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\exnumber{7212}
\heading{Adiabatic Transfer Along Chain (2019A)} 
\auname{Tomer Stolik and Noam Liron}

{\bf The problem:}

\Dn

In this problem we assume a chain with 5 states x=1,2,3,4,5 with jump frequencies between adjacent states are (a,b,a,b) where a=Csin($\theta$), b=Ccos($\theta$).\newline
The potential energy in all states is 0, beside the $2^{nd}$ state which is V.\newline There is no possibility to jump between states 1 and 5 directly (non-cyclic).\newline
The system starts with one particle in the first state and then the value of $\theta$ starts to change "slowly" from the value of $\theta=0$ to the value $\theta=\frac{\pi}{2}$

\Dn

(1) For V=0, write the matrix R that represents a gauge action so
$R^{-1}\mathcal{H}R=-\mathcal{H}$ \newline
Therefore, the Hemiltonian has an eigenstate with the eigenvalue of 0.\newline

(2) For V=0 write in the standard base the adiabatic state that is implied by the initial condition. Where will the particle be at the end of the adiabatic process? does the answer depend on the value of the potential barrier V?\newline

(3) For V=0 write the energies for $\theta=0$ and $\theta=\frac{\pi}{4}$.\newline
Hint: notice that for the $2^{nd}$ case, the problem is the same as a potential well in length of L=6.\newline

(4) Repeat section 3 for $V\gg C$\newline
In this case, the leading order answer will be enough.
This section demands a diagonalizing of a 3X3 matrix in order to set the energies. The correctness of the numerical coefficient in the result is important.\newline

(5) Draw a schematic graph of the energies as a function of $\theta$ for V=0.\newline
Draw a schematic graph of the energies as a function of V for $\theta=0$.\newline
Make sure that the graph will be clear.\newline

(6) What is the adiabatic condition on $\dot{\theta}$?\newline
solve for the cases  V=0 and the case $V\gg C$\newline
You can ignore numerical factors.\newline

Hint: without solving sections 1 and 2, you can still successfully handle with the other sections.

\Dn\Dn

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

\Dn
\text The Hamiltonian describing the system is: 

