A solved problem for spins

Source: Introduction to Quantum Mechanics,D. Griffiths,Problem 4-34.

Problem: 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:

(a)

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

(b)

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

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

${\dot {\alpha }}={\frac {i\mu B_{0}}{2}}Cos(\omega t)\alpha \Rightarrow {\frac {\mathrm {d} \alpha }{\alpha }}={\frac {i\mu B_{0}}{2}}Cos(\omega t)dt\Rightarrow Ln\alpha ={\frac {i\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}+constant.$

$\alpha (t)=Ae^{{\frac {i\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}};\alpha (0)=A={\frac {1}{\sqrt {2}}}$, so $\alpha (t)={\frac {1}{\sqrt {2}}}e^{{\frac {i\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}$

${\dot {\beta }}=-i{\frac {\mu B_{0}}{2}}Cos(\omega t)\beta \Rightarrow \beta (t)={\frac {1}{\sqrt {2}}}e^{-i{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}\Rightarrow \chi (t)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}e^{i{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}\\e^{-i{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}\end{pmatrix}}$

(c)

$c_{-}^{(x)}=\chi _{-}^{(x)T}\chi ={\frac {1}{2}}{\begin{pmatrix}1&-1\end{pmatrix}}{\begin{pmatrix}e^{i{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}\\e^{-i{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}\end{pmatrix}}={\frac {1}{2}}[e^{i{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}-e^{-i{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}}]=iSin[{\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }}]$

$P_{-}^{(x)}(t)=\left|c_{-}^{(x)}\right|^{2}=Sin^{2}({\frac {\mu B_{0}}{2}}{\frac {Sin(\omega t)}{\omega }})$

(d)

The argument of $Sin^{2}$ must reach ${\frac {\pi }{2}}$ (so P=1)$\Rightarrow {\frac {\mu B_{0}}{2\omega }}={\frac {\pi }{2}}$ , or $B_{0}={\frac {\pi \omega }{\mu }}.$