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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.