Prove that Laughlin's 3-electron states are a complete set of states In R. B. Laughlin's 1983 Physical Review B article, Quantized motion of three two-dimensional electrons in a strong magnetic field, Laughlin separates out the center of mass motion of the electrons, and writes down a set of states that he says form a "complete orthonormal basis". The states he writes down in equation (18) are:
$$|m,n\rangle = \frac{1}{\sqrt{2^{6m+4n+1}(3m+n)!n!\pi^2}}\biggr[ \frac{(z_a+iz_b)^{3m}-(z_a-iz_b)^{3m}}{2i} \biggr](z_a^2+z_b^2)^ne^{-(1/4)(|z_a|^2+|z_b|^2)}$$
where the internal coordinates $z_a$ and $z_b$ are defined as
$$z_a\equiv\sqrt{\frac{2}{3}}\biggr[\frac{z_1+z_2}{2}-z_3\biggr]$$
$$z_b=\frac{1}{\sqrt{2}}(z_1-z_2)$$
and $z_1$, $z_2$, and $z_3$ are the coordinates of the three electrons. My understanding is that Laughlin is claiming this is a complete orthonormal basis for the lowest Landau level. I can prove orthonormality.  I can also prove that this state obeys Fermi statistics. I also see how this state is constructed as a linear combination of products of single electron lowest Landau level states. However, I don't know how to prove that this set of states is complete. Does anyone know how to prove that this is a complete set of states for the lowest Landau level? Thanks!
 A: My reading of the paper was that it was a complete basis for the three particle system (modulo the center of mass) subject to the antisymmetry condition on the three electrons, which was originally $\begin{eqnarray} \Psi(z_1,z_2,z_3) &=&-\Psi(z_2,z_1,z_3) = \\
\Psi(z_2,z_3,z_1) &=&-\Psi(z_1,z_3,z_2) = \\
\Psi(z_3,z_1,z_2) &=&-\Psi(z_3,z_2,z_1) \end{eqnarray}$
but after you modulo by the center of mass (the above antisymmetry has no effect on the center of mass) but for the $z_a$ and $z_b$ coordinate system the above requires antisymmetry in $z_b$ (straightforward) and symmetry in rotations of $120^\circ$ in the $z_a,z_b$ plane (not straightforward).
In fact since higher values of $n$ and $m$ lead to higher total angular momentum $M=3m+2n,$ it is not clear that the set of all such $|m,n\rangle$ will span the lowest Landau level, I think it will span all the states with the right (anti)symmetry.  You could consider $M_k=\{|m,n\rangle: 3m+2n=k\}$ and if each $M_k$ spans the set of states with the right (anti)symmetry and total angular momentum $k,$ then I think you would be done.
