NMR population difference from a pulse of known flip angle I want to know how a radiofrequency pulse of known flip angle $\beta$ changes the population difference of a set of isolated spin 1/2 nuclei.

I know that the equilibrium magnetization is related to the population difference by:


  $M_0= \frac{1}{2}\gamma\hbar\Delta{n_{eq}}$


I would like to know how to calculate the x, y, z components of the magnetization after the pulse, and then calculate the population difference from these values.
immediately after the pulse:

$M_z = M_0cos(\beta )$

$M_x = M_0sin(\beta )$

$M_y = 0$

so for example after a 90 degree pulse, all the magnetization has been rotated into the x-y plane, and after a 180 pulse, the magnetization is in the -z direction.