New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here


by: Hugh Wilkinson


Hugh Wilkinson
GPA 3.56


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Mechanical and Aerospace Engineering

This 33 page Class Notes was uploaded by Hugh Wilkinson on Monday October 19, 2015. The Class Notes belongs to MANE 4410 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/224921/mane-4410-rensselaer-polytechnic-institute in Mechanical and Aerospace Engineering at Rensselaer Polytechnic Institute.

Similar to MANE 4410 at RPI

Popular in Mechanical and Aerospace Engineering




Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/19/15
MANE 44106930 Applied Atomic amp Nuclear Physics Fall 2006 Review of Lecture 25 Course website httpwwwrpieduliueTeachinghtml Chapter 11 Quantum Mechanical Description of Nuclei A Schrodinger Wave Equation B Bound States in One Dimensional Systems Particle in a Square Well C Bound States in Three Dimensional Systems Orbital Angular Momentum D Barrier Penetration A Schrodinger Wave Equation 1 Waves and Particles We will review some basic properties of waves and the concept of waveparticle duality In classical mechanics the equation for a onedimensional periodic disturbance x t is 625 625 BIZ 62 6x2 21 which has as a general solution 5xt ioe lk quot 22 where a27w is the circular frequency v the linear frequency and k is the wavenumber related to the wavelength 7 by k 27TH It was rst postulated by deBroglie 1924 that one can associate a particle of momentum p and total relativistic energy E with a group of waves wave packet which are characterized by a wavelength 7 and a frequency v with the relation 7 h p 25 v E h 26 and that moreover the motion of the particle is governed by the wave propagation of the wave packet Page 1 of19 MANE44106930 Fall 2006 The wave function of a group of two waves of slightly different wavelengths and frequencies 211 I xt Z 2cos dkx dwt2sinkx tut gt 1rn 14 Hg 1 Spatial variation of a sum of two waves of slightly different frequencies and wavenumbers showing the wave packet moves with velocity g which is distinct from the propagation phase velocity w from Eisberg p 144 Eq 211 shows the wave packet oscillates in space with a period of 21 k while its amplitude oscillates with a period of 21 dk see Fig 1 Notice that the latter oscillation has is own propagation velocity dd dk This velocity is in fact the speed with which the associated particle is moving Thus we identify g u d7 212 as the group velocity This velocity should not be confused with the propagation velocity of the wave packet which we can calculate from w ukvtEp cl1 cp2 213 Here mg is the rest mass ofthe particle and c the speed oflight Thus we see the wave packet moves with a velocity that is greater than 5 whereas the associated particle speed is necessarily less than c There is no contradiction here because the former is the phase velocity while the latter is the group velocity Go to Example 1 Page 2 cf 19 MANEVM1069307Fall2006 2 The Schrodinger Wave Equation We will write down the Schrodinger equation in its time dependent form for a particle in a potential eld Vr 2 m a l h vz Vr I 5t 214 at 2m The Hamiltonian H of the system is hi H v2 Vr 215 2m Its physical meaning is the total energy which consists of the kinetic part p22m and the potential part Vr The particle momentum 2 is an operator in con guration space and it is represented as p 7 139 So a I 6t 75 H I Q t 216 Consider a periodic solution to 216 of the form 1 U 44 D 6 217 where E is a constant the total energy Inserting this solution into 216 gives the time independent Schrodinger equation H W E W l 2 22 VwEl 218 m We see that 218 has the form of an eigenvalue problem with H being the operator E the eigenvalue and 1115 the eigenfunction 3 Which form to choose To answer this question we will consider 218 in one dimension for the sake of illustration Writing out the equation explicitly we have 61211 abc2 kllx 221 where k2 2 mE 7 Vx hz In general k2 is a function of x because of the potential energy Vx Page 3 of19 MANE44106930 Fall 2006 but for piecewise constant potential functions such as a rectangular well or barrier we can write a separate equation for each region whee Vx is constant and theeby treat 1 as a constant in 221 A general solution 221 is then 144 Az k Bow 222 whee A and B are constants to be deternined by appropriate boundary conditions Now suppose we a d alin quot 39 39 39 39 r then kbecomes mgbl 1 For E gt 0 k is real and 15 as given by 217 is seen to have the form of traveling plane waves 2 For E lt 0 k m is irnaginary then T e We and the solution has the form of a standing wave What this means is that for the description of scattering problems one should use pomvecmgv solutions these are called scaoeing