ch19 - CHAPTER 19 1 We have M fi 1 3 d re V i.r V(r If V(r...

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CHAPTER 19 1. We have M fi = 1 V d 3 re - i . r V ( r ) If V ( r) = V ( r ), that is, if the p9otential is central, we may work out the angular integration as follows: M fi = 1 V r 2 V ( r ) 0 d rd φ s in θ d θ e - i r cos θ 0 π 0 2 π with the choice of the vector as defining the z axis. The angular integration yields d φ s in θ d θ e - i r co s θ = 2 π d (co s θ ) e - i r co s θ - 1 1 = 4 π r 0 π 0 2 π s in r so that M fi = 1 V 4 π rdrV ( r )s in r 0 Note that this is an even function of that is, it is a function of 2 = ( p f - p i ) 2 / h 2 2. For the gaussian potential M fi = 1 V 4 π V 0 rdr s in r 0 e - r 2 / a 2 Note that the integrand is an even function of r . We may therefore rewrite it as rd r s in r 0 e - r 2 / a 2 = 1 2 rd r s in r -∞ e - r 2 / a 2 The integral on the right may be rewritten as 1 2 rd r s in r -∞ e - r 2 / a 2 = 1 4 i rd re - r 2 / a 2 + i r - c . c ) ( 29 -∞ Now 1 4 i rd r -∞ e - r 2 / a 2 + i r = 1 i ∂∆ d r -∞ e - r 2 / a 2 + i r =- i ∂∆ a π e - a 2 2 /4 = i a 3 π 2 e - a 2 2 /4 Subtracting the complex conjugate and dividing by 4 i gives

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M fi = 1 V a π (29 3 V 0 e - a 2 2 /4 The comparable matrix element for the Yukawa potential is M fi = 1 V 4 π V Y bd r 0 e - r / b s in r = 1 V 4 π V Y b 3 1 + b 2 2 We can easily check that the matrix elements and their derivatives with respect to 2 at = 0 will be equal if a = 2 b and V Y = 2 π V 0 . The differential cross section takes its simplest form if the scattering involves the same particles in the final state as in the initial state. The differential cross section is d σ d = μ 2 4 π 2 h 4 | U ( )| 2 where μ is the reduced mass and U ( ) = VM fi . We are interested in the comparison ( d σ / d ) gauss ( d σ / d ) Yukaw a = e - 2 b 2 2 (1 + b 2 2 ) - 2 = (1 + X ) 2 e - 2 X where we have introduced the notation X = b 2 x 2 . This ratio, as a function of X , starts out at X = 0 with the value of 1, and zero slope, but then it drops rapidly, reaching less than 1% of its initial value when X = 4, that is, at = 2/ b . 3. We use the hint to write d σ d = p 2 π h 2 d σ d 2 = μ 2 4 π 2 h 4 4 π V 0 b 3 1 + b 2 2 2 The total cross section may be obtained by integrating this over 2 with the range given by 0 ≤∆ 2 4 p 2 / h 2 , corresponding to the values of cos θ between –1 and + 1.. The integral can actually be done analytically. With the notation k 2 = p 2 / h 2 the integral is d 2 0 4 k 2 1 (1 + b 2 2 ) 2 = 1 b 2 dx (1 + x ) 2 0 4 k 2 b 2 = 4 k 2 1 + 4 k 2 b 2 This would immediately lead to the cross section if the particles were not identical. For identical particles, there are symmetry problems caused by the Pauli Exclusion Principle and the fact that the protons have spin 1/2. The matrix elements are not affected by the
spin because there is no spin-orbit coupling or any other spin dependence in the potential.

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