Spontaneous rotational symmetry breaking and roton like excitations in gauged sigma-model a

初一数学图形英语阅读理解20题1<背景文章>A circle is a very special shape. It is perfectly round and has no corners or edges. The distance around the circle is called the circumference. The line that goes through the center of the circle and connects two opposite points on the circle is called the diameter. The radius of a circle is half of the diameter.Circles have many interesting properties. For example, all points on the circumference of a circle are equidistant from the center. This means that if you measure the distance from any point on the circle to the center, it will always be the same. Another property of circles is that they have rotational symmetry. This means that if you rotate a circle around its center, it will always look the same.Circles are very common in our daily lives. We can see circles everywhere, such as wheels on cars and bicycles, plates, coins, and clocks. Circles are also used in art and design. For example, many paintings and sculptures use circles to create beautiful patterns.The area of a circle can be calculated using the formula A = πr², whereA is the area and r is the radius. The value of π is approximately 3.14.1. The distance around the circle is called the ___.A. radiusB. diameterC. circumferenceD. area答案：C。

关于角动量各分量相关表达在量子力学里，角动量算符(angular momentum operator)是一种算符，类比于经典的角动量。

在原子物理学涉及旋转对称性 (rotational symmetry)的理论里，角动量算符占有中心的角色。

角动量，动量，与能量是物体运动的三个基本特性[1]。

角动量促使在旋转方面的运动得以数量化。

在孤立系统里，如同能量和动量，角动量是守恒的。

在量子力学里，角动量算符的概念是必要的，因为角动量的计算实现于描述量子系统的波函数，而不是经典地实现于一点或一刚体。

在量子尺寸世界，分析的对象都是以波函数或量子幅来描述其概率性行为，而不是命定性 (deterministic)行为。

定义：在经典力学里，角动量定义为位置与动量的叉积：。

在量子力学里，对应的角动量算符定义为位置算符与动量算符的叉积：。

由于动量算符的形式为。

角动量算符的形式为。

其中，是梯度算符。

如果角动量分量的单位矢量用i ,j,k 表示，z y x p p p ,,，表示直角坐标系中个方向的动量，则可以有z y xp p p z yx k j i，如果相关其中一个分量与其它分量的关系可以用行列式关系。

关于z 轴方向的角动量可以用行列式x y y x z yp xp p p y xM -==角动量是厄米算符在量子力学里，每一个可观察量所对应的算符都是厄米算符。

角动量是一个可观察量，所以，角动量算符应该也是厄米算符。

让我们现在证明这一点，思考角动量算符的 x-分量 ：。

其伴随算符为。

由于 、 、 、 ，都是厄米算符，。

由于与之间、与之间分别相互对易，所以，。

因此，是一个厄米算符。

类似地，与都是厄米算符。

总结，角动量算符是厄米算符。

symmetry 英语解释《Symmetry: Unveiling the Beauty of Balance》Symmetry is a fascinating concept that touches upon the fundamental principles of balance, harmony, and beauty in the world around us. From the magnificent works of art to the intricate patterns in nature, symmetry plays a crucial role in creating visual appeal and evoking a sense of orderliness.Defined as a correspondence in size, shape, or arrangement on opposite sides of a dividing line or plane, symmetry is found in various forms throughout our existence. It exists in the realms of mathematics, biology, architecture, and even in the smallest particles that make up our universe. By examining the concept of symmetry, we can gain a deeper understanding of the fundamental organizing principles that shape our world.In nature, symmetry is a prevalent feature. From the petals of a flower to the branches of a tree, many living organisms exhibit a remarkable balance in their structures. The perfect radial symmetry of a sunflower, with its petals arranged uniformly around a central axis, is simply awe-inspiring. Similarly, the bilateral symmetry found in animals, where each side of the body mirrors the other, gives rise to the undeniable beauty of creatures such as butterflies and peacocks.In the realm of art and design, symmetry has always been highly valued. Throughout history, artists have utilized symmetry to create aesthetic masterpieces that captivate our senses. From the mesmerizing patterns on a Persian carpet to the architectural wonders of the Taj Mahal, symmetry is a key element in enhancing the sense of harmony and proportion. Michelangelo's famous painting on the ceiling of the Sistine Chapel is a testament to the power of symmetry in evoking a sense of grandeur.Moreover, symmetry plays a crucial role in mathematics and physics. Symmetrical shapes and patterns are deeply intertwined with geometric principles and mathematical formulas. This relationship can be seen in the symmetry of polygons, such as squares and circles, as well as in intricate fractal patterns. Even in the realm of physics, concepts like mirror symmetry and rotational symmetry are fundamental to understanding the laws that govern our universe.Symmetry not only captivates our senses but also holds profound philosophical implications. The quest for symmetry has driven scientists and researchers to uncover the fundamental truths behind the natural world. By seeking patterns and order, we strive to unlock the mysteries of the universe and gain a deeper appreciation for the beauty inherent in its balanced design.In conclusion, symmetry is a captivating concept that transcends disciplines and permeates every aspect of our lives. From the delicate patterns found in nature to the magnificent works of art, symmetry unveils the hidden beauty of balance and harmony. By embracing the power of symmetry, we can develop a greater appreciation for the world around us and gain insight into the fundamental principles that shape our existence.。

2015 AMC 12B竞赛真题Problem 1What is the value of ?Problem 2Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1:00 PM and finishes the second task at 2:40 PM. When does she finish the third task?Problem 3Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?Problem 4David, Hikmet, Jack, Marta, Rand, and Todd were in a 12-person race with 6 other people. Rand finished 6 places ahead of Hikmet. Marta finished 1 place behind Jack. David finished 2 places behind Hikmet. Jack finished 2 places behind Todd. Todd finished 1 place behind Rand. Marta finished in 6th place. Who finished in 8th place?Problem 5The Tigers beat the Sharks 2 out of the 3 times they played. They then played more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for ?Problem 6Back in 1930, Tillie had to memorize her multiplication facts fromto . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?Problem 7A regular 15-gon has lines of symmetry, and the smallest positive angle for which it has rotational symmetry is degrees. What is ?Problem 8What is the value of ?Problem 9Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a playerknocks the bottle off the ledge is , independently of what has happened before. What is the probability that Larry wins the game?Problem 10How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?Problem 11The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?Problem 12Let , , and be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation?Problem 13Quadrilateral is inscribed in a circle withand . What is ?Problem 14A circle of radius 2 is centered at . An equilateral triangle with side4 has a vertex at . What is the difference between the area of the regionthat lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?Problem 15At Rachelle's school an A counts 4 points, a B 3 points, a C 2 points, and a D 1 point. Her GPA on the four classes she is taking is computed as the total sum of points divided by 4. She is certain that she will get As in both Mathematics and Science, and at least a C in each of English and History.She thinks she has a chance of getting an A in English, and a chance of getting a B. In History, she has a chance of getting an A, and a chanceof getting a B, independently of what she gets in English. What is the probability that Rachelle will get a GPA of at least 3.5?Problem 16A regular hexagon with sides of length 6 has an isosceles triangle attached to each side. Each of these triangles has two sides of length 8. The isosceles triangles are folded to make a pyramid with the hexagon as the base of the pyramid. What is the volume of the pyramid?Problem 17An unfair coin lands on heads with a probability of . When tossed times,the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of ?Problem 18For every composite positive integer , define to be the sum of the factors in the prime factorization of . For example, because the prime factorization of is , and . What is the range of the function , ?Problem 19In , and . Squares and are constructed outside of the triangle. The points , , , and lie on a circle. What is the perimeter of the triangle?Problem 20For every positive integer , let be the remainder obtained when is divided by 5. Define a functionrecursively as follows:What is ?Problem 21Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of ?Problem 22Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same chair and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?Problem 23A rectangular box measures , where , , and are integers and. The volume and the surface area of the box are numerically equal. How many ordered triples are possible?Problem 24Four circles, no two of which are congruent, have centers at , , , and , and points and lie on all four circles. The radius of circleis times the radius of circle , and the radius of circle is timesthe radius of circle . Furthermore, and . Let be the midpoint of . What is ?Problem 25A bee starts flying from point . She flies inch due east to point . For , once the bee reaches point , she turns counterclockwise and then flies inches straight to point . When the bee reaches she is exactly inches away from , where , , andare positive integers and and are not divisible by the square of any prime. What is ?2015 AMC 12B竞赛真题答案1.C2.b3.a4.b5.b6.a7.d8.d9.c 10.c 11.e 12.d 13.b 14.d 15.d 16.c 17.d 18.d 19.c 20.b 21.d 22.d 23.b 24.d 25.b。