states while for boundrslale calculations one should work with negtm39veenergy solutions Fig 2 illustrates the behavior of the two types of solutions The condition at in nity x a too is that to is a plane wave in the scatteing problen and an exponentially decaying function in the boundstate problen 1n othe words outside the potential the exteior region the scatteing state should be a plane wave representing the presence of an incoming or outgoing particle while the bound state should be represented by an exponentially darnped wave signifying the localization of the particle inside the potential well Inside the potential the inteior region both solutions are seen to be oscillatory with the shorte peiod corresponding to highe kinetic enegy T E7 v Smiterim Fig 2 Traveling and standing wave functions as solutions to scatteing and boundstate problens respectively Pagerm MANEMIOE wil allJCOE Some properties of l or w which can be invoked as conditions for the solutions to be physically meaningful are i finite everywhere ii singlevalued and continuous everywhere iii first derivative continuous iv 1 a 0 when Va 00 Since a timeindependent potential cannot create or destroy particles the normalization condition Hwy52 1 223 cannot be applied to the boundstate solutions with integration limits extending to infinity However for scattering solutions one needs to specify an arbitrary volume Q for the normalization of a plane B Bound States in One Dimensional SystemsParticle in a Square Well Kx LZ Lz X X20 Fig 1 The square well potential centered at the origin with depth V0 and width L For a square well potential VX has the form Vx 7V0 L2 x L2 32 0 elsewhere as shown in Fig 1 For a quot 39 system the t39 A wave equation is im me x 7 E x 31 2m dx2 w w I Page 5 of 19 MANE441069307Fa11 2006 For the interior region the wave equation can be put into the standard form of a secondorder differential equation with constant coefficient d2Wxk2lx0 lxlSL2 33 dxl where we have introduced the wavenumber k such that k2 2mE V0 W is always positive and therefore k is always real For this to be true we are excluding solutions where 7E gt V0 For the exterior region the wave equation similarly can be put into the form d 2 x szx 0 x 2L2 34 where K2 2mE hz The solutions are LxAsinkx lxl EL 2 Be x gtL2 35 Ce x lt L2 The condition which we can now apply at the continuity conditions ii and iii in Lec2 At the interface x0 i L 2 the boundary conditions are 14 x0 Wm 99 36 dwinxx dlIMOC 3 7 dx XE dx XE 39 We therefore obtain C B B AsinkL 2eXpKL 2 and cotkL 2 K k 38 with the constant A determined by 223 Eq38 is the consequence of choosing the oddparity solution for the interior wave For the evenparity solutionzimx A cos kx the corresponding Page 6 of19 MANE44106930 Fall 2006 dJspersmn relaan ls mad 2 Kk 3 9 slnee bath sulununs are equally acceptable une has twu mum sets uf energy levels we 3 a and 3 3 and me cunespundmg wave funcuuns wmxAsmkx Dr A cuskx x ltLZ 3l3 WAX 52quot xgt L2 cgx xlt em 3 14 Whats the Energy levels are 3 15 The cunstznts B and c are determmed 39nm the Eunhnulty cundmuns at the lnterface A andA are m be xed bythenurmallzanun eenamen The dlscretevalues quhebuundrstate emerges k em are ubtamed 3 a and 3 3 ln Flg hm u the Sm exalted ats wnh eaa panty m 70 HammadMme Fig 3 Groundstate and first two excitedstate solutions from Cohen p 16 To obtain more explicit results it is worthwhile to consider an approximation to the boundary condition at the interface Instead of the continuity of 11 and its derivative at the interface one might assume that the penetration of the wave function into the external region can be neglected and require that 11 vanishes at x i L 2 or say V gt 00 then I gt 0 Applying this condition to 313 gives kL rm where n is any integer or equivalently n2 lhl 2mL2 En V0 nl2 316 This shows explicitly how the energy eigenvalue En varies with the level index n which is the quantum number The corresponding wave functions are zjnxAcosmeL nl3 AsinmeL n 2 4 317 The first solutions in this approximate calculation are also shown in Fig 3 We see that requiring the wave function to vanish at the interface is tantamount to assuming that the particle is con ned in a well of width L and inf39mitely steep walls the inf39mite well potential or limit of V0 C Bound States in Three Dimensional Systems Orbital Angular Momentum 3 D Spherical coordinates The problem we want to solve is to determine the boundstate energy levels and corresponding wave functions for a particle in a spherical well potential The potential is of the form Vr V0 r lt r0 0 otherwise 41 Here r is the radial position of the particle relative to the origin We again begin with the timeindependent wave equation h 2 Z VZ Vrl Ew 42 m Since the potential