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\exnumber{4853}
\heading{Dynamics Of Two Coupled Spins} 
\auname{Shimon Haver and Gideon Ilani}



{\bf The problem:}

\Dn

Two coupled spins are described by the Hamiltonian, ${ \mathcal{H} =\mu {\bf S}^A\cdot {\bf S} ^B+hS^{A}_{z} }$, when ${ {\bf S}=\frac{1}{2} {\boldsymbol \sigma} }$. The first spin was prepared in ${ Z }$ polarization.
We will focus on the movemnt of the second spin.
The spin state is described by a polarization vector ${ {\bf M}=(\langle\sigma^{B}_{x}\rangle,\langle\sigma^{B}_{y}\rangle,\langle\sigma^{B}_{z}\rangle) }$.
\Dn

(1) Write the Hamiltonian of the system in the standard basis.
\Dn

The second spin is prepared in $X$ polarization.
\Dn

(2) What will be his frequency in the limit $h\to\infty$ ?

(3) What will be his frequency in the limit $h\to0$ ?
\Dn

Now the second spin is prepared in $-Z$ polarization (spin down).
\Dn

(4) What will be his frequency ? $\mu,\ h$ are given.

(5) Write an expression for the polarization vector ${ {\bf{M}}(t) }$.

(6) Write a condition so that for all times ${ {\bf M}(t) \neq 0}$. 
\Dn\Dn

In section 5 it is possible to use known expressions for spin precession (Rabi formula).

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1)
Our problem is 4 dimensional, with the standard basis ${ |m^{A}\rangle \otimes |m^{B}\rangle }$.
\[
\left|\uparrow\uparrow \right\rangle=\begin{pmatrix} 1\\ 0\\ 0\\ 0 \end{pmatrix};\  \left|\uparrow\downarrow\right\rangle=\begin{pmatrix} 0\\ 1\\ 0\\ 0 \end{pmatrix};\  \left|\downarrow \uparrow\right\rangle=\begin{pmatrix} 0\\ 0\\ 1\\ 0 \end{pmatrix};\  
\left|\downarrow\downarrow \right\rangle=\begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix}
\]
 To find the Hamiltonian in the standard basis it's usefull to rewrite it as:
\[
 \mathcal{H} = \frac{ \mu}{2} (( {\bf S } ^A +{\bf S } ^B)^2-({\bf S } ^B)^2-({\bf S } ^A)^2)+hS ^{A}_{z}
\]
 which is more easy to handl,  since for ${ ( {\bf S } ^A +{\bf S } ^B)^2 = {\bf J}^{2}}$ we know the matrix (lecture notes 36.1), and ${ \left({\bf S } ^B\right) ^2,\ \left( {\bf S } ^A \right)^2 }$ , are still  diagonal in this basis.  in this basis the Hamiltonian is: 
\[
\mathcal{H} = \frac{\mu}{4}\cdot
\begin{pmatrix}
1 &  &  & \\
  & -1 & 2 & \\
  & 2 &-1 & \\
  &  &   & 1\end{pmatrix}
+\frac{h}{2}\cdot
\begin{pmatrix}
1 &  &  & \\
  & 1 &  & \\
 &   &-1 & \\
  &  &   & -1 \end{pmatrix}
\]


%%%%%%%%%%%%%%%
\Dn

(2)
Because the spin of A is up, when $h\to\infty$, it will not change during the system evolution. That means that all the evolution is occuring in the $2\times2$ subspace of ${ \left|\uparrow,m\right\rangle }$.
If we will set the energy zero to $\frac{h}{2}$  the effective Hamiltonian will be:
\[
\mathcal{H} =  \frac{ \mu}{4} \cdot \begin{pmatrix}
1& \\
 &-1 \end{pmatrix} 
\]
which can be written as $ \mathcal{H} = \frac{\mu}{2}\cdot S^{B}_{z}$. Any two levels system Hamiltonian can be put in the form $ \mathcal{H} = \boldsymbol{\Omega} \cdot {\bf {S}}$, so by comparsion to the latter expression we find that:
\[
\boldsymbol{\Omega} = \left( 0, 0, \frac{ \mu}{2} \right);\  \left| \boldsymbol{\Omega} \right|=\frac{\mu}{2}
\]
	
