# Fractional filling of Laughlin wavefunction

I am not clear about the following argument why Laughlin wavefunctions have $$1/m$$ filling.

The single-electron wavefunction in the zeroth Landau level is $$$$\psi_{m}(z)\sim z^m e^{-|z|^2/4l_B^2}$$$$

where $$z=x-iy$$. The probability density of $$\psi_m$$ peaks at a radius $$r_m=\sqrt{2m}l_B$$, so the highest $$m$$ state that is occupied satisfies $$A=\pi r_N^2=2\pi N l_B^2$$, where $$N$$ is the highest occupied $$m$$ state, and is also equal to the number of electrons. The degeneracy of the ground state per unit area is, therefore, $$N/A=1/2\pi l_B^2 =eB/h$$.

The many-body Laughlin wavefunction is $$$$\Psi_m(z_1,\ldots,z_N)=\prod_{i

for $$m$$ odd. The argument for filling fraction goes as follows: the highest exponent of $$z_k$$ in the polynomial for any $$k$$ is $$m(N-1)$$, so the radius of the Laughlin wavefunction is $$r_{max}=\sqrt{2m(N-1)}l_B$$ as in the one-body case. The number of electrons is constrained by the area of the sample as $$A=\pi r_{max}^2=2\pi m(N-1)l_B^2=m(N-1)h/eB$$. The degeneracy of the Laughlin state is $$N/A \sim(eB/h) \times 1/m$$ in the limit of large $$N$$, so the filling fraction is $$\nu=1/m$$.

My question is: How can we take the highest exponent of an electron in the Laughlin wavefunction and say that the many-body state will be confined within radius $$r_{max}$$ like in the single-electron wavefunction? Is there a more rigorous way to derive the filling fraction for the Laughlin state?

Any clarification regarding this is appreciated.

Consider the Laughlin $$m=3$$ state for two particles \begin{align} \psi(z_1,z_2) &\sim (z_1-z_2)^3 e^{-|z_1|^2/4 - |z_2|^2/4 } \\ &\sim (z_1^3 z_2^0 - 3 z_1^2 z_2^1 + 3 z_1^1 z_2^2 - z_1^0 z_2^3) e^{-|z_1|^2/4 - |z_2|^2/4 } \end{align} Can be thought of as a superposition in occupation number representation in the angular momentum basis $$\psi \sim \vert 1,0,0,1,0,0,0,0,\dots\rangle + \vert 0,1,1,0,0,0,0,0,\dots\rangle$$ You can show that the single particle angular momentum state with $$L$$ (equivalent to the exponent of $$z$$) as being localized at a radius $$r(L) = \sqrt{2 L} l_B$$, where $$l_B$$ is the magnetic length $$=1$$ in this representation. Written in this angular momentum basis, it is manifest that the electron density corresponding to the many-body Laughlin wavefunction decays rapidly beyond $$r(L)$$, where $$L$$ is the highest exponent in the wavefunction.