function has spherical symmetry it is natural for us to carry out the analysis in the Page 8 of19 MANE44106930 Fall 2006 spherical coordinate system rather than the Cartesian system A position vector p then is speci ed by the radial coordinate r and two angular coordinates a and a the polar and azimuthal angles respectively see Fig 1 Fig 1 The spherical coordinate system A point in space is located by the radial coordinate r and polar and azimuthal angles a and a 2 In this coordinate system theLaplacian operator v is ofthe form 43 where D is an operator involving the radial coordinate l a a 02 72 44 72 87 97 2 and the operator L involves only the angular coordinates 2 2 74441 are 467 45 h smeaa as sin ea In terms ofthese operators the wave equation 42 becomes Pages of 19 MANEMIOEMOiI allJDOE 2 2 h Df L 2 Vr yngp Eyrgqp 46 2m 27m 2 2 For any potential Vr the angular variation of 11 is always determined by the operator L 2mr 2 Therefore one can study the operator L separately and then use its properties to simplify the solution of 46 2 We summarize the basic properties of L before discussing any physical interpretation It 2 can be shown that the eigenfunction of L are the spherical harmonics functions Y m 9 a LZYZ39quot t9 0 306 lYquot t9 0 47 It is clear from 47 that Ym a is an eigenfunction of L2 with corresponding eigenvalue Z lh2 Since the angular momentum of the particle like its energy is quantized the index X can take on only positive integral values or zero 01 23 Similarly the index m can have integral values from to X m 1 l 01 l Z For a given I there can be 2 1 1 values of m The significance of m can be seen from the property of L2 the projection of the orbital angular momentum vector L along a certain direction in space in the absence of any external field this choice is up to the observer LzYzm 9 0 MhYzm 9 0 The indices X and m are called quantum numbers We should regard the particle as existing in various states which are specified by a unique set of quantum numbers each one is associated with a certain orbital angular momentum which has a definite magnitude and orientation with respect to our chosen direction along the zaxis The particular angular momentum state is described by the In function Y ago with Z known as the orbital angular momentum quantum number and m the magnetic quantum number It is useful to keep in mind that Y I 9 go is actually a rather simple function for low order indices For example the first four spherical harmonics are Page 10 of19 MANE44106930 Fall 2006 Y00 1 l47z39Y1 14387z39e q sin0110 V347z39cos0Y11V38aelwsin0 One property we need to know is orthonormality I 27 j sin we I WY a gym a a 5225 4 11 0 0 m Then we may use this property to cancel out Y ago while we solve wave equation of the total system Returning to the wave equation 46 we look for a solution as an expansion of the wave function in spherical harmonics series WOW 2R1 0 950 412 2 Because of 47 the L operator in 46 can be replaced by the factor 1h2 In view of 411 we can eliminate the angular part of the problem by multiplying the wave equation by the compleX conjugate of a spherical harmonic and integrating over all solid angles recall an element of solid angle is sin Bd dgo obtaining m r W D MW0 RrERZr 413 2 2m This is an equation in one variable the radial coordinate r although we are treating a threedimensional problem We can make this equation look like a onedimensional problem by transforming the dependent variable RI Define the radial function MN RN 414 Inserting this into 413 we get h2 d2ur zz1h2 M dr 2er VrurEur 415 We will call 415 the radial wave equation It is the basic starting point of threedimensional problems involving a particle interacting with a central potential field It is now just a 1 D problem again The wave function describing the state of orbital angular momentum Z is often called the th partial Page 11 of19 MANE44106930 Fall 2006 wave lIz 9W R1 0W 9 lt0 4 16 The parity of Y 9 a is 1 In other words the parity of a state with a de nite orbital angular momentum is even if X is even and odd if X is odd All eigenfunctions of the Hamiltonian with a spherically symmetric potential are therefore either even or odd in parity 3 D Cartesian coordinates There are situations where it will be more appropriate to work in another coordinate system As a simple example of a boundstate problem we can consider the system of a free particle contained in a cubical box of dimension L along each side In this case it is clearly more convenient to write the wave equation in Cartesian coordinates r12 a a 6x2 6y2 6 2 2Ixyz E lIxyz 417 m 62 0 ltx y z lt L The boundary conditions are 1p 0 whenever x y or z is 0 or L Since both the equation and the boundary conditions are separable in the three coordinates