Parallel grippers HGPL, heavy-duty with long strokeKey features At a glanceSpace-saving and suitable for high forces–Two parallel and opposingpistons move the gripper jawsdirectly and without loss of force Reliable–A pinion that synchronises themovement of both gripper jawsensures controlled, precise andcentred gripping–The space-saving design of theparallel gripper jaws permits along guide length for the gripperjaws Sturdy–The T-slot in combination witha long guide length allows thegripper jaws to withstand highforces and torquesFlexible range of applications–Double-acting gripper suitablefor external and internal gripping –Versatile mounting options andcompressed air connections–Opening stroke can be adjustedto optimise time-H-Note Sizing software for gripper selection è 1Gripper jaw2Synchronising element3Piston with magnet4Moment compensatorWide range of supply ports Mounting optionsDirect Via adapter plate Direct mountingfrom the front from underneath from above from underneath1Supply ports2Centring sleeves3O-rings1Mounting screws2Centring sleeves-H-NoteThese grippers are not suitable forthe following or similar applicationexamples:Aggressive mediaMachiningWelding splashesSubject to change – 2018/05 2è Internet: /catalog/...2018/05 – Subject to change 3è Internet: /catalog/...Peripherals overview and type codes Peripherals overviewSystem product for handling and assembly technologyAccessoriesType Size Descriptionè Page/Internet 1Centring sleeve ZBH14 … 63 For centring gripper jaw blanks/gripper fingers on the gripper jaws 4 included in the scope of delivery 172Proximity sensor SME/SMT-1014 … 40 For sensing the piston position 18Proximity sensor SME/SMT-863 For sensing the piston position 3Position transmitter SME/SMT-863 For sensing the piston position184Push-in fitting QS14 … 63For connecting compressed air tubing with standard O.D.qs 5Blanking plug B14 … 63For sealing supply ports when using ports at the front 176Centring sleeve ZBH14 … 63 For centring the gripper during mounting 2 included in the scope of delivery 177Stroke reducing kit HGPL-HR-...14 … 63For reducing the opening stroke 168Adapter kitDHAA, HMSV, HMVA, HAPG 14 … 63Drive/gripper connections13–Gripper jaw blank BUB-HGPL14 (63)Blank specially matched to the gripper jaws for custom production of gripper fingersLEERER MERKERType codesHGPL—14—40—AType HGPL Parallel gripperSize Stroke [mm]Position sensing AVia proximity sensorSubject to change – 2018/054è Internet: /catalog/...Technical data Function Double-acting HGPL-…-A-N-Size14 … 63 mm -T-Stroke40 … 300 mm-W- Wearing parts kits èpage 12General technical data Size 14254063DesignSynchronised pneumatic pistons Force-guided motion sequence Mode of operationDouble-acting T-shape Gripper functionParallel Number of gripper jaws 2Stroke per gripper jaw [mm]20, 40, 60, 8020, 40, 60, 8020, 40, 60, 80, 10060, 100, 150Pneumatic connectionM5G1/8Max. load per external gripper finger 1)[g]80250420940Repetition accuracy 2)[mm]< 0.03Max. interchangeability [mm]< 0.2Max. operating frequency [Hz]< 1Rotational symmetry [mm]< ∅ 0.2Position sensing Via proximity sensor –Position transmitterType of mounting Via through-holes and centring sleeves Via female thread and centring sleeves Mounting positionAny1)Valid for unthrottled operation2)End-position drift under constant conditions of use with 100 consecutive strokes in the direction of movement of the gripper jawsOperating and environmental conditions Operating pressure [bar]3 (8)Operating mediumCompressed air in accordance with ISO 8573­1:2010 [7:4:4]Note on operating/pilot medium Operation with lubricated medium possible (in which case lubricated operation will always be required)Ambient temperature 1)[°C]+5 … +60Corrosion resistance class CRC 2)21)Note operating range of proximity sensors2)Corrosion resistance class CRC 2 to Festo standard FN 940070Moderate corrosion stress. Indoor applications in which condensation may occur. External visible parts with primarily decorative requirements for the surface and which are in direct contact with the ambient atmo­sphere typical for industrial applications.Weight [g]Size14254063Stroke per gripper jaw20 mm 3051015 2560 –40 mm 4401400 3300 –60 mm 5951780 4165 10460 80 mm 7202200 4800 –100 mm ––5340 13800 150 mm–––181002018/05 – Subject to change 5è Internet: /catalog/...Technical data Materials Sectional viewParallel gripper 1Gripper jaw Steel, nitrided 2Piston High-alloy steel3Housing Wrought aluminium alloy, hard-anodised –SealsNitrile rubber, polyurethane –Note on materialsFree of copper and PTFE RoHS-compliantMeasured gripping force [N] at 6 bar (èalso see graphs starting on page 7)Size14254063Gripping force per gripper jaw Opening 632065191233 Closing792566081371Total gripping force Opening 1264121038 2466 Closing1585121216 2742Characteristic load values at the gripper jawsThe indicated permissible forces and torques apply to a single gripper jaw.They include the lever arm, additional applied loads due to the workpiece or external gripper fingers and acceleration forces occurring during movement. The zero coordinate line (guide groove of the gripper jaws)must be taken into consideration for the calculation of torques.Size14254063Max. permissible force F z [N]5001500 2500 9000 Max. permissible torque M x [Nm]35100125300Max. permissible torque M y [Nm]356080200Max. permissible torque M z[Nm]3570100250Technical dataMass moment of inertia [kgm2x10-4]Mass moment of inertia of the parallelgripper in relation to the central axis,without external gripper fingers,without load.Size14254063 Stroke per gripper jaw20 mm 1.4011.9827.60–40 mm 6.6918.8866.83–60 mm11.4339.95118.30470.0780 mm21.9378.70198.87–100 mm––318.251018.17150 mm–––2247.54Opening and closing times [ms] at 6 barWithout external gripper fingersWith external gripper fingersThe indicated opening and closingtimes [ms] were measured at roomtemperature at an operating pressureof 6 bar with horizontally mountedgrippers without additional gripperfingers. The grippers must be throttledfor greater loads [g]. Opening andclosing times must then be adjustedaccordingly.Size14254063Stroke mm20406080204060802040608010060100150 Without external gripper fingersOpening times1201712702861702253704231902384304146204106501020 Closing times110163230270150230370418180205430438690330600850 Max. permissible opening and closing times with external gripper fingers (as a function of the load per gripper finger)Load of the gripper fingers100 g123108257243––––––––––––200 g174136364343––––––––––––300 g213167445420164210405401––––––––400 g246192514485190243468463––––––––500 g––––212272523518196260469478676–––600 g––––––––215284514524741–––700 g––––––––232307555565800–––800 g––––––––248328593604856–––900 g–––––––––––––3235878321000 g–––––––––––––3406198771100 g–––––––––––––3576499191200 g–––––––––––––373678960Subject to change – 2018/05 6è Internet: /catalog/...2018/05 – Subject to change 7è Internet: /catalog/...Technical dataGripping force F h per gripper jaw as a function of operating pressure and lever arm x The gripping forces as a function of operating pressure and lever arm can be determined from the followinggraphs.Gripping force F h per gripper jaw as a function of operating pressure and lever arm x External gripping (closing)HGPL-14HGPL-25HGPL-40HGPL-633 bar 6 bar8 barSubject to change – 2018/058è Internet: /catalog/...Technical dataGripping force F h per gripper jaw as a function of operating pressure and lever arm x Internal gripping (opening)HGPL-14HGPL-25HGPL-40HGPL-633 bar 6 bar8 bar2018/05 – Subject to change 9è Internet: /catalog/...Technical dataGripping force F h per gripper jaw at 6 bar as a function of lever arm x and eccentricity a and b The following formula must be used to calculate the lever arm x with eccentric gripping:The gripping force F h can be read from the graphs (è page 7) using the calculated value x.x +a 2)b 2ǸCalculation example Given:Distance a = 45 mm Distance b = 40 mm To be calculated:The gripping force at 6 bar,with an HGPL-25,used as an external gripperProcedure:Calculating the lever arm x x = 60 mmx +452)402ǸThe graph (è page 7) shows a value of F h= 225 N for the gripping force.Technical data10è Internet: /catalog/...Subject to change – 2018/052018/05 – Subject to change 11è Internet: /catalog/...Technical data TypeB1±0.05B2±0.1B3±0.1B6±0.01D1D2∅+0.1D3∅H8/h7D4D5∅H8/h7D8∅H13D9D10EEE1HGPL-144834.53722M5 4.25M397.46M3M5M3HGPL-2580606538M6 5.17M59108M5M5M5HGPL-40106708750M108.59M615158M6M5M5HGPL-6315411613078M1210.415M81516.515M10G1/8G1/8TypeH1H2±0.1H3±0.1H4–0.3H5–0.3H6±0.1H7±0.1H8±0.1H9±0.1L1±0.1L2±0.021)±0.12)L3±0.021)±0.12)L4±0.5HGPL-14-20302911 1.9 1.2–10121873.6–3662HGPL-14-40113.660102HGPL-14-60153.6100142HGPL-14-80193.6182HGPL-25-20504918 1.9 1.4–18203086–6064HGPL-25-40126104HGPL-25-60166100144HGPL-25-80206184HGPL-40-208078.56 2.9 1.94617.54050.596–6670HGPL-40-40136100110HGPL-40-60176150HGPL-40-80216180190HGPL-40-100256200230HGPL-63-60121.512014 2.9 2.960305874.8190.8–100160HGPL-63-100270.8200240HGPL-63-150370.8300340TypeL5±0.5L7±0.1L8±0.1L9±0.2L10±0.021)±0.12)L11±0.5L12±0.1T1min.T2+0.1T3+0.1T4min.T5min.T7+0.1T9HGPL-14-202241416.88916.812.5 1.3 2.1 5.5 6.5101HGPL-14-40HGPL-14-60HGPL-14-80HGPL-25-20241114201017.52012.5 1.6 2.18.57171HGPL-25-40HGPL-25-60HGPL-25-80HGPL-40-20301332–1520.5–15.52.13.110.58301HGPL-40-4021HGPL-40-60HGPL-40-80HGPL-40-100HGPL-63-604028.530–2229–18 3.1 3.117.58451HGPL-63-100HGPL-63-1501)For centring 2)For through-holeTechnical dataOrdering dataSize Stroke Double-acting without compression spring[mm][mm]Part No.Type1420567820HGPL-14-20-A40535852HGPL-14-40-A60567821HGPL-14-60-A80535853HGPL-14-80-A2520567822HGPL-25-20-A40535854HGPL-25-40-A60567823HGPL-25-60-A80535855HGPL-25-80-A4020567824HGPL-40-20-A40535856HGPL-40-40-A60567825HGPL-40-60-A80535857HGPL-40-80-A100567826HGPL-40-100-A6360567827HGPL-63-60-A100567828HGPL-63-100-A150567829HGPL-63-150-AOrdering data – Sets of wearing partsSize[mm]Part No.Type14701585HGPL-14-A25701586HGPL-25-A40701587HGPL-40-A63752917HGPL-63-A12è Internet: /catalog/...Subject to change – 2018/05AccessoriesAdapter kitDHAA, HAPG, HMSV, HMVA Material:Wrought aluminium alloyFree of copper and PTFERoHS-compliant-H-NoteThe kit includes the individualmounting interface as well as thenecessary mounting material.1)Corrosion resistance class CRC 2 to Festo standard FN 940070Moderate corrosion stress. Indoor applications in which condensation may occur. External visible parts with primarily decorative requirements for the surface and which are in direct contact with the ambient atmo­sphere typical for industrial applications.2018/05 – Subject to change13è Internet: /catalog/...AccessoriesAdapter kit DHAA Material:Wrought aluminium alloyFree of copper and PTFERoHS-compliant-H-NoteThe kit includes the individualmounting interface as well as thenecessary mounting material.1)Corrosion resistance class CRC 2 to Festo standard FN 940070Moderate corrosion stress. Indoor applications in which condensation may occur. External visible parts with primarily decorative requirements for the surface and which are in direct contact with the ambient atmo­sphere typical for industrial applications.Subject to change – 2018/05 14è Internet: /catalog/...AccessoriesAdapter kit DHAA, HAPG Material:Wrought aluminium alloyFree of copper and PTFERoHS-compliant-H-NoteThe kit includes the individualmounting interface as well as thenecessary mounting material.1)Corrosion resistance class CRC 2 to Festo standard FN 940070Moderate corrosion stress. Indoor applications in which condensation may occur. External visible parts with primarily decorative requirements for the surface and which are in direct contact with the ambient atmo­sphere typical for industrial applications.2018/05 – Subject to change15è Internet: /catalog/...AccessoriesDimensions and ordering dataFor size [mm]B1±0.1D1D2H1±0.1H2±0.1H3±0.1H4±0.1H5±0.1149M6M327.523.517.59.512 2512M8M547.537.529.517.520 4018M12M67763501740 6319M14M8118.594.574.529.558For size [mm]L1±1L2±0.1L3±0.1L4±0.1ß1ß2Weight[g]Part No.Type146145371110345539092HGPL-HR-14 2561776519134150539093HGPL-HR-25 40611038725196455539094HGPL-HR-40 6381151130392261060 567831HGPL-HR-63Subject to change – 2018/05 16è Internet: /catalog/...2018/05 – Subject to change 17è Internet: /catalog/...AccessoriesGripper jaw blank BUB-HGPL (scope of delivery: 2 pieces)Material:AluminiumFree of copper and PTFEDimensions and ordering data For size B1B2B3D1D2H1H2∅∅[mm]±0.1+0.02+0.1H8±0.1142584 3.2580112535108 5.371201********* 6.49150186368221210.51520020For size L1L2L3T1Weight per blank Part No.Type[mm]±0.1+0.1+0.1+0.1[g]1420.58 3.3 1.375537316BUB-HGPL-142536125 1.6295537317BUB-HGPL-254049.516.58 2.1720537318BUB-HGPL-40637727123.11960567830BUB-HGPL-631)Packaging unitAccessoriesProximity sensors for size 14 (40)Proximity sensors for size 6318è Internet: /catalog/...Subject to change – 2018/052018/05 – Subject to change 19è Internet: /catalog/...AccessoriesProximity sensors for size 63Position transmitterThe position transmitter continuously senses the position of the piston.It has an analogue output with an out­put signal in proportion to the piston position.Festo - Your Partner in AutomationConnect with us/socialmedia 1Festo Inc.2Festo Pneumatic 3Festo Corporation 4Regional Service Center 5300 Explorer DriveMississauga, ON L4W 5G4CanadaAv. Ceylán 3,Col. Tequesquináhuac 54020 Tlalnepantla, Estado de México1377 Motor Parkway Suite 310Islandia, NY 117497777 Columbia Road Mason, OH 45040Festo Customer Interaction CenterTel:187****3786Fax:187****3786Email:*****************************Multinational Contact Center 01 800 337 8669***********************Festo Customer Interaction Center180****3786180****3786*****************************S u b j e c t t o c h a n g e。