\begin{equation*}
\mathcal{H}\mapsto
\begin{pmatrix}
0&Csin(\theta)&0&0&0 \\
Csin(\theta)&V&Ccos(\theta)&0&0 \\
0&Ccos(\theta)&0&Csin(\theta)&0 \\
0&0&Csin(\theta)&0&Ccos(\theta) \\
0&0&0&Ccos(\theta)&0 \\
\end{pmatrix}
\end{equation}
\newline
\newline
(1) For V=0 we are looking for a matrix R that will:\newline
multiply (by -1) the $2^{nd}$ and $4^{th}$ rows, when multiplying from the left side (the inverse), and also multiply (by -1) the $2^{nd}$ and $4^{th}$ columns, when multiplying from the right side.\newline
We will choose the following R:\newline
\begin{equation*}
R\mapsto
\begin{pmatrix}
1&0&0&0&0 \\
0&-1&0&0&0 \\
0&0&1&0&0 \\
0&0&0&-1&0 \\
0&0&0&0&1 \\
\end{pmatrix}
\end{equation}
Note that $R=R^{-1}$ and $R^{-1}\mathcal{H}R=-\mathcal{H}$ as requested.
\newline
Therefore, $\mathcal{H}$ and $-\mathcal{H}$ has the same eigenvalues, so for each eigenvalue E of $\mathcal{H}$, there is a complete eigenvalue -E. Since there are odd number of eigenvalues, at least one of them must be 0.
\newline
\newline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(2) In adiabatic processes the particles remain with the same initial energy.\newline
For V=0, and the initial condition ( $\Psi=|1\rangle$ and $\theta =0$ ) we get:
\begin{equation*}
\mathcal{H}
\cdot
\Psi
=
  \left(\begin{array}{@{}ccccc@{}}
0&0&0&0&0 \\
0&0&C&0&0 \\
0&C&0&0&0 \\
0&0&0&0&C \\
0&0&0&C&0 \\
  \end{array}\right)
  \cdot
  \begin{pmatrix}
1 \\0 \\0 \\0 \\0 \\
\end{pmatrix}
=0\cdot \Psi
\end {equation*}
Therefore, the state $|1\rangle$ in time t=0 is an eigenstate of $\mathcal{H}$ with the eigenvalue of E=0.\newline
In order to find the adiabatic state we must demand that the energy remains 0:
\begin{equation*}
\mathcal{H}\Psi=
\begin{pmatrix}
0&Csin(\theta)&0&0&0 \\
Csin(\theta)&0&Ccos(\theta)&0&0 \\
0&Ccos(\theta)&0&Csin(\theta)&0 \\
0&0&Csin(\theta)&0&Ccos(\theta) \\
0&0&0&Ccos(\theta)&0 \\
\end{pmatrix}
\cdot\begin{pmatrix}
a \\b \\c \\d \\e \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\0 \\0 \\0 \\0 \\
\end{pmatrix}
\end{equation}
\newline
\end {equation}
We get 5 equations that must be true for each value of $\theta$ \newline
i.  b$\cdot$ sin($\theta$)=0\newline
ii.  a$\cdot$ sin($\theta$)+c$\cdot$ cos($\theta$)=0\newline
iii.  b$\cdot$ cos($\theta$)+d$\cdot$ sin($\theta$)=0\newline
iv.  c$\cdot$ sin($\theta$)+e$\cdot$ cos($\theta$)=0\newline
v.  d$\cdot$ cos($\theta$)=0\newline
With the choose of a=1 (for non-normalized state) we get:
\begin{equation*}
|\Psi(\theta)\rangle \mapsto
\begin{pmatrix}
1 \\0 \\-tg(\theta) \\0 \\tg^2(\theta) \\
\end{pmatrix}
\end {equation}
\newline
Note that the state $|2\rangle$ is always 0, so the partical has 0 probability to get there. The answer for $\Psi(\theta)$ stays the same for any value of V.
\newline
At the end of the process $\theta=\frac{\pi}{2}$:
\begin{equation*}
\mathcal{H}\mapsto
\begin{pmatrix}
0&C&0&0&0 \\
C&0&0&0&0 \\
0&0&0&C&0 \\
0&0&C&0&0 \\
0&0&0&0&0 \\
\end{pmatrix}
\end{equation}
For V=0 the $\mathcal{H}(\theta=\frac{\pi}{2})$ is symetric to $\mathcal{H}(\theta=0)$, so it is easy to see that state $|5\rangle$ has the eigenenergy of 0. Therefore, at the end of the adiabatic process, the particle is at state $|5\rangle$.
\newline
From another point of view, when we normalized $\Psi(\theta)$ and take the limit $\theta \longrightarrow \frac{\pi}{2}$ we get that $\Psi(\theta) = |5\rangle$.
\newline
The value of V has no effect here since the particle dose not get there during the process.
\newline
\newline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(3) For V=0 and $\theta=0$ we get:
\newline
\begin{equation*}
\mathcal{H}\mapsto
  \left(\begin{array}{@{}c|cc|cc@{}}
0&0&0&0&0 \\ \hline
0&0&C&0&0 \\
0&C&0&0&0 \\ \hline
0&0&0&0&C \\
0&0&0&C&0 \\
  \end{array}\right)
\end {equation*}

