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# Introduction to Modern Physics PHY 361

Syracuse

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This 26 page Class Notes was uploaded by Clement Bernier on Wednesday October 21, 2015. The Class Notes belongs to PHY 361 at Syracuse University taught by Staff in Fall. Since its upload, it has received 32 views. For similar materials see /class/225638/phy-361-syracuse-university in Physics 2 at Syracuse University.

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Date Created: 10/21/15

Lecture Notes Physics 361 Re ection and transmission by barrlers Tunnellng Additional readings on the Scanning Tunneling Microscope STM o Gerd Binnig and Heinrich Roher The Scanning Tunneling Microscope Scienti c American August 1985 page 50 0 Calvin F Quate Vacuum tunneling a new technique for microscopy Physics Today August 1986 page 26 o For an STM Image Gallery see httpwwwalmadenibmcomvisstmgalleryhtml Our discussion so far has been limited to particles of energy E in potential wells of depth U0 for E lt U0 This corresponds to bound states where the wavefunction goes to zero at in nity Next we want to consider what happens when the particle is unbound For instance consider particles moving in one dimension in the presence of a potential step as shown below The step occurs at z 0 where the particles suddenly move from a region of constant here zero potential energy to a region of higher but still constant potential energy Such a potential step for electrons could for instance be obtained by a sudden change in voltage difference as sketched below from the book by R Harris Figure 51 We understand well what happens for classical particles Imagine a steady state situation where a constant stream of particles comes in from the left with E lt U0 Of course all particles bounce against the step and we have complete reflection If on the other hand the particles have E gt U0 no particles are re ected but rather they all keep travelling towards the right although they slow down in the region z gt 07 as required by energy conser vation This is complete transmission We will see below that the behavior of quantum particles is very different 8 777777777777777 7 7 0 VOItS V0V01ts l U eV l U0 E59 0 X Until now we have not considered the time dependence of the Schrodinger equation or of its solutions We have looked only at time independent so lutions that describe bound states7 rather than particles travelling to the right or to the left Now we recall that if 714 represents a time indepen dent solution of the Schrodinger equation corresponding to energy E7 the corresponding tirne dependent solution is 1967 t W96 1 where the frequency w is given by the de Broglie relation7 E 7 i 2 w h lt gt see Appendix A for a quick review of the complex notation cm In partic ular7 free particles are described by plane wave solutions of the Schrodinger equation These are given by k gt 0 Tl7 t Ae kw m right 7 moving 3 which represents a free particle of momentum pm hk pointing in the positive x direction and energy E ha 12k22m7 ie7 a free particle moving towards x A particle with momentum in the 7s direction ie7 a particle moving towards 7x is described by Tl7 t Ae ikm m left 7 moving 4 Notice that this choice is simply a convention The bound states of particles in a well that we have discussed so far correspond to wavefunctions given by combinations of cos kx and sin kz which are in turn superposition of right and left moving states For instance 1 sin kx Zelkm 7 6 7 5 ie7 the bound state sin kx is the sum or superposition of right and left moving states Notice that both right and left moving states have the same time dependence It is their spatial dependence that tells us whether they are right or left moving We are interested in describing steady state situations7 where we have a constant ux of particles moving to the right or to the left For this purpose we can drop the time dependence ofthe wavefunctions again and simply consider 7 Aeikw right 7 moving 6 This is an in nitely long7 right moving wave train7 describing a beam of mo noenergetic particles of xed momentum moving towards m These states strictly speaking cannot be normalized as lzblz MP is constant everywhere for 700 S x 3 00 The right way to think of this is that we have a con tinuous ow of right moving particles from 700 to 00 and that M N of particles per unit length7 ie7 MP is proportional to the intensity of the beam of particles Similarly7 7 Be ik left 7 moving 7 is an in nitely long left moving wave train describing a beam of monoener getic particles of xed momentum moving towards 7x In general we may have a wavefunction of the form A617 Be ik 8 describing a beam of monoenergetic particles some moving to the right and some to the left The rst term represents particles moving to the right and MP is the intensity of the wave moving towards x The second term represents particles moving to the left and B is the intensity of the wave moving towards 