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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2010 HOMEWORK ASSIGNMENT 11: Solutions Topics covered: Scattering amplitude, differential cross-section, scattering probabilities. 1. [5 pts] Using only the definition, G = ( E- H + i )- 1 , show that the free-space Green’s function is the solution to E + ~ 2 2 M ∇ 2 ~ r G ( ~ r,~ r ) = δ 3 ( ~ r- ~ r ) . (1) The purpose of this problem is just to establish the equivalence between our operator-based approach, and the standard Green’s function formalism encountered, e.g., in classical EM. According to it’s definition, we must have: [ E- H + i ] G = I. (2) Hitting from the left with < ~ r | and from the right with | ~ r i then gives: h ~ r | [ E- H + i ] G | ~ r i = h ~ r | ~ r i . (3) Using H = 1 2 M P 2 and taking → 0 then gives E + ~ 2 2 M ∇ 2 ~ r G ( ~ r,~ r ) = δ 3 ( ~ r- ~ r ) . (4) 1 2. If we define the operator F via f ( ~ k , ~ k ) = h ~ k | F | ~ k i , then it follows that F =- (2 π ) 2 M ~ 2 T , where T is the T-matrix operator. In principle, one would like to deduce the form of the potential V from scattering data. First, derive an expression for the operator V in terms of the operators G and T only. In preparation for problem 11.4, use this expression for V to prove that the full Green’s function, G = ( E- H- V + i )- 1 is related to the background Green’s function, G via the simple relation: G = G + G TG . (5) (Hint: don’t forget that order matters in operator inversion ( AB )- 1 = B- 1 A- 1 .) The relationship between T , V , and G is T = (1- V G )- 1 V. (6) Operating from the left with (1- V G ) then gives (1- V G ) T = V. (7) Multiply out the l.h.s. to get T- V G T = V (8) Putting all terms containing V on the r.h.s. gives T = V + V G T = V (1 + G T ) . (9) Operate from the right with (1 +...
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- Spring '11
- Work, lim, Trigraph, Pallavolo Modena, bound state, G0