text:
Капитолина Григорьевна Барко́вская (1920—2010) — советская российская актриса. Заслуженная артистка РСФСР (1985). Лауреат Государственной премии РСФСР им. К. С. Станиславского (1985).

Биография
На профессиональную сцену пришла из самодеятельности.

В 1936—1946 годах работала в драматических театрах Алдана (Якутская АССР), Моздока, Элисты, Читы,

в 1946—1956 годы — актриса Омского ТЮЗа. Среди ролей; Беатриче («Слуга двух господ» К. Гольдони), Джема («Овод» Э. Войнич), Агафья Тихоновна («Женитьба» Н. Гоголя) и другие.

В 1956—1993 годы — одна из ведущих актрис Омского академического театра драмы. В её репертуаре как характерные, так и драматические роли: Ксения («В старой Москве» В. Пановой), Монахова («Варвары» М. Горького), пани Завальска («Сверчок» Т. Кожушника), Ракитина («Дали неоглядные» Н. Вирты), тётя Тони («Проснись и пой» М. Дьярфаша), Королева Елизавета («Мария Стюарт» Ф. Шиллера), Регана («Король Лир» У. Шекспира), Анна («Прошлым летом в Чулимске» А. Вампилова), Вера Сергеевна («Энергичные люди» В. Шукшина), Зыбкина («Правда — хорошо, а счастье — лучше» А. Островского), Алла Васильевна («С вечера до полудня» В. Розова), Кукушкина («Доходное место» А. Островского)., Морозова («У войны — не женское лицо» С. Алексиевич).

В 1994—2004 годы — на сцене Омского государственного драматического «Пятого театра». Играла в спектаклях «Фиктивный брак» В. Войновича, «Светит, да не греет» А. Н. Островского.

Исполнила главную роль Марии Морозовой в фильме-телеспектакле Геннадия Тростянецкого и Ольги Соковых «У войны — не женское лицо» (1988).

Как режиссёр возглавляла любительский драматический коллектив медицинского института.

В 1970-е годы — депутатом Омского городского совета шести созывов, возглавляла комиссию по культуре.

Похоронена в Омске на Старо-Северном кладбище.

Театральные работы

Омский драматический театр

 Ксения - «В старой Москве», В. Панова (реж. С. Владычанский);
 Пани Завальска, владелица галантерейной лавочки – «Сверчок», Т. Кожушник (реж. И. Копытман);
 Нина Сергеевна – "Четыре капли", В. Розов (реж. В. Тумилович)
 Клава – «Хочу быть честным», В. Войнович (реж. И. Копытман);
 Калугина Людмила Прокофьевна – «Сослуживцы», Э. Рязанов, Э. Брагинский (реж. Я, Киржнер, А. Хайкин);
 Рябинина – «Человек со стороны», И. Дворецкий (реж. Я. Киржнер);
 Звягинцева – «Ясная Поляна», Д. Орлов (реж. Я. Киржнер);
 Хозяйка – «Преступление и наказание», Ф. Достоевский (реж. Я. Киржнер);
 Кукушкина - «Доходное место», А. Островский (реж. А. Хайкин);
 Людмила Петровна – «Проходной балл», В. Константинов, Б. Рацер (реж. Л. Шварц);
 Алла Васильевна – «С вечера до полудня», В. Розов (реж. А. Хайкин);
 Репникова – "Прощание в июне", А. Вампилов (реж. Л. Шалов)
 Мать Алёши – "Муж и жена снимут комнату", М. Рощин (реж. Л. Шалов)
 Алена – "Беседы при ясной луне", В. Шукшин (реж. Н. Мокин)
 Комариха – "Деньги для Марии", В. Распутин (реж. В. Симановский)
 Вера Сергеевна - «Энергичные люди», В. Шукшин, (реж. Я. Киржнер);
 Анна Васильевна Хороших, мать Пашки, буфетчица – «Прошлым летом в Чулимске», А. Вампилов (реж. А. Хайкин)
 Надежда Монахова - «Варвары», М. Горький, (реж. Ю. Альховский);
 Ракитина - «Дали неоглядные», Н. Вирта, (реж. Ю. Альховский);
 тётя Тони - «Проснись и пой», М. Дьярфаш (реж. Л. Дурасов);
 Королева Елизавета - «Мария Стюарт», Ф. Шиллер, (реж. Ф. Шейн);
 Девица – «Старик» М. Горький (реж. Б. Малкин)
 Воронкова – «Ретро» А. Галин (реж. В. Пази)
 Морозова Мария Ивановна, ефрейтор, снайпер стрелковой дивизии – «У войны не женское лицо», С. Алексиевич (реж. Г. Тростянецкий)
 Зыбкина - «Правда — хорошо, а счастье — лучше», А. Островский (реж. А. Хайкин)
 Марфа Петровна, бригадир-овощевод – «Всего три дня», Н. Анкилов (реж. А. Хайкин);
 Евгения Семеновна – "Тревога", Н. Анкилов (реж. С. Бульба)
 Обноскина Анфиса Петровна – "Село Степанчиково и его обитатели", Ф. Достоевский (реж. А. Хайкин)
 Прасковья Маслова – "Солдатская вдова", Н. Анкилов (реж. Я. Киржнер)
 Гертруда, королева Датская, мать Гамлета – «Гамлет», У. Шекспир (реж. Я. Киржнер);
 Бьюла Биннингс – "Орфей спускается в ад", Т. Уильямс (реж. А. Хайкин)
 Софья Игнатьнвна – "Смотрите, кто пришел", В. Арро (реж. Г. Тростянецкий, С. Коромщиков)
 Мать Валентины – «Валентин и Валентина», М. Рощин (реж. А. Хайкин);
 Регана - «Король Лир», У. Шекспир (реж. Г. Тростянецкий).

Награды и звания
 Государственная премия РСФСР имени К. С. Станиславского (1985) — за исполнение роли Марии Морозовой в спектакле «У войны не женское лицо» С. А. Алексиевич
 Премия «Легенда омской сцены» (2001)
 Орден Трудового Красного Знамени (1971)
 Медаль «За доблестный труд в Великой Отечественной войне 1941—1945 гг.» (1992)

Ссылки
http://www.gtrk-omsk.ru/news/detail.php?IBLOCK_ID=3&ID=49636

Персоналии:Омский академический театр драмы
Похороненные на Старо-Северном кладбище


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Another Minnesota lawmaker accused of sexual harassment
Tony Cornish
A Thursday, March 10, 2016 photo shows Representative Dan Schoen,at the Minnesota State Office Building, Thursday, March 10, 2016 in St. Paul, Minn. Senate Minority Leader Tom Bakk, Rep. Erin Murphy and state Auditor Rebecca Otto all issued statements Wednesday ... more >
By KYLE POTTER - Associated Press - Friday, November 10, 2017
ST. PAUL, Minn. (AP) - Minnesota's House speaker suspended the committee chairmanship of state Rep. Tony Cornish after two women accused him of sexual harassment, the second allegation against a sitting lawmaker in as many days.
Rep. Erin Maye Quade shared text messages with the Star Tribune from Cornish, a Republican, who said he "got busted" staring at her on the House floor in May, saying it was her fault: "Look too … good."
And a longtime female lobbyist, who asked the newspaper not to publish her name due to fear it would hurt her career, said Cornish has propositioned her for sex dozens of times and once pushed her against a wall in his office while trying to kiss her. In a 2010 incident, the lobbyist said Cornish used crude language to point out that he had an erection and said, "You can't leave."
In a string of text messages with The Associated Press, Cornish said his exchange with Maye Quade was "obviously a congenial conversation" and one of many text messages they shared over the past year. But he called the lobbyist's accusations completely false.
"There was no interaction that even resembled that in my office," Cornish said. "But with the mistrust in politicians today naturally some people will believe it."
Republican House Speaker Kurt Daudt issued a statement late Thursday announcing he had suspended Cornish's chairmanship of the Public Safety and Crime Prevention committee, where Cornish has been the face of efforts to expand gun laws and crack down on protesters who have shut down highways in the last several years. Daudt also asked the House's nonpartisan human resources staff to investigate the allegations.
Daudt called the accusations "extremely troubling."
"I spoke with Rep. Cornish and told him that his reported actions were inappropriate and unacceptable as a member of our caucus and the Legislature," Daudt said in a statement.
Asked if he had been pushed to resign, Cornish said: "I have not been asked by anyone to resign nor has it even been hinted to me by anyone. Definitely not."
The allegations against Cornish came as several of Minnesota's top Democrats urged state Sen. Dan Schoen to resign after accusations that he sexually harassed women on the campaign trail, including grabbing a woman's buttocks. The Democrat from St. Paul Park said the allegations are false or that his actions were taken out of context, and suggested he has no plans to quit.
Schoen informed Senate Democrats he had hired an attorney, a spokeswoman for Senate Democrats said late Thursday, a sign he plans to fight a potential ethics complaint.
Schoen has served in the Legislature since 2013 and also works as a police officer in the Minneapolis suburb of Cottage Grove. Cornish is a retired police officer who has served in the Legislature since 2002.
The allegations against the two Minnesota lawmakers come amid a tide of sexual harassment and assault allegations against powerful figures in politics, media and entertainment, including movie producer Harvey Weinstein.
The incidents prompted several female victims of alleged sexual harassment - including Maye Quade - to call upon Gov. Mark Dayton and other legislative leaders to establish a new state task force. They recommended the state implement a new reporting system, overhaul sexual harassment training and create a new process for lawmakers to investigate claims - one separate from the current House and Senate ethics committees, where complaints often deadlock on partisan votes.
"Minnesota state government should set the standard for zero tolerance for sexual harassment in the workplace," their statement said.
Sign up for the AP's weekly newsletter showcasing our best reporting from the Midwest and Texas: http://apne.ws/2u1RMfv

redpajama_set_name: RedPajamaCommonCrawl


text:
Zoe Saldana Says These Instagram Hashtags Helped Her Lose the Baby Weight
Cortney Clift
If there is one unsung celeb hero on Instagram, it might just be Zoe Saldana. Scroll through the actress's page and you'll find photos of her super handsome husband Marco Perego and adorable twin boys, as well as some behind-the-scenes, very real mommy moments.
As with many mommy celebs, we have to admit we're a bit curious about how she whipped herself into killer shape after giving birth to her sons Bowie and Cy back in 2014. Anne Hathaway recently shared her experience of hitting the gym post-baby, and now Zoe has also chimed in to share her secret to success.
I've got nothing to hide. #NationalUnderwearDay @calvinklein
A post shared by Zoe Saldana (@zoesaldana) on Aug 5, 2015 at 2:13pm PDT
In an interview with Allure, Zoe said, "After I had the boys, I was so inspired by the mommies [on social media] using their babies as weights, doing all their exercises in between naps and feeding times. I never went to a single gym — I exercised in my backyard! I went on YouTube. Social media isn't just there to feed into your narcissism. Search #workoutmoms, #bouncingback, whatever and you'll be flooded with women who will inspire you."
Well damn, we did not see that answer coming. While a personal trainer is the most common way to go about the post-baby bod in Hollywood, we're totally into this DIY alternative. In fact, Chris Hemsworth's wife Elsa Pataky also does the same sort of workout. Back in the beginning of 2016, she shared a video of her using her stroller to work out in her backyard. Looks like those mommy and me classes just got a lot more interesting.
How did you get in shape after having a baby? Share with us on Twitter @BritandCo.
(Photos via Frazer Harrison/Jason Merritt/Getty)
Cortney is the senior writer and special projects editor at Brit + Co. She typically covers topics related to women's issues, politics, and lifestyle. When she's not buried behind her computer you'll find her exploring New York City or planning her next big adventure.