7x 1 Potential Step Now lets go back to our potential step lmagine sending a continuous ow of monoenergetic particles from the left ie the wave function starts off as 7 A0617 If E lt U0 we expect a wavefunction as shown below In the region z lt 0 the wavefunction is a superposition of right and left moving waves For x gt 0 there is a bit of leakage but the wavefunction dies off exponentially Esercz39se How does the penetration length depend on U0 7 E and on the mass of the particles Ask yourself how does 6 change when U0 7 E gets smaller or when the mass of the particles is doubled Sketch the wavefunction for different values of U0 7 E and of m to show the change In the steady state there is no transrnission because the exponential wave function in the region z gt 0 does not describe particles that can propagate So we get complete re ection The situation is very similar to total internal re ection of light from a glass boundary Light is completely re ected7 but there is actually a tiny exponential leakage77 of electric and magnetic eld outside the glass These elds die out exponentially do not propagate and in steady state all light bounces When E gt U07 however7 we get something radically new Unlike what happens classically7 we will see that in this case part of the wave is transmit ted and part is re ected In general we need to write down the solution of the Schrodinger equation in each of the regions z lt 0 where Vz 0 and z gt 0 where Vz U0 In each region the solution is a combination of sines and cosines or7 equivalently7 right and left rnoving waves7 z lt 0 ww Aeik1m Be iklm Z39A 7 B sin 161 A B cos kg with k1 2mEh2 9 where z gt 0 714 06 De ka 20 7 D sin 162 C D cos 162 with k2 2mE i U0h2 10 and A7 B7 C7 and D are constants to be determined from the continuity and the normalization condition for the wavefunction A typical solution may look like that sketched below 9L1 ZTE k1 EgtU0 9V22 TEkz The squared magnitude of each coef cient A B C and D gives the in tensity of the corresponding component of the wave In particular MP is proportional to the intensity of the wave incident on the step from the left lBlZ is proportional to the intensity of the re ected wave and 1012 to the intensity of the transmitted wave lDlZ is proportional to the intensity of particles moving to the left in the region z gt 0 Such particles can only be present if we have a source of a beam of particle coming in from the right from x 00 To describe the physical situation where particles come in from the left and are either re ected or transmitted by the step we choose D 0 We de ne particles transrnittedsec T transrnission coef cent 11 particles incidentsec 7 particles re ectedsec 7 t f t 12 particles incidentsec re 80 Ion COG Gen Clearly we must have R T 1 13 If one does the calculation one nds 4k1k2 T 14 k1 k22 R 1 7 T 15 This is not the classical result as 0 lt R lt 1 In other words some of the particles are re ected even for E gt U0 This is kind of like light hitting a plate of glass 7 some re ects I should stress that T is not equal to lBlzlAlZ To understand why we need to make our de nitions more precise MP is the density of particles7 ie7 the number of particles per unit length volume if we were in three dimensions that are found in the region x S 0 travelling from the far left towards z 0 The intensity of such a beam of particles is the number of particles that travel across any given point z per second This is determined by both the density of particles and their velocity The momentum of these particles is hkl So the intensity of this stream of particles incident on the barrier at z 0 from the left is proportional to lAlzhkl Similarly the intensity of the stream of particles that are transmitted7 ie7 that travel towards the right in the region z gt 07 is proportional to the density 0 of such particles in this region times their momentum hkg In other words particles incident from leftsec N lAlzhkl particles transmitted to the rightsec lClzhkg particles re ected to the leftsec lBlzhkl 16 so that particles transmittedsec lClzhkg particles incidentsec lAlzhkl particles re ectedsec lBlzhkl lBlZ particles incidentsec lAlzhkl T 17 When we work out the coef cients B and C using the conditions of continuity of the wavefunction7 we get the result given in Eq 14 Notice that the magnitude of momentum of the particles travelling to the left in the region z lt 07 ie7 of the re ected particles7 is hkl and is equal in magnitude to the momentum of the incident particles7 so that these cancel in the second of Eqs 17 2 Potential Barrier and Tunneling If we now consider a potential energy barrier rather than a step7 we nd that quantum particles can tunnel trough the barrier and get out on the other side 7 X x0 xL Suppose particles with energy E lt U0 are incident from the left Clas sically all such particles would be re ected In the quantum case we expect solutions ofthe forrn shown below 7 oscillatory solutions in the region z lt 0 incident and re ected waves