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This 8 page Class Notes was uploaded by Dr. Carissa Rowe on Tuesday October 20, 2015. The Class Notes belongs to PHYS410 at San Diego State University taught by Staff in Fall. Since its upload, it has received 26 views. For similar materials see /class/225321/phys410-san-diego-state-university in Physics 2 at San Diego State University.
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Date Created: 10/20/15
Lecture 14 Outline TWO rings to bind o Solvay talk Brillouin The free particle contd 24 The wavepacket 24 Bound and scattering states 251 The nite square well E lt Oeven 26 Fourier Transforms Problem given xnf 0 determine using Fourier analysis 2713912 fme fx 6 ikxdx hk 1 00 2 Mm E kelltkx 2mtgtdk M f2 00 We Example 26 xnf O A for a lt a lt a thus fsin 1m 7139 1m and the other limit a gt 00 has sinusoidal 06 oc 6 4kxd1 limit a gt O m mnka ka Planewave vs Wavepacket Velocity 0 SO llx7t A ik9 wkt and k iv272nE 0 Where wk 73622771 wavelength A 27Tlc Speed coef cient of If ever a coef cient wk vquam um k 2m hk E 2m 2 classy but the classical speed is given by E mv 2E velassy W 2vquam um Picture 211 wave packet with phase ripples 2 hkt IJLUt VLZT 00 eikx dk Phase velocity vs Group velocity 0 Dispersion relation was 73622771 0 Assume was wo w6c k0 let u k 160 1 00 x11 932 s k elltkx Wktgtdk m LO g 1 00 I IJLU t e iwo kow6t 00 k0 iukox w6tdu 1 V271quot 1 00 11 9 O u k e ltuk0 du lt gt m 00 lt 0 x t eiw0k0w6t x 00615 O 0 ie wavepacket moves at vgmup wg dwdk hism 0 Compare with individual Uphase wk his 2m Bound and Scattering States Two forms of our linear combination of stationary states mm t V gt0 we eiltkwgtdk 7T 00 ms t Z cnwncr eXp iEnth To understand the physical difference consider the 4 forms of gure 212 if E lt V oo AND E lt Voo bound state if E gt V oo AND E gt Voo scattering state oo well and SHO only have bound states Vioo 00 free particle only has scattering states Finite Square Well outside Can have bound E lt O and scattering E gt 0 states Vx a l V0 Consider bound states rst V0 lt E lt 0 outside well E2 d2 d2 E d 2 2m dzr2 w an dzr2 H w Where H V ZmEh gt O E R General solution is z x Aexp x B eXp r r lt a BexpOex and r gt a FeXp x Finite Square Well inside Vx V 0 Consider the bound states V0 lt E lt 0 inside well E2 d2 d2 2 W VW E and W W o Where k x2mEV0h gt O E R 0 General solution is x C39sinUm D ooskx Finite Square Well even funk Still Considering the bound states V0 lt E lt 0 since Vx symmetry even or odd solns about a O Dcoscx V0 lt a lt a 7M1 Fez m V a gt a x V a lt O D F Via boundary cond xxa and d xdxxa Fez H D cosca and ltFe cD sinca Dividing one by the other H ktanca Note KLE V ZmEh and 2mE V0h Introducing z 1m zo axZmVOh then we can show that tanz zOz2 1