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\heading{ Ex5662: Landau Levels in Grafene} 
\auname{Shmu'el Greenwald and Tom Weiss}


{\bf The problem:}

\Dn

The effective Hemiltonian of an electron in a 2-D layer of Grafene is
 $H=v_{0}\sigma(p-eA)$

when $\sigma=(\sigma_{x},\sigma_{y},\sigma_{z})$ and $A$ is the vector potential for a pependicular magnetic field $B$ in Landau
gauge. Be advised that the standard base for representing the electron is $\mid x,y,m>$, when
$m=\downarrow,\uparrow$.

\Dn

(1) In the lack of a magnetic field, and for a given momentum $p=(p_{x},p_{y})$
what are the eigen energies of the particle?

(2) In Landau gauge $\widehat{Y}=-(1/(eB))\widehat{p_{x}}$ is a constant
of motion.
Write the Hemiltonian $H^{Y}=H_{m,m^{,}}(p_{y},y-Y)$ of after the
seperation of variables. 
Please note that a linear combination of canonical coordinates $a=(sQ+iP/s)/\sqrt{2}$
is a ladder operator, and write the Hemiltonian in the alternative way $H_{m,m^{,}}(a,a^{\dagger})$

 
(3)  Define the operator $C=(H^{Y})^{2}$ and calculate the eigenvalues
$\lambda_{n}=0,1,2...$of $C$. Please note that all the eiegenvalue
except $n$=0 are degenerated. 

(4) Since $C$ is a constant of motion it is possible to perform a
second seperation of variables.Write the $2x2$ matrix representing
$H^{Y,n}$.

(5) Find the energy levels $E_{Y,n,\pm}$ for $n>0$

(6) Write the matrix form of the operator $\hat{I}(t)$ and represent it as a sum of Pauli Matrices. Find the eigenstates and write them in the standard base. Be advised
that the quantum state in the standard base in represented by $\mathbf{\mathcal{\bar{\Psi}}}\rightarrow(\mathcal{\Psi}_{\uparrow}(x,y),\Psi_{\downarrow}(x,y))$.
It is possible to make use of $\varphi^{n}()$ for the eigen functions
of a 1-D harmonic oscilator.


\Dn\Dn


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

\Dn

Preface: 
\Dn
$H=V_{0}\sigma\cdot(p-e\bar{A})$ , $\bar{B}=B\hat{z}\Longrightarrow\{Lnadau-Gauge\}\Longrightarrow\bar{A}=(-By,0,0)$

$\sigma=(\sigma_{x},\sigma_{y})$ , $p=(p_{x},p_{y})\Longrightarrow\sigma\cdot p=\sigma_{x}p_{x}+\sigma_{y}p_{y}$
, $\sigma\cdot e\bar{A}=-eBy\sigma_{x}$

$\Longrightarrow H=v_{0}[\sigma_{x}(p_{x}+eBy)+\sigma_{y}p_{y})=v_{0}\left(\begin{array}{cc}
0 & (p_{x}+eBy-ip_{y})\\
(p_{x}+eBy+ip_{y)} & 0\end{array}\right)$

$\Longrightarrow E=\pm v_{0}(p_{x}^{2}+2p_{x}eBy+(eBy)^{2}+p_{y}^{2})^{1/2}=\pm v_{0}(\bar{p}^{2}+2p_{x}eBy+(eBy)^{2})^{1/2}$

\Dn\Dn

(1) 
for $B=0$$\Longrightarrow E=\pm v_{0}\left|\bar{p}\right|$

%%%%%%%%%%%%%%%
\Dn\Dn
(2)
.$H=V_{0}\sigma(\bar{p}+eBy\hat{x})$ , Let us define $\widehat{Y}=-\widehat{p_{x}}/(eB)$
so that $\left[H,Y\right]=0$

$\Longrightarrow H=v_{0}[\sigma_{x}eB(-Y+y)+\sigma_{y}p_{y})=v_{0}eB\left(\begin{array}{cc}
0 & y-Y-ip_{y}/eB\\
y-Y+ip_{y}/eB & 0\end{array}\right)$

Let us define $Q$ and $P$ as canonical conjugates in the following
manner:

$Q=y-Y$, $P=p_{y}$

and so we can define ladder oppeators $a$ and $a^{\dagger}$in the
following way:

$a=(sQ+iP/s)/\sqrt{2}$ 

$a^{\dagger}=(sQ-iP/s)/\sqrt{2}$

when $s=\sqrt{2eB}$ and the Hemiltonian will take the form: 

$H=v_{0}\sqrt{2eB}\left(\begin{array}{cc}
0 & a^{\dagger}\\
a & 0\end{array}\right)$

\Dn\Dn

(3)
 $[a,a^{\dagger}]=aa^{\dagger}-a^{\dagger}a=1\Longrightarrow aa^{\dagger}=1+a^{\dagger}a$