\newline
We can diagonalize each 2X2 block separately according to the symmetric and a-symmetric eigenvectors:\newline
\begin{equation*}
|+\rangle \mapsto \frac1{\sqrt{2}} 	
\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}
\text{, }
|-\rangle \mapsto \frac1{\sqrt{2}} 	
\begin{pmatrix}
1 \\
-1 \\
\end{pmatrix}
\end {equation}
And then we get the full eigenvectors base:
\newline
\begin{equation*}
|1\rangle \mapsto
\begin{pmatrix}
1 \\0 \\0 \\0 \\0 \\
\end{pmatrix}
\text{, }
|2\rangle \mapsto
\frac1{\sqrt{2}} 	
\begin{pmatrix}
0 \\1 \\-1 \\0 \\0 \\
\end{pmatrix}
\text{, }
|3\rangle \mapsto
\frac1{\sqrt{2}} 	
\begin{pmatrix}
0 \\1 \\1 \\0 \\0 \\
\end{pmatrix}
\text{, }
|4\rangle \mapsto
\frac1{\sqrt{2}} 	
\begin{pmatrix}
0 \\0 \\0 \\1 \\-1 \\
\end{pmatrix}
\text{, }
|5\rangle \mapsto
\frac1{\sqrt{2}} 	
\begin{pmatrix}
0 \\0 \\0 \\1 \\1 \\
\end{pmatrix}
\end {equation}
With the eigenvalues: $E_1=0$, $E_2=-C$, $E_3=C$, $E_4=-C$, $E_5=C$
\newline
\newline
For V=0 and $\theta= \frac\pi{4}$ we get:\newline
\begin{equation*}
\mathcal{H}\mapsto\frac C{\sqrt{2}}
\begin{pmatrix}
0&1&0&0&0 \\
1&0&1&0&0 \\
0&1&0&1&0 \\
0&0&1&0&1 \\
0&0&0&1&0 \\
\end{pmatrix}
\end{equation}
By diagonalizing the matrix we get:
\newline
\begin{equation*}
|1\rangle \mapsto
\frac1{\sqrt{3}} 	
\begin{pmatrix}
1 \\0 \\-1 \\0 \\1 \\
\end{pmatrix}
\text{, }
|2\rangle \mapsto
\frac1{3} 	
\begin{pmatrix}
1 \\ \sqrt{3} \\2 \\ \sqrt{3} \\1 \\
\end{pmatrix}
\text{, }
|3\rangle \mapsto
\frac1{3} 	
\begin{pmatrix}
1 \\ -\sqrt{3} \\2 \\ -\sqrt{3} \\1 \\
\end{pmatrix}
\text{, }
|4\rangle \mapsto
\frac1{2} 	
\begin{pmatrix}
-1 \\-1 \\0 \\1 \\1 \\
\end{pmatrix}
\text{, }
|5\rangle \mapsto
\frac1{2} 	
\begin{pmatrix}
-1 \\1 \\0 \\-1 \\1 \\
\end{pmatrix}
\end {equation}
With the eigenvalues:\newline
$E_1=0$ , $E_2=\sqrt{\frac3{2}}C$, $E_3=-\sqrt{\frac3{2}}C$, $E_4=\sqrt{\frac1{2}}C$, $E_5 = -\sqrt{\frac1{2}}C$
\newline \newline
Another option to solve this question is to notice that we are looking on a discrete version of a potential well with L=6. The 5 values of each vector represent 5 samples of the continues state with equal length between each 2 samples.
\newline
Reminder: the eiganstates of a potential well between in the range [-L/2,L/2] are $sin(\frac{2\pi n}{L}x)$ and $cos(\frac{2\pi (n+\frac{1}{2})}{L}x)$.\newline
Therefore, the non-normalized eigenstate will be the samples of those wave functions in the points x= 2,1,0,-1,-2. For each continues wave function we sampled the function and got:
\begin{equation*}
cos(\frac{\pi}{6}x) \mapsto
\begin{pmatrix}
\frac{1}{2} \\\frac{\sqrt{3}}{2} \\1 \\\frac{\sqrt{3}}{2} \\\frac{1}{2} \\
\end{pmatrix}
\text{, }
cos(\frac{3\pi}{6}x) \mapsto
\begin{pmatrix}
-1\\0\\1\\0\\-1
\end{pmatrix}
\text{, }
cos(\frac{5\pi}{6}x) \mapsto
\begin{pmatrix}
\frac{1}{2} \\\frac{-\sqrt{3}}{2} \\1 \\\frac{-\sqrt{3}}{2} \\\frac{1}{2} \\
\end{pmatrix}
\text{, }
sin(\frac{2\pi}{6}x) \mapsto
\begin{pmatrix}
\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}\\0\\\frac{-\sqrt{3}}{2}\\\frac{-\sqrt{3}}{2}\\
\end{pmatrix}
\text{, }
sin(\frac{4\pi}{6}x) \mapsto
\begin{pmatrix}
\frac{-\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}\\0\\\frac{-\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}\\
\end{pmatrix}
\end {equation}
Note: for higher orders of n, we will get the same answers as we got, since there are no more that 5 orthogonal states in a $5^{th}$ order matrix.\newline
As you can see, each state from the potential well states, has a matching state from the normalized eigenstates. So, they have the same eigenvalues.\newline
Note: it is possible to solve it also for x in range [0,L] and use only sin() functions.
\newline