exponentials in the region 0 lt z lt L and oscillatory solutions again in the region z gt L transrnitted wave The wavelength A0 27rk0 with k0 2mEh is the same on both sides of the barrier but the amplitude of the transmitted wave is much smaller than that of the incident wave The particles are never observed inside the barrier where they would have negative kinetic energy but are observed beyond the barrier k021t k 0 k021t k 0 x0 xL X This phenomenon is called tunneling and we discussed it already on our rst day of classes The transmission coef cient T is also called the tun neling probability or tunneling rate For wide barriers7 where the tunneling probability is very small7 it is given by E E T2167177 4 18 UOlt we ltgt where 1 h 5 19 7 is the penetration length introduced when we discussed the penetration of quantum particles in a step potential This expression for T applies when the barrier is wide7 in the sense that 6 ltlt L The tunneling probability depends exponentially on the thickness of the barrier L7 as well as on the energy difference U0 7 E For7 say7 electrons with E 004eV attempting to penetrate a barrier of height U0 010eV7 the penetration length is 6 08nm If the barrier has width L 15nm7 the barrier is indeed wide as 1Note that the notation here is different from that of your textbook by Beiseri If you compare Eqs 18 and 19 to Eqs 5 60 and 561 of Beiser you see that the quantity that Beiser calls 162 corresponds to the quantity 16 of these notes7 iiei7 legals lSi Note that 6 is a length7 while legals has dimensions of llengthi L gtgt 6 and we can use the formula above to calculate T The tunneling rate is a very sensitive function of L This fact is used in the STM that we talked about at the beginning of the semester 3 Application We discussed some of these in class Read the articles given at the top for a description of the SMT Here I will just list a few applications 0 Scanning Tunneling Microscope 0 Alpha decay o Tunnel diode o Ammonia molecule 1 will add a few words about the last one 31 Inversion of the ammonia molecule A sketch of the ammonia molecule NHg is shown below The molecule has two equilibrium con gurations with the same energy The two equilibrium con gurations correspond to the N atom being located 10 on either side of the plane containing the three H atoms The Coulomb repulsion of the hydrogen atom generates a potential barrier that the N atom has to surmount to move from one equilibrium position to the other The potential energy of the nitrogen atom is shown below as a function of its distance from the plane containing the three hydrogen atoms Here x 0 is the location of this plane potential energy of nitrogen potential barrier l distance Li 0 ii from lane equilibrium positions 0f hy rogen In the ground state the N atom has an energy indicated by the dashed line in the gure This energy is lower that the height of the barrier between the two minima7 corresponding to the two equilibrium con gurations of the molecule Classically7 the molecule would remain in whichever of the two con gurations happened to be formed initially In other words7 the nitrogen atom would always be on the same side of the three hydrogen plane as to get to the other side it has to surmount an energy barrier In a classical world the only way an object can get over an energy barrier is by haVing an energy higher than the barrier This extra energy could in principle come from thermal energy However7 the height of the barrier at z 0 is about 026 eV7 while at room temperature kBT 26 meV 0026 eV7 ie7 much smaller In a quantum world7 however7 the nitrogen atom can tunnel through the barrier and it will indeed do so As a result7 you can think of the nitrogen as oscillating between the two equivalent equilibrium positions The frequency of these oscillations is f 23786 gtlt 10 l0HZ7 which is rather low compared 11 to the frequencies of most molecular Vibrations The N atom makes more than 1010 oscillations per seconds The tunneling frequency of the ammonia molecule was used as the standard of the rst atomic clocks7 which provide the worlds standard for precision time keeping 4 Appendix A You will need to recall some properties of complex numbers A complex number 2 is written as 2 z z y where z Rez is the real part and y lmz is the imaginary part x and y are ordinary real numbers Also7 239 7 1 The magnitude of a complex number is de ned as lzl z2y2 20 The complex conjugate of z is denoted by 2 and is the number that has the same real part as 2 and opposite imaginary part7 ie7 2 z 7 2y Since 2392 71 we nd that 22 lle 2 y2 One can represents complex numbers as points in the xy plane The real part is measured along the z axis and the imaginary part along the y axis You can see then that the magnitude of the complex number is simply the length of its distance from the origin The gure below shows for instance the number 2 3z394 lts magnitude is m 5 Try drawing on the gure 3 the complex conjugate of z 3z394 It is clear from the gure that the