
Reading F. C.
 Más tarjetas rojas: 6'''
 Derby County F. C.
West Bromwich Albion F.C.

Véase también
 Premier League 2021-22
 League One 2021-22
 League Two 2021-22
 National League 2021-22
 National League North 2021-22
 National League South 2021-22
 FA Cup 2021-22
 EFL Cup 2021-22
 English Football League Championship

Referencias

2022
Fútbol en 2021
Fútbol en 2022
Deporte en Reino Unido en 2021
Deporte en Reino Unido en 2022

redpajama_set_name: RedPajamaWikipedia