the solution is of the product form nyz wn x wny y an Z 2 L 2 sinnx nx Lsinnyny Lsinnz n2 L 418 where m ny nz are positive integers excluding zero and the energy becomes a sum of three contributions EM z Enx Eny Enz T1702 2mL2 71 n 71 43919 We see that the wave functions and corresponding energy levels are specified by the set of three quantum numbers m m 712 While each state of the system is described by a unique set of quantum numbers there can be more than one state at a particular energy level Whenever this happens the level is said to be degenerate For example 112 121 and 211 are three different 2 states but they are all at the same energy so the level at 605112 2mL is triply degenerate Page 12 of19 MANE44106930 Fall 2006 2 The energy unit is seen to be AE 71102 2mL We can use this expression to estimate the 28 magnitude of the energy levels for electrons in an atom for which m 91 X10 g and L N 3 X 8 24 10 cm and for nucleons in a nucleus for which m 16X10 g and L N 5 F The energies come out to be N30 eV and 6 MeV respectively values which are typical in atomic and nuclear physics 10 Notice that if an electron were in a nucleus then it would have energies of the order 10 eV C Barrier Penetration We have previously observed that one can look for different types of solutions to the wave equation Other than bounded system problems barrier penetration is treated as scattering problem In this case one looks for positiveenergy solutions as in a scattering problem We consider a onedimensional system where a particle with mass m and energy E is incident on a potential barrier with width L and height V0 that is greater than E Fig 1 shows that with the particle approaching from the left the problem separates into three regions left of the barrier region 1 inside the barrier region 11 and right of the barrier region 111 Fig 1 Particle with energy E penetrating a square barrier of height V0 V0 gt E and width L In regions 1 and 111 the potential is zero so the wave equation 31 is of the form d 2 x k2yxx0 kZ 2mEhz 51 where k2 is positive The wave functions in these two regions are therefore 14 ale k blew E 11 m 52 Page 13 of 19 MANE441069307Fa11 2006 1 W3 aselkx bseikx E W3 53 where we have set b3 0 by imposing the boundary condition that there is no particle in region 111 traveling to the left since there is nothing in this region that can re ect the particle In region 11 the wave equation is Z xivx 0 K2 2mV0 Eh2 54 So we write the solution in the form KX KX ll612 bge 55 2 Notice that in region 11 the kinetic energy E 7 V0 is negative so the wavenumber is imaginary in a propagating wave another way of saying the wave function is monotonically decaying rather than oscillatory What this means is that there is no wavelike solution in this region By introducing K we can think of it as the wavenumber of a hypothetical particle whose kinetic energy is positive V0 7 E We recall that given the wave function 1p we know immediately the particle density number of 3 particles per unit volume or the probability of the finding the particle in an element of volume d r about r W 0 2 and the net current is related to the wave function by the expression 1 w Zipwa 56 2mz Using the wave functions in regions I and III we obtain Z Z 11xva b J 57 Z 13x Vlasl 58 where v hk m is the particle speed From here on we can regard a1 b1 and a3 as the amplitudes of the incident re ected and transmitted waves respectively With this interpretation we define 59 and Page 14 of19 MANE44106930 Fall 2006 TR1 510 is always satisfied as one would expect The transmission coefficient is sometimes also called the Penetration Factor and denoted as P To calculate a1 and a3 we apply the boundary conditions at the interfaces x 0 and x L d d way2 1 2 x0 511 d d gumx3 xL 512 This result then leads to lasl2 a3 2 1 2 a 2 P 514 lall 1 sinh2 KL with sinhx e iei 2 Using the leading expression of sinhx for small and large arguments one can readily obtain simpler expressions for P in the limit of thin and thick barriers 2 2 PN1V 0 Z 1 V0L 2 m z ltlt1 515 4EV0 E 4E hi PNE 1 87M KLgtgtl 516 V0 V0 Thus the transmission coef cient decreases monotonically with increasing V0 or L relatively slowly for thin barriers and more rapidly for thick barriers Which limit is more appropriate for our interest Consider a 5 MeV proton incident upon a barrier 12 of height 10 MeV and width 10 F This gives K N 5 x 10 cm39l or KL N 5 Using 516 we find Pl6gtltlgtltlgtlte3910 2gtlt10394 2 2 As a further simplification one sometimes even ignores the prefactor in 516 and takes P e 517 with Page 15 of19 MANE44106930 Fall 2006 Y2KL il Lszan 7E 518 We show in Fig 3 a schematic of the Wave function in each region In regions I and III 14 is complex so We plot its real or imaginary