《《物理學(xué)教學(xué)課件》26introductiontoquantummecha》由會(huì)員分享，可在線閱讀，更多相關(guān)《《物理學(xué)教學(xué)課件》26introductiontoquantummecha（36頁(yè)珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。

1、Introduction to quantum mechanicsChapter 2 6 2 6 .1 Operator and physical quantity2 6 .2 Schrdinger equation2 6 .3 One dimensional potential well2 6 .4 Hydrogen atom 2 6 .5 Quantum numbers and atomical orbitals Operator represents certain kind of mathematical operation which applies on one function

2、and leads to another function in result. 2 6 .1 Operator and physical quantity vuF Example: differential operator vdxduudxd )( vux coordinate operator vuF wwF C The eigen equation of operator Feigen function eigen value In quantum mechanics, all the physical quantities are expressed by operators. co

3、ordinate operatorxx xhipx 2 yhip y 2 zhipz 2 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 2 2mpEk kinetic energy operator 222222222 8)4(21 2

4、dxdmhdxdhmmpEk classical energy )( 2 2 xUmpEEE Pk Energy operator in quantum mechanics is usually called Hamiltonian operator: )(8)( 2222 xUdxdmhxUEH k )(8)( 2222 xUdxdmhxUEH k 2 2 2 2 2 2 2d d dH ( ) U(x,y,z)2m dx dy dz 2hin 3 dimensional coordinates:where: 2 6 .2 Schrdinger equationThe eigen equat

5、ion for energy operator is called Schrdinger equation (Sch. eq.): EH Hamiltonian wavefunction 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 ca

6、n easily derived after wavefunction is obtained. EH Steps to solve a physical system in quantum mechanics: 1dV2 2 is 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)

7、(1 PPPP PPPP PP , , , , , , , , 4321 )(4)(3)(2)(1 xxxx xxxx xx 2 6 .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

8、 (0 xa) )(8)( 2222 xUdxdmhxUEH k EH 0)(8dd 2222 xUEh mx 0 x aU(x) mSchrdinger equation 0)(8dd 2222 xUEh mxoutside box region: U(x) (x0 , xa)0The probability of finding particle outside box is ZERO. 0)(8dd 2222 xUEh mxinside box region: U(x) = 0 (0 xa)08dd 2222 Eh mx 0dd 222 kx 2228 kEh m with 0dd 22

9、2 kx BcoskxAsinkx(x) solution:where constant A and B should be determined by boundary conditions.0(0) 0(a) 2228 kEh m with nka Asinkx(x) B=0n = 1 ,2 ,3 , ank 2228 kEh m ank 2228 nmahE Energy quantization !Energy level: E1 , E2 , E3 , E4 , n - quantum number 1d)(sin0 22 a xxanA 1dV2 Asinkx(x) ank xan

10、ax sin2)( aA 2 EH 0 x aU(x) mA series of En and are obtained, with n = 1 ,2 ,3 , n xansina2(x)n 1Enn8mahE 2222n xansina2(x)n 1Enn8mahE 2222n 2 6 .4 Hydrogen atom r4eU(r) 02+e -er v 2 2 2 22 2 2d d dH ( ) U(r)2m dx dy dz 2 2 2 22 2 2d d dH ( ) U(r)2m dx dy dz Cartesian coordinates spherical coordinat

11、es)(r, ,+e -er v z)y,(x, 2 2 2 22 2 2d d dH ( ) U(r)2m dx dy dz ),(r,H 2 ( ) ( ) 222 2 2 220 1 1 1 1r sin2m r r r r sin sine 4 r 2 6 .5 Quantum numbers and atomic orbitals E),(r,H ),(r,H 2 ( ) ( ) 222 2 2 220 1 1 1 1r sin2m r r r r sin sine 4 r - How to solve this complicated Sch. eq.? E),(r,H The t

12、rick to solve this Sch. eq. :(1 ) variables separation(2 ) solve 3 equations successivelyequation for r equation for equation for )()(R(r),(r, separated into 3 equations3 quantum numbers appearquantum number n quantum number l quantum number lm E),(r,H equation for r equation for quantum number n l

13、lmequation for nE energy R(r) )( )(quantum number quantum numberwavefunction wavefunction wavefunctionorbital angular momentum 1)l(lL Z-component of angular momentum lz mL In addition, electrons have spin angular momentum, and the z-component of which has two possible values: sz mS 21 smwithto as sp

14、in up and spin down.which are often referred 4 quantum numbers are required to mark each orbital.sl mmln lmln Exe -1 -An electron is contained in a one-dimensional box of length 0 .1 0 0 nm. (a) Draw an energy-level diagram for the electron for levels up to n=4 .(b) Find the wavelengths of all photo

15、ns 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 possible for a hydrogen atom for which (a) n=1 , (b) n=2 , (c) n=3 , (d) n=4 , and (e) n=5 ? Check your results to show that they agree with the general rule that the number of sets of quantum numbers for a shell is equal to 2 n2 .