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\heading{E1472: Three site system in a row, Dynamics} 
\auname{Arnon Ben-Meir, Tomer Ygael}


{\bf The problem:}

\Dn

A particle in a three site system is described using the standard basis: $|1 \rangle , |2 \rangle , |3 \rangle $, with the Hamiltonian   
\[
\mathcal{H} = \pppmatrix{ 0 & c_a & 0 \cr c_a & 0 & c_b \cr 0 & c_b & 0} ,   \qquad c_a=c\cos(\theta), \;  c_b=c\sin(\theta)
\]

The particle is prepared in the first site at t=0. \\
Express your answer with $ \theta, c $. \\
To solve clause (1) find the roots of the polynomial equation (very simple formula).

\Dn

(1) Find the eigenvalues of the Hamiltonian: $ E_0,E_+, E_-$.

\Dn

(2) Using the standad basis write the eigenstates $|0 \rangle , |+ \rangle , |- \rangle $.

\Dn

(3) Write the intial state in the energy basis.

\Dn

(4) Write an expression for $ P(t) $, the probability of finding the particle in the first site after a time period of t.

\Dn

(5) Write and expression for $ I(t) $, the expectation vaule of the current which is measured between sites $ |2 \rangle $  and  $ |3 \rangle $ at time t.



\Dn\Dn


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{\bf The solution:}

\Dn

(1) Using diagonalization, we solve the equation,
\[ det(\mathcal{H}-\lambda \mathbb{I} )=0 \]
thus finding the eigenvalues:
\[E_0=0 \qquad 	E_\pm =\pm c  \]

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\Dn

(2)
Solving $ (\mathcal{H}-\lambda_i\mathbb{I}) $ we find that the eigenstates of the Hamiltonian are: 
\[|0\rangle = \pppmatrix {\sin\theta \cr 0 \cr -\cos\theta}, \qquad |+\rangle = \frac{1}{\sqrt{2}}\pppmatrix{\cos\theta \cr 1 \cr \sin\theta}, \qquad |+\rangle = \frac{1}{\sqrt{2}}\pppmatrix{\cos\theta \cr -1 \cr \sin\theta}\]

and written in the dirac notation:

\[ |0\rangle  =  (\sin\theta|1 \rangle - \cos\theta|3\rangle )                   \]
\[ |+ \rangle =  \frac{1}{\sqrt{2}} (\cos\theta|1 \rangle +  |2 \rangle + \sin\theta|3\rangle ) \]
\[ |- \rangle=   \frac{1}{2} (\cos\theta|1 \rangle - |2 \rangle + \sin\theta|3\rangle ) \]


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\Dn\Dn

(3)
Expressing the initial state in the energy basis we get:

\[|1\rangle = \sin\theta|0\rangle + \frac{1}{\sqrt{2}}\cos\theta(|+\rangle + |-\rangle) \]

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\Dn

(4)
To find P(t) we use the formula : $ P(t) = |\langle 1 | \psi(t) \rangle|^2 $

\Dn

First we need to define $\psi(t)$

\[\Phi_t = |1\rangle \eexp{-iE_1t} = \sin\theta |0\rangle + \frac{1}{\sqrt{2}} \cos\theta(|+\rangle \eexp{-ict} + |-\rangle \eexp{ict}) \]

then $\psi(t) = \pppmatrix{\langle 1| \Phi_t\rangle \cr \langle 2| \Phi_t\rangle \cr \langle 3| \Phi_t\rangle \cr} $ giving us:

\[\psi(t) = \pppmatrix{\sin^2\theta + \cos^2\theta \cos(ct) \cr \cos\theta (-i\sin(ct)) \cr -\sin\theta \cos\theta (1-\cos(ct)) } \]

\[ P(t) = | \sin^2\theta +\frac{1}{2} \cos^2\theta  \eexp{-ict} +\frac{1}{2 }\cos^2\theta \eexp{ict} |^2 = |\sin^2\theta + \cos^2\theta \cos(ct)|^2 \]


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

(5)
In order to find I(t) we first must find the operator $\hat{I}$ we can do that in two ways :
\begin{enumerate}
\item The current operator is defined as $\hat{I} = \pd{\hat{\mathcal{H}}}{\phi}|_{\phi = 0} $, in order to do that we must place a test flux between sites $ |2\rangle \, and \, |3\rangle $ giving us 

\[\hat{\mathcal{H}} = \pppmatrix{ 0 & c_a & 0 \cr c_a & 0 & c_b\eexp{i\phi} \cr 0 & c_b\eexp{-i\phi} & 0} \]

Using this $ \mathcal{H} \:we\,can\,find\: \hat{I} $:

\[\hat{I} = \pd{\hat{\mathcal{H}}}{\phi}|_{\phi = 0} = i\pppmatrix{ 0 & 0 & 0 \cr 0 & 0 & c_b \cr 0 & -c_b & 0} \] 

\item Using the general site model:

\[
\hat{I}_{i\rightarrow j} = -i \left[\hat{\mathcal{H}}, \hat{P}^i \right] = -i[|j\rangle \mathcal{H}_{ji}\langle i| - |i\rangle \mathcal{H}_{ij}\langle j|] 
\]
\[\hat{I}_{2\rightarrow 3}= i \pppmatrix { 0 & 0 & 0 \cr 0 & 0 & c_b \cr 0 & -c_b & 0}
\]
\end{enumerate}

To find the expectation value of $\hat{I}_{2\rightarrow 3}(t) $:

\[ \langle \psi (t) |\hat{I}| \psi (t) \rangle = \pppmatrix{\psi _1 , & \psi_2 , & \psi_3} \cdot \pppmatrix{ 0 & 0 & 0 \cr 0 & 0 & i c_b \cr 0 & -i c_b & 0} \cdot \pppmatrix{\psi _1  \cr \psi_2  \cr \psi_3} = 2C_b \cos^2\theta \sin\theta \sin(ct) (1-\cos(ct))  \]

With the use of a simple trigonometric identity we finally get:

\[ \langle I(t) \rangle = \frac{1}{2} c |\sin\theta |^2 \sin(ct) (1-\cos(ct)) \]


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