%%%%%%%%%%%%%%
\Dn

(3)
By observing the Hamiltonian matrix form we find that only two states are coupled $\left| \uparrow,\downarrow \right\rangle$, $\left| \downarrow,\uparrow \right\rangle$, we can write the effective Hamiltonian for this subspace as:
\[
\mathcal{H} =- \mathbb{I} +\mu \cdot S_x
\]
notice that here ${   S_x }$ is a ${ 2 \times2 }$  matrix defined on our subspace.
The system frequency will be:

$\left| \boldsymbol{\Omega} \right|= \mu$


%%%%%%%%%%%%%
\Dn

(4)
From the Hamiltonian it is clear that the initial  state of the system ${ \left| \uparrow \downarrow \right\rangle }$ is still coupled only to ${ \left| \downarrow \uparrow \right\rangle }$, so considering the subspace of these two states the Hamiltonian is now (after redefining the energy floor):
\[
 \mathcal{H} = \mu S_{x} +h S_z
\]
The frequency of the system is ${ \boldsymbol{\Omega} = \left(  \mu , 0 ,h \right) ;\   \left|  \boldsymbol{\Omega} \right| = \sqrt{ \mu^2+h^2 }  }$.

%%%%%%%%%%%%%
\Dn

(5)
We are dealing with a two level system so the time evolution can be written with the help of Rabi formula for the survival probabilty ${P(t)=1-\sin^{2}(\theta)\sin^{2}(\frac{\Omega t}{2}) }$, when ${ \theta = \arctan( \frac{ \mu }{h}) }$, and ${ \Omega}$ is the system frequncy that  was found  in the previuos section:
\[
|\psi_0\rangle=|\uparrow \downarrow\rangle \implies |\psi_t\rangle = P(t)|\uparrow \downarrow\rangle+(1-P(t))| \downarrow \uparrow \rangle
\]
Now we can calculate the polarization vector in the usual way. we can simplify the algebra by observing that by definition ${  \boldsymbol{\sigma} ^{B} }$ is acting only on ${\left| m^{B}\right\rangle  }$:
\[
 \sigma^{B}_{x} | \psi_{t} \rangle=  P(t)|\uparrow \uparrow \rangle+(1-P(t))| \downarrow \downarrow \rangle
\]
so obviously:
\[
\langle \psi_{t} | \sigma^{B}_{x} | \psi_{t} \rangle = 0
\]
with similar calculation one find's that ${ \langle \psi_{t} | \sigma^{B}_{y} | \psi_{t} \rangle = 0 }$. for ${ \sigma^{B}_{z} }$:
\[
\sigma^{B}_{z} |\psi_{t}\rangle =  -P(t)|\uparrow \downarrow\rangle+(1-P(t))| \downarrow \uparrow \rangle
\]
so:
\[
\langle \psi_{t} | \sigma^{B}_{z} | \psi_{t} \rangle =(1-P(t))^2-P(t)^2=1-2P(t)
\]
so the polarization vector is:
\[
{\bf{M}} (t) = (0,0,1-2P(t))
\]
%%%%%%%%%%%%%
\Dn

(6)
In order of the spin not to get to 0 we demand that the polariztion vector wil not get null which means:
\[
1-2P(t) < 0 \implies P(t) > \frac{1}{2}
\]
this is true because ${ M_{z} (0) = -1 }$ and it changes continuosly. this condition implies that:
\[
1-\sin^{2}(\theta)\sin^{2}(\frac{\Omega t}{2})  >\frac{1}{2}
\] 
and because ${ \sin^{2}(\frac{\Omega t}{2}) \in \{0,1\} }$ this means that:
\[
| \sin(\theta)| < \frac{1}{\sqrt{2}}; |\theta| < \frac{ \pi}{4}
\]
inserting the expression for ${ \theta }$ the condition is:  \[  |\arctan(\frac{ \mu}{h})| <  \frac{ \pi}{4}  \] wihch means:
\[
\left| \frac{\mu}{h} \right| <1 \implies | \mu| < |h|
\] 
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\end{document}
