# Electron positions in the lowest Landau level

In the lowest Landau level, the position operators $\hat{x}$ and $\hat{y}$ do not commute. So in writing e.g. the Laughlin wave function $\Psi\left(z_1,...z_N\right) = \prod_{i<j} \left(z_i -z_j\right)^{2p+1} e^{-\frac{1}{4} \sum_i |z_i|^2}$ with $p$ integer*, the x,y components of the coordinates $z=x+iy$ cannot be simultaneously sharply defined.

Now $\hat{x}$ and $\hat{y}$ have a very similar commutator as $\hat{x}$ and momentum $\hat{p}$ for a particle in 1d. So, thinking of the Laughlin wave function as \begin{equation} \Psi\left(z_1,...,z_N\right) = \langle z_1,z_2,...,z_N\big| \Psi\rangle = \langle \left(x_1,y_1\right),...,\left(x_N,y_N\right)\big| \Psi\rangle \end{equation} this seems similar to writing a 1d wave function in an "intermediate" representation $\Psi\left(x,p\right)$ instead of in the position ($\Psi\left(x\right)$) or momentum ($\Psi\left(p\right)$) representation. This seems odd; can I think of such a wave function as a probability amplitude? How should I make sense of the Laughlin wave function in this regard?

Also, how does a statement like "the wave function vanishes with a power $2p+1$ when two electrons are taken to the same position" make sense when their positions are "fuzzy"?

*p could also be zero, in which case this wave function describes a fully filled lowest Landau level; in that case my question still applies

The Laughlin wavefunction is a solution to some replusive-interaction many-body Hamiltonian in the entire Hilbert space, in which $x$ and $y$ still commute. It is constructed out of single particle-wave functions that are in the lowest-Landau-level (LLL), but when $x$, $y$ act as operators by multiplying LLL wavefunctions $\psi(x,y)= f(x+iy) \exp\{-(x^2+y^2)/4\}$ they take you out of the lowest Landau level. The operators that don't commute are $PxP$, $PyP$ where $P$ is the projector onto the lowest Landau level.