READ THESE INSTRUCTIONS FIRSTWrite your Centre number, candidate number and name on all the work you hand in.Write in dark blue or black pen.You may use a pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fl uid.DO NOT WRITE IN ANY BARCODES.Answer all questions.If working is needed for any question it must be shown below that question.Electronic calculators should be used.If the degree of accuracy is not speci fi ed in the question, and if the answer is not exact, give the answer to three signi fi cant fi gures. Give answers in degrees to one decimal place.For π, use either your calculator value or 3.142.At the end of the examination, fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.The total of the marks for this paper is 70.MATHEMATICS0580/23Paper 2 (Extended)May/June 20131 hour 30 minutesCandidates answer on the Question Paper.Additional Materials:Electronic calculator Geometrical instrumentsTracing paper (optional)UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certi fi cate of Secondary EducationThis document consists of 12printed pages.[Turn overIB13 06_0580_23/4RP © UCLES 20130580/23/M/J/13© UCLES 2013ForExaminer ′s Use1 Sheila can pay her hotel bill in Euros (€) or Pounds (£).T he bill was €425 or £365 when the exchange rate was £1 = €1.14 . In which currency was the bill cheaper?S how all your working. Answer ............................................... [2]_____________________________________________________________________________________2 The Ocean View Hotel has 300 rooms numbered from 100 to 399.A room is chosen at random.F ind the probability that the room number ends in zero. Answer ............................................... [2]_____________________________________________________________________________________3 The time in Lisbon is the same as the time in Funchal.A plane left Lisbon at 08 30 and arrived in Funchal at 10 20.I t then left Funchal at 12 55 and returned to Lisbon.T he return journey took 15 minutes more.W hat time did the plane arrive in Lisbon? Answer ............................................... [2]_____________________________________________________________________________________ForExaminer′sUse4U se a calculator to fi nd(a)Answer(a) (1)(b)°407cos.Answer(b) (1)_____________________________________________________________________________________ 5W rite the following in order of size, smallestfi rst..152^h 1.52c m 1.5-2c m32-2c mAnswer ................... < ................... < ................... < . (2)_____________________________________________________________________________________6T he volumes of two similar cones are 36πcm3 and 288πcm3.T he base radius of the smaller cone is 3cm.C alculate the base radius of the larger cone.Answer ......................................... cm [3]_____________________________________________________________________________________0580/23/M/J/13© UCLES 2013[Turn over0580/23/M/J/13© UCLES 2013For Examiner ′s Use74 cmNOT TO SCALEThe shaded shape has rotational symmetry of order 2.W ork out the shaded area. Answer ........................................ cm 2 [3]_____________________________________________________________________________________8 T he mass, m , of a sphere varies directly with the cube of its radius, r .m = 160 when r = 2.F ind m when r = 5.Answer m = ............................................... [3]_____________________________________________________________________________________For9C alculate, giving your answers in standard form,Examiner′sUse(a)2× (5.5 × 104) ,Answer(a) (2)(b) (5.5 × 104) – (5 × 104) .Answer(b) (2)_____________________________________________________________________________________10F ind the value of 2x + y for the simultaneous equations.3x + 5y = 482x – y = 19Answer 2x + y = (4)_____________________________________________________________________________________© UCLES 2013[Turn over0580/23/M/J/130580/23/M/J/13© UCLES 2013ForExaminer ′s Use11 T he sum of the prime numbers less than 8 is equal to 17.(a) F ind the sum of the prime numbers less than 21.Answer(a) (2)(b) T he sum of the prime numbers less than x is 58.F ind an integer value for x . Answer(b) x = ............................................... [2]_____________________________________________________________________________________12 T wo spinners have sections numbered from 1 to 5. E ach is spun once and each number is equally likely.T he possibility diagram is shown below.1123452345SecondspinnerFirst spinnerF ind the probability that(a) b oth spinners show the same number,Answer(a) (2)(b) t he sum of the numbers shown on the two spinners is 7.Answer(b) ............................................... [2]_____________________________________________________________________________________0580/23/M/J/13© UCLES 2013[Turn overForExaminer ′s Use13 W rite as a single fraction in its simplest form.3x x 1+--- Answer ............................................... [4]_____________________________________________________________________________________14 (a) S olve 3n + 23 < n + 41.Answer(a) (2)(b) F actorise completely ab + bc + ad + cd .Answer(b) ............................................... [2]_____________________________________________________________________________________0580/23/M/J/13© UCLES 2013For Examiner ′s Use15ABCyx12 – x15 – x20 – x14138T he Venn diagram shows the number of elements in sets A , B and C .(a) n (A ∪ B ∪ C ) = 74F ind x .Answer(a) x = (2)(b) n () = 100F ind y .Answer(b) y = (1)(c) F ind the value of n((A ∪ B )' ∩ C ).Answer(c) ............................................... [1]_____________________________________________________________________________________0580/23/M/J/13© UCLES 2013[Turn overFor Examiner ′s Use16 f(x ) = x + x 2– 3, x ¸ 0 g(x ) = 2x– 5Find(a) f g(18),Answer(a) (2)(b) g –1(x ).Answer(b) g –1(x ) = ............................................... [2]_____________________________________________________________________________________17 M = 2336e o N = 211752eo(a) W ork out MN .Answer(a) [2](b) F ind M –1, the inverse of M .Answer(b) [2]_____________________________________________________________________________________0580/23/M/J/13© UCLES 2013For Examiner ′s Use18NOT TO SCALEA andB lie on a circle centre O , radius 5 cm. A ngle AOB = 120°.F ind the area of the shaded segment. Answer ........................................ cm 2 [4]_____________________________________________________________________________________110580/23/M/J/13© UCLES 2013[Turn over For Examiner ′sUse19OE DOABCDE is a regular polygon.(a) W rite down the geometrical name for this polygon.Answer(a) (1)(b)Ob c .F ind, in terms of b and c, in their simplest form, (i)Answer(b)(ii)Answer(b) (iii) t he position vector of E . Answer(b)(iii) (1)_____________________________________________________________________________________Question 20 is printed on the next page.畅享冠军纯正特产？快来绵羊国正品代购！120580/23/M/J/13Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Everyreasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included thepublisher will be pleased to make amends at the earliest possible opportunity.University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.© UCLES 2013For Examiner ′s Use20 (a)yF ind y when x = 2.G ive your answer correct to 4 decimal places.Answer(a) y = ............................................... [2] (b) R earrangey x the subject.Answer(b) x = (4)畅享冠军纯正特产？快来绵羊国正品代购！。