complex number can also be represented by its magnitude and the angle 6 that the radius to the origin makes with the z axis The angle 6 is given by 6 arctany In other words one can write 2 cos6 sin 6 If you like7 this is nothing but changing from cartesian to polar coordinates Now consider complex numbers of magnitude 1 These are all the numbers in the zy plane that lie on a circle of radius one Any complex number of magnitude 1 can be written as M 1 z cos6zsin6 E 619 21 The last equality is just the de nition of a new notation The complex function 619 is de ned by the equality 6w cos 6 z sin 6 22 The complex conjugate of 619 is 6 67w cos 6 7 z sin 6 23 We can then verify immediately that 61967 cos2 6 sin2 6 1 24 Lecture Notes Physics 361 Re ection and transmission by barrlers Tunnelmg Our discussion so far has been limited to particles of energy E in potential wells of depth U for E lt Ug This corresponds to bound states where the wavefunction goes to zero at in nity Next we want to consider what happens when the particle is unbound For instance consider particles moving in one dimension in the presence of a potential step as shown below The step occurs at C 0 where the particles suddenly move from a region of constant here zero potential energy to a region of higher but still constant potential energy Such a potential step for electrons could for instance be obtained by a sudden change in voltage difference as sketched below from the book by R Harris Figure 51 We understand well what happens for classical particles Imagine a steady state situation where a constant stream of particles comes in from the left with E lt Um Of course all particles bounce against the step and we have complete re ection If on the other hand the particles have E gt Ug no particles are re ected but rather they all keep traveling towards the right although they slow down in the region C gt 0 as required by energy conser vation This is complete transmission We will see below that the behavior of quantum particles is very different Until now we have not considered the time dependence of the Schrodinger equation or of its solutions We have looked only at time independent solu tions that describe bound states rather than particles traveling to the right or to the left Now we recall that if lm represents a time independent solution of the Schrodinger equation corresponding to energy E the corresponding time dependent solution is min t WW4 1 where the frequency w is given by the de Broglie relation I E dg 2 see Appendix A for a quick review of the complex notation cm In partic ular free particles are described by plane wave solutions of the Schrodinger equation These are given by gt 0 t Ae Likx w right moving 3 which represents a free particle of momentum pI Tali pointing in the positive C direction and energy E he h2k22m ie a free particle moving towards 1 A particle with momentum in the C direction ie a particle moving towards C is described by t Ae ikx m left moving 4 Notice that this choice is simply a convention The bound states of particles in a well that we have discussed so far correspond to wavefunctions given by combinations of cos and sin which are in turn superposition of right and left moving states For instance 1 sin e e W ie the bound state sin is the sum or superposition of right and left moving states Notice that both right and left moving states have the same time dependence It is their spatial dependence that tells us whether they are right or left moving We are interested in describing steady state situations where we have a constant ux of particles moving to the right or to the left For this purpose we can drop the time dependence of the wavefunctions again and simply consider 28m Aei right moving 6 This is an in nitely long right moving wave train describing a beam of mo noenergetic particles of xed momentum moving towards 1 These states strictly speaking cannot be normalized as 1ZI 2 A 2 is constant everywhere for oo 5 C 5 00 The right way to think of this is that we have a con tinuous ow of right moving particles from oo to 00 and that A 2 of particles per unit length ie A 2 is proportional to the intensity of the beam of particles Similarly lm Be i left moving 7 is an in nitely long left moving wave train describing a beam of monoener getic particles of xed momentum moving towards C In general we may have a wavefunction of the form lm Adi B641 8 describing a beam of monoenergetic particles some moving to the right and some to the left The rst term represents particles moving to the right and A 2 is the intensity of the wave moving towards 1 The second term represents particles moving to the left and B is the intensity of the wave moving towards C 1 Potential Step Now lets go back to our potential step Imagine sending a continuous ow of monoenergetic particles from the left ie the wave function starts off as ram 1406quot If E lt Ug we expect a wavefunction as shown below In the region C lt 0 the wavefunction is a superposition of right