\newline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(4) For $V\gg C$ and $\theta=0$: \newline
\[
    \mathcal{H}\mapsto
  \left(\begin{array}{@{}c|cc|cc@{}}
0&0&0&0&0 \\ \hline
0&V&C&0&0 \\
0&C&0&0&0 \\ \hline
0&0&0&0&C \\
0&0&0&C&0 \\
  \end{array}\right)
\]
Let us observe each block separately: \newline
The lower block can be diagonalize with the standard  $|+\rangle$, $|-\rangle$ states:
\newline
\[
base:
\begin{pmatrix}
1&0&0&0&0\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\
0&0&0&\frac1{\sqrt{2}}&-\frac1{\sqrt{2}}\\
\end{pmatrix}
\Longrightarrow
\mathcal{H}\mapsto
\begin{pmatrix}
0&0&0&0&0 \\
0&V&C&0&0 \\
0&C&0&0&0 \\
0&0&0&C&0 \\
0&0&0&0&-C \\
\end{pmatrix} \]
The upper block, state $|1\rangle$ is not coupled with any other state.\newline
For the middle block, we get the eigenvalues using perturbation theory: \newline
$E_1 = 0$ \newline
$E_2 = V^{[0]}+0^{[1]}+\frac{C^2}{V-0}^{[2]} \approx V$  \newline
$E_3 = 0^{[0]}+0^{[1]}+\frac{C^2}{0-V}^{[2]} \approx \frac{-C^2}{V}$  \newline
$E_4 = C$  \newline
$E_5 = -C$ \newline
\newline
For $V\gg C$ and $\theta$=$\frac{\pi}{4}$:
\newline \newline
$\mathcal{H}\mapsto$ 
\begin{pmatrix}
0&\frac{C}{\sqrt{2}}&0&0&0\\
\frac{C}{\sqrt{2}}&V&\frac{C}{\sqrt{2}}&0&0\\
0&\frac{C}{\sqrt{2}}&0&\frac{C}{\sqrt{2}}&0\\
0&0&\frac{C}{\sqrt{2}}&0&\frac{C}{\sqrt{2}}\\
0&0&0&\frac{C}{\sqrt{2}}&0\\
\end{pmatrix}
\newline
The first step is to choose the base where the matrix
\begin{pmatrix}
0&\frac{C}{\sqrt{2}}&0\\
\frac{C}{\sqrt{2}}&0&\frac{C}{\sqrt{2}}\\
0&\frac{C}{\sqrt{2}}&0\\
\end{pmatrix}
will be diagonolized.
\newline
We can see that this matrix is equal to C\cdot $S_{x}$. We are familiar with the basis $|\Uparrow \rangle$,$|\Updownarrow \rangle$,$|\Downarrow \rangle$ where $S_{z}$ is diagonal, and we can rotate these vectors by $90^o$ around y axis to get the basis where $S_{x}$ would be diagonal: \newline
$|3\rangle$=U($\frac{\pi}{2}$ $n_y$)$|\Uparrow \rangle$ = $\frac{1}{2}$
\begin{pmatrix}
1&-\sqrt{2}&1\\
\sqrt{2}&0&-\sqrt{2}\\
1&\sqrt{2}&1\\
\end{pmatrix}
\begin{pmatrix}
1\\0\\0\\
\end{pmatrix}
=$\frac{1}{2}$
\begin{pmatrix}
1\\$\sqrt{2}$\\1\\
\end{pmatrix}
\newline
$|4\rangle$=U($\frac{\pi}{2}$ $n_y$)$|\Updownarrow \rangle$ = $\frac{1}{2}$
\begin{pmatrix}
1&-\sqrt{2}&1\\
\sqrt{2}&0&-\sqrt{2}\\
1&\sqrt{2}&1\\
\end{pmatrix}
\begin{pmatrix}
0\\1\\0\\
\end{pmatrix}
=$\frac{1}{\sqrt{2}}$
\begin{pmatrix}
-1\\0\\1\\
\end{pmatrix}
\newline
$|5\rangle$=U($\frac{\pi}{2}$ $n_y$)$|\Downarrow \rangle$ = $\frac{1}{2}$
\begin{pmatrix}
1&-\sqrt{2}&1\\
\sqrt{2}&0&-\sqrt{2}\\
1&\sqrt{2}&1\\
\end{pmatrix}
\begin{pmatrix}
0\\0\\1\\
\end{pmatrix}
=$\frac{1}{2}$
\begin{pmatrix}
1\\-$\sqrt{2}$\\1\\
\end{pmatrix}
\newline
By working with the following basis, we get: \newline
\newline
$\mathcal{H}$\mapsto 
\begin{pmatrix}
1&0&0&0&0\\
0&1&0&0&0\\
0&0&\frac{1}{2}&-\frac{1}{\sqrt{2}}&\frac{1}{2}\\
0&0&\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\\
0&0&\frac{1}{2}&\frac{1}{\sqrt{2}}&\frac{1}{2}\\
\end{pmatrix}^T
\begin{pmatrix}
0&\frac{C}{\sqrt{2}}&0&0&0\\
\frac{C}{\sqrt{2}}&V&\frac{C}{\sqrt{2}}&0&0\\
0&\frac{C}{\sqrt{2}}&0&\frac{C}{\sqrt{2}}&0\\
0&0&\frac{C}{\sqrt{2}}&0&\frac{C}{\sqrt{2}}\\
0&0&0&\frac{C}{\sqrt{2}}&0\\
\end{pmatrix}
\begin{pmatrix}
1&0&0&0&0\\
0&1&0&0&0\\
0&0&\frac{1}{2}&-\frac{1}{\sqrt{2}}&\frac{1}{2}\\
0&0&\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\\
0&0&\frac{1}{2}&\frac{1}{\sqrt{2}}&\frac{1}{2}\\
\end{pmatrix}
\newline
=
\begin{pmatrix}
0&\frac{C}{\sqrt{2}}&0&0&0\\
\frac{C}{\sqrt{2}}&V&\frac{C}{2\sqrt{2}}&\frac{-C}{2}&\frac{C}{2\sqrt{2}}\\
0&\frac{C}{2\sqrt{2}}&C&0&0\\
0&\frac{-C}{2}&0&0&0\\
0&\frac{C}{2\sqrt{2}}&0&0&-C\\
\end{pmatrix}
\newline