text:
'use strict';

/**
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*
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*
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*
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*
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redpajama_set_name: RedPajamaGithub


text:
What Is The Difference Between A Screwdriver And A Screwdriver
Response to tool operation at full power
Number of charge and discharge cycles
Battery selection based on application
What does the battery voltage say?
What characteristics are important when choosing a battery for a screwdriver
Pros and cons of the impact wrench
The difference between a screwdriver and a nutrunner
The difference between a wrench and a screwdriver
How to make a wrench from a screwdriver
What are they
Comparing screwdriver and drill
Find out how a screwdriver differs from a drill
What is the difference
What is the difference between a wrench and a screwdriver: 3 main criteria
Which is better wrench or screwdriver?
How to convert a screwdriver to a wrench?
What is the difference between a wrench and a screwdriver
What is a screwdriver
What is the difference between tools
A few words about the impact wrench
Is it possible to independently make a nutrunner from a screwdriver
Electric screwdriver: differences from a screwdriver and selection criteria
Screwdriver and electric screwdriver: differences
Criteria for choosing an electric screwdriver
When you need to drill a hole in a dense material or screw a thick and long self-tapping screw, you have to use the tool at full power. In this case, active charge consumption begins, and the ability to complete the process qualitatively without a drawdown of revolutions depends on the battery's reaction to this phenomenon.
In view of this, modern Li-Ion batteries are significantly inferior to their earlier counterparts. The most powerful in this regard are NiCd and NiMh. When maximum energy output is required, then they are chosen.
All batteries have one drawback. sooner or later they run out of energy and the device must be put on charge. Depending on the type of battery, there is an average statistical limit on the number of charging cycles after which the battery will no longer be able to store current and will have to be replaced.
For screwdrivers with NiCd batteries this figure is 1000 cycles, and the tool often continues to work after reaching this figure. This makes it a leader when compared in this characteristic.
A common type of Li-Ion in screwdrivers is capable of being charged and discharged over 600 times. The advanced technology of the last century NiMh is significantly inferior to the previous ones and the battery will withstand only 300-500 cycles.
One of the new generations of batteries, which first appeared in 1991. Japanese radio engineering companies began to actively use them, and subsequently they managed to increase their power and use them with power tools.
It is this type that is used in mobile phones, laptops and electric vehicles. Their cathode and anode are made of aluminum and copper foil, between which a perforated separator is installed. A valve may be present to release internal pressure.
So, having thoroughly examined all the key factors, we will discuss with which battery to choose a screwdriver for home and professional activities. This will help you not to overpay and have a reliable tool at hand.
For cordless models, the power and rotation speed of the screwdriver depends on the battery voltage, which must correspond to the parameters of the tool:
For screwdrivers, the maximum indicator is 36 V. These are the most powerful devices for working with metal and hard wood species.
12-18 V. average value, suitable for most processes in repair and installation.
3-10 V. refer to low-power, and are used only for assembling soft materials or small diameter fasteners.
Installing a battery with a higher capacity on a weaker model will sooner damage the latter. A 12 V battery on an 18 V device will, on the contrary, limit the power of the tool, so the voltage indicator of the drive must exactly match the value on the screwdriver.
Keep in mind that if you take two batteries with the same capacity, the longer operation of the tool will be provided by the one with the higher voltage.
Considering the advantages and disadvantages of different types of batteries, you can choose the one suitable for specific operating conditions. Now let's look at two key characteristics that all types of batteries have that affect the choice of a screwdriver.
A significant advantage of the nutrunner lies in its ability to tighten fasteners with increased force and untwist rusted joints. This tool becomes an excellent assistant when working with large equipment. Unfortunately, the increased power of the instrument makes it more noisy. If you can work quietly with a screwdriver after rebound, then the wrench will quickly make itself felt.
To twist small elements with a wrench, you need to have a certain skill, otherwise there is a high risk of material damage. With a screwdriver in this regard, everything is much more convenient, albeit longer. The tuned ratchet mechanism will not allow you to twist the self-tapping screw, setting its head exactly on the surface of the material.
Summing up, we can say that the wrench is a powerful and noisy tool designed for tightening large fasteners. Suitable for domestic needs, but not everyone needs it. The screwdriver works at a lower speed. At the same time, it is very easy to use and does not require special skills and adapters.
Conclusion: The wrench is needed for those who know exactly where they will use it. The screwdriver is suitable for most household jobs.
Nowadays there are many tools for tightening fasteners, and one of the most popular representatives of this class is the screwdriver. Almost any housework will be easier if you have this wonderful assistant in your arsenal. But despite all its versatility, the screwdriver has a very similar competitor. The impact wrench has a narrower focus, but is much better suited for certain jobs. Let's figure out how these tools differ and how you can make a wrench from a screwdriver with your own hands.
The electric screwdriver covers a wide range of jobs. It is used both for tightening (screws, self-tapping screws, screws) and for drilling low-density materials. The tool has a speed regulator and can be equipped with various types of chucks (keyless or cam). Suitable for home and industrial work.
With a wrench, things are different. This tool is widely used in production for quick and high-quality fastening of parts, but more and more often it begins to appear in home kits of motorists. Available with a hex chuck or square with a friction ring. Impact wrenches are considered the most effective, although impulse-type models are very common. The main thing that distinguishes a wrench from a screwdriver is the increased power of the former. A wrench can tighten self-tapping screws, but it requires a fair amount of skill for this, since, with extra effort, it can tear off the caps or drive the fasteners deep into the board. The main everyday use of the nutrunner is to unscrew the wheels of the car and fasten the overall threaded structures. For this business, he is perfect.
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The next difference. This is the patron. Unlike a screwdriver's clamping mechanism, adapters are required to use drills or bits on a wrench.
For various reasons in everyday life, you may need to unscrew a couple of dozen nuts. It is rather expensive to buy a wrench for such a fleeting need, but you don't want to turn it manually when you have a good old screwdriver at your disposal. It performs the main rotational function, it remains to remake it for unscrewing the nuts. Everything is extremely simple here.
Humanity has long been using drills and other tools for making holes in various works.
Over time, a drill appeared, which made it possible to mechanize the manual labor of drillers.
After the invention of the electric motor, a drill was invented, in which the drill rotated with the help of such a motor.
In various sectors of the national economy and in everyday life, the most different types of drills are used, making holes in materials of any hardness.
In addition, attachments attached to them began to be used, allowing them to perform grinding and cutting work. Drills began to be used for loosening and tightening screws.
Today, pneumatic, electric and even gasoline drills are common. The most popular are electric drills, which are used by both professionals and for various household tasks.
They usually differ:
Pistol-shaped body
An electric motor located inside it
Impact mechanism for drilling
The presence of a switch, power regulator
Chuck for fixing drills and various accessories.
Outwardly similar to a drill is a screwdriver. To understand how a screwdriver differs from a drill, it is enough to determine their purpose. If the drill functionally serves to make holes, then the screwdriver is specially designed to quickly tighten or unscrew screws, screws and other screw fasteners.
In a screwdriver, one of the most important parameters is the maximum torque.
No high rotation speed required.
Under normal conditions, two to three hundred revolutions per minute are sufficient.
Sometimes a gearbox is installed to reduce speed and increase torque.
This is done using a special regulator.
They are given the force required to tighten a specific type of screw or screw, to avoid their deformation.
The screwdriver usually has a reverse to unscrew screws or free from jamming.
It can have multiple speeds, overload protection. There are screwdrivers on sale with a drilling function. The main thing that distinguishes a screwdriver from a drill-screwdriver is the presence of a release clutch, which regulates the twisting force, it does not allow deeper than you need to tighten the self-tapping screw.
The difference is also in the holder. If the screwdriver involves the use of a bit of the desired size with a hex shank, then the drill uses a keyless chuck for a round in diameter drill.
The drill, regardless of the presence of adjustment, cannot ensure the accuracy of tightening, as a result, the head of the fastener or the screwing nozzle itself may be damaged.
Most common electric screwdrivers and drills can be powered by mains power and a battery. powerful drills use mains electricity. Especially when it comes to a tool with an impact function, where rotary and reciprocating movements are combined.
When using cordless power tools, the advantage of screwdrivers is obvious. With the same charge and voltage, they show an advantage over a drill. This is determined by their technical parameters, primarily the maximum torque.
Most often, household products have nickel-cadmium batteries. Their peculiarity is that they need to be discharged "to zero" and charged until fully charged.
For many works in construction, in production and at home, you need to know how a screwdriver differs from a drill.
This affects the correct organization of work and its efficiency.
Sometimes, these mechanized tools perform operations that, together, ensure the strength and aesthetics of the structure.
Both types of tools are necessary in professional activities and everyday life.
Answering the question, what is the difference between a screwdriver and a cordless drill, it should be noted that the screwdriver has a specific purpose. to tighten and unscrew screws, screws and self-tapping screws.
It is convenient for assembling and disassembling various structures, furniture, wood products, drywall, etc.
Its design allows you to do it quickly and accurately.
The cordless drill also allows for a quick and accurate drilling function.
Both of these tools are good for the tasks for which they were created.
However, with the advent of efficient battery drills with the functions of a screwdriver and a screwdriver with the ability to drill, the difference between them decreases.
Cordless drill requires a torque of up to 20 Nm and at least 1500 rpm.
When choosing the right tool, you should take into account:
Battery type and spare
Infinitely variable speed adjustment for drilling and twisting
Information about the torque that determines the purpose of the tool
Convenient regulation of optimum torque.
If we are talking about a professional tool, the screwdrivers can work continuously for more time. They wear out less.
The comparison of impact and non-impact screwdrivers is presented on:
Found a mistake? Select it and press CtrlEnter to tell us.
In some everyday situations, it is quite difficult to get by with a wrench. Then devices come to the rescue that help to assemble and disassemble threaded devices. Some of these are a screwdriver and a wrench. They have one task. Assistance when working with fasteners. But there are also a lot of differences. We propose to consider the main differences between the devices, find out which is more convenient to use, and how you can upgrade an old screwdriver to a nutrunner.
An impact wrench is a mechanical tool for screwing in and out nuts and bolts. You cannot do without it when you need to unscrew a rusty nut in hard-to-reach places, and a simple wrench is powerless. The impact wrench can be confused with a drill. But an experienced owner will see the difference: the attachment installed in the holder. It is this nozzle that compresses the nut and holds it while the engine twists it in or unscrews it.
