Projection operator (relative angular momentum) in FQHE Toy hamiltonian

Question

I am working on Fractional Quantum Hall Effect and reading these lecture notes http://www.damtp.cam.ac.uk/user/tong/qhe/qhe.pdf. As all others sources I have found, none of them precisely define the projecto operator $ P_{m'}(ij)$ mentioned here on p84 or give a mathematical argument on why is thus $V_m \Phi = 0 $ (so why are the Laughlin states ground state of the toy hamiltonian).

The best 'attempt' or beginning of a explicit proof I have found is on the p11 of the following : http://www.phys.virginia.edu/Files/fetch.asp?EXT=Seminars:1752:SlideShow

But I am still a bit confused on how to get the explicit expression for $\phi_c^{rel}$, or said otherwise how to go from line 2 to 3 ?

\begin{equation} H = \sum_{m'=1}^{\inf} \sum_{i<j} v_{m'}P_{m'}(ij) \end{equation}

\begin{equation} \phi(z_i) = \prod_{i<j}(z_i-z_j)^m e^{-\sum _i \mid z_i \mid ^2 / 4l^2_B} \end{equation} So ho do we move from here to a decomposition in relative angular momentum ? \begin{equation} \phi(z_i) = \sum_a \sum_{b,c} C_a^{b,c} \phi_b^{CM}(\frac{1}{2}(z_1+z_2)) \phi_c^{rel} (z_1-z_2) e^{-\sum _i \mid z_i \mid ^2 / 4l^2_B} \Phi(z_2,z_3, ...,z_N) \end{equation}

Answer 1

I don't know if you're still interested in the answer, but I think I found something. Since we're working in two dimensions (i.e. x-y plane), angular momentum is defined as $$J = i\hbar\, (x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}).$$ This transformed in $z = x-iy,\,\overline{z}=x+iy$, coordinates reads (fairly easy to check) $$J = \hbar\, (z\,\frac{\partial}{\partial z}-\overline{z}\,\frac{\partial}{\partial{\overline{z}}}).$$ From here, it means that relative angular momentum operator reads (neglecting $\hbar$) $$J_{ij} = (z_i-z_j)\,\frac{\partial}{\partial (z_i-z_j)}-(\overline{z}_i - \overline{z}_j)\,\frac{\partial}{\partial{(\overline{z}_i - \overline{z}_j)}}.$$ Now, projection operators are projecting on eigenstates of $J_{ij}$ with eigenvalue $m'$ which are $\phi_{m'}^{ij} = (z_i-z_j)^{m'} e^{-|z_i-z_j|^2/4}$. This isn't hard to check (remember that $|z|^2 = z\cdot \overline{z}$).

I still have to figure out how to show that Laughlin state is then ground state of Toy Hamiltonian composed of these projection operators, but as soon as I do I will post it.