Now, since we are only looking for the leading order, we can clearly see that 3 energies will be V, C, -C (C and -C are coupled only with V so the next correction will be proportional to $\frac{C^2}{V-C}$ and much smaller than C). For the other 2 energies, we can observe only the relevant parts of the matrix:\newline
\begin{pmatrix}
0&\frac{C}{\sqrt{2}}&0\\
\frac{C}{\sqrt{2}}&V&-\frac{C}{2}\\
0&-\frac{C}{2}&0\\
\end{pmatrix}
\newline
The $3^{rd}$ and the $5^{th}$ rows and columns are not interesting since they are not related to the energies we are looking for. \newline
We must notice that using perturbation theory will not be accurate enough because we have degenerated energies:
\newline
\begin{figure}[htp]
    \centering
    \includegraphics[width=10cm]{pic1.jpeg}
\end{figure}
\newline
\newline
\newline
\newline

Even though the two degenerated states are not directly coupled, they are indirectly coupled through a third state with energy V.\newline
There are various ways to solve the problem such as choosing a different basis or considering the contribution of the indirect hopping (the latter is out of the scope of this course). We will follow the guidance in the question and diagonalize the 3X3 matrix to get: 
\newline
$E_1$=0 \newline
$E_2$=$\frac{V}{2} +\frac{V}{2} \sqrt{1+3\frac{C^2}{V^2}} \approx V$ \newline
$E_3$=$\frac{V}{2} -\frac{V}{2} \sqrt{1+3\frac{C^2}{V^2}} \approx -\frac{3}{4}\frac{C^2}{V}$ \newline
Finall answer: E = $V, C, -C, -\frac{3}{4}\frac{C^2}{V}, 0$
\newline
\newline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newline (5) E($\theta$) for V=0:
\newline
We saw that for $\theta=0$ and $\theta=\frac{\pi}{2}$ we get a symmetric Hemiltonian, and thus the same energies. \newline
In fact the Hemiltonian is symmetric around $\mathcal{H}(\theta=\frac{\pi}{4}$) and there is no degeneracy, so one can infer:\newline
\newline
\begin{figure}[htp]
    \centering
    \includegraphics[width=10cm]{graph1.jpeg}
\end{figure}
\newline
\newline
\newline
for $\theta=0$ and E(V):\newline
The exact solution for the eigenvalues can be achieved by diagonalizing
\begin{pmatrix}
V&C\\
C&0\\
\end{pmatrix}
and we get:
$E_1 =0$, $E_2 =\frac{V}{2}+\sqrt{\frac{V^2}{4}+C^2}$, $E_3 =\frac{V}{2}-\sqrt{\frac{V^2}{4}+C^2}$, $E_4 =C$, $E_5 =-C$\newline