READ Stihl Chain 180 File Which
Tip: DeWalt DW292 is considered a convenient option. The model consumes 710 watts of power.
Screwdriver. a device for screwing and unscrewing screws, screws, self-tapping screws and other fasteners. Its peculiarity. Work at different speeds. Expands the range of possibilities. The device drives screws, drills, has an impact function.
Tip: Bosch has also established itself as a market leader in this niche. Especially popular are the models Bosch GSR 6-45 TE, Bosch GSR 14.4-2-LI Plus.
A wrench, unlike a screwdriver, is considered a more professional tool. According to the method of application, the mechanism for nuts can be: gasoline, pneumatic, hydraulic and work from the network. It is designed for screwing in or out bolts or nuts. Unlike a screwdriver, a cartridge is not attached to a nutrunner. In its place, a rod with a nut head is installed.
The second difference. In terms of tightening power, the impact wrench has an advantage here. With him you can "master" any crafts.
Another difference lies in the volume and weight of the mechanisms, for example, not every person can lift some wrenches. Also, screwdrivers are able to regulate their work process, while another representative of mechanics does not have this opportunity. The nutrunner copes with complex tasks no faster than the screwdriver, which creates a lot of noise during operation. This fact is the key difference. In such a situation, equipped with a 3-speed gearbox, the DeWalt DCF895M2 screwdriver in light tasks can replace any device for tightening nuts.
In terms of operational capabilities, the nut mechanism has serious limitations. It is good to use drills when drilling. So, for soft wood, it is not necessary to use a professional mechanism, you can get by with the BlackDecker EPC12CAB screwdriving tool.
It may also be interesting: How to choose a drill
Given the many functions of a screwdriver, it performs a wider range of work, which is what wins over a nutrunner. In many cases, it replaces the nut device, performs the simple work of a wrench. The screwdriver works even more accurately: the tangential impact of a professional nutrunner, although it works quickly, but the screws do not withstand the stress and can fly out of the building material.
Tip: From the outside, the nut mechanism looks like a conventional drill, and its holder is equipped with an extension attachment.
The impact wrench becomes indispensable during large construction works, when working with cars, ships, airplanes and other units. A device for screwing in self-tapping screws and screws is often used at home for furniture repair. Since such a mechanism is free from mains power, the use of the impact wrench is.
Consider how to convert an unusable screwdriver into a new nutrunner. For such a design you will need:
Adapter set.
Device battery.
Control indicator.
Button for wrench.
Tip: Store the devices in the case, so the chance of breakage if dropped is reduced to a minimum. It is important to remember that attachments for mechanisms are very fragile and require careful maintenance.
For the device, you need to set the horn for the brakes. Then you need to connect and solder all the connectors of the indicators. If you connect the components correctly, you get a screwdriver. Watch your voltage. It is difficult for the battery to tolerate changes. Assembly of parts will take no more than 20 minutes; you can also convert the device to its original state. Such a device will be able to screw in the screws, but for this you need to put a suitable nozzle.
In summary, it is not necessary to buy new home appliances. Practical expert advice will help you build your wrench. At home, it can be easily replaced with a screwdriver. For more serious construction, the wrench is irreplaceable. We hope our material helped to understand the main differences between devices for screws and nuts.
See clearly in how to make a new nutrunner from an old screwdriver: Turning a screwdriver into a nutrunner
During work, difficult situations are often encountered when using a wrench alone becomes clearly insufficient. Then you have to pick up special tools: they help in working with parts that have a thread. Among them are a screwdriver and a wrench. These technical devices serve one purpose. to speed up work associated with a large number of nuts, bolts and screws used. At the same time, despite the similarity of the tasks being solved, they also have a number of significant differences. You can find out about this and which of the tools is better to use from the review below.
A screwdriver is an electric tool, the main purpose of which is to work on screwing in or out of various screws, bolts and screws. You can only imagine how long it will take to screw in a few hundred screws by hand when assembling, for example, a conventional built-in cabinet for things or shelves for tools. With this device, all work will take a few minutes. In addition, this power tool has some other additional features:
Use as a cordless drill;
Replacement with special nozzles for conventional hex keys;
Backlight function for poor visibility.
A wrench is considered to be a tool for professionals, while a screwdriver is mainly used for smaller household jobs. The main differences between a nutrunner and a screwdriver are their following characteristics.
The wrench is more cumbersome than a screwdriver and does not have a chuck. At its end there is a special rod on which the nut head is attached.
The power of a wrench applied to an object is many times greater than that of a screwdriver. With the help of the first one can solve any "problems".
There is a significant difference in the weight and size of these mechanisms. Wrenches are usually larger and heavier.
Screwdrivers have several operating modes that can be adjusted. Unlike them, no other special tools have this capability.
Thus, impact wrenches are used more often for heavy mechanical work, while its competitor is more often used for simpler domestic repairs. In addition, for simple tasks, you can try to make the wrench yourself using ordinary nozzles.
Which tool should you choose? The answer to this question will depend on the type of work where these tools are needed. Considering that the impact wrench has a tangential impact during operation, it cannot be used to tighten the screws, as it will simply tear off their caps. However, if you ensure that there is no stop, then he can make such a mount that even the most powerful screwdriver cannot create. If there is a small amount of work to be done, which includes only screwing in screws, then an ordinary screwdriver is enough for this. Therefore, it is necessary to decide what to choose for work. a screwdriver or a wrench, based on the specific circumstances.
READ Alteration Battery Screwdriver On Li Ion
An industrial or homemade wrench is used to install and remove bolts and nuts. Some may accidentally confuse it with a drill, but experienced craftsmen will immediately point to a special attachment in the holder, which a conventional drill does not have. You will definitely not be able to do without it if you need to unscrew the old nut, which is so rusted that you simply cannot unscrew it with your hands or with a key. The tool is produced in the following versions:
Pneumatic;
Gasoline;
Hydraulic;
Powered by electricity.
Of course it is possible, with only one important caveat: it is unlikely that it will be possible to obtain the same power as a tool made in an industrial environment. But, very often too much power is not required. For example, if you have to change tires on a car. Many people know how long the process of unscrewing wheel nuts is. Of course, there is an opportunity to contact specialists, but it will cost money. If you have a screwdriver at hand, then it can be quickly converted into a nutrunner: this requires only an adapter with a head of a certain size, which must completely match the size of the wheel wrench. It remains to fix this adapter in the chuck and install the head. Wrench from a screwdriver completely ready.
Professionals know that the expansion of functions and the universalization of the tool impairs its capabilities when carrying out individual, highly specialized manufacturing operations. This fully applies to power tools, where universal screwdrivers and specialized electric screwdrivers are allocated to separate categories.
Just like a locksmith or assembly screwdriver, an electric screwdriver is designed for a narrow range of operations. She specializes in fast screwing or unscrewing of screws on finished threads, as well as assembly screwing of screws and self-tapping screws into soft materials. In industrial applications, such properties of an electric screwdriver make it possible to effectively use it when assembling furniture, in screwdriver assembly of electronic devices or in electronics repairs.
The basic functions are provided by the simplest household models of electric screwdrivers with a minimum set of interchangeable devices. screwdriver bits. In most cases, the tool does not require high power, therefore all its mechanisms (electric motor, gearbox, power source) are compact and located in a small handle.
The power of an electric screwdriver is not a key indicator and is usually not indicated in the technical specifications. The main working indicator is the number of revolutions, which rarely exceeds 200-240 revolutions per minute. For comparison, in a professional screwdriver, the maximum rotational speed can reach 400-1300 rpm. such speeds and high torque can be harmful when mounting operations on soft materials or brittle plastics.
Due to the unnecessary design of the electric screwdriver, there are no cam and collet clamps, precise centering devices, an impact mechanism, which a screwdriver needs for drilling, as well as screwing self-tapping screws into hard materials. For the same reason, electric screwdrivers are not equipped with additional power handles, which are typical for powerful screwdrivers.
Cordless screwdriver Einhell Blue BT-SD 4.8 F
The absence of the need to perform power operations and drilling with an electric screwdriver leads to a simple design for fastening replaceable tools, which uses simple spindles for a hexagonal rod, quick-release fasteners or low-power magnetic holders.
Low speed indicators make a powerful electric motor and mains supply unnecessary. Most electric screwdrivers are of a cordless design. During continuous operation, recharging of the batteries is provided by a special stand with a built-in charger, on which the electric screwdriver is placed in the intervals between operations.
The shape of the electric screwdriver body can be different, unlike the screwdriver, for which most manufacturers prefer to choose the pistol grip configuration. However, many household electric screwdrivers have the same shape.
Cordless screwdriver Bosch IXO Medium
For domestic use, the pistol grip with elastic pads is comfortable enough. For specialized use (work in narrow and hard-to-reach mounting places), a tool with a straight or T-shaped handle is required.
The most convenient for universal use are electric screwdrivers with a variable handle shape. such models are produced by many manufacturers of power tools.
When choosing an electric screwdriver with an eye on the brand or the reputation of the manufacturer, it should be borne in mind that many reputable European companies have moved production to China and Southeast Asia. Retaining design features, the power tool has somewhat lost in the quality of assembly and durability of the plastics of the case.
An indirect indicator of quality can be called the availability of additional features that well-known manufacturers of power tools supply their products. These include:
Battery charge indicator;
Ratchet torque adjustment;
Direction of rotation indication (forward, reverse).
Cordless screwdriver Intertool DT-0301
Recently, manufacturers of inexpensive tools have also achieved good quality. Their models use durable plastics, reliable electric motors, high-quality steel for replacement bits.
When using the tool in many ways, the minimum set of bits may not be enough. For advertising purposes, many manufacturers produce attachments for household use, such as a replaceable electric screwdriver or a barbecue attachment. Such attachments are not suitable for professional use.
A quality power tool is always sold in a special case and comes with a significant number of replacement bits. The set includes bits for straight flat slots, cruciform, hexagonal configurations, "stars", "trefoils" used for furniture fittings and assembly hardware. Extra long bits included for special applications.
The choice of battery type (nickel-cadmium or lithium-ion) depends on the application of the electric screwdriver. Lithium-ion batteries have practically no memory effect of the accumulator, have a high capacity, have a low self-discharge, but do not tolerate negative temperatures, therefore, they are not suitable for outdoor work in winter.
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Articles difference, electric screwdriver, electric screwdrivers, impact wrench, screwdriver