and left moving waves For C gt 0 there is a bit of leakage but the wavefunction dies off exponentially Exercise How does the penetration length depend on Ug E and on the mass of the particles Ask yourself how does 6 change when Ug E gets smaller or when the mass of the particles is doubled Sketch the wavefunction for different values of U E and of m to show the change In the steady state there is no transmission because the exponential wave function in the region C gt 0 does not describe particles that can propagate So we get complete re ection The situation is very similar to total internal re ection of light from a glass boundary Light is completely re ected but there is actually a tiny exponential leakage of electric and magnetic eld outside the glass These elds die out exponentially do not propagate and in steady state all light bounces When E gt U however we get something radically new Unlike what happens classically we will see that in this case part of the wave is transmit ted and part is re ected In general we need to write down the solution of the Schrodinger equation in each of the regions C lt 0 where 0 and C gt 0 where Ug In each region the solution is a combination of sines and cosines or equivalently right and left moving waves 8 lt 0 lm 146 Be i m iA B sin km A B cos km with k1 V2mE7z2 9 where C gt 0 11 06 De m iC D sin 115238 C D cos 1121 with k2 2mE Ug7L2 10 and A B C and D are constants to be determined from the continuity and the normalization condition for the wavefunction A typical solution may C21 look like that sketched below 9M1 275 k1 EgtU0 9M22 TEkz The squared magnitude of each coef cient A3 By C and D gives the in tensity of the corresponding component of the wave In particular A 2 is proportional to the intensity of the wave incident on the step from the left B is proportional to the intensity of the re ected wave and C to the intensity of the transmitted wave D is proportional to the intensity of particles moving to the left in the region C gt 0 Such particles can only be present if we have a source of a beam of particle coming in from the right from C 00 To describe the physical situation where particles come in from the left and are either re ected or transmitted by the step we choose D 0 We de ne W transmission coef cent 11 particles incidentsec particles re ected sec t39 f t 12 particles incidentsec r0 0C Ion C00 Gen Clearly we must have R T 1 13 If one does the calculation one nds 41111115 T 14 k1 k22 R 1 T 15 This is not the classical result as 0 lt R lt 1 In other words some of the particles are re ected even for E gt Ug This is kind of like light hitting a plate of glass 7 some re ects I should stress that T is not equal to B 2 A 2 To understand why we need to make our de nitions more precise 3A is the density of particles ie the number of particles per unit length volume if we were in three dimensions that are found in the region C S 0 traveling from the far left towards C 0 The intensity of such a beam of particles is the number of particles that travel across any given point C per second This is determined by both the density of particles and their velocity The momentum of these particles is him So the intensity of this stream of particles incident on the barrier at C 0 from the left is proportional to A 27zk1 Similarly the intensity of the stream of particles that are transmitted ie that travel towards the right in the region C gt 0 is proportional to the density C 2 of such particles in this region times their momentum Mtg In other words particles incident from left sec A 2hkt1 particles transmitted to the right sec C 2fzkt2 particles re ected to the leftsec B 2hkt1 16 so that particles transmitted sec C 2hkt2 particles incidentsec A 2fzkt1 particles re ectedsec B 27zkt1 B 2 7 17 particles incidentsec A 2fzkt1 A 2 When we work out the coef cients B and C using the conditions of continuity of the wavefunction we get the result given in Eq Notice that the magnitude of momentum of the particles traveling to the left in the region C lt 0 ie of the re ected particles is his and is equal in magnitude to the momentum of the incident particles so that these cancel in the second of Eqs 2 Potential Barrier and Tunneling If we now consider a potential energy barrier rather than a step we nd that quantum particles can tunnel trough the barrier and get out on the other 7 side x0 xL X Suppose particles with energy E lt Ug are incident from the left Clas sically all such particles would be re ected In the quantum case we expect solutions of the form shown below 7 oscillatory solutions in the region C lt 0 incident and re ected wavesa exponentials in the region 0 lt C lt L and oscillatory solutions again in the region C gt L transmitted wave The wavelength A0 27rktga with k0 V2mEha is the same on both sides of the barrier but the amplitude of the transmitted wave is much smaller than that of the incident wave The particles are never observed inside the barrier where they would have negative kinetic