A solved problem for spins

An electron is at rest in an oscillating magnetic field

$B=B_{0}Cos\left(\omega t\right){\hat {k}}$

where $B_{0}$ and $omega$ are constants.

(a) Construct the Hamiltonian matrix for this system.

(b) The electron starts out (at t = 0) in the spin-up state with respect to the x-axis [that is,$\chi ^{0}=\chi _{+}^{(x)})$]. Determine $\chi (t)$ at any subsequent time. Beware.' This is a time-dependent Hamiltonian, so you cannot get $\chi (t)$ in the usual way from stationary states. Fortunately, in this case you can solve the time-dependent Schr/Sdinger equation directly.

(c) Find the probability of getting $-\hbar /2$ if you measure $S_{x}$

(d) What is the minimum field $(B_{0})$ required to force a complete flip in $S_{x}$?

Solution:

$H=-\mu \mathbf {B} .\mathbf {S} =-\mu B_{0}Cos(\omega t)S_{z}=-{\frac {\mu B_{0}\hbar }{2}}Cos(\omega t){\begin{pmatrix}1&0\\0&-1\end{pmatrix}}$

(b) $\chi (t)={\begin{pmatrix}\alpha (t)\\\beta (t))\end{pmatrix}}$ with $\alpha (0)=\beta (0)={\frac {1}{\sqrt {2}}}$

$i\hbar {\frac {\partial \chi }{\partial t}}=i\hbar {\begin{pmatrix}{\dot {\alpha }}\\{\dot {\beta }}\end{pmatrix}}=\mathbf {H} \chi =-{\frac {\mu B_{0}\hbar }{2}}Cos(\omega t){\begin{pmatrix}1&0\\0&-1\end{pmatrix}}$