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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2010 HOMEWORK ASSIGNMENT 11: Solutions Topics covered: Scattering amplitude, differential crosssection, scattering probabilities. 1. [5 pts] Using only the definition, G = ( E H + i ) 1 , show that the freespace 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 operatorbased 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 Tmatrix 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
 Moore
 Work, lim, Trigraph, Pallavolo Modena, bound state, G0

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