\newline
\begin{figure}[htp]
    \centering
    \includegraphics[width=10cm]{graph2.jpeg}
\end{figure}
\newline
\newline
\newline\newline
\newline
\newline\newline
\newline
\newline
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(6) The adiabatic condition as learned in class is:
\begin{equation*}
\dot{\theta}\ll \frac{\Delta^2}{V_{nm}}
\end{equation*}

Where $\Delta$ is the minimal distance between any 2 energy levels and $V_{nm}$ is the corresponding matrix element. \newline
For V=0:\newline
The energy difference $\Delta\sim$ C so:\newline
\begin{equation*}
\dot{\theta}\ll \frac{C^2}{C}=C
\end{equation}

For V\gg C:

We saw that after losing the degeneracy we get $\Delta\sim\frac{C^2}{V}$ for any value of $\theta$. In order to find $V_{nm}$ we can remember that $V_{nm}=(\frac{d\mathcal{H}}{d\theta})_{nm}$. So, we can observe $\mathcal{H}(\theta=\frac{\pi}{4})$ in the base that diagonalized $\mathcal{H}(\theta=0)$ (or choose any other 2 times) and that will be an approximation for $V_{nm}$.
\newline
Finally, we get $V_{nm}\sim \frac{C^2}{V}$, and:
\newline
\begin{equation*}
\dot{\theta}\ll \frac{C^2}{V}
\end{equation}
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\end{document} 