Chapter5Chiral Dynamics5.1What is spontaneous symmetry breaking?Symmetries and their breakings are important part of modern physics.Spacetime symmetry and its supersymmetric extensions are the basis for building quantumfield theories.Internal symmetries, such as isospin(proton and neutron,up and down quark symmetry),flavor,color etc.,form the fundamental structure of the standard model.On the other hand,studying symmetry breakings is as interesting as studying symmetries themselves.As far as we know,there are three ways to break a symmetry:explicit breaking,spontaneous breaking,andfinally anomalous breaking.In this part of the lectures we will concern ourselves with thefirst two types of breakings of the so-called chiral symmetry,the exact meaning of which will become clear later.We will come to the anomalous symmetry breaking towards the end of the course.In quantum mechnics,a symmetry of a hamiltonian is usually reflected in its energy spectrum. For instance,the rotational symmetry of a three-dimensional system often leads to a2ℓ+1-fold degeneracy of the spectrum.This standard realization of a symmetry is called Wigner-Weyl mode. On the other hand,in the late50’s Nambu and Goldstone discovered a new way through which a symmetry of a system can manifest itself:spontaneous breaking of the symmetry.This realization of a symmetry is called Nambu-Goldstone mode.To understand the Nambu-Goldstone realization of a symmetry,let us recall a related problem in statistical mechanics:second-order phase transitions.We have many examples of the second-order phase transitions in which a continuous change of order parameters happens.Consider a piece of magnetic material.Its hamiltonian is certainly rotationally symmetric and therefore normally one would expect its ground state wave function is also rotationally symmetric.This apparently is not the case below a certain critical temperature at which a spontaneous magnetization occurs. The magnitization vector points to a certain direction in space,and hence the rotational symmetry is lost.We say in this case that the rotational symmetry is spontaneously broken.Likewise,for a conductor below a certain temperature,the electromagnetic U(1)symmetry is spontaneously broken and the wavefunction of the Cooper pairs developes certain classical value.A useful mathematical formulation of the SSB is the concept of the effective action.Let us introduce thisfirst.Consider a scalarfield theory with lagrangian density L(φ).We define the green’s function8182CHAPTER5.CHIRAL DYNAMICS functional or generating functional Z(j)asZ(j)=∞i=0i n[Dφ]e i d4x L(x).(5.2)We define the connected green’s function G(n)c throughW(j)=∞i=1i nδj(x),(5.4)from which one can solve j(x)as a functional ofφ(x).Perform now the Legendre transformation,Γ(φ)= W− d4xj(x)φ(x) |j=j(φ)(5.5) ThenΓ(φ)is the generating functional for the one-particle irreducible Green’s functionsΓ(n)(x1,···,x n),Γ(φ)= n=11δφ(x).(5.7)Effective action can be computed through the shift offield in the lagrangianφ→φ+φc,and calculating the1PI contribution to the effective W.There are two popular usage of the effective action formalism:First,the effective action containsall the1PI which are the target for renormalization study.The renormalization condition can5.1.WHAT IS SPONTANEOUS SYMMETRY BREAKING?83 easily expressed in terms of1PI,like the mass of the particles and coupling constants.Moreover, the symmetry of these1PI can be expressed in terms of the Ward-Takahashi identities which can be summarized in terms of a simple equation for the effective action.This equation can be used to prove the Goldstone theorem.Second,the effective action can be used as a thermodynamic function with natural variableφc which diagnoses the phase structure of the system.For instance, according to Colemann-Weinberg,the natural phase of the massless scalar electrodynamics is the Higgs phase in which the vector and scalar particles aquire mass through radiative corrections. Another use of the effective action is in cosmology.The spontanous symmetry breaking happens only if there is a degeneracy in the vacuum.This degeneracy can arise from certain symmetry of the original lagrangian.Consider a symmetry transformation offields,φi(x)→φ′i(x)= j L ijφj(x),(5.8)here we have assumed multiplefields with i=1,...,n.If the action and measure are both invariant, then the effective action is invariant under a similar transformation of the classicalfieldsΓ[φ]=Γ[Lφ].(5.9) As we mentioned before,the vaccum state is a solution¯φof−Γ[φ]at its minimum.If the solution is invariant L¯φ=¯φ,i.e.the vacuum is invariant under the symmetry transformation,the vacuum is unique.On the other hand,if L¯φ=¯φ,the solution is not.Then we have many degenerate vacua which are all physically equivalent.By choosing a particular barφas the true vaccum,we have a spontaneous symmetry breaking.According to the above discussion,the key condition for SSB is there are multiple,equivalent vacua.Although it is easy tofind ground state degeneracies in the classical systems,in quantum systems it is difficult to have multiple vacuum.For instance,in a potential with a double well,the ground state is a non-degenerate symmetrical state.In other words,the real vacuum is a linear combination of the various classical vacua.The same thing happens for a rotationally symmetric system in which the ground state has J=0,i.e.,allθangles are equally probable.There are special cases in quantum mechanics in which the ground state may be degenerate. For instace,in an atom with a ground state J=0,the state can be prepared in the eigenstates of J2and J z.However,there is no SSB because the states of different J z are not equivalent vacua in the sense that they blong to the same Hilbert space and are easily connected through a transitions operators.Therefore,the spontaneous symmetry breaking happens only if the volume of the system is approaching infinity and the transition rate between the degenerate states goes to zero.In this case,it turns out that the vacuum states are not representations of the symmetry generators. Rather they are eigenstates of the conjugating coordinate operators and are superposition of states with symmetry quantum numbers.Any perturbation which causes the transition between different vacua have exponentially small matrix elements.On the other hand,the diagonal matrix elements of the perturbation is much larger than the off-diagonal matrix elements.In other words,the vacuum states are those with definite¯φ,or in the rotationally symmetric system,definiteθ.So in the limit of infinit volume,the states with definite¯φbecome the exact vacua.It can be shows that with local hamiltonian and operators,different vacua obey the super-selection rule.Assume the degenerate vacua are|v i andv i|v j =δij(5.10)84CHAPTER5.CHIRAL DYNAMICS By considering the matrix element of v i|A( x)B(0)|v j in the limit of x→∞,it can be shownu i|A(0)|u j =δij a i.