I think that a good children's book is one that tells an engaging story while encouraging creativity and mental exercise and this book definitely does both. It was entertaining to solve puzzles right along with Zane, and a relief that one could just plow ahead for the answer if one was feeling lazy (or lost, lol) about trying to figure out what was being asked. The combination of mental and physical challenges make me wish these games really existed because I think that they would be both fun and educational, and would encourage children to exercise both sets of muscles (mental and physical). The underlying story is all too relevant a subject, as Zane has to face the potential repercussions of brain injury resulting from head trauma, but the message that being well-rounded is a bonus is artfully conveyed. I was pleased that a gamut of attitudes toward gamesmanship and the spirit of competition were described (with delightfully satisfying results for some attitudes) even as I was enthralled by the games themselves. The asides that deal with the machinations of the adults weren't quite as necessary as far as I was concerned, but I don't know whether this is an ongoing series that addresses the sabotage. This was a fun read with delightful illustrations that I think will be appreciated by both adults and youngsters, and I would love to read more by this author.
"The Gollywhopper Games: Friend or Foe", written by Jody Feldman and illustrated by Victoria Jamieson, is a children's story that follows the adventures of Zane Braycott as he competes in the annual contest. The series of tasks that he must complete requires excelling in both mental and physical skills as well as a competitive spirit that combines teamwork and strategy as well as self-confidence. Only one person can win the Gollywhopper Games, but what will it take to be that person and is Zane willing to focus on something other than his beloved football long enough to compete?
Chock-full of puzzles, riddles, and challenges for the reader to solve along with the main characters, this companion to the first two Gollywhopper Games books offers readers plenty of action and fun. The perfect choice for fans of Escape from Mr. Lemoncello's Library, Charlie and the Chocolate Factory, and Chasing Vermeer.
Zane is not that interested in the Gollywhopper Games. He'd rather play football and is sure that he's headed straight for the NFL. But when he gets his second concussion, his parents tell him, "No football for a year." Instead, to his surprise, he gets a chance to compete in the Gollywhopper Games. Zane's sense of strategy, his physical strength, and his competitive edge are all assets, and so is his ability to motivate his teammates and get them to work together. Zane becomes particularly close to Elijah, a young and scrawny genius who is friendly, awkward, and funny—Zane's polar opposite. These two unlikely friends end up head-to-head in the final challenge, where Elijah's quick thinking and Zane's physical strength make it a tough fight. This is a fun and fast-paced and interactive read for fans of The Mysterious Benedict Society.


text:
Hosokawa Morihiro
(born 1938). A political reformer who broke with the country's traditional ruling party, Hosokawa Morihiro served as prime minister of Japan in 1993–94. Early in his political career, he was supported by the powerful Liberal Democratic Party (LDP), which had dominated Japanese politics for decades. Hosokawa eventually became opposed to one-party rule. He founded the conservative Japan New Party as an alternative to the Liberal Democrats.
Hosokawa was born on January 14, 1938, in Kyushu, Japan. He was the grandson of Konoe Fumimaro, who had twice served as prime minister of Japan before World War II. Hosokawa attended Sophia University, graduating in 1963. Intent on a career in journalism, he took a job with the liberal newspaper Asahi shimbun. His run in 1969 for election to the lower house of the Diet (parliament) was unsuccessful. However, in 1971, with strong support from the ruling LDP, Hosokawa was elected to the less powerful upper house. He went on to serve in the upper house of the Diet for 12 years.
In 1983 Hosokawa was elected governor of Kumamoto prefecture on Kyushu island. While serving as governor, he pursued an aggressive economic policy and strengthened environmental laws. He was often frustrated, however, by the powerful bureaucracy of the central government. In 1992 Hosokawa formed the Japan New Party. He called for reform of the electoral system and an end to political corruption and one-party rule. The new party was quickly embraced by people around the country who were tired of the scandals and internal disagreements plaguing the Liberal Democrats.
In 1993 a coalition of seven dissident LDP factions and opposition parties in the House of Representatives elected Hosokawa prime minister. He was the first non-LDP prime minister of Japan since 1955. Once in office, Hosokawa got a bill passed to restructure the electoral system. It was passed in an effort to limit political corruption and increase the relative voting strength of urban areas. Hosokawa resigned in April 1994, however, when the LDP accused him of financial wrongdoing. He retired from politics in 1998 and devoted himself to making pottery.
After 16 years of retirement, Hosokawa reentered politics in 2014 to run for the governor of Tokyo. He ran on a platform of ending in Tokyo the use of electricity generated by nuclear power, in the wake of a serious accident at a Japanese nuclear power plant (see Japan earthquake and tsunami of 2011). Despite being endorsed by former prime minister Koizumi Junichiro, Hosokawa lost the election.


text:
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Why You Should Visit Canada Now
Between its natural beauty and cultural offerings, there are more than enough reasons to make the trip north.
By Cailey Rizzo on August 22, 2018
Even if Canada isn't your home and native land, there are more than enough reasons to visit. In addition to boasting the world's longest coastline and largest number of lakes, the country has a fast-growing culinary scene, a well-preserved history, and an astonishingly friendly culture. Read on for why you should book a trip as soon as possible.
Chase Dekker Wild-Life Images/Getty Images
Since 1885, Canada's National Park system has grown to include more than 40 different entities. While that number may not seem impressive, consider that the national park system makes up about three percent of Canada's entire landmass. From the impressive Banff, the snowboarding destination in Alberta that is also the nation's first national park, to Quttinirpaaq, one of the northernmost places in the country, the diversity of Canada's national parks means that you can spend a lifetime exploring them all.
It's More Affordable
The currency conversion rate is one very practical reason for Americans to visit: The Canadian dollar is now worth about 75 cents stateside. Splurge on luxury hotels, indulge in fine cuisine, and know that you're paying less for these experiences than you would back home.
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Canada's bilingual system is fascinating to watch in action, particularly in the Quebec province. A visit to Montreal or Quebec City might feel more like a trip to Europe than anywhere else: About 7.3 million people (or 80 percent of Quebec) claim French as their first language to this day.
Canadian cuisine is finally getting the recognition it deserves—and yes, there's more to it than maple syrup and poutine. The country's growing food scene takes inspiration from indigenous foraging techniques and borrows from the culture of modern immigrants, like those from Syria. Some of the most exciting places to dine right now include Canoe in Toronto (for dishes like rabbit or fresh Atlantic lobster) and Toqué in Montreal (for an elevated seven-course tasting menu centered on Quebecois classics).
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Some of the world's finest festivals call Canada home. Every year, the Toronto International Film Festival (TIFF) premieres groundbreaking new films and brings the industry's biggest names up north. On the other side of the country, the Edmonton Folk Fest brings thousands of people to North Saskatchewan River valley for four days of music. This year's lineup included heavy-hitters (like The Decemberists and Regina Spektor) alongside exciting up-and-comers like This Is The Kit.
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There has never been a better time for thrill-seekers to visit Niagara Falls. Earlier this year, a zipline opened over the Canadian side of the Falls, allowing visitors to experience the mist from 220 feet above the water. The zipline extends 2,200 feet and sends visitors up to 40 miles per hour to their landing over Horseshoe Falls. While the entire experience only lasts about 30 seconds, it's certainly the most thrilling way to see the natural wonder.
Preserved History
Last year, Canada celebrated its 150th year of independence. But those who wish to explore Canada's history throughout the ages will find plenty of well-preserved buildings that date back hundreds of years. Visit Old Quebec City (dating back to the 1600s), Inuksuk Point in Nunavut (structures estimated to be 2,000 years old) and Dawson City (preserved from the 1890s Gold Rush in Yukon) for a firsthand look at Canadian history.
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Canada claims to have the best spot in the world for viewing the Northern lights. In Yellowknife (Northwest Territories), visitors have a 90 percent chance of seeing the aurora borealis on most nights from mid-November through April. Aurora Village, located just outside of Yellowknife, is the most popular place in Canada for viewing the phenomenon. But don't be afraid to venture beyond the most popular sites. From Yukon to Nova Scotia, there are plenty of locations where it's possible to get a glimpse of the lights.
Craft Beer + Wine
Canadian beer is much more than Labatt Blue and Molson. Today, there's a thriving microbrewery scene for beer fanatics, plus many vintners producing unique wines. For the most variety in terms of beer, make a stop at Bar Hop or the Craft Brasserie, both in Toronto, for the best of what the country has to offer. Wine fanatics should book a trip through the Okanagan Valley or the Niagara region of Ontario.
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Morocco is unlike any place I have been to before and it is such a culture shock for me. As soon as I got off the plane I realized I was in for so much more than I expected. The streets in the Medina are busy and full of life with traffic running in both directions and narrow roads full of merchants and donkeys.
With all of this going on it is easy to feel unsafe and to be worried especially since I stand out like a sore thumb in my western clothes compared to the traditional garb of the Muslim women.
The form of currency used here is the dirham and the current exchange rate is 1 USD to 8.41 MAD. With that being said, everything was pretty cheap for me. After checking into the Riad (apartment or flat) I headed into the town to get some food.
I decided to eat at a nearby restaurant outside of the main square (Jemaa Al Fna) as I was recommended by my driver not to eat the food in the square, in fear of food poisioning. When visiting here it is best to not eat american fast food (McDonald's etc.) or market food.
Also don't be surprised by the stares from the locals. They are not used to our dress and will look out of curiousity. In regards to the local dress, we have to make sure we do not stare at them because this makes them uncomfortable.
I took very few pictures today because I didn't want to make any of the locals feel uncomfortable and I was also still feeling out the city.
The weather was lovely, around the 70's all day until it hits dusk and steadily dropped down to the 50's which is very common for desert climates.
The door to my balcony.
Atlas Mountains view from Marrakech.
PS-When traveling throughout Jamma El Fna (the center square in the old city) be on the look out for merchants trying to get money out of you. They may say they will give you a service and even if you tell them you don't want the service they may do it anyway and expect payment. This is what happened with two Moroccan ladies in the square. We told them we were in a rush and would come back to get henna yet they insisted they give us a little "free" design until we get back. Once she did our hands she demanded that we pay 500 dirhams even though we told her we had no change. The exchange ended when they realized the couldn't get money from us and cursed us out.
All in all today was quite an experience here and I learned the importance of being on guard at all times when traveling to foreign countries. Despite the natural beauty of Morocco, the inhabitants are working class and will try to take advantage of tourists to make money. Feel free to triple check prices and bargain before buying anything. Also don't allow someone to start a service or insist on helping you, they are most likely looking to get paid.
I can't wait to head to the Atlas Mountains tomorrow and the Ourika waterfalls to see more of the natural beauty of Marrakech.