energy but are observed beyond the barrier k021t k 0 k021t k 0 x0 xL X This phenomenon is called tunneling and we discussed it already on our rst day of classes The transmission coef cient T is also called the tun neling probability or tunneling rate For wide barriers where the tunneling probability is very small it is given by E E T 2 167 1 7 HI6 18 U0 U0 lt gt where i fa O x2m 0 E is the penetration length introduced when we discussed the penetration of quantum particles in a step potential This expression for T applies when the barrier is wide in the sense that 6 ltlt L The tunneling probability depends exponentially on the thickness of the barrier L as well as on the energy difference U E For say electrons with E 004eV attempting to penetrate a barrier of height U 010eV the penetration length is 6 08nm If the barrier has width L 15nm the barrier is indeed wide as L gtgt 6 and we can use the formula above to calculate T The tunneling rate is a very sensitive function of L This fact is used in the STM that we talked about at the beginning of the semester 19 3 Application For applications read Krane We already discussed some of these in class Let me list a few 0 Scanning Tunneling Microscope 0 Alpha decay o Tunnel diode o Ammonia molecule I will add a few words about the last one 31 Inversion of the ammonia molecule A sketch of the ammonia molecule NHg is shown below The molecule has two equilibrium con gurations with the same energy The two equilibrium con gurations correspond to the N atom being located on either side of the plane containing the three H atoms The Coulomb repulsion of the hydrogen atom generates a potential barrier that the N atom has to surmount to move from one equilibrium position to the other The potential energy of the nitrogen atom is shown below as a function of its distance from the plane containing the three hydrogen atoms Here C 0 is the location of this plane potential energy of nitrogen potential barrier distance Ll 0 ii from lane equilibrium positions 0f hy rogen In the ground state the N atom has an energy indicated by the dashed line in the gure This energy is lower that the height of the barrier between the two minima corresponding to the two equilibrium con gurations of the molecule Classically the molecule would remain in whichever of the two con gurations happened to be formed initially In other words the nitrogen atom would always be on the same side of the three hydrogen plane as to get to the other side it has to surmount an energy barrier In a classical world the only way an object can get over an energy barrier is by having an energy higher than the barrier This extra energy could in principle come from thermal energy However the height of the barrier at C 0 is about 026 eV while at room temperature ktBT 26 meV 0026 eV ie much smaller In a quantum world however the nitrogen atom can tunnel through the barrier and it will indeed do so As a result you can think of the nitrogen as oscillating between the two equivalent equilibrium positions The frequency of these oscillations is f 23786 x 1010Hz which is rather low compared to the frequencies of most molecular vibrations The N atom makes more than 1010 oscillations per seconds The tunneling frequency of the ammonia molecule was used as the standard of the rst atomic clocks which provide the worlds standard for precision time keeping 11 4 Appendix A You will need to recall some properties of complex numbers A complex number z is written as z C where C Rez is the real part and y Imz is the imaginary part and y are ordinary real numbers Also i 1 The magnitude of a complex number is de ned as izi we 212 20 The complex conjugate of z is denoted by z and is the number that has the same real part as z and opposite imaginary part ie z C Since i2 1 we nd that zz z 2 2 312 One can represents complex numbers as points in the 31 plane The real part is measured along the 3C axis and the imaginary part along the y axis You can see then that the magnitude of the complex number is simply the length of its distance from the origin The gure below shows for instance the number z 3i4 Its magnitude is z 5 Try drawing on the gure the complex conjugate ofz 3i4 It is clear from the gure that the complex number can also be represented by its magnitude z and the angle 6 that the radius to the origin makes with the C axis The angle 6 is given by 6 arctanyC In other words one can write z z cos sin 6 If you like this is nothing but changing from carthesian to polar coordinates 12 Now consider complex numbers of magnitude 1 These are all the numbers in the 31 plane that lie on a circle of radius one Any complex number of magnitude 1 can be written as z 1 z cost isin 6 E a 21 The last equality is just the de nition of a new notation The complex function a is de ned by the equality em cos 6 isin 6 22 The complex conjugate of 6quot is CWTk 6 cost isin 6 23 We can then verify immediately that ewe w cos2 6 sin2 6 1 24

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