\hat{\bf y}$. Now our Hamiltonian is $$H=\frac{1}{2m}\left[p_x^2+\pi_y^2\right]=\frac{1}{2m}\left[p_x^2+(p_y+eBx)^2\right]$$ where $\pi_y$ is the kinetic momentum operator in the $\boldsymbol{\hat{y}}$ direction. Immediately, we see that $[H,p_y]=0$, so we can separate our eigenstate $\Psi$ as $$\Psi(x,y)\propto e^{ik_yy}\psi(x),$$ where the operator $p_y$ is replaced by its eigenvalue $\hbar k_y$. Substituting this into the Schrödinger Equation leads to the one dimensional differential equation $$\left[\frac{-\hbar^2}{2m}\partial_x^2+\frac{1}{2}m\omega_c^2(x+k_yl_B^2)^2\right]\psi(x)=E\psi(x),$$ which is the usual harmonic oscillator (albeit not centred at zero) equation. Our solution is therefore $$\Psi_{n,k_y}(x,y)=e^{ik_yy}\frac{1}{\sqrt{2^{n}n!
    \sqrt{\pi}l_B}}e^{-\frac{(x+k_yl_B^2)^2}{2l_B^2}}H_{n}\left[\frac{x+k_yl_B^2}{l_B}\right]
    \label{psi_landau}$$ where $H_n(z)$ are the Hermite polynomials. This method of solution has introduced two quantum numbers, $k_y$, a real number, and $n$, a non-negative integer. The corresponding energies are as before, in Eq. (\[energy\_landau\_level\]). Note that the wave functions are one-dimensional-like and certainly don’t resemble those obtained previously. The reason this is possible is because of the degeneracy present (energy does not depend on $k_y$), which we now discuss further.

[Degeneracy and the LLL Wavefunction in the Landau Gauge]{}
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In the Landau gauge, the $LLL$ wavefunction must have $n=0$ for any $k_y\in\mathbb{R}$, giving us: $$\Psi_{LLL}(x,y)=e^{ik_yy}\frac{e^{-\frac{(x+k_yl_B^2)^2}{2l_B^2}}}{\sqrt{\sqrt{\pi}l_B}}.$$ Usually periodic boundary conditions are imposed in the $y$-direction, resulting in the requirement that $k_y = 2\pi m_y/a$, where $a$ is the length of the sample (i.e. “ribbon”) 