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text:
\section{Introduction and main results}

The aim  of this paper is to construct a sequence of
eigenvalues, and a corresponding sequence of eigenfunctions,
for the  2D isotropic quantum harmonic oscillator
\begin{equation}
\widehat
H:= -\Delta + x^2 + y^2\,,
\end{equation} with exactly two nodal domains. Similar
results were stated  in Antonie Stern' PhD thesis (with R. Courant
as advisor), for the Dirichlet problem for the Laplacian in the
square, and in the case of the Laplace-Beltrami  operator on the
sphere $\mathbb S^2$ \cite{St}. This is only much later that H. Lewy
\cite{Lew} in the case of the sphere (see also \cite{BeHe2}), and
the authors \cite{BeHe} in the case of the square with Dirichlet
conditions (see also Gauthier-Shalom--Przybytkowski \cite{GSP}),
have proposed complete proofs of her statements.

Coming back to the isotropic harmonic oscillator,   an
orthogonal basis of eigenfunctions is given by
\begin{equation}
\phi_{m,n} (x,y) = H_m(x) H_n (y) \exp( - \frac{x^2+y^2}{2})\,,
\end{equation}
 for $(m,n)\in \mathbb N^2$, where $H_n(x)$ denotes the
Hermite polynomial of degree $n$.

The eigenfunction $\phi_{m,n}$ corresponds to the eigenvalue
$2(m +n+1)\,,$
\begin{equation}
\widehat H \phi_{m,n} = 2(m+n+1)\, \phi_{m,n}\, .
\end{equation}
Here we use the definitions and notation of Szeg\"o \cite[\S
5.5]{Sz}.

The eigenspace $\mathcal{E}_{\ell}$ associated with the eigenvalue
$\hat{\lambda}(\ell)= 2(\ell +1)$ has dimension $(\ell + 1)$, and is
generated by the eigenfunctions $\phi_{\ell,0}$, $\phi_{\ell-1,1}$,
\dots, $\phi_{0,\ell}$.

For $\theta \in [0,\pi[$, we shall consider the families of
eigenfunctions,
\begin{equation}
\Phi^{\theta}_{n} := \cos\theta \, \phi_{n,0} + \sin\theta \,
\phi_{0,n}\,,
\end{equation}
corresponding to the eigenvalue $2(n+1)$.

Our aim is to prove the following  theorems.

\begin{theorem}\label{prop1}   Assume that  $n$ is odd. Then,
there exists an open interval $I_{\frac{\pi}{4}}$ containing
$\frac{\pi}{4}$, and an open interval $I_{\frac{3\pi}{4}}$,
containing $\frac{3\pi}{4}$, such that for
$$
\theta \in I_{\frac{\pi}{4}} \cup I_{\frac{3\pi}{4}} \setminus
\{\frac{\pi}{4},\frac{3\pi}{4}\}\,,
$$
the nodal set $N(\Phi^{\theta}_n)$ is a connected simple regular
curve, and the eigenfunction $\Phi^{\theta}_n$ has two nodal domains
in ${\mathbb R}^2$.
\end{theorem}

\begin{theorem}\label{th2}
 Assume that $n$ is odd. Then, there exists $\theta_c >0$ such
that, for $0 < \theta < \theta_c$, the nodal set
$N(\Phi^{\theta}_n)$ is a connected simple regular curve, and the
eigenfunction $\Phi^{\theta}_n$ has two nodal domains in ${\mathbb R}^2$.
\end{theorem}

As in the case of the square,  to prove Theorem~\ref{prop1}, we
begin by a symmetry argument to reduce to a neighborhood of either
$\pi/4$ or $3\pi/4$, say $3\pi/4$. The first step in the proof is to
analyze the zero set when $\theta = \frac{3\pi}{4}$, in particular
the points where the zero set is critical, and to show that this
only occurs on the diagonal.

The second  step is then to show that the double crossings on
the diagonal disappear by perturbation, for $\theta$ close to
and different from $\frac{3\pi}{4}$\,. Using the local nodal
patterns and some barrier lemmas, one can then show that the nodal
set becomes a connected simple curve, asymptotic to $x=y$ at $\pm
\infty\,$. The local stability of the nodal set under perturbation,
then gives an explicit interval containing $3 \pi/4$ in which the
phenomenon occurs (for $\theta \neq \frac{3\pi}{4}$ of course).

The proof of Theorem~\ref{th2} follows similar lines.

\newpage

\textbf{Remarks}.\vspace{-3mm}
\begin{enumerate}
\item In the case of the square the same kind of analysis is also
interesting for determining when the number of nodal domains of an
$n$-th eigenfunction  is equal to $n$ (the so called Courant sharp
situation). Pleijel \cite{Pl} observed that in the case of the
square (Dirichlet condition) this  only occurs for the first, the
second and the fourth eigenfunctions (see \cite{BeHe} for a complete
argument). In the case of the sphere, as a consequence of the
analysis of Leydold \cite{Ley}, this  only occurs for the first and
the second eigenfunctions. It is natural to investigate a similar
question in the case of the isotropic harmonic oscillator, and more
generally, the validity of Pleijel's theorem in this case.  In
the last section, we will give a Leydold's like proof of the fact
that the only Courant sharp eigenvalues of the harmonic oscillator
are $\hat{\lambda }(\ell)$, for $\ell = 0, 1$ and $2$. As
communicated by I. Polterovich, this question will be analyzed from
a different point of view in \cite{Cha}.
\item A connected question is to analyze the zero set when $\theta$
is a random variable. We refer to \cite{HZZ} for  results in
this direction.
\item These questions are related to the question of spectral minimal
partitions \cite{HHOT}. In the case of the harmonic oscillator
similar questions appear in the analysis of the properties of
ultracold atoms (see for example \cite{RL}).
\end{enumerate}

 Theorems~\ref{prop1} and \ref{th2} concern the eigenspace
$\mathcal{E}_n$ of the harmonic oscillator, with $n$ odd. When $n$ is even,
the picture is different. Some nodal sets have compact connected
components, with or without critical zeros, some have both compact
and non-compact components. Other examples can be analyzed as well.
This will be analyzed in the future.

{\bf Acknowledgements.}\\
The second author would like to thank D. Jakobson, I. Polterovich
and M. Persson-Sundqvist for useful discussions, transmission of
information or computations.

\section{A  reminder on Hermite polynomials}

We use the definition, normalization, and notations of Szeg\"{o}'s book
\cite{Sz}. With these choices, $H_n$ has the following properties,
\cite[\S~5.5 and Theorem~6.32]{Sz}.\vspace{-3mm}

\begin{enumerate}
\item  $H_n$ satisfies the differential equation
$$
y''(t) - 2 t \,y'(t) + 2n\, y(t) = 0\,.
$$
\item  $H_n(t)$ is a polynomial of degree $n$ which is even (\emph{resp. } odd) for $n$
even (\emph{resp. } odd).
\item  $H_n(t) = 2 t\, H_{n-1}(t) - 2 (n-1) \,
H_{n-2}(t)\,, ~n\ge 2\,, ~~H_0(t)= 1\,, ~~H_1(t) = 2t\,.$
\item $H_n$ has $n$ simple zeros $t_{n,1} < t_{n,2} < \cdots <
t_{n,n}\,$.
\item
$$
 H_n(t) =  2 t\, H_{n-1}(t) - H'_{n-1}(t)\,.
$$
\item \begin{equation}
H'_n(t) = 2n H_{n-1} (t)\,.
\end{equation}
\item The coefficient of $t^n$ in  $H_n$ is $2^n$.
\item
$$
\int _{-\infty}^{+\infty} e^{-t^2} |H_n(t)| ^2\, dt = \pi^\frac 12 \,2^n\, n! \,.
$$
\item The  first zero $t_{n,1}$ of $H_n$ satisfies
\begin{equation}
 t_{n,1} = (2n+1)^\frac 12 - 6^{-\frac 12} (2n+1)^{-\frac 16}
(i_1+\epsilon_n)\,,
\end{equation}
where $i_1$ is the first positive real zero of the Airy function,
and $\lim_{n\rightarrow +\infty} \epsilon_n =0\,$.
\end{enumerate}

The following result (Theorem 7.6.1 in Szeg\"o's book \cite{Sz}) will
also be useful:

\begin{lemma}\label{L-leh}
The successive relative maxima of $|H_n(t)|$ form an increasing
sequence for $t \geq 0$\,.
\end{lemma}
{\bf Proof.}
\\
It is enough to observe that the function
$$
\Theta_n (t):= 2 n H_n(t)^2 + H_n'(t)^2
$$
satisfies
$$
\Theta_n'(t) = 4 t \, (H'_n(t)) ^2 \,.
$$\hfill $\square$

\section{Stern-like constructions for the harmonic oscillator, case $n$-odd}

\subsection{The case of the square}

 Consider the square  $[0,\pi]^2$, with Dirichlet boundary
conditions, and the following families of eigenfunctions associated
with the eigenvalues $\hat{\lambda}(1,2r) := 1+ 4r^2$, where $r$ is
a positive integer, and $\theta \in [0,\pi/4]$,
$$
(x,y) \mapsto \cos \theta \, \sin x \, \sin(2ry)  + \sin \theta \,
\sin(2rx) \, \sin y\,.
$$

According to \cite{St}, for any given $r$, the typical evolution of
the nodal sets when $\theta$ varies is similar to the case $r=4$
shown in Figure~\ref{sq-1} \cite[Figure~6.9]{BeHe}: generally
speaking, the nodal sets deform continuously, except for finitely
many values of $\theta$, for which crossings appear or disappear.
\begin{figure}[!ht]
\begin{center}
\includegraphics[width=14cm]{square-1-R-R8-plate-4x4-1600x1600.png}
\caption{Evolution of the nodal set in the case of the square.} \label{sq-1}
\end{center}
\end{figure}

We would like to get similar results for the isotropic quantum
harmonic oscillator.

\subsection{Symmetries}

 Recall the notation,
\begin{equation}\label{P2}
\Phi^{\theta}_n(x,y) :=: \Phi_n(x,y,\theta) := \cos\theta \,
\phi_{n,0} + \sin\theta \, \phi_{0,n}\,.
\end{equation}
Since $\Phi^{\theta + \pi}_n = - \Phi^{\theta}_n$, it suffices to
vary the parameter $\theta$ in the interval $[0,\pi[$.

\emph{Since $n$ is odd}, we have the following symmetries.

\begin{equation}\label{P4}
\left\lbrace
\begin{array}{ll}
\Phi^{\theta}_n(-x,y) & = \Phi^{\pi-\theta}_n(x,y)\,,\\[4pt]
\Phi^{\theta}_n(x,-y) & = - \Phi^{\pi-\theta}_n(x,y)\,,\\[4pt]
\Phi^{\theta}_n(y,x) & = \Phi^{\frac{\pi}{2}-\theta}_n(x,y)\,.\\[4pt]
\end{array}
\right.
\end{equation}

When $n$ is odd, it therefore suffices to vary the parameter
$\theta$ in the interval $[0,\frac{\pi}{4}]$. The case $\theta = 0$ is
particular, so that we shall mainly consider $\theta \in ]0,\frac{\pi}{4}]$.

