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Green Function for Delta Potential and One Loop Correction to Quantum Electrodynamic Scattering Process

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Green Function for Delta Potential and One Loop Correction to Quantum Electrodynamic Scattering Process GREEN FUNCTION FOR DELTA POTENTIAL AND ONE LOOP CORRECTION TO QUANTUM ELECTRODYNAMIC SCATTERING PROCESS A project report for partial fulfillment of requirements for degree of Integrated Master of Science in Physics Submitted by Raj Kishore 410PH5084 Supervised by Dr. Biplab Ganguli Department of Physics and Astronomy National Institute of Technology Rourkela Abstract Greens function a technique used to solve in general non homogeneous dif- ferential equations. It is basically a correlation function. Its application in high energy physics in finding the propagators has discussed here. In this paper, its application in quantum physics to find Greens functions for quan- tum operators and its solutions has the main focus. By knowing the Greens function we can calculate density of states. In this paper, contains a detail calculations to find the Greens function for single and double delta function potential and then analysis of the bound state. In particle physics one of the main observable concern for the experiment is the scattering cross-section. Physicists used the perturbation theory to approximate results of the scattering process, as we do not have the ex- act knowledge of scattering process. In the higher order corrections, due to loops, the amplitude expression for the Feynman diagram diverges. We apply a theory of renormalization to get a physically possible results. This paper start with the basic introduction to renormalization theory then its application to quantum electrodynamics scattering process. I have briefly discussed all three types of loops found in QED and focused mainly on the vacuum polarization (photon self energy) corrections which give the charge renormalization. Certificate This is to certify that the work done in thesis entitled, “Green’s function for delta potential” and “one loop correction to quantum electrodynamic scattering process”, are submittedby Raj Kishore towards partial fulfillment of the requirements for the award of Master of Science in Physics degree by National Institute of Technology Rourkela. It is a record of the work done by him, under my supervision. The published results have been partly reproduced. Dr. Biplab Ganguli Department of Physics and Astronomy National Institute of Technology Rourkela India Acknowledgment I would like to express my gratitude to my project advisor Dr. Biplab Ganguli and as my co-advisor Prof. Sasmita Mishra for all that I have learned from them during my project. It has been a wonderful experience working with them. Raj Kishore National Institute of Technology, Rourkela, India Email:[email protected] 1 Contents 1 Definition of Green's function 4 2 Green's function in Quantum Physics 6 3 Green's Functions For Single Delta Function Potentials 8 3.1 Bound States . 9 4 Green's Function For Double Delta Function Potential 11 4.1 Conclusion . 13 5 Some Necessary Mathematics Notations and Conventions 15 6 Baseline Concepts 17 7 Relativistic Quantum Mechanics 18 7.1 Klein-Gordon Equation . 18 7.2 Dirac Equation . 18 8 Scattering Amplitude Calculation from Feynman Diagram 20 9 Renormalization in Quantum mechanics 24 9.1 In One -Dimension . 24 9.2 In two-dimension . 28 10 Ultraviolet Divergences 30 10.1 Classification of ultraviolet divergences appear in QFT . 30 11 Radiative Correction 32 11.1 Vacuum Polarization . 34 11.2 One-loop Vertex Correction . 