Time Development of a Spin \bgroup\color{black}${1\over 2}$\egroup State in a B field

Assume that we are in an arbitrary spin state \bgroup\color{black}$\chi(t=0)=\pmatrix{a\cr b}$\egroup and we have chosen the z axis to be in the field direction. The upper component of the vector (a) is the amplitude to have spin up along the z direction, and the lower component (b) is the amplitude to have spin down. Because of our choice of axes, the spin up and spin down states are also the energy eigenstates with energy eigenvalues of \bgroup\color{black}$\mu_BB$\egroup and \bgroup\color{black}$-\mu_BB$\egroup respectively. We know that the energy eigenstates evolve with time quite simply (recall the separation of the Schrödinger equation where \bgroup\color{black}$T(t)=e^{-iEt/\hbar}$\egroup). So its simple to write down the time evolved state vector.

\begin{displaymath}\bgroup\color{black}\chi(t)=\pmatrix{ae^{-i\mu_BBt/\hbar}\cr ...
=\pmatrix{ae^{-i\omega t}\cr be^{i\omega t}} \egroup\end{displaymath}

where \bgroup\color{black}$\omega={\mu_BB\over\hbar}$\egroup.

So let's say we start out in the state with spin up along the x axis, \bgroup\color{black}$\chi(0)=\pmatrix{{1\over\sqrt{2}}\cr {1\over\sqrt{2}}}$\egroup. We then have

\chi(t) & = & \pmatrix{{1\over\sqrt{2}}e^{-i\omega t}\cr {1\ov...
...+e^{-2i\omega t}\right)
={\hbar\over 2}\cos(2\mu_BBt/\hbar) \\

So again the spin precesses around the magnetic field. Because \bgroup\color{black}$g=2$\egroup the rate is twice as high as for \bgroup\color{black}$\ell=1$\egroup.

Jim Branson 2013-04-22