in Equation 15.5.6 and expanding to see what you get? It looks like you should get the result they obtain if you work at it using that relation.
#3
jhon
21
0
the second eq II ?? it's equale zero why??
#4
jdwood983
382
0
If you have expanded Equation 15.5.6 using the substitution given, try separating the resulting equation into the real and imaginary components. It looks to me like Equation I in 15.5.7 is the imaginary part and Equation II of 15.5.7 is the real part after substituting for \sin[x].
I don't think that the two parts of 15.5.7 are the real and imaginary parts of 15.5.6, but rather they result from the fact that e^{ikr} and e^{-ikr} are orthogonal functions of r, and hence the relation \alpha e^{ikr}+\beta e^{-ikr}=0[/itex] can only be true for all r if \alpha=\beta=0.