part In region 11 14 is not oscillatory Although the Wave mction in region 11 is nonzero it does not appear in either the transmission or the re ection coef cient Fig 3 Panicle penetration through a square barrier of height VB and Width L at energy E E lt V schematic behavior of Wave functions in the three regions Process of solving waveifunctjon problems 1 Look at potential carefully and Write down Vx or Vr for each region N Write down the appropriate timeindependent or timedependent 1D or 3D mction for each region 3 Borrow solutions for each region under that speci c Wave lnction 4 Write down boundary conditions 11 112 anddTIV dTIlf at each boundary 5 Solve boundary conditions to link the pammeters 6 Apply normalization condition to mher determine the pammeters Page 15 of 19 MAN39EA4106930Fallm06 Example 2 Consider a onedimensional Wave equation with the potential Vo 7L1 5 x 5 11 region 1 Vx Vl 7L2 lt x 5 fly L1 5 x 5 L2 region 2 0 otherwise region 3 a Find the Xdependence of the wave function in each of the 3 regions for E lt 0 b What are the boundary conditions to be applied at the interface Y on are asked to state the boinldary conditions but not to apply them Page 17 uf19 MAN39E7441069307Fall 2005 Example 3 Consider a beanr of particles each of mass nr and having kinetic energy E incident from the left upon a square barrier of height Va and width L wit 1 E lt V0 Inside the barrier the wave function is of the form in 110 we 179 where you are given the result I 1irK 4d 8 171 K wrth hlk22111E and Elle 2m VD 7E 15 a Using the information given on yx inside the barrier derive an approximate expression for the transmission coef cient T for the case of thick barriers 10 b Sketch qualitatively the absolute square of the wave function I1 1 I2 everywhere and indicate the spatial dependence of I 1101quot wherever it is lcuown Does IllX 1 vanish at any point Page 18 of 19 MANEMrOssswrall 2005 Example 4 Suppose you are given the result for the transmission coef cient T for the barrier penetration problem onedimensional baurier of height VO extending from X0 to XL 71 T l sinthL 4EVl7 E where K2 2711070 E7 12 is positive E lt V0 a From the expression given deduce T for the case E gt V 0 without solving the wave equation again b Deduce T for the case of a square well potential from the result fora square barrier Page 19 of 19 MANE44106930 F311 2006 MAN39E44106930 Applied Atomic amp Nuclear Physics Fall 2006 Appendix B Cross Section Calculation A Method of Phase shin Wevcillstudy v lt i m t t t a L 4 w v V quot blem39 39 quot F hhina Ulllel dlced massmo mgina quotVr 39 39 39 39 39 two colliding particles We will solve the 511 dmgm39wave equauou for the spatial distnburionof 39 39 wd needed to deterinme the angular differentral cross section 56 For a discussion of the concepts ofcruss sections see AppendixA The Scattering Amplitude M 11 standard ms t Iquot A rs 4 U6ill 111150fh K i 1 t V aw t 39 1 equation Forthe rst step 39 39 39 I a HI a quot potential eld Vtr centered at me on gm LCML as shown In kg 51 lhe incident beam is represented by a traveling plane wave Wm be m B 1 k my the energy of the effective particle E r k 2 mt11e relative ensign ofthe colliding particles For the scattered wave which results from the interaction in the region of the potential Vr we will write it in the form of an outgoing spherical wave Ian e 139 f9b 132 r where f6 which has the dimension of length denotes the amplitude of scattering in a direction indicated by the polar angle 6 relative to the direction of incidence see Fig Bl It is clear that by representing the scattered wave in the form of B2 our intention is to work in spherical coordinates Once we have expressions for the incident and scattered waves the corresponding current or ux can be obtained from the relation 71 1 P NAME 1 133 2m The incident current is J vb2 where v hku is the speed ofthe effective particle Fig31 Scattering of an incoming plane wave by a potential eld Vr resulting in spherical outgoing wave The scattered current crossing an element of surface area d9 about the direction 9 is used to de ne the angular differential cross section dadQ 2 66 where the scattering angle 6 is the angle between the direction of incidence and direction of scattering For the number of particles per sec scattered through an element of surface area dQ about the direction 9 on a unit sphere we have 19de vft92 do 134 The angular differential cross section for scattering through dQ about Q is therefore see Appendix A 1 9 06 J If t92 35 This is the fundamental expression relating the scattering amplitude to the cross section it has an analog in the analysis of potential scattering in classical mechanics Method of Paltial Waves To calculate f 6 from the Schrodinger wave equation