《物理学教学》26 introduction to quantum mecha_第1页
《物理学教学》26 introduction to quantum mecha_第2页
《物理学教学》26 introduction to quantum mecha_第3页
《物理学教学》26 introduction to quantum mecha_第4页
《物理学教学》26 introduction to quantum mecha_第5页
已阅读5页,还剩31页未读 继续免费阅读

下载本文档

版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领

文档简介

1、Introduction to quantum mechanicsChapter 26.26.1 Operator and physical quantity26.2 Schrdinger equation26.3 One dimensional potential well26.4 Hydrogen atom 26.5 Quantum numbers and atomical orbitals.Operator represents certain kind of mathematical operation which applies on one function and leads t

2、o another function in result. 26.1 Operator and physical quantityvuF Example: differential operatorvdxduudxd )(vux coordinate operator.vuF wwF C The eigen equation of operator Feigen functioneigen value.In quantum mechanics, all the physical quantities are expressed by operators. coordinate operator

3、xx xhipx2 yhipy2 zhipz2 momentum operator.If there is a classical expression of quantity F: F=F(x,p), then the corresponding operator expression in quantum mechanics can be written as: ) , (pxFF example: classical kinetic energy 22mpEkkinetic energy operator 2222222228)4(21 2dxdmhdxdhmmpEk.classical

4、 energy)( 22xUmpEEEPkEnergy operator in quantum mechanics is usually called Hamiltonian operator: )(8) (2222xUdxdmhxUEHk.)(8) (2222xUdxdmhxUEHk2222222dddH() U(x,y,z)2m dxdydz 2hin 3 dimensional coordinates:where:.26.2 Schrdinger equationThe eigen equation for energy operator is called Schrdinger equ

5、ation (Sch. eq.): EHHamiltonianwavefunction .1) derive Hamiltonian for the system according to different conditions;2) set up Schrdinger equation;3) solve Schrdinger equation to get energies and wavefunctions; 4) all other physical informations can easily derived after wavefunction is obtained. EHSt

6、eps to solve a physical system in quantum mechanics:.1dV22is the probability of finding particle in the unit volume about the certain point.The probability of finding particle in all possible space should be one. , , , , , , , 4321)(4)(3)(2)(1PPPPPPPP PP. , , , , , , , 4321)(4)(3)(2)(1xxxxxxxx xx.26

7、.3 One dimensional potential wellOne particle is confined in an one dimensional box with a width “a”. It can move free within the box, but can NOT go outside. 0 x aU(x)mU(x) (x0, xa)potential inside boxpotential outside boxU(x) = 0 (0 xa).)(8) (2222xUdxdmhxUEHk EH0)(8dd2222xUEhmx 0 x aU(x)mSchrdinge

8、r equation.0)(8dd2222xUEhmxoutside box region:U(x) (x0, xa)0The probability of finding particle outside box is ZERO.0)(8dd2222xUEhmxinside box region:U(x) = 0 (0 xa)08dd2222 Ehmx0dd222 kx2228kEhm with.0dd222 kxBcoskxAsinkx(x)solution:where constant A and B should be determined by boundary conditions

9、.0(0) 0(a) 2228kEhm withnka Asinkx(x) B=0n = 1,2,3,ank.2228kEhm ank2228nmahE Energy quantization !Energy level: E1 , E2 , E3 , E4 , n - quantum number.1d)(sin022 axxanA1dV2Asinkx(x) ankxanaxsin2)( aA2 . EH 0 x aU(x)mA series of En and are obtained, with n = 1,2,3,nxansina2(x)n1Enn8mahE2222n.xansina2

10、(x)n1Enn8mahE2222n.26.4 Hydrogen atom r4eU(r)02+e-erv2222222dddH() U(r)2m dxdydz .2222222dddH() U(r)2m dxdydz Cartesian coordinatesspherical coordinates)(r,+e-ervz)y,(x,.2222222dddH() U(r)2m dxdydz ),(r,H2()()222222201111rsin2m rrrrsinsine 4r .26.5 Quantum numbers and atomic orbitalsE),(r,H ),(r,H2(

11、)()222222201111rsin2m rrrrsinsine 4r - How to solve this complicated Sch. eq.?.E),(r,H The trick to solve this Sch. eq. :(1) variables separation(2) solve 3 equations successivelyequation for requation for equation for )()(R(r),(r,separated into 3 equations3 quantum numbers appearquantum numbernquan

12、tum numberlquantum numberlm.E),(r,H equation for requation for quantum numbernllmequation for nE energy R(r)()(quantum numberquantum numberwavefunctionwavefunctionwavefunctionorbital angular momentum 1)l(lLZ-component of angular momentum lzmL.In addition, electrons have spin angular momentum, and th

13、e z-component of which has two possible values:szmS21smwithto as spin up and spin down.which are often referred.4 quantum numbers are required to mark each orbital.slmmln . lmln.Exe -1-An electron is contained in a one-dimensional box of length 0.100nm. (a) Draw an energy-level diagram for the electron for levels up to n=4.(b) Find the wavelengths of all photons that can be emitted by the electron in making downward transitions that could eventually carry it from the n=4 state to the n=1 state. .Exe -2-How many sets of quantum numbers are possib

温馨提示

  • 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
  • 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
  • 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
  • 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
  • 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
  • 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
  • 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。

评论

0/150

提交评论