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\exnumber{5662}
\heading{Landau levels in Graphene} 
\auname{Gilad Sade and Sahar Malka}

{\bf The problem:}

\Dn

The effective Hamiltonian of an electron in a 2-D layer of Graphene is $H=v_0\bm{\sigma}(\bm{p}-e\bm{A})$ while $\bm{\sigma}=(\sigma_x,\sigma_y,0)$ and $\bm{A}$ is the vector potential for a perpendicular magnetic field $\bm{B}$ within the Landau gauge.\newline
Notice that the standard basis for an electron is $\ket{x,y,m}$, $m=\downarrow,\uparrow.$

\Dn

(1)In the absence of a magnetic field, for a given momentum $\bm{p}=(p_x,p_y)$, what are the eigenenergies of the particle?

(2)Within the Landau gauge $\hat{Y}=-(\frac{1}{eB})\hat{p_x}$ is a constant of motion. Write down the Hamiltonian $H^{(Y)}=H_{m,m'}(p_y,y-Y)$ after the variable separation. Note that the linear combination of canonical coordinates $a=\frac{(sQ+\frac{iP}{s})}{\sqrt{2}}$ is a ladder operator, and write down the Hamiltonian derived in the alternative notation $H_{m,m'}(a,a^\dagger)$ 

(3)Define the operator $C=(H^{(Y)})^2$.\newline
Calculate the eigenvalues $\lambda_{n=0,1,2...}$ of $C$,\newline
notice, the eigenvalues are degenerated except $n=0$.

(4)Since $C$ is a constant of motion it is possible to use variable separation once more.\newline
Write down the $2X2$ matrix representing $H^{(Y,n)}$. 

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

(6)Find the eigenstates and write them in the standard basis.\newline
Notice that the quantum state in the
standard basis in represented by $\Psi\mapsto (\psi_\uparrow(x,y),\psi_\downarrow(x,y))$.\newline
You may use the notation $\varphi^n()$ for the eigenfunctions of a 1-D harmonic oscillator. 

\Dn\Dn

{\bf The solution:}

\Dn

(1)The Hamiltonian without a magnetic field is: $$H=v_0\bm{\sigma}\cdot\bm{p}=v_0(\sigma_xp_x+\sigma_yp_y)$$$$=v_0
\begin{pmatrix}
0 & p_x-ip_y\\
p_x+ip_y & 0
\end{pmatrix}=
v_0|\bm{p}|\begin{pmatrix}
0 & e^{-i\phi}\\
e^{i\phi} & 0
\end{pmatrix}$$
where $\bm{p}=(p_x,p_y)$ is given and $tg(\phi)=\frac{p_y}{p_x}$.\newline
Diagonalization of this Hamiltonian provides the following eigenenergies: $E_{\pm}=\pm v_0|\bm{p}|$.
\Dn

(2)The vector potential within the Landau gauge is: $$\bm{A}=B(-y,0,0)$$
Therefore the Hamiltonian is:$$H=v_0[\sigma_x(p_x+eB_y)+\sigma_yp_y]$$
plugging the definition of Y into H yields:$$H=v_0[eB\sigma_x(y-Y)+\sigma_yp_y]$$
One should notice that $[Y,H]=0$, thus the Hamiltonian is separable into blocks of the form:$$H^{(Y)}=v_0[eB\sigma_x(y-Y)+\sigma_yp_y]=v_0\sqrt{eB}[\sqrt{eB}\sigma_x(y-Y)+\frac{1}{\sqrt{eB}}\sigma_yp_y]$$
We define:$$Q=y-Y, P=p_y, s=\sqrt{eB}$$
So:
$$H^{(Y)}(Q,P)=v_0[\sigma_x\sqrt{eB}Q+\frac{1}{\sqrt{eB}}\sigma_yP]$$
Using the ladder operator given:$$a=\frac{1}{\sqrt{2}}(Q\sqrt{eB}+\frac{i}{\sqrt{eB}}P)$$$$\Downarrow$$$$Q=\frac{1}{\sqrt{2eB}}(a^\dagger+a),\space P=\frac{\sqrt{eB}}{i\sqrt{2}}(a-a^\dagger)$$
plugging into the Hamiltonian:$$H^{(Y)}(a,a^\dagger)=v_0\sqrt{2eB}[a(\sigma_x-i\sigma_y)+a^\dagger(\sigma_x+i\sigma_y)]$$
$$H^{(Y)}\mapsto v_0\sqrt{2eB}\begin{pmatrix}
0 & a^\dagger\\
a & 0
\end{pmatrix}$$  
\Dn

(3)$$[a,a^\dagger]=1\Rightarrow aa^\dagger=1+a^\dagger a$$ $$C=(H^{(Y)})^2\mapsto 2eBv_0^2\begin{pmatrix}
a^\dagger a & 0\\
0 & a a^\dagger
\end{pmatrix}=2eBv_0^2\begin{pmatrix}
a^\dagger a & 0\\
0 & a^\dagger a+1
\end{pmatrix}$$
The eigenstates of $a^\dagger a$ satisfy:(1D harmonic oscillator) $$a^\dagger a\ket{n}=n\ket{n},(n=0,1,2\dots)$$ The eigenvectors of $C$ are: $\begin{pmatrix}
\ket{n}\\
0
\end{pmatrix},\begin{pmatrix}
0\\
\ket{n-1}
\end{pmatrix}$$$C^{(n)}\mapsto2eBv_0^2\begin{pmatrix}
n & 0\\
0 & n
\end{pmatrix}\Rightarrow \lambda_n=2eBv_0^2n$$\Dn

(4) $[C,H^{(Y)}]=0$, therefore $H^{(Y)}$ is separable into blocks.$$H^{(Y)}\begin{pmatrix}
\ket{n}\\
0
\end{pmatrix}=v_0\sqrt{2eB}\begin{pmatrix}
0\\
a\ket{n}
\end{pmatrix}=v_0\sqrt{2eBn}\begin{pmatrix}
0\\
\ket{n-1}
\end{pmatrix}$$
$$H^{(Y)}\begin{pmatrix}
0\\
\ket{n-1}
\end{pmatrix}=v_0\sqrt{2eB}\begin{pmatrix}
a^{\dagger}\ket{n-1}\\
0
\end{pmatrix}=v_0\sqrt{2eBn}\begin{pmatrix}
\ket{n}\\
0
\end{pmatrix}$$$$\Downarrow$$$$H^{(Y,n)}\mapsto v_0\sqrt{2eB}\begin{pmatrix}
0 & \sqrt{n}\\
\sqrt{n} & 0
\end{pmatrix}$$
\Dn

(5)$$H^{(Y,n)}= v_0\sqrt{2eBn}\sigma_x\Rightarrow E_{Y,n,\pm}=\pm v_0\sqrt{2eBn}, (n>0)$$
\Dn

(6) The eigenstates of $H^{(Y)}$ are (up to a normalization factor): $$\begin{pmatrix}
\ket{n}\\
\pm\ket{n-1}
\end{pmatrix}$$ the representation of these eigenstates in the standard basis is: $$\begin{pmatrix}
\varphi^n(y)\\
\pm \varphi^{n-1}(y)
\end{pmatrix}$$remembering that $Y$ shares the same eigenstates with $p_x$ yields the following eigenstates:$$e^{-ieBYx}\begin{pmatrix}
\varphi^n(y)\\
\pm \varphi^{n-1}(y)
\end{pmatrix}$$ 








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