(5.11) Therefore the local operators have nofinite matrix elements between different vacuum states. 5.1.1SSB and Space(-time)DimensionsIn afinite quantum mechanical system,there is no SSB.For discrete symmetry,such as Z2symmetry (σi→−σi)in the Ising model,it cannot be broken in one-dimensional(0+1)system.This is known in1938to Peierls.But,it can be broken in two-dimensional(1+1)system.For example, the Onsagar solution contains a spontaneous magnetization for a two-dimensional Ising model.For continuous symmetry,it cannot be spontaneously broken in two-dimensional system.This is called the Mermin-Wagner-Coleman theorem.For example,the classical Heisenberg model consists of interactions of spins living on a n-dimensional sphere.The system has O(n)symmetry.This model has spontaneous symmetry breaking only in3D.To see the MWC theorem,let’s assume there is a SSB in2D.Then we have massless Goldstone bosons.The correlation of these massless Goldstone bosons reads0|φ(x)φ(0)|0 = d2k2πk1cos(k1x1)e ik1x0(5.12)which is hopelessly infrared divergent.This strongfluctation will destroy any long-range order. In a two-dimensional classical Heisenberg model,an disordered phase has as much weight as an ordered one.5.2SSB of the continuous symmetry and Goldstone TheoremIn the case of the spontaneous breaking of a continuous symmetry,a theorem can be proved.The theorem says that the spectrum of physical particles must contain one particle of zero mass and spin for each broken symmetry generator.Those particles are called Goldstone bosons.Consider an infinitesimal transformationφi→φi+iǫa(t aφ)i.(5.13) The same transformation leaves the effective action invariantijd4xδΓ∂φit a ijφj=0,(5.15)This relation is true independent ofφ.Differentiate the above equation with respect toφk and take φ=¯φin a vacuum,∂2V(φ)5.2.SSB OF THE CONTINUOUS SYMMETRY AND GOLDSTONE THEOREM85According to the definition of the effective potential,we have∂2V (φ)2∂µφi ∂µφi −14(φi φi )2(5.19)In the tree approximation Γ=V 3L ,we haveV =14(φi φi )2(5.20)If M 2is negative,we have¯φi ¯φi =−M 2/g(5.21)We can choose a solution as ¯φi =(0, 0∂φi ∂φj |φ=¯φ=2g ¯φi ¯φj =(0,...,0,2|M 2|)(5.22)Thus the last particle now has mass √86CHAPTER5.CHIRAL DYNAMICS we have the effective classical hamiltonianH eff= N1∂φi 1···φi N (5.27)From equations derived earlier,it is easy to see that the amplitude for a zero-momentum Goldstone boson disappearing into the vacuum is zero.The amplitude for a zero-momentum goldstone boson to make transition to another boson is zero.Finally,the amplitude for three massless Goldstone bosons to make transtion is zero.This is in fact true to all orders.Let us consider now the interactions of Goldstone bosons with other massive particles.The following approach assumes exact symmetry.To calculate the process ofα→β+B a,we start from the matrix element with the corresponding conserved currentβ|Jµa|α .(5.28) The current supports a momentum transfer q=pα−pβ.Clearly the most important contributionto matrix element comes from the Goldstone boson pole which has the following structureiF qµMβB,αFqµNµβ+J,α.(5.30)This is a form of Ward identity.If Nµhas no pole,then the process of emitting a Goldstone boson vanishes as q→0.This is called the Adler zero.The most important contribution in the regular term comes from the Feynman diagrams in which J acting on the external line.In this case,there is a heavy-particle pole which enhance the contribution.The pole contribution can often be calculated or extracted from experimental data,from example,the nucleon pole contribution is related to the neutron beta decay constant g A.Knowing g A,we can calculate the meson-nucleon interaction as we shall do in the next section.The above result can also be derived from a theory with explicit breaking of the symmetry.This approach is called PCAC.In this case,the masses of the Goldstone bosons are not exactly zero, butfinite.They are called pseudo-Goldstone bosons.Let us consider the SSB of an approximate symmetry.In this case,the vacuum is no longer degenerate,and strictly speaking,there is no spontaneous symmetry breaking.This is very much like a magnet in an external magneticfield(first order phase transition).In the following we would like tofind the constraint on the vacuum from the symmetry breaking effects;we also want to derive the masses of the pseudo-Goldstone bosons.5.3.PION AS GOLDSTONE BOSON,PCAC87Now the effective potential has two terms V(φ)=V0(φ)+V1(φ).The real solution isφ=φ0+φ1 which is no longer degenerate.The condition onφ0andφ1is contained in the expanded version of ∂V(φ)/∂φi|φ=¯φ=0∂2V0=0(5.31)∂φiUsing the equation we found early,we have∂V1(φ0)(t aφ0)i(5.33)∂φi∂φjwhich vanish to the zeroth order.To the frist order,wefindM2ab=− cd F−1ac F−1bd 0|[T a,[T b,H1]]|0 (5.34) where T a is the quantum generator of the symmetry group.5.3pion as goldstone boson,PCACOne of the most interesting examples of SSB is exhibited by fundamental strong interactions:quan-tum chromodynamics.Consider the QCD lagrangian.The only parameters with mass dimension are quark masses.For ordinary matter,we just consider up and down quarkflavors.The QCD scale ΛQCD is about200MeV,which is much larger than the up and down quark masses(5to9MeV). Therefore,to a good approximation,we can negelect the quark masses in the QCD lagrangian. Then the QCD lagrangian has the U(2)×U(2)chiral symmetry.Recall the chiral projection operators P L=(1−γ5)/2and P R=(1+γ5)/2,whereγ5is diag (-1,1),which project out the left-handed and right-handed quarkfields,ψL,R=P L,Rψ.(5.35) Then the QCD lagrangian we can be written in terms ofL=ψR(i D)ψR−188CHAPTER5.CHIRAL DYNAMICS where U L,R are unitary matrices in the two-dimensionalflavor space.Since U(2)=U(1)×SU(2),we have two U(1)symmtries.From now on,we focus on the two SU(2)symmetries,leaving the U(1) symmetries to later discussion.According to Noether’s theorem,the SU L(2)×SU R(2)chiral symmetry leads to the the following conserved currents,jµL,R=¯ψL,R t aγµψL,R,(5.38) where t a=τa/2andτa is the usual Pauli matrices.We have the vector and axial vector currents from the linear combinations,j aµV=¯ψt aγµψ=jµL+jµRj bµA=¯ψt aγµγ5ψ=jµR−jµL.(5.39) From the above currents,we can define the charges Q a and Q a5in the usual way.And it is easy to see that the charges obey the following algebra:[Q a,Q b]=iǫabc Q c;[Q5a,Q b]=iǫabc Q5c;[Q5a,Q5b]=iǫabc Q c.