we note that since this is not a timedependent problem we can look for a separable solution in space and time kMgr yQIO with 11 expGilE h The Schrodinger equation to be solved then is ofthe form if 2 e V V l Elll 136 2 For twobody scattering through a central potential this is the wave equation for an effective particle with mass equal to the reduced mass u mlmz ml m2 and energy E equal to the sum of the kinetic energies of the two particles in CMCS or equivalently E uv2 2 with v being the relative speed The reduction of the twobody problem to the effective onebody problem B6 is a useful exercise although quite standard For those in need of a review a discussion of the reduction in classical as well as quantum mechanics is given at the end of this Appendix As is well known there are two kinds of solutions to B6 boundstate solutions for E lt 0 and scattering solutions for E gt 0 We are concerned with the latter situation In view of B2 and Fig B 1 it is conventional to look for a particular solution to B6 subject to the boundary condition 110 xkz 8 ML e f 97 B7 where Va is the range of force Vr 0 for r gt r0 The subscript k is a reminder that the entire analysis is carried out at constant k or at fixed incoming energy E hzkz 2 u It also means that f 6 depends on E although this is commonly not indicated explicitly For simplicity of notation we will suppress this subscript henceforth According to B7 at distances far away from the region of the scattering potential the wave function is a superposition of an incident plane wave and a spherical outgoing scattered wave In the faraway region the wave equation is therefore that of a free particle since Vr 0 The freeparticle solution to is what we want to match up with B7 The form of the solution that is most convenient for this purpose is the expansion of yD into a set of partial waves Since we are considering central potentials interactions which are spherically symmetric or V depends only on the separation distance magnitude of 5 of the two colliding particles the natural coordinate system in which to find the solution is spherical coordinates 5 a r6ltp The azimuthal angle p is an ignorable coordinate in this case as the wave function depends only on r and 6 The partial wave expansion is yr iRlrPlcos9 B8 where Rcos 8 is the Legendre polynomial of order I Each term in the sum is a partial wave of a definite orbital angular momentum with I being the quantum number The set of functions 3 x is known to be orthogonal and complete on the interval 1 1 Some of the properties of Bx are 1 2 idxeltxmvltxgtm6uv 1311 137171 B9 PoxlPlxx Px3xhi2Px5x3 73902 Inserting B8 into B6 and making a change of the dependent variable to put the 3D problem into 1D form u r rR r we obtain cgd rk27 fVr7ulr0 rltr0 B10 This result is called the radial wave equation for rather obvious reasons it is a one dimensional equation whose solution determines the scattering process in three dimensions made possible by the properties of the central potential Vr Unless Vr has a special form that admits analytic solutions it is often more effective to integrate B 10 numerically However we will not be concerned with such calculations since our interest is not to solve the most general scattering problem EqB 10 describes the wave function in the region of the interaction r lt r0 where Vr 0 r gt r0 Its solution clearly depends on the form of Vr On the other hand outside of the interaction region r gt r0 EqB 10 reduces to the radial wave equation for a free particle Since this equation is universal in that it applies to all scattering problems where the interaction potential has a finite range r0 it is worthwhile to discuss a particular form of its solution Writing Eq B 10 for the exterior region this time we have d k27 Dulr0 1311 dr r which is in the standard form of a secondorder differential equation whose general solutions are spherical Bessel functions Thus ulrBlrjlkrClmlkr B12 where B and C are integration constants to be determined by boundary conditions and j and n are spherical Bessel and Neumann functions respectively The latter are tabulated functions for our purposes it is sufficient to note the following properties Lxsinxx nax7cosxx sinx cosx cosx sinx 11x nx7 7 x x x x x 1352 71 39 x a n x a Bl3 M quot 1352z1 0 x l l j x at sinx7 It 2 n x at i cosx 7 It 2 x x The Phase Shift 5 Using the asymptotic expressions for j and n we rewrite the general solution 1312 as My Blksinkr7 7r27Clkcoskr7 7r2 krgtgt1 alksinkr7 7r25l B14 The second step in B 14 deserves special attention Notice that we have replaced the two integration constant B and C by two other constants a and 5 the latter being introduced as a phase shift The signi cance of the phase shift will be apparent as we proceed further