Fractional filling of Laughlin wavefunction

Question

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 \begin{equation} \psi_{m}(z)\sim z^m e^{-|z|^2/4l_B^2} \end{equation}

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 \begin{equation} \Psi_m(z_1,\ldots,z_N)=\prod_{i<j}^N (z_i-z_j)^me^{-\sum_i^N |z_i|^2/4l_B^2} \end{equation}

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.

Answer 1

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.