# False solution of Landau Hamiltonian

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?

## 1 Answer

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.