in discussing how one can calculate the angular differential cross section through B5 In Fig B2 below we give a simple physical explanation of how the sign of the phase shift depends on whether the interaction is attractive positive phase shift or repulsive negative phase shift Combining B 14 with B8 we have the partialwave expansion of the wave function in the asymptotic region c056 B 15 llr 9 Akrgtgt1 2 at t This is the lefthand side of B7 Our intent is to match this with the righthand side of B7 also expanded in partial waves and thereby relate the scattering amplitude to the phase shift Both terms on the righthand side are seen to depend on the scattering angle6 Since the scattering amplitude is still unknown we can simply expand it in terms of partial waves f 9 Z Bcos9 B 16 where the coefficients jj are the quantities to be determined in the present cross section calculation The other term in B7 is the incident plane wave It can be written as Emma Z Iquot 2 1 krB cos 9 I em 2 2 DWMCOSB B 17 l Inserting both B 16 and B 17 into the righthand side of B7 we see that terms on both sides are proportional to either expikr or expikr If B7 is to hold in general the coefficients of each exponential have to be equal This gives 1 z 5 7 t f 71 ale 1 2 1 13 18 a 0 1w B 19 EqB18 is the desired relation between the 2 th component of the scattering amplitude and the 2 th order phase shift Combining it with B 16 we have the scattering amplitude expressed as a sum of partialwave components f9 0102 2 1W sin 53cos 9 1320 0 This expression more than any other shows why the present method of calculating the cross section is called the method of partial waves Now the angular differential cross section B5 becomes w 2 79 X2 203 le 5 sin5Pcos9 B21 0 where i 1 k is the reduced wavelength Correspondingly the total cross section is 039 j 61909 4an 2 1 sinz 5 322 EqsB21 and B22 are very well known results in the quantum theory of potential scattering They are quite general in that there are no restrictions on the incident energy Since we are mostly interested in calculating neutron cross sections in the lowenergy regime kr0 ltlt 1 it is only necessary to take the leading term in the partialwave sum The Z 0 term in the partialwave expansion is called the swave One can make a simple semiclassical argument to show that at a given incident energy E hzk2 2u only those partial waves with Z lt kr0 make signi cant contributions to the scattering If it happens that furthermore kr0 ltlt 1 then only the Z 0 term matters In this argument one considers an incoming particle incident on a potential at an impact parameter b The angular momentum in this interaction is h pb where p hk is the linear momentum of the particle Now one argues that there is appreciable interaction only when b lt r0 the range of interaction in other words only the 2 values satisfying b Z k lt r0 will have signi cant contriubution to the scattering The condition for a partial wave to contribute is therefore I lt kr0 S wave scattering We have seen that if kr0 is appreciably less than unity then only the Z 0 term contributes in B21 and B22 What does this mean for neutron scattering at energies around kBT N 0025 eV Suppose we take a typical value for r0 at N 2 X 10 3912 cm then we find that for thermal neutrons kr0 N 105 So one is safely under the condition of low energy scattering kr0 ltlt 1 in which case only the swave contribution to the cross section needs to be considered The differential and total scattering cross sections become 79 X2 sinZ 130k B23 039 4an sin2 130k B24 It is important to notice that swave scattering is spherically symmetric in that 70 is manifestly independent of the scattering angle this comes from the property P0X 1 One should also keep in mind that while this is so in CMCS it is not true in LCS In both B23 and B24 we have indicated that swave phase shift 5D depends on the incoming energy E From B 18 we see that f0 65quot sin ok Since the cross section must be nite at low energies as k a 0 f0 has to remain nite or 50 k a 0 Thus we can set Emma sin 600 60 iak 325 where the constant a is called the scattering length Thus for lowenergy scattering the differential and total cross sections depend only on knowing the scattering length of the target nucleus 79 a2 B26 039 47mZ B27 We will see in the next lecture on neutronproton scattering that the large scattering cross section of hydrogen arises because the scattering length depends on the relative orientation of the neturon and proton spins Physical signi cance of sign of scattering length One can ask about the physical meaning of the scattering length a Although the sign of a does not affect the magnitude of the cross sections it carries information about the scattering potential This is