The Landau Hamiltonian in 2D is given (in natural units $q=c=2m=1$) by $$ \hat{H} = (\hat{\vec{p}}-\vec{A}(\hat{\vec{x}}))^2 \,,$$ where $\vec{A}$ is the magnetic vector potential field.

We know that the momentum operator $\hat{\vec{p}}$ may be shifted using the position operator, that is, if $f$ is any scalar field, then $$\exp(if(\hat{\vec{x}}))\hat{\vec{p}}\exp(-if(\hat{\vec{x}})) = \hat{\vec{p}}-(\nabla f)(\hat{\vec{x}}) \,.$$

Hence we may re-write $\hat{H}$ as \begin{align} \hat{H} &= \exp(i\int_{\vec{x_0}}^{\hat{\vec{x}}}A(\vec{l})\cdot d\vec{l})\,\,\,\,\,\hat{\vec{p}}^2\,\exp(-i\int_{\vec{x_0}}^{\hat{\vec{x}}}A(\vec{l})\cdot d\vec{l})\label{eq:one}\end{align}where $\vec{x_0}$ is any arbitrarily chosen reference point.

Since we know the (non-normlizable) eigenstates of $\hat{\vec{p}}^2$, namely, for any $\vec{k}$, we have $\psi_k^{\mathrm{A=0}}(x)=\exp(\pm i\vec{k}\cdot\vec{x})$, we may now write down the eigenstates of $\hat{H}$ as $$ \psi_k^{\mathrm{A\neq0}}(x)=\exp(i\int_{\vec{x_0}}^{\vec{x}}A(\vec{l})\cdot d\vec{l})\exp(\pm i\vec{k}\cdot\vec{x}) \,. $$

This is of course false, as it doesn't give the famous quantization of the the Landau energy levels.

My question is: what is the mistake I made?


You've assumed that $\vec{A}$ can be described by a function $f$ such that $\nabla f=\vec{A}$. You've also attempted to write down an explicit formula for $f$, namely $f=\int_{x_0}^{x}\vec{A}\cdot d\vec{\ell}$. This all works fine, assuming $\vec{A}$ has no curl. If $\vec{A}$ has curl, you cannot find a function $f$ such that $\nabla f=\vec{A}$, and your formula $f=\int_{x_0}^{x}\vec{A}\cdot d\vec{\ell}$ is not well-defined because it depends on the path $x_0\rightarrow x$. Of course, if $\vec{A}$ has no curl, it describes a system with zero magnetic field, and you don't get Landau quantization. You are interested in precisely when $\vec{A}$ HAS curl, which is exactly when your argument fails.


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