36 12 Renormalization of Electric Charge 39 13 References 42 2 PART-I 3 Chapter 1 Definition of Green's function For any arbitrary linear differential operator L, in Rn (Euclidean space), the Green's function G(r; r0) is defined by the solution of equation 0 0 L · G(r; r ) = δ(r0 − r ) (1.1) where, r; r0S where S is a surface domain and δ in the Dirac delta function. By this definition of green function, we can have the solution of any inhomogeneous differential eqn. of the form Lφ(r) = f(r) (1.2) and the solution is given by Z φ(r) = G(r; r0)f(r0)dr (1.3) Poisson's equation −ρ r2 = (1.4) 0 Now the Green function corresponding to this differential eq. is given by 2 0 0 r G(r; r ) = +δ(r0 − r ) (1.5) but from Gauss' theorem Z 1 Z 1 r2( dτ) = r( · d~σ) (1.6) r r Since, 1 −~r r^ r( ) = = − (1.7) r r3 r2 So, Z 1 Z r^ · d~σ r( ) · d~σ = − (1.8) r r2 4πr2 Z r^ · ndS^ = (1.9) r2 0 4πr2 1 Z = − dS (1.10) r2 0 4 So, Z 1 0 if if origin is excluded r2( )dτ = (1.11) r −4π if origin is included or 1 r2 = −4πδ(r) (1.12) r Hence, 1 r2( ) = −4πδ(r − r0) (1.13) jr − r0j Now, from eqn (1.5) and (1.13) 1 G(r; r0) = − (1.14) 4πjr − r0j So, the solution of eqn. (1.4) can be 1 Z ρ(r0) (r) = 0 dr (1.15) 4π0 jr − r j 5 Chapter 2 Green's function in Quantum Physics The time independent Gree's function can be defined as the solution of inhomogeneous differential eqn of (z − C(r))G(r; r0; z) = δ(r − r0) (2.1) on some boundary condition for r and r', and z = λ + ιs and L(r) is a time independent linear Hermitian differential operator. Let φn(r) be the complete set of eigen functions of the differential operator L(r) then C(r)φn(r) = λnφn(r) (2.2) 0 and φn(r) also satisfy the same boundary conditions as G(r; r ; z). Since, eigen functions φn corresponding to difficulty eigen values are orthogonal. So, φn(r) can be considered as orthogonal. So, Z ∗ φn(r)φm(r)dr = δnm (2.3) from completeness condition X jφni hφnj = 1 (2.4) n and X ∗ 0 X 0 φn(r)φ (r ) = hrjφni hφnjr i (2.5) n n X 0 = jφni hφnj hrjr i (2.6) n = hrjr0i (2.7) = δ(r − r0) (2.8) (2.9) and G(r; r0; z) ≡ hrj G(z) jr0i (2.10) δ(r − r0)L ≡ hrj L jr0i (2.11) hrjr0i ≡ δ(r − r0) (2.12) 6 where jri is the eigenvector of position operator. In new notations, the eqn. (1.1) can be written as (z − L)G(z) = 1 (2.13) Now, if all eigenvalues of z − L are non-zero i.e. z 6= λn then, 1 G(z) = (2.14) z − L P jφni jφni G(z) = n (2.15) z − L but eigenvalues of L can have both discrete and continuous. So, X Z = Σ0 + dc (2.16) n Z 0 jφni hφnj jφci jφci G(z) = Σn + dc z − λn z − λc 7 Chapter 3 Green's Functions For Single Delta Function Potentials Green's function G(x; y; E) associated with the Hamiltonian H is the solution to the eqn. (E − H)G(x; y; E) = δ(x − y) (3.1) Satisfying the boundary conditions lim G(x; y; E) = 0 jx−yj!1 here, x and y are points in D-dimensional Euclidean space and correspondingly δ(x − y) is a D-dimensional delta function. Let us suppose f n(x)g be the eigenstates of H ∗ So, G(x; y; E) = P n(x) n(y) E−En n Let H = H0 + λδ(x) Where H0 is the Hamiltonian of free particle. so, the Hamiltonian H contains one delta potential. Now, we know the Green's function associated with H0 i.e. for the free particle. Let G0(x; y; E) be the Green's function associated with H0. Then, −1 1 G0(E) = (E − H0) = E − H0 Now, 1 G(E) = (3.2) E − H −1 = (E − H0 − λδ(x)) (3.3) −1 = G0(E) f1 − G0(E)λδ(x)g (3.4) = G0(E) + G0(E)λδ(x)(G0(E) + G0(E)λδ(x)G0(E) + E + ::: ) (3.5) G(E) = G0(E) + G0(E)λδ(x)G(E) In x; y representation, Z D G(x; y; E) = G0(x; y; E) + d zG0(x; z; E)λδ(z)xG(z; y; E) 8 G(x; y; E) = G0(x; y; E) + λG0(x; 0; E)G(0; y; E) Putting x = 0 is in the above eqn. G(0; y; E) = G0(0:y; E) + λG0(0; 0; E)G(0; y; E) G (0; y; E) G(0; y; E) = 0 1 − λG0(0; y; E) now, λG0(x; 0; E)G0(0; y; E) G(x; y; E) = G0(x; y; E) + 1 − λG0(0; 0; E) G0(x; 0; E)G0(0; y; E) G(x; y; E) = G0(x; y; E) + 1 λ − G0(0; 0; E) This is the Green's function associated with Hamiltonian H. 3.1 Bound States Bound states of the Hamiltonian H with H0, the Hamiltonian of the free particle. We know that poles of the Greens function gives the eigenvalues of the eigenvalues of the Hamiltonian. So, poles of G(x; y; E) in the above eqn. are the energy levels in bound states. Since, there are no bound state in free particle problem, so from the above eqn. only poles occur if 1 − G (0; 0; E) = 0 λ 0 2 Suppose, H0 = −∇ then, Z dDk eιk(x − y) G 0(x; y; E) = 0 (2π)D E − k2 So, in order to find the energy of bound states, we must solve the equation 1 Z dDk 1 + = 0 (3.6) λ (2π)D K2 + k2 (where k2 = −E) now, for 1-dimension 1 Z dk 1 (2π) K2 + k2 −∞ Let k = K tan θ, then dk = K sec2 θdθ 1 π=2 Z dk 1 Z π = dθ = ; k > 0 K2 + k2 K K −∞ −π=2 So, 1 Z 1 dk 1 = 2π K2 + k2 2K −∞ 9 This means, 1 1 + = 0 2k λ which gives, λ k = − 2 λ2 E = − B 4 since, k > 0, so, λ < 0.
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