readily seen from the geometric significance of a Fig 42 shows two sine waves one is the reference wave sinkr which has not had any X SirLlu lil x a r rm 9 x1 Fig 42 Comparison of unscattered and scattered waves showing a phase shift 6 in the asymptotic region as a result of the scattering interaction unscattered and the other one is the wave sinkr 6 which has suffered a phase shi by virtue of the scattering The entire effect of the scattering is seen to be represented by the phase shift 6 or equivalently the scattering length through B25 Thus in swave scattering the angular differential and total scattering cross sections depend only on knowing the scattering length a The scattered wave written in the form of uO N sin kr r a suggests a simple but revealing geometric construction In the vicinity of the potential we can take kro to be small this is agin the condition of low energy scattering so that uo N kr r a in which case a becomes the distance at which the wave function extrapolates to zero from its value and slope at r r0 There are two ways in which this extrapolation can take place depending on the value of kro As shown in Fig B3 when krO gt 7r 2 the wave function has reached more than a quarter of its wavelength at r re So its slope is downward and the extrapolation gives a distance a which is positive If on the other hand krO lt 7 2 then the extrapolation gives a distance a which is negative The signi cance is that a gt 0 means the potential is such that it can have a bound state whereas a lt 0 means that the potential can only give rise to a virtual state LL 39239 gt TVz W3 W2 A l r 1 r L r r 070 110 Fig 133 Geometric interpretation of positive and negative scattering lengths as the distance of extrapolation of the wave function at the interface between interior and exterior solutions for potentials which can have a bound state and which can only virtual state respectively Reduction of twobody collision to an effective onebody problem We conclude this Appendix with a supplemental discussion on how the problem of twobody collision through a central force is reduced in both classical and quantum mechanics to the problem of scattering of an effective one particle by a potential eld Vr Meyerhof pp 21 By central force we mean the interaction potential is only a function of the separation distance between the colliding particles We will rst go through the argument in classical mechanics The equation describing the motion of particle 1 moving under the in uence of particle 2 is the Newton39s equation of motion mi L 328 where g is the position of particle l and F2 is the force on particle l exerted by particle 2 Similarly the motion of motion for particle 2 is mi F2 5 1329 where we have noted that the force exerted on particle 2 by particle l is exactly the opposite of F2 Now we transform from laboratory coordinate system to the centerof mass coordinate system by de ning the centerofmass and relative positions mlll mZLZ 6 L 1172 1330 m m2 Solving for 51 and 52 we have m m 7 Z 7 1 1172 l 127 L 1331 m m2 m m2 We can add and subtract B28 and B29 to obtain equations of motion for L and 5 One nds ml mJi 0 B32 2 12 dVrdz 1333 with u mimz m m2 being the reduced mass Thus the centerofmass moves in a straightline trajectory like a free particle while the relative position satis es the equation of an effective particle with mass u moving under the force generated by the potential Vr EqB33 is the desired result of our reduction It is manifestly the onebody problem of an effective particle scattered by a potential field Far from the interaction field the particle has the kinetic energy Mg 2 The quantum mechanical analogue of this reduction proceeds from the Schrodinger equation for the system of two particles h2 h2 7 2m 712P 1 2 E1 E2 T 1912 1 2 Transforming the Laplacian operator V2 from operating on r1 52 to operating on If 5 we find Ah V ih W Vr PLL E Em B35 2m1 m2 2 Since the Hamiltonian is now a sum of two parts each involving either the centerof mass position or the relative position the problem is separable Anticipating this we have also divided the total energy previously the sum of the kinetic energies of the two particles into a sum of centerofmass and relative energies Therefore we can write the wave function as a product TQNZ yQJyg so that B35 reduces to two separate problems 7 v2 E B36 2mlm2 JIJL MAL eh 2V2 V l EllZ 1337 2 It is clear that B36 and B37 are the quantum mechanical analogues of B32 and B33 The problem of interest is to solve either B33 or B37 As we have been discussing in this Appendix we are concerned with the solution of B37


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Janice Dongeun University of Washington

"I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.