\subsection{Critical zeros}

A \emph{critical zero} of $\Phi^{\theta}_n$ is a point $(x,y) \in
{\mathbb R}^2$ such that both $\Phi^{\theta}_n$ and its differential
$d\Phi^{\theta}_n$ vanish at $(x,y)$. The critical zeros of
$\Phi^{\theta}_n$ satisfy the following equations.

\begin{equation}\label{P6}
\left\lbrace
\begin{array}{ll}
\cos\theta \, H_n(x) + \sin\theta \, H_n(y) &= 0\,,\\[4pt]
\cos\theta \, H'_n(x) & = 0 \,,\\[4pt]
\sin\theta \, H'_n(y) &= 0\,.\\
\end{array}
\right.
\end{equation}

Equivalently, using the properties of the Hermite polynomials, a
point $(x,y)$ is a critical zero of $\Phi^{\theta}_n$ if and only if

\begin{equation}\label{P8}
\left\lbrace
\begin{array}{ll}
\cos\theta \, H_n(x) + \sin\theta \, H_n(y) &= 0\,,\\[4pt]
\cos\theta \, H_{n-1}(x) & = 0 \,,\\[4pt]
\sin\theta \, H_{n-1}(y) &= 0\,.\\
\end{array}
\right.
\end{equation}

The only possible critical zeros of the eigenfunction
$\Phi^{\theta}_n$ are the points $(t_{n-1,i}\,,t_{n-1,j})$ for $1
\le i,j \le (n-1)$, where the coordinates are the zeros of the
Hermite polynomial $H_{n-1}$. The point $(t_{n-1,i}\,,t_{n-1,j})$ is
a critical zero of $\Phi^{\theta}_n$ if and only if $\theta =
\theta(i,j)$, where $\theta(i,j) \in ]0,\pi[$ is uniquely defined by
the equation,
\begin{equation}\label{P10}
\cos\left( \theta(i,j) \right) \, H_n(t_{n-1,i}) + \sin\left(
\theta(i,j) \right) \, H_n(t_{n-1,j}) = 0\,.
\end{equation}
Here we have used the fact that $H_n$ and $H'_n$ have no common
zeros. We have proved the following lemma.

\begin{lemma}\label{L-cz}
For $\theta \in [0,\pi[$, the eigenfunction $\Phi^{\theta}_n$ has no
critical zero, unless $\theta$ is one of the $\theta(i,j)$ defined
by equation \eqref{P10}. In particular $\Phi^{\theta}_n$ has no
critical zero, except for finitely many values of the parameter
$\theta \in [0,\pi[$. Let $\theta_0 = \theta(i_0,j_0)$,
defined by some $(t_{n-1,i_0}\,, t_{n-1,j_0})$. The function
$\Phi^{\theta_0}_n$ has finitely many critical zeros, namely the
points $ (t_{n-1,i}\,, t_{n-1,j})$ which satisfy
\begin{equation}
\cos\theta_0 \, H_n(t_{n-1,i}) + \sin\theta_0 \, H_n(t_{n-1,j}) =
0\,,
\end{equation}
among them the point  $(t_{n-1,i_0}\,, t_{n-1,j_0})$.
\end{lemma}

\textbf{Remarks}.\\
 From the general properties of nodal lines
\cite[Properties~5.2]{BeHe}, we  derive the following facts.
\vspace{-3mm}
\begin{enumerate}
    \item When $\theta \not \in \left\lbrace \theta(i,j) ~|~ 1 \le i,j \le n-1
    \right\rbrace$, the nodal set of the eigenfunction $\Phi^{\theta}_n$, denoted by
    $N(\Phi^{\theta}_n)$, is a smooth $1$-dimensional submanifold of ${\mathbb R}^2$.
    \item When $\ \theta  \in \left\lbrace \theta(i,j)~|~ 1 \le i,j \le n-1
    \right\rbrace$, the nodal set $N(\Phi^{\theta}_n)$ has finitely
    many singularities which are double crossings. Indeed, the Hessian of the
    function $\Phi^{\theta}_n$ at a critical zero $(t_{n-1,i},t_{n-1,j})$ is
    given by
    %
    \begin{equation*}\label{hessian}
    \mathrm{Hess}_{(t_{n-1,i},t_{n-1,j})}\Phi^{\theta}_n =
    \exp{(-\frac{t_{n-1,i}^2 + t_{n-1,j}^2}{2})}\,
    \begin{pmatrix}
      \cos\theta \, H''_n(t_{n-1,i}) & 0 \\
      0 & \sin\theta \, H''_n(t_{n-1,j}) \\
    \end{pmatrix}\,,
    \end{equation*}
    %
and the assertion follows from the fact that $H_{n-1}$ has simple
zeros.
\end{enumerate}

\subsection{General properties of the nodal set $N(\Phi^{\theta}_n)$}

Denote by $\mathcal L$ the finite lattice
\begin{equation}\label{P22}
\mathcal L := \left\lbrace (t_{n,i}\,,t_{n,j}) ~|~ 1 \le i,j \le n
\right\rbrace \subset {\mathbb R}^2 \,,
\end{equation}
consisting of points whose coordinates are the zeros of the Hermite
polynomial $H_n$. Since we can assume that $\theta \in ]0,\frac{\pi}{4}]$, we
have the following inclusions for the nodal set,

\begin{equation}\label{P24}
\mathcal L \subset N(\Phi^{\theta}_n) \subset \mathcal L \cup \left\lbrace (x,y)
\in {\mathbb R}^2 ~|~ H_n(x) \, H_n(y) < 0 \right\rbrace \,.
\end{equation}

\textbf{Remarks}. Assume that $\theta \in ]0,\frac{\pi}{4}]$.\\
(i) The nodal set $N(\Phi^{\theta}_n)$ cannot meet the vertical
lines $\{x=t_{n,i}\}$, or the horizontal lines $\{y=t_{n,i}\}$ away
from the set $\mathcal L$. \\
(ii) The lattice point $(t_{n,i},t_{n,j})$ is not a
critical zero of $\Phi^{\theta}_n$ (because $H_n$ and $H'_n$
have no common zero). As a matter of fact, near a lattice point, the
nodal set $N(\Phi^{\theta}_n)$ is a single arc through the lattice
point, with a tangent which is neither horizontal, nor
vertical.\medskip

Figure~\ref{FSch-1} shows the evolution of the nodal set of
$\Phi^{\theta}_n$ when $\theta$ varies in the interval $]0,\frac{\pi}{4}]$.
The values of $\theta$ with two digits are regular values (\emph{i.e. }
correspond to an eigenfunction without critical zeros), the values
of $\theta$ with at least three digits are critical values (\emph{i.e. }
correspond to an eigenfunction with critical zeros). The form of the
nodal set is stable between two consecutive critical values of the
parameter $\theta$. In the figures, the grey lines correspond to the
zeros of $H_7$. The blue lines correspond to the zeros of $H'_7$,
\emph{i.e. } to the zeros of $H_6$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=10cm]{FSchroedinger-0-7-bifurc.png}
\caption{Evolution of the nodal set $N(\Phi^{\theta}_n)$, for
$\theta \in ]0,\frac{\pi}{4}]$.} \label{FSch-1}
\end{center}
\end{figure}\medskip

We now describe the nodal set $N(\Phi^{\theta}_n)$ outside a large
enough square which contains the lattice $\mathcal L$.  For this
purpose, we give two \emph{barrier lemmas}.

\begin{lemma}\label{L-ext1}
Assume that $\theta \in ]0,\frac{\pi}{4}]$.  For $n\geq 1$, define
$t_{n-1,0}$ to be the unique point in $]-\infty, t_{n,1}[$ such that
$H_n(t_{n-1,0}) = - H_n(t_{n-1,1})$. Then,\vspace{-3mm}
\begin{enumerate}
    \item $\forall t \le t_{n,1}$, the function $y \mapsto
    \Phi^{\theta}_n(t,y)$ has exactly one zero in the interval
    $[t_{n,n},+\infty[$\,;
    \item $\forall t < t_{n-1,0}$, the function $y \mapsto
    \Phi^{\theta}_n(t,y)$ has exactly one zero in the interval\break
    $]-\infty,+\infty[$\,.
\end{enumerate}
\end{lemma}

\textbf{Proof}.\\
Let $v(y) := \exp(\frac{t^2+y^2}{2})\, \Phi^{\theta}_n(t,y)$. In
$]t_{n,n},+\infty[$, $v'(y)$ is positive, and $v(t_{n,n})\le 0\,$.
The first assertion follows. The local extrema of $v$ occur at the
points $t_{n-1,j}\,$, for $1 \le j \le (n-1)\,$. The second assertion
follows from the definition of $t_{n-1,0}\,$, and from the
inequalities,
\begin{equation*}
\begin{split}
\cos\theta \, H_n(t) + \sin\theta \, H_n(t_{n-1,j}) & \le
\frac{1}{\sqrt{2}}\Big( H_n(t) - |H_n|(t_{n-1,j})\Big)\\
& < -\, \frac{1}{\sqrt{2}} \Big( H_n(t_{n-1,1}) -
|H_n|(t_{n-1,j})\Big) \le 0\,,
\end{split}
\end{equation*}
where we have used Lemma~\ref{L-leh}.\hfill $\square$\medskip

\textbf{Remark}. Using the symmetry with respect to the vertical
line $\{x=0\}$, one has similar statements for $t \ge t_{n,n}$ and
for $t > - t_{n-1,0}\,$.

\begin{lemma}\label{L-ext2} Let $\theta \in ]0,\frac{\pi}{4}]$. Define
$t_{n-1,n}^{\theta} \in ]t_{n,n}\,,\,\infty[$ to be the unique point
such that $\tan\theta \, H_n(t_{n-1,n}^{\theta}) = H_n(t_{n-1,1})$.
Then,\vspace{-3mm}
\begin{enumerate}
    \item $\forall t \ge t_{n,n}$, the function $x \mapsto
    \Phi^{\theta}_n(x,t)$ has exactly one zero in the interval
    $]-\infty,t_{n,1}]$\,;
    \item $\forall t > t_{n-1,n}^{\theta}$, the function $x \mapsto
    \Phi^{\theta}_n(x,t)$ has exactly one zero in the interval
    $]-\infty,\infty[$\,.
    \item For $\theta_2 > \theta_1$, we have
    $t_{n-1,n}^{\theta_2} < t_{n-1,n}^{\theta_1}\,$.
\end{enumerate}
\end{lemma}