(5.40)From the above,wefind that Q a forms a subgroup of the chiral symmetry group and is called the isospin group.From the experimental hadron spectrum,wefind that the isospin subgroup is realized in Wigner-Weyl mode.For instance,the pion comes in with three charge states and near degenerate mass.The proton and neutron also have nearly degenerate mass.However,the spectrum does not show the full chiral symmetry.For instance,the three pion states do not form an irreducible reps of the chiral group.Together a scalar particleσ,they form(1/2,1/2)reps. Therefore,if the chiral symmetry is realized fully in Wigner-Weyl mode,there must be a scalar particle with the same mass as the pion.We do not see such a particle in Nature.Thus,the chiral group SU L(2)×SU(2)R must break spontaneously to the isospin subgroup SU(2).Thus the QCD vacuum|0 satisfiesQ a|0 =0,Q5a|0 =0.(5.41) According to Goldstone’s theorem,there are three massless spin-0pseudo-scalar bosons.They are pseudoscalars because Q5a changes sign under parity transformation.Of course,in the real world,we don’t have massless pseudoscalars.We have pions.The pion masses are indeed much smaller than a typical hadron mass.For instance,the rho meson has mass 770MeV.The nucleon mass is940MeV.And the pion mass is140MeV.The pions are called pseudo-Goldstone bosons because the chiral symmetry is not exact.It is broken by thefinite up and down quark masses.H1=m u¯u u+m d¯dd.(5.42) If we write u in terms of left and right-handedfields,we haveH1=m u(¯u L u R+¯u R u L)+m d(¯d L d R+¯d R d L).(5.43)Therefore the left and right-handedfields are now coupled through the mass terms.The mass operator transforms as the components of(1/2,1/2)representations of the chiral group.Using the relation we found earlier,we can calculate the pion mass,m2π=−(m u+m d) 0|¯u u+¯dd|0 /f2π(5.44)5.3.PION AS GOLDSTONE BOSON,PCAC89 where 0|¯u u+¯dd|0 is the chiral condensate.Since¯u u is a part of the representation(1/2,1/2),it vacuum expectation value vanishes ordinarily because of the chiral symmetry.However,it has a vacuum expectation value because of the vacuum is no longer chirally symmetric(chiral singlet). In fact,the vacuum contains all(k,k)type of representations because the vacuum has zero isospin. Any chiral tensor of type(k,k)has non-zero vacuum expectation value.The pion decay constant fπis defined from0|jµa(x)|πb =ipµδab fπe−ip·x.(5.45) It can be measured from the semi-leptonic weak decayπ+→µ+νµrateG2F m2µf2π(m2π−m2µ)2Γ=U(p′)t a[g A(q2)γµγ5+g p(q2)qµγ5]U(p),(5.48) where q=p−p′and U’s are the on-shell Dirac spinors of the nucleon states.Multiplying qµto both sides of the equation and using current conservation and Direc equation(p−M)U(p)=0, we have−2Mg A(q2)+q2g P(q2)=0.(5.49) g A(q2)in the limit of q2→0is just the neutron decay constant(the axial current is part of the weak interaction current)and has been measured accuratelyg A(0)=1.257.(5.50) Thus according to the above equation g P(q)must have a pole in1/q2.This pole corresponds to the intermediate massless pion contribution to the interaction between the the axial current and the nucleon.If we introduce the pion-nucleon interaction vertex gπNN¯Niγ5τa Nπa,the contribution to the axial current matrix element is(ifπqµ).(5.51)i2gπNNq2In the limit of q2→0,wefind the following celebrated Goldberger-Treiman relationg A(0)M=gπNN fπ.(5.52) Using gπNN from experimental data(g2πNN/4π=14.6),wefind that the above relation is obeyed at better than10%level.90CHAPTER5.CHIRAL DYNAMICS According to the recipe derived from the previous section,we calculate the interactions between the soft pion and the nucleon system as follows.First use a vertex iqµ/fπconnecting the Goldstone boson to the axial current.Then the non-singular part of the axial current interaction with the nu-cleon is approximated through the g Aγµγ5vertex.This yields the effective pion-nucleon interaction vertex i qγ5/fπ.This is a peudo-vector interaction.Another way to study the interactions among the pions and with other particles is through what is called the PCAC(partially-conserved axial-vector current),in which we assume there is a small explicit symmetry breaking through nonvanishing quark masses.Applying the derivative operator to the current matrix between the vacuum and the pion,we have0|∂µjµa|πb =m2πδab fπ.(5.53) The right-hand side is proportional to the pion mass squared.This motivates the assumption that∂µj aµ=m2πfππa,(5.54)whereπa is a pion interpolatingfield.Of course,the above relation is in some sense empty because any pseudo-scalar operator can be used as an interpolatingfield for pion.The content of the PCAC is that axial current at zero momentum transfer(this is the place where we know how to calculate the matrix element)is dominated by the pion contribution at q2=m2π.In other words, the variation of the matrix elements of the axial current from q2=0to m2πis smooth.In fact,we can derive the Goldberger-Treiman relation using PCAC andfind now one has to use g A(0)instead of g A(m2π).The content of PCAC is that the variation of this small.Therefore,when the pion energy is small,we can calculate using PCAC.PCAC can be used to study the multi-pion interactions.For instance,consider the amplitudeT abµν= d4xe iqx H(p2)|T A aµ(x)A bν(0)|H(p1) (5.55) Applying differentical operators to the above quantity,we derive a Ward ing PCAC, one can calculate the pion-nucleon scattering amplitude at low-energy.However,it turns out that it is much easier to get the predictions using the low-energy effective theory.5.4the linearσmodelMany of the essential physics exhibited in spontaneous breaking of the chiral symmetry can be illustrated by a simple phenomenological model.This is very similar to the Ginsburg-Landau theory for second-order phase transitions.This model isfirst introduced by Gell-Mann and Levy, and is called the linearσmodel.The lagrangian is,L=L S+cσ,L S=¯ψ[i∂+g(σ+i π·τγ5)]ψ+12(σ2+ π2)−λ5.5.EFFECTIVE FIELD THEORY:CHIRAL PERTURBATION THEORY WITH PIONS91 term cσ,the lagrangian is clearly symmetric under the chiral SU L(2)×SU R(2),and the correspond-ing vector and axial vector current isτajµa=¯ψγµψ+(σ∂µπa−πa∂µσ)(5.57)2After introducing the symmetry breaking term,the axial vector current is no longer conserved.We have instead∂µA aµ=−cπa(5.58) according to the equation of motion.The above has the form of PCAC.Whenµ2<0,the spontaneous symmetry breaking happens.The potential has its minimum not atπa=σ=0but atπ2+σ2=v2,where v2=−µ2/λ.Thus,the shape of the potential is a Mexican hat.There are infinite many degenerate minima.We need to choose a particular direction as our vacuum state.If we want to keep the isospin group intact,we takeσ =v.