2 Particles in a Landau level interacting via a Central Potential I am studying Robert Laughlin's paper about the Franctional Quantum Hall Effect http://gtwlx.jpkc.fudan.edu.cn/reference/FQHE-T.pdf . In it, while he is setting up his motivation for the Laughlin wavefunctions, he mentions that for 2 particles in a Landau level interacting via a central potential, the wavefunction must be of the form
$$\psi_{n,m} = (z_1-z_2)^m (z_1+z_2)^n \exp{(-\frac{1}{4}(|z_1|^2+|z_2|^2))} $$
I am having trouble seeing why this is so. I know that the single particle Lowest Landau Level states are given (in the symmetric gauge) by 
$$\psi_m = z^m \exp{(-\frac{1}{4}|z|^2)} $$
(ignoring normalization in both cases) In both of the equations, a complex number $z$ is introduced to encode the coordinates 
$$z = x-iy $$
I am having trouble seeing the path for going from the second equation to the first. I have read that the first equation is also the only way that we can form this wavefunction, i.e. it is unique. Is there a formal way to show this? 
I think the main problem I am having is introducing the CoM and relative coordinates using the complex notation. 
 A: An argument to justify this solution is given in the review by David Tong. However, his argument is very brief, and misses a part related to the Peierls substitution. Here I bring a full explanation: 
In the symmetric gauge, the quantum Hamiltonian of two identical particles located on the plane in the points $z_1$, $z_2$in the Landau problem is given by:
$$H = \hbar{\omega}(a_1^{\dagger} a_1+\frac{1}{2}) + \hbar{\omega}(a_2^{\dagger} a_2+\frac{1}{2}) + V(|z_1-z_2|),$$
where $\omega = \frac{eB}{mc}$ is the cyclotron frequency and $l = \sqrt{\frac{\hbar c}{eB}}$ is the magnetic length
$$a_i = -i \sqrt{2}(l\bar{\partial_i}-\frac{z_i}{4l})$$
Defining
$$a_R = \frac{1}{\sqrt{2}}(a_1+a_2)$$
and 
$$a_r = \frac{1}{\sqrt{2}}(a_1-a_2)$$
Then the Hamiltonian has the form:
$$H = \hbar{\omega}(a_R^{\dagger} a_R+\frac{1}{2}) + \hbar{\omega}(a_r^{\dagger} a_r+\frac{1}{2}) + V(|z|),$$
The first part of the Hamiltonian depends only on the center of mass coordinates $Z = \frac{1}{2}(z_1+z_2)$ and momentum, because:
$$a_R = -i \sqrt{2}(l\bar{\partial_Z}-\frac{Z}{2l})$$
and 
$$a_r = -i \sqrt{2}(l\bar{\partial_z}-\frac{z}{2l})$$
The Hamiltonian separates into the center of mass Hamiltonian:
$$H_{R} = \hbar{\omega}(a_R^{\dagger} a_R+\frac{1}{2}) $$
and the relative Hamiltonian
$$H_{r} = \hbar{\omega}(a_r^{\dagger} a_r+\frac{1}{2}) +  V(|z|) $$
The lowest Landau levels of the center of mass problem will take the form:
$$\Psi_m(Z) = Z^m e^{-\frac{\bar{Z}{Z}}{2 l^2}}$$
These are mutual functions of the Hamiltonian and the center of mass angular momentum: $L_Z = \hbar (Z \partial_Z-\bar{Z} \bar{\partial}_Z)$
The Hamiltonian of the relative motion is not free, but it also commutes with the relative angular momentum $L_z = \hbar (z \partial_z-\bar{z} \bar{\partial}_z)$
because:
$$[L_z, V(|z|)] = \frac{dV(|z|)}{d|z|}(z\bar{z}-z\bar{z}) = 0$$
Thus the eigenstates of the relative Hamiltonian can be taken as eigenstates of the angular momentum. In the lowest landau level, these are given by:
$$\psi_n(z) = z^n e^{-\frac{\bar{z}z}{2 l^2}}$$
However, by inspection one can see that these are not eigenfunctions of the relative Hamiltonian with the interaction. They cannot be eigenfunctions of the Schrödinger operator with an arbitrary central potential. The full solution, which was not emphasized by Tong, must mix different Landau levels of the free Hamiltonian with the same angular momentum. Thus, it will be of a definite angular momentum, but its pre-factor will not be holomorphic.
Thus strictly speaking Laughlin's two-body wave function is not a solution of the Landau problem with a central potential. However, when we restrict the motion to the Lowest landau level and this can be achieved by increasing the magnetic field to a very large value, then the Peierl's substitution becomes valid. This substitution basically entails the replacement $\bar{z}$ in the interaction term by $l^2 \frac{\partial}{\partial z}$ and performing normal ordering. Please see the above reference by Horváthy for the details in the case of a polynomial potential in $\bar{z}z$.
Thus instead of the potential $ V(z\bar{z})$ , under the Peierl's substitution, we will wind up with a potential of the form
$$ e^{-\frac{\bar{z}{z}}{2 l^2}}\tilde{V} (l^2 z \frac{\partial}{\partial z}) e^{\frac{\bar{z}{z}}{2 l^2}}$$
In this approximation, i.e., when we restrict the motion to the lowest Landau level, then the angular momentum eigenstates become also eigenstates of the Hamiltonian, because they are also eigenstates of the potential term (which becomes a differential operator!) in this case:
$$ e^{-\frac{\bar{z}{z}}{2 l^2}}\tilde{V} (l^2 z \frac{\partial}{\partial z}) e^{\frac{\bar{z}{z}}{2 l^2}}\psi_n(z) = \tilde{V} (l^2 n) \psi_n(z)$$
