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# ADVANCED QUANTUM THEORY PHYS 7280

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This 6 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 7280 at University of Colorado at Boulder taught by Victor Gurarie in Fall. Since its upload, it has received 29 views. For similar materials see /class/232096/phys-7280-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.

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Date Created: 10/30/15

Quantum Many Body Theory Victor Gumm39e Week 2 2 Scattering Theory with Green s Functions 21 Relation between the scattering amplitude and the T matrix Consider a particle moving in a potential Vz in three dimensions thus the Hamiltonian is 1 H727AVz H E 21 m In quantum mechanics it is shown that the wave function of this particle can have the following asymptotic at large distance r behavior 9 W ym am r Here pZ2m E z is one of the coordinates of the 3D space along which an incoming wave cm is propagating ft9 is the scattering amplitude which depends only on the angle 6 between F and 2 but not on the magnitude of F denoted as r The problem how does one nd ft9 given the Greens function GR This is done in the following way One way to look for a solution of 21 is by creating initially the incoming wave 6quot E Ei Here is a vector directed along the z coordinate The incoming wavefunction is not an eigenfunction of the Hamiltonian Then we let it evolve in time according to the time dependent Schrodinger equation Eventually it stops depending on time except for the usual factor e iEt and develops the asymptotics 22 describing the incoming and the outgoing wave Denote H H0V where H0 iiA G5 will denote the retarded Green7s function of H0 while GR is reserved for the full Green7s function By the de nition of the time dependent retarded Green7s function mm 2 dgr GRFFte W 239 y GRFWE5 W 39E 23 In the Fourier space the initial wave function 71200 6quot is nothing but 71200 603713 We go into Fourier space and recall that the Fourier transform of the noninteracting Green7s function is given by C35 lt12 E 2390 7 g However for brevity this delta function is often omitted and the following notation is introduced 1 GR kE 25 a gt Ewi ltgt 2m Now recall the de nition ofthe T matrix C3 0 0T 0 With its help this becomes E d H d kdE Wt 39 g 651 7 3 7GB k E T I r E GR E WM 26 2714 O 7 7 7 0p7 This is the solution to the problem of representing the scattering problem in terms of Green7s functions We can also write it in a different way by denoting the Fourier time transform of the wave function by 3 71FE 2051Eei TISG EET1EG0pEei 7 27 The rst term in this expression is trivial it represents the initial wave propagating in time And indeed recalling that G5 1E ipz2m 20 we can take the integral in the rst term of 26 over E by residues to get H p2 eW ZW 28 The second term represents the scattered wave To evaluate it we need to know the T matriX which depends in some unknown way on the potential V So it would appear that until we gure out T for a speci c problem no further progress is possible This is true however if all we are interested in is a large 7 asymptotics of the wave function in the far future this can be simpli ed further The integral over 1 consists of angular integration and integration over the magnitude of k The angular integration involves the term 617 ammo where 6 is the angle of I counted from the direction of 15 direction of the incoming wave The integral itself goes over sin6d6 T0115 E also depends on 6 in some unknown way making taking this integral impossible However GWCOW is a fast oscillating function of 6 if r is large We will thus employ the steepest descend method which says that only a vicinity of the point where the rst derivative of cos6 vanishes which is important There are two such points 6 0 and 6 7139 Take the rst such point There we write sin6eik 059 00 6d6 eWO WZ all 29 0 zkr Similarly the second point gives sin6e ik 059 00 6d6 e ikf1 922 76711 210 0 zkr 2 Thus we nd as far as the second term in 27 is concerned Z kzdk eikTTkp 6 0 E 7 e Tkp 6 7139 E 1 1 0 4712 z kr E 0ikz2mE 07p22m39 211 We observe that the second term can be interpreted as the integral over negative k Com bining the terms together thus gives Zltgt kdk eikTTkpE 1 1 212 70047172 2397quot E 07kZ2mE 07p22m39 39 Next step is to do the integral over k We do it by residues closing the k contour over the upper half plane the only pole which contributes is k 2mE 20 This gives m ei39rxZmE i39iT k E 213 22m 07 Ez3907p22m Finally this gives for the wave function m dE 6i sz m p2 i 7T k E lEt iiT k E 2 2 717 17 Z27r27rr p E 0p22m5 27W 7107 10 me 2 e 214 To do the integral over E we assumed that the only pole in the lower half plane is E pZ2m 720 This is in fact not completely true there could be poles of Tkp E in the lower half plane However those will give contributions at energy which is not pZ2m which we will ignore because we would like to study scattering at E pZ2m We emphasize that the T matrix which sits in this expression must be evaluated with an incoming momentum 15 and the outgoing momentum 1 whose length is p and whose direction parallels F Comparison with 22 gives nally m f6 7T6 215 This is the relation we need which connects the T matrix to the scattering amplitude Here 6 is the angle between the direction of the incoming wave and the direction of F just as in 22 22 Scattering in a deltafunction potential in 3D Let us solve the problem of scattering in 3D in a delta function potential A 1 H 77A A6 216 2m a lt gt 3 We need to compute the T matrix The calculation proceeds completely equivalent to the calculation of T matrix in the 1D case7 which was studied in Week 1 notes The only dsk 1 n73 k2 217 27139 E 7 20 which now must be done in 3D7 not in 1D difference is in the integral This integral is divergent and does not make sense This re ects the following well known fact the delta function potential in 3D is not a well de ned potential However7 let us imagine that this potential is similar to a delta function7 but has a nite extend in space Then its Fourier transform will be A at low momenta7 and quickly go to zero when momenta exceed A7 where A is the inverse extend of the potential Then the integral becomes A kzdk 1 A dk k2 i 2mE A dk 2mE 2 2 2 39 0 2W2E72k7n20 0 2W2E721 im20 0 2W2E721 im20 This gives A A dk 1 n 77 2mE 219 7r 0 27r Ei 20 The integral is convergent and can be extended to in nity as long as E lt AZ2p It can also be extended to minus in nity7 at the expense of introducing an extra factor of 2 With the help of 139 of Week 17 we now nd mA 2mE m 117777 7T2 2W 772E 220 The expression 140 of Week 1 gives 1 221 1 mA mE m X f 7 f T m Notice that it is completely independent ofthe angle 0 and depends only on the scattering energy E Substituting E pZ2m gives 1 724 7 2A mA f7 T 27139 772p 222 Thus the scattering is angle independent fully s wave and proceeds with the scattering length 7 71 2 23 a i m W lndeed7 the scattering length is de ned as a 7fp 0 The poles of the T matrix re ect the bound states7 as usual Examining 221 we nd that the poles are possible only if 7T2 A lt 7 224 That is7 the delta function must be strong enough and attractive for the bound state to appear When A WZmA the scattering length is equal to in nity and the scattering amplitude is equal to f 7E 225 This is called the unitary limit of scattering

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