\textbf{Proof}. Let $h(x) := \exp(\frac{x^2+t^2}{2})\,
\Phi^{\theta}_n(x,t)$. In the interval $]-\infty,t_{n,1}]$, the
derivative $h'(x)$ is positive, $h(t_{n,1}) > 0$, and
$\lim_{x\rightarrow  -\infty} h(x) =-\infty$, since $n$ is odd. The
first assertion follows. The local extrema of $h$ are achieved at
the points $t_{n-1,j}$. Using Lemma~\ref{L-leh}, for $t \geq
t_{n-1,n}^{\theta}\,$, we have the inequalities,
\begin{equation*}
\begin{split}
H_n(t_{n-1,j}) + \tan\theta \, H_n(t) &\ge
\tan\theta \, H_n(t_{n-1,n}^{\theta}) - |H_n(t_{n-1,j})|\\
& = H_n(t_{n-1,1}) - |H_n(t_{n-1,j})| \ge 0\,.
\end{split}
\end{equation*}
\hfill $\square$\medskip

\textbf{Remark}. Using the symmetry with respect to the horizontal
line $\{y=0\}$, one has similar statements for $t \le t_{n,1}$ and
for $t < - t_{n-1,n}^{\theta}\,$.\medskip

As a consequence of the above lemmas, we have the following
description of the nodal set far enough from $(0,0)$.

\begin{proposition}\label{P-ext}
Let $\theta \in ]0,\frac{\pi}{4}]$. In the set ${\mathbb R}^2 \setminus
]-t_{n-1,n}^{\theta},t_{n-1,n}^{\theta}[\times ]t_{n-1,0},
|t_{n-1,0}|[$, the nodal set $N(\Phi^{\theta}_n)$ consists of two
regular arcs. The first arc is a graph $y(x)$ over the
interval\break $]-\infty,t_{n,1}]$, starting from the point
$(t_{n,1},t_{n,n})$ and escaping to infinity with,
$$ \lim_{x\to -\infty}\frac{y(x)}{x} = - \sqrt[n]{\cot\theta}\,.$$
The second arc is the image of the first one under the symmetry with
respect to $(0,0)$ in ${\mathbb R}^2$.
\end{proposition}

\subsection{Local nodal patterns}\label{SS-lnp}

As in the case of the square, we study the possible local nodal
patterns taking into account the fact that  the nodal set contains the
lattice points $\mathcal L$, can only intersect the connected components of
the set $\{H_n(x) \, H_n(y) < 0\}$, and consists of a simple arc at
the lattice points. The following figure summarized the possible
nodal patterns in the interior of the square
\cite[Figure~6.4]{BeHe},

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=10cm]{FSchr-local-nodal-patterns.png}
\caption{Local nodal patterns.} \label{FSchr-local-nodal-patterns}
\end{center}
\end{figure}

Except for nodal arcs which escape to infinity, the local nodal
patterns for the quantum harmonic oscillator are the same (note that
in the present case, the connected components of the set $\{H_n(x)
\, H_n(y) < 0\}$ are rectangles, no longer equal squares).

Case (C) occurs near a critical zero. Following the same ideas as in
the case of the square, in order to decide between cases (A) and
(B), we use the barrier lemmas, Lemma~\ref{L-ext1} or~\ref{L-ext2},
the vertical lines $\{x=t_{n-1,j}\}$, or the horizontal lines
$\{y=t_{n-1,j}\}$.

\newpage

\section{Proof of Theorem \ref{prop1}}

Note that
$$\phi_{n,0}(x,y) - \phi_{0,n}(x,y) = - \Phi^{\frac{3\pi}{4}}_n(x,y) = -
\Phi^{\frac{\pi}{4}}_n(y,x)\,.
$$

Hence it is the same to work with $\theta =\frac \pi 4$ and the
anti-diagonal, or to work with $\theta = \frac{3\pi}{4}$ and the
diagonal. From now on, we work near $\frac{3\pi}{4}$.

\subsection{ The nodal set of $\Phi_n^{\frac{3\pi}{4}}$}

 \begin{proposition}\label{prop2} Let $\{t_{n-1,i}\,, 1 \le i \le
n-1\}$ denote the zeroes of $H_{n-1}$. For $n$ odd, the nodal set of
$\phi_{n,0} - \phi_{0,n}$ consists of the diagonal $x=y$, and of
$\frac{n-1}{2}$ disjoint simple closed curves crossing the diagonal
at the $(n-1)$ points $(t_{n-1,i}\,,t_{n-1,i})$, and the antidiagonal
at the $(n-1)$ points $(t_{n,i}\,, - t_{n,i})$.
\end{proposition}

To prove Proposition \ref{prop2}, we first observe that it is enough
to analyze the zero set of
$$
(x,y) \mapsto \Psi_{n}(x,y)  :=H_n(x) - H_n(y)\,.
$$
The critical points  of $\Psi_n$ are determined by
$$
H_n'(x)=0\,,\, H_n'(y)= 0\,.
$$

 Hence, the critical points of $\Psi_{n}$  consist of the
$(n-1)^2$ points $(t_{n-1,i}\,,t_{n-1,j})$, for \break $1 \le i,j \le (n-1)$,
where $t_{n-1,i}$ is the $i$-th zero of the polynomial $H_{n-1}\,$.

The zero set  of $\Psi_{n}$  contains the diagonal $\{x=y\}\,$.
 Since $n$ is odd, there  are only $n$ points belonging
to the zero set on the anti-diagonal $\{x+y=0\}$.

On the diagonal, there are $(n-1)$ critical points.  We claim
that there are no critical zeros outside the diagonal.
Indeed, let $(t_{n-1,i}\,,t_{n-1,j})$ be a critical zero. Then,
$H_n(t_{n-1,i}) = H_n(t_{n-1,j})$. Using Lemma~\ref{L-leh} and the
parity properties of Hermite polynomials, we see that
$|H_n(t_{n-1,i})| = |H_n(t_{n-1,j})|$ occurs if and only if
$t_{n-1,i} = \pm t_{n-1,j}\,$. Since $n$ is odd, we can conclude that
$H_n(t_{n-1,i}) = H_n(t_{n-1,j})$ occurs if and only if
$t_{n-1,i}=t_{n-1,j}\,$.

\subsection{ Existence of disjoint simple closed curves in the nodal
set of $\Phi_n^{\frac{3\pi}{4}}$}

The second part in the proof of the proposition follows closely the
proof in the case of the square (see Section 5 in \cite{BeHe}).
Essentially, the  Chebyshev polynomials are replaced by the Hermite
polynomials. Note however that the checkerboard is no more with
equal squares, and that  the square $[0,\pi]^2$ has to be
replaced in the argument by the rectangle  $[t_{n-1,0}, -t_{n-1,0}] \times
[-t_{n-1,n}^{\theta},t_{n-1,n}^{\theta}]$, for some $\theta$ such
that $0 < \theta < \frac{3\pi}{4}$, see Lemmas~\ref{L-ext1} and~\ref{L-ext2}.

The checkerboard argument holds, see \eqref{P24}.

The separation lemmas of our previous  paper \cite{BeHe} must
be substituted by Lemmas~\ref{L-ext1} and~\ref{L-ext2}, and similar
statements with the lines $\{x=t_{n-1,j}\}$ and $\{y=t_{n-1,j}\}$,
for $1 \le j \le (n-1)$.

One  needs to control what is going on at infinity.  As a matter of
fact, outside a specific rectangle  centered at the origin, the zero set
is the diagonal $\{x=y\}$, see Proposition~\ref{P-ext}.

Hence in this way (like for the square), we get that inside the zero
set, we have the diagonal and $\frac{n-1}{2}$  disjoint simple
closed lines turning around the origin.

\subsection{No other closed curve in the nodal set of
$\Phi_n^{\frac{3\pi}{4}}$}\label{nocc}

It remains to show that there are no other closed curves which do
not cross the diagonal. The ``energy'' considerations of our
previous papers work in the following way.

Assume there is a connected component of the nodal set which does
not meet the lattice $\mathcal L$. Using Proposition~\ref{P-ext}, we see
that this component must be contained in some large coordinate
square centered at $(0,0)$, call it $C$. Since the nodal set cannot
meet the vertical or horizontal lines defined by the zeros of $H_n$,
we would have a nodal domain $\omega$ contained in $C$, hence in one
of the bounded connected components of $\{H_n(x)\, H_n(y) < 0\}$,
and hence also in some infinite rectangle $R$ between two
consecutive zeros of $H_n$. We can compute the energy for $\omega$
by applying Green's formula in $\omega$. We can compute the energy
of the infinite rectangle $R$ by applying Green's formula first in a
finite rectangle, and then taking the limit (using the decaying
exponential factor). We have that the first Dirichlet eigenvalues
$\lambda_1$ satisfy $\lambda_1(\omega) = \lambda_1(R) = 2(n+1)$. On
the other hand, taking some $\omega_1$ such that $\omega \subset
\omega_1 \subset R$, with strict inclusions, we have
$\lambda_1(\omega) > \lambda_1(\omega_1) \ge \lambda_1(R)$, a
contradiction.

A simple alternative argument is the following. We look at
the line $y=\alpha x $ for some $\alpha \neq 1$. The intersection of
the zero set with this line corresponds to the zeroes of the
polynomial $x\mapsto H_n(x) - H_n (\alpha x)$ which has at most $n$
zeroes. But in our previous construction, we get at least $n$
zeroes. So the presence of extra  curves would lead to a
contradiction for some $\alpha$. This  argument  solves the
problem at  infinity as well.

\subsection{Perturbation argument}

 Figure~\ref{FSch-2} shows the desingularization of the nodal
set $N(\Phi^{\frac{3\pi}{4}}_n)$, from below and from above. The
picture is the same as in the case of the square (see Figure
\ref{sq-1}), all the critical points disappear at the same time and
in the same manner, \emph{i.e. } all the double crossings open up
horizontally or vertically depending whether $\theta$ is less than
or bigger than $\frac{3\pi}{4}$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=12cm]{FSchroedinger-0-7-bifurc-2.png}
\caption{The nodal set of $N(\Phi^{\theta}_n)$ near
$\frac{3\pi}{4}$.} \label{FSch-2}
\end{center}
\end{figure}

 As in the case of the square, in order to show that the nodal
set can be desingularized under small perturbation, we look at the
signs of the eigenfunction $\Phi_n^{\frac{3\pi}{4}}$ near the
critical zeros. We use the cases [i] and [ii] which appear in
Figure~\ref{FSchr-local-signs-1} below (see also
\cite[Figure~6.7]{BeHe}).