(5.59) The pion excitation corresponds to the motion along the minima and therefore has zero energy unless the wavelength isfinite.Theσmass corresponds to the curvature in theσdirection and is 2λv2.The nucleon also get its mass from spontaneous symmetry breaking and is−gv.From the PCAC,wefind that fπ=−v.When the symmetry breaking term is introduced,the Mexican hat is tilted.In this case,the minimum of the potential is unique and the pion excitations do have mass.5.5effectivefield theory:Chiral Perturbation theory with pionsCurrent algebra and Ward identity approach were popular in the60’s for calculating Goldstone boson interactions.However,they are tedious.In1967,Weinberg used the nonlinearly-transformed effective lagrangian to study the Goldstone boson interactions.This is the precursor of effective field theory approach which is popular today.The key observation is that when the Goldstone boson energy is small,the coupling is weak. Therefore their interactions must be calculable in perturbation theory.However,in the strong interactions,we also have the usual QCD or hadron(rho meson or nucleon)mass scale.The physics at these two different scales have to be separated before one can apply chiral perturbation theory.The physics at QCD or hadron mass scale can be parametrized in terms of various low-enegy constants which can be determined from experimental data.Through a particular model,we demonstrate the separation of physics through nonlinear trans-formations.Wefirst perform a symmetry transformation at every point of the spacetime to get rid of the Goldstone boson degrees of freedom.We then re-introduce them through the spacetime-dependent symmetry transformation.When the Goldstone-bosonfields are constant,the transfor-mation is the usual chiral tranformation;and the Goldstone bosonfields disappear.Therefore,in the new lagrangian,the Goldstone boson interaction must have derivative-type interactions.Consider the linear sigma model.Let us introduce a(1/2,1/2)2×2matrixU=σ+i π·τ(5.60)92CHAPTER5.CHIRAL DYNAMICSUnder the chiral transformation,we haveU→U L UU†R(5.61) We can write the linear sigma model asL=14Tr[UU†]−λπ2+σ2.We reintroduce back the goldstone boson by parametrizing the U including the axial transformation parameters,U=σe i πa(x)·τa/fπ(5.64) whereπa=fπθa A is now the Goldstone bosonfield.For the convenience,we call the exponential factorΣ.Now substituting U=σΣinto the original lagrangian,we get,L=14σ2Tr[∂µΣ∂µΣ†]−14σ4.(5.65) Now the Goldstone bosonfields contain derivatives and therefore the above lagrangian will pro-duce appropriate Goldstone boson interactions.Since theσparticle has a typical hadronic mass, its effects can be integrated out completely and theσis then replaced by its expectation value. Therefore,the effective intereaction lagrangian for pion is justL(2)ππ=f25.5.EFFECTIVE FIELD THEORY:CHIRAL PERTURBATION THEORY WITH PIONS93 Using L=I− i V i+1,we haveν= i V i(d i−2)+2L+2.(5.69)Therefore the lowest power of Q in any pion process is2.We can use the above leading order lagrangian to calculate the interactions between the pions. Expand in1/fπto to the second order,we have[(∂µ π· π)2− π2(∂µ π)2]+...(5.70) L(2)ππ=16f2πThe second term can be used to calculate the S-matrix element between pion scattering.Assume the incoming pions with momenta p A and p B and isospin indices a and b and the outgoing pions with momenta p C and p D and isospin indices c and d.We have the following leading-order invariant amplitude(S=1−iM),M=−f−2π(δabδcd s+δacδbd t+δadδbc u)(5.71) where s=(p A+p b)2,t=(p A−p C)2and u=(p A−p D)2are called Mandelstam variables.5.5.1Scalar and Pseudoscalar SourcesWe can include the quark mass effects at this order.The quark mass term transforms like(1/2,1/2) under chiral transformations.In general,let us introduce s and p source in the QCD lagrangianL sp=−¯ψs(x)ψ+¯ψiγ5p(x)ψ=−¯ψR(s+ip)ψL−¯ψL(s−ip)ψR(5.72) Call s−ip=χand s+ip=χ†.Then the interaction is invariant ifχ→LχR;χ†→Rχ†L†(5.73) Without the p source,χ∼χ†∼s∼m q,which counts as second order in momentum.The effective lagrangian then containχas a O(p2)external source.The lowest order isL(2m)ππ=B Tr(Σχ†+Σ†χ).(5.74) When expanded to the leading order,the above gives the pion mass contribution if B=f2π/4and χ=m2π.The next-order contribution ism2π94CHAPTER5.CHIRAL DYNAMICSAt the threshold where s=4m2π,t=u=0,we haveM=−m2πf−2π[3δabδcd−δacδbd−δadδbc].(5.77) The scattering amplitude f=−M/8π√16f4π −1µ2−1µ2−1µ2 −14c2(t2+u2) +crossing(5.78)where c1and c2are constants which must be determined from experimental data.In fact,there are also pion mass contribution at this order which we will not go into.The p4-order mass term include the followingL4Tr(DµΣ†DµΣ)Tr(χ†Σ+χΣ†)+L5Tr(DµΣ†DµΣ)(χ†Σ+χΣ†)+L6(Tr(χ†Σ+χΣ†))2+L7(Tr(χ†Σ−χΣ†))2+L8Tr(χ†Σχ†Σ+χΣ†χΣ†)+H2Tr(χ†χ)(5.79) where H2is pointless because there is no meson depedence.5.5.2Electromagnetic and Axial InteractionsWhen there are electromagnetic and weak interactions with the Goldstone boson system,we need to construct a gauge theory in which the effective theory is gauge invariant under gauge transfor-mations.Introduce the the following coupling the QCD lagrangianL=¯ψ(γµvµ(x)+γµγ5aµ(x))ψ=¯ψLγµ(vµ−aµ)ψL+¯ψRγµ(vµ+aµ)ψR(5.80) If vµand aµare gaugefields,under gauge transformation,they must transform in the following way,vµ−aµ→L(vµ−aµ)L†+iL∂µL†vµ+aµ→R(vµ−aµ)R†+iR∂µR†(5.81) The above equation means that these gaugefields have to appear together withΣin the following formDµΣ=∂µΣ−i(v−a)µΣ+iΣ(v+a)µ(5.82)5.6.BANKS-CASHER FORMULA AND VAFA-WITTEN THEOREM95 Then all the partial derivatives will be replaced by the above covariant derivatives.For example,consider the electromagnetic interaction of the pions.In this case,we replace vµ=−ie(τ3/2+1/6)Aµwhere e is the charge of a proton andτ3is the isospin and1/6is the hypercharge.Then the partial derivative becomes,DµΣ=∂µΣ+ieAµ[τ3∂x1+∂A2∂x3+∂A496CHAPTER5.CHIRAL DYNAMICS The electricfield E in the Euclidean space is the imaginary of that in the Minkowski space and so E2→−E2,and FµνFµν=−2(E2−B2)→2(B2+E2)=FµνFµν.We also define the Euclidean version of theγmatrix withγE4=γ0andγE i=−iγi and the commutators now become{γEµ,γEν}=2δµν(5.91) The newγmatrices are hermitian.The QCD lagrangian is nowL QCD=−4FµνFµν (5.92)Notice thatγµDµis now an antihermitian operator.We can define the Euclidean L to absorb the minus sign.Consider now the exponential factor exp(iS)in the path integral.After rotation,the integral d4x becomes−i d4x.The−i here cancels the i in front of the iS and define the Eulidean action asS E=− d4x L(5.93) Therefore the integration meansure becomes exp(−S E)Let us see how the spontaneous symmetry breaking takes place in QCD.To this goal,we need to introduce an explicit breaking of the symmetry.For example,we give a small mass to quarks. Consider the expectation value of ¯u u .We writeV4d4x u(x)¯u(x)=− [DA]e−S Y M Det(D+M)1D+m u].(5.94)where Tr is over spatial,color,and spin indices.Now consider the eigenstates of D.Because it is an anti-hermitian operator,we haveD|λ =iλ|λ ,(5.95) whereλis real.The different|λ are orthogonal and therefore we haveTr[1iλi+m u.(5.96)On the other hand,we have Tr(D+M)=Tr(−D+M)because(γ5)2=1.We get thenTr 1iλi+m u+1m2u+λ2i(5.97)Intrduce now aδ(λ−λi)and integration overλ.We have then¯u u =−dλρ(λ)m uZV4 [DA]exp(−S Y M)Det(D+M)i2δ(λ−λi)(5.99)。