Question 11.4: Calculate the differential cross section in the first Born a...

As seen in Chapter 7, the spin states of two identical particles with spin s_1=s_2=1  consist of a total of nine states: a quintuplet |2,m〉 (i.e.,|2,\pm 2〉,|2,\pm 1〉,|2,0〉)  and a singlet |0,0〉,   which are symmetric, and a triplet |1,m〉 (i.e.,|1,\pm 1〉,|1,0〉),  which are antisymmetric under particle permutation. That is, while the six spin states corresponding to S = 2 and S = 0 are symmetric, the three S = 1  states are antisymmetric. Thus, if the scattering particles are unpolarized, the differential cross section is

\frac{d\sigma }{d\Omega }=\frac{5}{9}\frac{d\sigma _S}{d\Omega }+\frac{1}{9}\frac{d\sigma _S}{d\Omega }+\frac{3}{9}\frac{d\sigma _A}{d\Omega }=\frac{2}{3}\frac{d\sigma _S}{d\Omega }+\frac{1}{3}\frac{d\sigma _A}{d\Omega },               (11.128)

where

\frac{d\sigma _S}{d\Omega } =\left|f(\theta )+f(\pi -\theta )\right|^2,    \frac{d\sigma _A}{d\Omega } =\left|f(\theta )-f(\pi -\theta )\right|^2.             (11.129)

The scattering amplitude is given in the Born approximation by (11.69) f(\theta )=-\frac{2\mu }{\hbar ^2q}\int_{0}^{\infty }{r^\prime }V(r^\prime )\sin (qr^\prime )dr^\prime , :

=-\frac{4V_0\mu a}{\hbar ^2}\frac{1}{(a^2+q^2)^2} =-\frac{4V_0\mu a }{\hbar ^2}\frac{1}{(a^2+4k^2\sin ^2(\theta /2))^2},               (11.130)

where we have used q=2k\sin (\theta /2),  with  \mu =m/2.   Since \sin [(\pi -\theta )/2]=\cos (\theta /2),   we have

\frac{d\sigma _S}{d\Omega }=\frac{16V^2_0\mu ^2a^2}{\hbar ^4}\left[\frac{1}{(a^2+4k^2\sin ^2(\theta /2))^2}+\frac{1}{(a^2+4k^2\cos ^2(\theta /2))^2} \right] ^2 ,               (11.131)

\frac{d\sigma _A}{d\Omega }=\frac{16V^2_0\mu ^2a^2}{\hbar ^4}\left[\frac{1}{(a^2+4k^2\sin ^2(\theta /2))^2}-\frac{1}{(a^2+4k^2\cos ^2(\theta /2))^2} \right] ^2 .              (11.132)