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I am playing with the Landau Level problem and Algebraic solutions to it. I am given $$a=\frac{l_{b}}{\sqrt{2}\hbar}(\pi_{x}-i\pi_{y}) ~~~~~~~~\text{and}~~~~~~~~~a^{\dagger}=\frac{l_{b}}{\sqrt{2}\hbar}(\pi_{x}+i\pi_{y})$$

The creation and annihilation operators. I am told that I can express these in terms of $x$, $\frac{\partial}{\partial x}$ and $k$, but am unsure how to proceed. Would I just expand every term and substitute? That seems like a lot of work that may be unnecessary. $\pi$ is the kinetic momentum.

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For context, I am studying the quantum Hall Effect, integer and fractional, and this came up in my instructors online notes. I am unsure why you would rewrite this, and further what is the best way to approach it. To expand on my idea, I was thinking to expand $\pi$ to $\vec{\pi}=m\vec{v}=\vec{p}-\frac{q}{c}\vec{A}$ and perhaps try from there. – Dylan Sabulsky Feb 17 '13 at 17:03
Perhaps you know something about the commutation relations that could help you rewrite these? How are the momentum and position operators related? – KDN Feb 17 '13 at 17:04
Great call, hadn't even considered it in my foolishness. $$[p_{i},r_{j}]=[p_{i},r_{j}]=-i\hbar \delta_{ij}$$ $$[R_{i},\pi_{j}]=0$$ $$[\pi_{i},\pi_{j}]=-i \epsilon_{ij} m \hbar \omega_{c}=-i \epsilon_{ij} \frac{\hbar^{2}}{l_{b}^{2}}$$ $$[R_{i},R_{j}]=i \epsilon_{ij} l_{b}^{2}$$ $$[\rho_{i},\rho_{j}]=-i \epsilon_{ij} l_{b}^{2}$$ $$[\rho_{i},\pi_{j}]=i \hbar \delta_{ij}$$ $$[p_{i},\pi_{j}]=-i \hbar \frac{e}{c} \frac{\partial A_{j}}{\partial r_{i}}$$ $$[R_{i},r_{j}]=i \epsilon_{ij} l_{b}^{2}$$ $$[\rho_{i},r_{j}]=-i \epsilon_{ij} l_{b}^{2}$$ – Dylan Sabulsky Feb 17 '13 at 17:18
Do you know anything about the commutation of $a$ and $a^\dagger$? – KDN Feb 17 '13 at 17:35
let us continue this discussion in chat – KDN Feb 17 '13 at 17:38

1 Answer 1

up vote 3 down vote accepted

First, we choose a gauge. We will use the Landau gauage, so that \begin{equation} \mathbf{A}=B_z x \mathbf{\hat{y}} \end{equation} corrseponding to a magnetic field $\mathbf{B}=B_z \mathbf{\hat{z}}$.

With this choice of gauge, the kinetic momentum operators can be rewritten in terms of the canonical momentum operators, as \begin{equation} \begin{aligned} \pi_x &= p_x - \frac{e}{c} A_x = p_x\\ \pi_y &= p_y - \frac{e}{c} A_y = p_y - \frac{e B_z x}{c} \end{aligned} \end{equation} If we write down the Hamiltonian in these terms, \begin{equation} \mathcal{H} = \frac{1}{2 m}\left(\mathbf{p}-\frac{e}{c}\mathbf{A}\right)^2 \end{equation} we see that, since $\mathbf{A}$ has no $y$ dependence, the canonical $p_y$ operator commutes with the Hamiltonian. We may therefore replace it with its eigenvalue, $\hbar k_y$. Rewriting the momentum operator in terms of the position operator, the creation operator may be written \begin{equation} a^\dagger=\frac{l_b}{\sqrt{2} \hbar}(\pi_x+i\pi_y)=\frac{l_b}{\sqrt{2} \hbar}\left( \frac{\hbar}{i}\frac{\partial}{\partial x} + i\hbar k_y - \frac{e}{c}B_z x\right), \end{equation} which rewrites this operator in terms of your chosen variables.

Different gauge choices can be made, preserving translational or rotational invariance. This particular gauge choice preserves translation invariance along the $y$-axis. For a more detailed discussion of this topic, see these lecture notes.

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Thanks so much for your help KDN. Awesome – Dylan Sabulsky Feb 17 '13 at 18:52

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