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\exnumber{4622}
\heading{Precession caused by a series of pulses} 
\auname{Guy Zisling and Noam Wunch}


{\bf The problem:}

\Dn

A series of magnetic pulses are applied to a spin ${\frac{1}{2}}$ particle. The Hamiltonian is given by
$\bf{ h(t){\cdot}S}$. The duration of each pulse is ${\tau}$ and the amplitude is ${h}$. The field is in the ${\hat{z}}$ direction during the first pulse and in the ${\hat{y}}$ direction during the second pulse, and so on and so forth. As a result of these pulses we get a precession with frequency ${\Omega}$ about an axis ${\bf n}$. 

\Dn\Dn

(1) Write an expression for the precession frequency ${{\Omega}}$.

(2) Find the rotation axis ${\bf n}$ (unit vector). 

(3) Find ${{\Omega}\rightarrow{\Omega_0}}$ in the limit where the driving force is a series of pulses with $ \tau\rightarrow 0$ .

(4) Find ${ \bf n \rightarrow n_0}$ in the limit where the driving force is a series of pulses with $\tau\rightarrow 0$.

(5) Write an expression for ${\tan{\theta}}$ of the angle between ${\bf n}$ and ${\bf n_0}$.

\Dn\Dn

Remarks:

The precession frequency is defined as ${\Omega=\frac{\phi}{2\tau}}$. Where ${\phi}$ is the equivalent rotation during two consecutive pulses. The answers to questions 3,4 are trivial and can be guessed. The expressions in question 2,5 after simplification include only a "tan" function. 

\Dn\Dn


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{\bf The solution:}

\Dn
(1) 
The evolution operator from this Hamiltonian is a product of alternating rotations around the ${\hat{y}}$ and ${\hat{z}}$ axis. We can group each pair of consecutive rotations and examine their combined rotation.
\Dn
\begin{align*}
U(2\tau)&=\eexp{-ih\tau S_z}\eexp{-ih\tau S_y}=\left[\cos{\left(\frac{h\tau}{2}\right)}\cdot\hat{1}-i\sin{\left(\frac{h\tau}{2}\right)}\cdot\sigma_z\right]\cdot\left[\cos{\left(\frac{h\tau}{2}\right)}\cdot\hat{1}-i\sin{\left(\frac{h\tau}{2}\right)}\cdot\sigma_y\right]\\
&=
\cos^2{\left(\frac{h\tau}{2}\right)}\cdot\hat{1}-i\left[\sin^2{(\frac{h\tau}{2})}\cdot\sigma_x+\cos{(\frac{h\tau}{2})}\sin{(\frac{h\tau}{2})}\cdot(\sigma_y+\sigma_z)\right]\\
&=
\cos{\left(\frac{\phi}{2}\right)}\cdot\hat{1}-i\sin{\left(\frac{\phi}{2}\right)}\cdot{\bf\sigma_n}
\end{align*}
\Dn

We used the fact that any rotation can be expressed as a rotation with angle ${\phi}$ around some axis ${\bf n}$. Examining the coefficient for the unit matrix on both sides of the equation above yields:
\Dn
\[
\phi=2\arccos{\left(\cos^2{\left(\frac{h\tau}{2}\right)}\right)}\rightarrow  
\Omega=\frac{\phi}{2\tau}=\frac{1}{\tau}\arccos{\left(\cos^2{\left(\frac{h\tau}{2}\right)}\right)}
\]
\Dn

(2) The rotation axis is similarly found from identifying ${\bf\hat{n}}$ in the left hand side of the equation above:
\Dn
\[
{\bf\hat{n}}=\frac{1}{\sqrt{\tan^2{(\frac{h\tau}{2})}+2}}
\left(\tan{\left(\frac{h\tau}{2}\right)},1,1\right)
\]
\Dn
\newpage

(3) Small rotations commute and so in the limit ${\tau\rightarrow0}$ the combined rotation simplifies to:
\Dn
\[
U(2\tau)=\eexp{-i\left(h{\tau}S_z+h{\tau}S_y\right)}=\eexp{-i\sqrt{2}h{\tau}\frac{(0,1,1)}{\sqrt{2}}\cdot\bf{S}}=\eexp{-i\phi_0\bf{n_0\cdot S}}
\Dn
\]
\Dn
And so it follows that ${\phi_0=\sqrt{2}h\rightarrow\Omega_0=\frac{1}{\sqrt{2}}h}$.
\Dn

(4) The rotation axis is also found from the expression above:
\Dn
\[
{\bf\hat{n}_0}=\frac{1}{\sqrt{2}}{(0,1,1)}
\Dn
\]
\Dn

(5) The tangent of the angle between two unit vectors is given by the ratio between their cross product and dot product.
\Dn
\[
\Dn
tan{(\theta)}=\frac{{\bf{n_0}\times{n}}}{\bf{n_0{\cdot}n}}=\frac{1}{\sqrt{2}}\tan{\left(\frac{h\tau}{2}\right)}
\]
\Dn


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