$C=\left(H^{Y}\right)^{2}=V_{0}^{2}\cdot2eB\left(\begin{array}{cc}
a^{\dagger}a & 0\\
0 & aa^{\dagger}\end{array}\right)=V_{0}^{2}\cdot2eB\left(\begin{array}{cc}
a^{\dagger}a & 0\\
0 & a^{\dagger}a+1\end{array}\right)$ 

Let us define: $N=a^{\dagger}a$ so that $N\mid n>=n\mid n>\Longrightarrow$

$C=V_{0}^{2}\cdot2eB\left(\begin{array}{cc}
n & 0\\
0 & n+1\end{array}\right)$

$u_{n}=\left(\begin{array}{c}
n\\
0\end{array}\right)$ , $u_{n+1}=\left(\begin{array}{c}
0\\
n+1\end{array}\right)$ , $\lambda_{1}=n$ , $\lambda_{2}=n+1$

$E_{n}^{2}=V_{0}^{2}\cdot2eBn$ , $\lambda_{n}=$$n=0,1,2....$ and
$E_{n}$ is of degeneracy 2 for every $n$ but $n=0$

\Dn\Dn

(4)
 $H^{Y}=V_{0}\cdot\sqrt{2eB}\left(\begin{array}{cc}
0 & a\\
a^{\dagger} & 0\end{array}\right)$ , $C=\left(H^{Y}\right)^{2}=V_{0}^{2}\cdot2eB\left(\begin{array}{cc}
a^{\dagger}a & 0\\
0 & a^{\dagger}a+1\end{array}\right)$ so $\left[H,C\right]=0$ 

so $C$ and $H$ share the same eigen states;

$u_{1}=\left(\begin{array}{c}
\shortmid n>\\
0\end{array}\right)$ , $u_{2}=\left(\begin{array}{c}
0\\
\shortmid n+1>\end{array}\right)$

we can rearrange the order of the base elements so that

$C=\left(H^{Y}\right)^{2}=V_{0}^{2}\cdot2eB\left(\begin{array}{cc}
n & 0\\
0 & n+1-1\end{array}\right)=V_{0}^{2}\cdot2eB\left(\begin{array}{cc}
n & 0\\
0 & n\end{array}\right)$

$u_{1}=\left(\begin{array}{c}
\shortmid n>\\
0\end{array}\right)$ , $u_{2}=\left(\begin{array}{c}
0\\
\shortmid n>\end{array}\right)$

$C=\left(H^{Y}\right)^{2}\Longrightarrow H^{Y}=\sqrt{C}=v_{0}\sqrt{2eB}\left(\begin{array}{cc}
0 & \sqrt{n}\\
\sqrt{n} & 0\end{array}\right)$ 

we can rearrange the order of the base elements so that 

$\sqrt{C}=H^{Y,n}=v_{0}\sqrt{2eB}\left(\begin{array}{cc}
0 & \sqrt{n}\\
\sqrt{n} & 0\end{array}\right)$$\Longrightarrow$

\Dn\Dn

(5)
$H^{Y,n}=V_{0}\cdot\sqrt{2eBn}\left(\begin{array}{cc}
0 & 1\\
1 & 0\end{array}\right)$ , $E_{Y,n}=\pm V_{0}\cdot\sqrt{2eBn}$

\Dn\Dn

(6) In the first form of the Y blocks, the eigen vectors of the matrix:

$H=v_{0}\sqrt{2eB}\left(\begin{array}{cc}
0 & a^{\dagger}\\
a & 0\end{array}\right)$ is $u_{n}^{+}=\left(\begin{array}{c}
\shortmid n>\\
\shortmid n+1>\end{array}\right)$ , $u_{n}^{-}=\left(\begin{array}{c}
\shortmid n>\\
-\shortmid n+1>\end{array}\right)$

Using $Q$ and $P$ we see that the given Hemiltonian is analogous
to one of a 

harmonic oscilator, therefore we can rewrite the eigenvectors in terms
of the 

eigen functions of a harmonic oscilator;

$u_{n}^{+}=\left(\begin{array}{c}
\shortmid\varphi^{n}(y)>\\
\shortmid\varphi^{n-1}(y)>\end{array}\right)$ , $u_{n}^{-}=\left(\begin{array}{c}
\shortmid\varphi^{n}(y)>\\
-\shortmid\varphi^{n-1}(y)>\end{array}\right)$

Though since the oscilator is shifted we need to add a phase factor
of the form $e^{-ieBYx}$

so the eigen states of the given Hemiltonian will be:

$\overline{u_{n}^{+}}=\left(\begin{array}{c}
\shortmid\varphi^{n}(y)>\\
\shortmid\varphi^{n-1}(y)>\end{array}\right)e^{-ieBYx}$ , $\overline{u_{n}^{-}}=\left(\begin{array}{c}
\shortmid\varphi^{n}(y)>\\
-\shortmid\varphi^{n-1}(y)>\end{array}\right)$$e^{-ieBYx}$

{*} if we concider the eigen functions to be normlized, another normlizing
factor should 

be added



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