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Rotating Field

This is defined as for constant field in \bgroup\color{col1}$ z$\egroup direction and an oscillating field in the \bgroup\color{col1}$ x$\egroup- \bgroup\color{col1}$ y$\egroup plane,
$\displaystyle B_z(t)=B_0={\rm const},\quad B_{\Vert}(t) = B_1e^{i\omega t}.$     (2.10)

Our equation for \bgroup\color{col1}$ \psi_2$\egroup thus becomes
$\displaystyle i \ddot{\psi}_2$ $\displaystyle =$ $\displaystyle i\frac{\dot{B_{\Vert}}}{B_{\Vert}} \dot{\psi}_2 + \left[ \frac{\d...
...}}{B_{\Vert}} B_z
-i B_z^2 -i \vert B_{\Vert}\vert^2 - \dot{B}_z \right] \psi_2$  
  $\displaystyle =$ $\displaystyle -\omega \dot{\psi}_2 + i[\omega B_0 - B_0^2 - \vert B_1\vert^2 ]\psi_2$  
$\displaystyle \leadsto \ddot{\psi}_2$ $\displaystyle -$ $\displaystyle i\omega \dot{\psi}_2 + [B_0^2 + \vert B_1\vert^2 - \omega B_0]\psi_2=0.$ (2.11)

This can be solved using the exponential ansatz method \bgroup\color{col1}$ \psi_2(t)= ce^{-izt}$\egroup which yields a quadratic equation for \bgroup\color{col1}$ z$\egroup,
    $\displaystyle z^2 - \omega z + [\omega B_0 -B_0^2 - \vert B_1\vert^2] =0$  
$\displaystyle z_{\pm}$ $\displaystyle =$ $\displaystyle \frac{\omega}{2}\pm \frac{1}{2}\sqrt{\omega^2 + 4B_0^2 -4\omega B_0 + 4 \vert B_1\vert^2}
= \frac{\omega}{2}\pm \frac{1}{2} \Omega_R$  
$\displaystyle \Omega_R$ $\displaystyle \equiv$ $\displaystyle \sqrt{ (\omega-2B_0)^2+ 4 \vert B_1\vert^2}$   Rabi-frequency$\displaystyle .$ (2.12)

Note that the term \bgroup\color{col1}$ 2B_0$\egroup in the Rabi-frequency is determined by the level-splitting \bgroup\color{col1}$ \Delta\equiv 2B_0$\egroup in absence of the time-dependent field \bgroup\color{col1}$ B_{\Vert}(t)$\egroup. The solution for \bgroup\color{col1}$ \psi_2(t)$\egroup (from which \bgroup\color{col1}$ \psi_1(t)$\egroup follows immediately) therefore is
$\displaystyle \psi_2(t)$ $\displaystyle =$ $\displaystyle c_1 e^{i\left( \frac{\omega}{2}+ \frac{\Omega_R}{2}\right)t}
+ c_2 e^{i\left( \frac{\omega}{2}- \frac{\Omega_R}{2}\right)t}$  
  $\displaystyle =$ $\displaystyle e^{i\frac{\omega}{2}t}\left[ c'_1 \cos \frac{\Omega_R}{2} t +
c'_2 \sin \frac{\Omega_R}{2} t\right].$ (2.13)

We can choose, e.g., the initial condition \bgroup\color{col1}$ \psi_2(0)=1$\egroup from which follows
$\displaystyle \psi_2(t)$ $\displaystyle =$ $\displaystyle e^{i\frac{\omega}{2}t} \left[\cos \frac{\Omega_R}{2} t
+ c'_2 \sin \frac{\Omega_R}{2} t\right]$  
$\displaystyle 0=\psi_1(0)$ $\displaystyle =$ $\displaystyle \left.\frac{i\dot{\psi}_2 + B_z \psi_2}{ B_{\Vert}}\right\Vert _{t=0}=
\frac{-\frac{\omega}{2} + i\frac{\Omega_R}{2}c'_2 + B_0 }{B_1}$  
$\displaystyle \leadsto c'_2$ $\displaystyle =$ $\displaystyle -i \frac{\omega-2B_0}{\Omega_R}$ (2.14)

This leads to
$\displaystyle \vert \psi_2(t)\vert^2$ $\displaystyle =$ $\displaystyle \cos ^2\frac{\Omega_R}{2} t + \frac{(\omega-2B_0)^2}{\Omega_R^2}
\sin ^2\frac{\Omega_R}{2} t$ (2.15)
  $\displaystyle =$ $\displaystyle \frac{(\omega-2B_0)^2}{\Omega_R^2} + \frac{4\vert B_1\vert^2}{\Omega_R^2}
\cos ^2\frac{\Omega_R}{2} t$   Rabi-Oscillations$\displaystyle .$  

Note that the quantum-mechanical oscillations at constant \bgroup\color{col1}$ {\bf B}$\egroup (e.g., the case \bgroup\color{col1}$ {\bf B}=(T_c,0,0)$\egroup in Eq. (VI.1.18)) occur for a time-independent Hamiltonian. The Rabi-oscillations occur in a time-dependent Hamiltonian containing a time-dependent term (`time-dependent field'). These two often get mixed up in the literature.


next up previous contents index
Next: Landau-Zener-Rosen problem Up: Spin in Magnetic Field Previous: Constant   Contents   Index
Tobias Brandes 2005-04-26