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=8cm]{FSchr-local-signs-1.png}
\caption{Signs near the critical zeros.} \label{FSchr-local-signs-1}
\end{center}
\end{figure}

The sign configuration for $\phi_{n,0}(x,y) - \phi_{0,n}(x,y)$ near
the critical zero $(t_{n-1,i},t_{n-1,i})$ is that of
\begin{equation*}\label{sign}
\left\lbrace
\begin{array}{l}
\text{case [i], ~if~} i \text{~is even},\\
\text{case [ii],~if~} i \text{~is odd}.\\
\end{array}\right.
\end{equation*}

Looking at the intersection of the nodal set with the vertical line
$\{y=t_{n-1,i}\}$, we have that
$$(-1)^i \left( H_n(t) - H_n(t_{n-1,i})\right) \ge 0, \text{~for~}
t \in ]t_{n,i},t_{n,i+1}[ \,.
$$

For positive $ \epsilon$ small, we write

$$(-1)^i \left( H_n(t) - (1+\epsilon) H_n(t_{n-1,i})\right) =
(-1)^i \left( H_n(t) - H_n(t_{n-1,i})\right) + \epsilon (-1)^{i+1}\,
H_n(t_{n-1,i})\,,
$$

so that

$$(-1)^i \left( H_n(t) - (1+\epsilon) H_n(t_{n-1,i})\right) \ge 0,
\text{~for~} t \in ]t_{n,i},t_{n,i+1}[ \,.
$$

 A similar statement can be written for horizontal line
$\{x=t_{n-1,i}\}$ and $-\epsilon$, with $\epsilon > 0$, small
enough. These inequalities describe how the crossings open up all at
the same time, and in the same manner, vertically (case I) or
horizontally (case II), see Figure~\ref{FSchr-local-signs-2}, as in
the case of the square \cite[Figure~6.8]{BeHe}.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=8cm]{FSchr-local-signs-2.png}
\caption{Desingularization at a critical zero.}
\label{FSchr-local-signs-2}
\end{center}
\end{figure}

We can then conclude as in the case of the square, using the local
nodal patterns, Section~\ref{SS-lnp}.\medskip

 \textbf{Remark}. Because the local nodal patterns can only
change when $\theta$ passes through one of the values $\theta(i,j)$
defined in \eqref{P10}, the above arguments work for $\theta \in
J\setminus \{\frac{3\pi}{4}\}$, for any interval $J$ containing
$\frac{3\pi}{4}$ and no other value $\theta(i,j)$.

\section{Proof of Theorem \ref{th2}}

\begin{proposition}\label{P-1}
The conclusion of Theorem \ref{th2} holds with
\begin{equation}\label{P26}
\theta_c := \inf \left\{ \theta(i,j) ~|~ 1 \le i,j \le n-1 \right\}
\,.
\end{equation}
\end{proposition}

\textbf{Proof}. The proof consists in the following steps. For
simplicity, we call $N$ the nodal set $N(\Phi^{\theta}_n)$.
\vspace{-3mm}
\begin{itemize}
    \item Step 1.~ By Proposition~\ref{P-ext}, the structure of the nodal
    set $N$ is known  outside a large coordinate rectangle centered at $(0,0)$
    whose sides are defined by the ad hoc numbers in
    Lemmas~\ref{L-ext1} and \ref{L-ext2}. Notice that the sides of
    the rectangle  serve as barriers for the arguments using the local
    nodal patterns as in our paper for the square.
    \item Step 2.~ For $1\le j \le n-1$, the line $\{x=t_{n-1,j}\}$
    intersects the set $N$ at exactly one point $(t_{n-1,j}, y_j)$,
    with $y_j > t_{n,n}$ when $j$ is odd, \emph{resp. } with $y_j < t_{n,1}$
    when $j$ is even. The proof is given below, and is similar to the proofs of
    Lemmas~\ref{L-ext1} or \ref{L-ext2}.
    \item Step 3.~ Any connected component of $N$
    has at least one point in common with the set $\mathcal L$. This
    follows from the  argument with $y = \alpha x$ or from the energy argument
    (see Subsection \ref{nocc}).
    \item Step 4.~ Follow the nodal set from the point
    $(t_{n,1},t_{n,n})$ to the point $(t_{n,n},t_{n,1})$, using the
    analysis of the local nodal patterns as in the case of the
    square.
\end{itemize}

\textbf{Proof of Step~2}. For $1 \le j \le (n-1)$, define the
function $v_j$ by
$$
v_j(y) := \cos\theta \, H_n(t_{n-1,j}) + \sin\theta \, H_n(y)\,.
$$

The local extrema of $v_j$ are achieved at the points $t_{n-1,i}$,
for $1 \le i \le (n-1)$, and we have
$$
v_j(t_{n-1,i} ) = \cos\theta \, H_n(t_{n-1,j}) + \sin\theta \,
H_n(t_{n-1,i})\,,
$$
which can be rewritten, using \eqref{P10}, as
$$
v_j(t_{n-1,i}) = \frac{H_n(t_{n-1,j})}{\sin\theta(j,i)}\,
\sin\left(\theta(j,i) -\theta \right).
$$

The first term in the right-hand side has the sign of $(-1)^{j+1}$
and the second term is positive provided that $0 < \theta <
\theta_c$. Under this last assumption, we have
\begin{equation}\label{P-32}
(-1)^{j+1}\, v_j(t_{n-1,i}) > 0, ~\forall i, ~1 \le i \le (n-1)\,.
\end{equation}

The assertion follows. \hfill $\square$

\section{Courant's theorem for the $2D$ quantum harmonic
oscillator}

Recall that $\mathcal{E}_{\ell}$ is the eigenspace of $\hat{H}$
associated with the eigenvalue $\hat{\lambda}(\ell) := 2(\ell + 1)$.
This eigenspace is generated by the eigenfunctions
$\phi_{\ell-j,j}$, for $0 \le j \le \ell$. It has dimension $(\ell +
1)$. The functions in $\mathcal{E}_{\ell}$ are even (\emph{resp. } odd) under the
map $a : (x,y) \mapsto (-x,-y)$ when $\ell$ is even (\emph{resp. } odd).

Since $\dim(\bigoplus_{j=0}^{\ell-1}\mathcal{E}_j) = \frac{\ell(\ell
+1)}{2}$, Courant's theorem gives the following estimate for the
number $\mu(u)$ of nodal domains of an eigenfunction $u \in
\mathcal{E}_{\ell}$,
\begin{equation}\label{courant1}
\mu (u) \le \frac{\ell(\ell +1)}{2} + 1 =: \mu_C(\ell)\,.
\end{equation}

Using the symmetry or anti-symmetry with respect to $a$, one can
improve Courant's estimate.

\begin{proposition}\label{P-courant2}
Let $u \in \mathcal{E}_{\ell}$. Then, the number $\mu(u)$ of nodal domains
of $u$ satisfies the inequalities,
\begin{equation}\label{courant2}
\mu(u) \le \mu_L(\ell) := \left\lbrace
\begin{array}{ll}
2(r^2+1) & \text{~if~} \ell = 2r\,,\\
2r(r+1)+2 & \text{~if~} \ell = 2r+1\,.\\
\end{array}
\right.
\end{equation}
In particular, we have that $\mu_L(\ell) < \mu_C(\ell)$ provided
that $\ell \ge 3$, and $\mu_L(\ell) = \mu_C(\ell)$, when $\ell = 2$.
\end{proposition}

\begin{corollary}\label{C-courant2}
The only Courant sharp eigenvalues of the quantum harmonic
oscillator are the eigenvalues,
\begin{equation}\label{courant-s}
\left\lbrace
\begin{aligned}
& \hat{\lambda}(0) = 2, \text{~with~} \mu_C(0)=1\,,\\
& \hat{\lambda}(1) = 4, \text{~with~} \mu_C(1)=2\,,\\
& \hat{\lambda}(2) = 6, \text{~with~} \mu_C(2)=4\,.\\
\end{aligned}
\right.
\end{equation}
\end{corollary}

\textbf{Proof of the corollary}. The first two assertions are clear.
The last one follows from the fact than the nodal set of an
eigenfunction in $\mathcal{E}_2$ is a hyperbola, the union of two lines
which intersect, or an ellipse. \hfill $\square$

\textbf{Proof of the proposition}. We use Leydold's argument in
\cite{Ley}, namely the symmetry properties of the eigenfunctions
with respect to $a$, the fact that an odd eigenfunction is always
orthogonal to an even one, and Courant's proof.

\noindent $\bullet $~ Assume that $u \in \mathcal{E}_{\ell}$ with $\ell = 2r$. We have
$$\dim(\bigoplus_{j=0}^{r-1}\mathcal{E}_{2j}) = r^2\,.
$$
There are $k_i$ nodal domains of $u$ which are $a$ invariant,
$a(\omega)=\omega$, and $2 k_a$ nodal domains which are not
invariant, $a(\omega) \cap \omega = \emptyset$. Assume that $k_i+k_a
\ge r^2+2$. Define functions $u_j$ such that $u_j = u|_{\omega_j}$,
and $0$ elsewhere, for each invariant domain $\omega_j$, $1 \le j
\le k_i-1$, and $u_j=u|_{\omega_p \cup a(\omega_p)}$, and $0$
elsewhere, for the $k_a$ non-invariant domains. This gives us
$k_a+k_i-1 \ge r^2+1$ independent functions. We can find a linear
combination $v$ of these functions such that $\|v\|_{L^2} = 1$, $v
\perp \bigoplus_{j=0}^{r-1}\mathcal{E}_{2j}$, and $\mathcal{Q}(v)) = 2(\ell
+1)$, where $\mathcal{Q}$ is the quadratic form associated with
$\hat{H}$. The function $v$ is even by construction so that it is
orthogonal to any odd eigenfunction. It follows that $v \in
\mathcal{E}_{\ell}$ which leads to a contradiction since $v$ vanishes on an
open set. It follows that $k_i+k_a \le r^2+1$ and hence that $k_i+
2k_a \le 2(r^2+1)$. This proves the first assertion.

\noindent $\bullet $~ Assume that $u \in \mathcal{E}_{\ell}$ with $\ell = 2r+1$. The proof
is similar. We have that $$\dim(\bigoplus_{j=0}^{r-1}\mathcal{E}_{2j+1}) =
r(r+1)\,.$$
For an odd eigenfunction $u$, the nodal domains satisfy
$a(\omega) \cap \omega = \emptyset$, so that $\mu(u) = 2k$, and we
can construct $k$ linearly independent functions $u_j=u|_{a(\omega)
\cup \omega}$, and we can proceed as above. \hfill $\square$

 \textbf{Remark}. In the above proof, we used Courant's proof
which is based on energy estimates, using Green's formula for the
eigenfunction $u$. That this can be done in the case of the quantum
harmonic operator follows from the following argument. At infinity,
the nodal set of $u$ is a regular submanifold. It consists of arcs
asymptotic to lines determined by the homogeneous higher order terms
in $\exp(\frac {x^2+y^2}{2})\, u(x,y))$. We can apply Green's formula to
the intersections of the nodal domains of $u$ with balls $B(0,r)$.
When $r$ tends to infinity, the boundary terms involving the ball
tend to zero due to the presence of the exponential factors.

redpajama_set_name: RedPajamaArXiv


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