why the Laughlin's wave function is an incompressible quantum state? some comments about the meaning of an incompressible quantum liquid are posted here: 
Incompressible quantum liquid
In the same context, the Laughlin's wave function for a filling factor of 1/3 describes an incompressible quantum state. Jain in his book (composite fermions; page  35) suggested we may see this property  by looking at the wave function itself in the Vandermonde determinant. I don't find this conclusion as obvious as in his book. My question is: Is there a precise mathematical way to see the in-compressibility in the Laughlin's wave function or any wave function? 
 A: I think we can understand Jain's comments as follows: when you look at the $n=1$ Landau level fully filled wavefunction $\psi(z_1,z_2,...,z_N)$as a function of one particle, say $z_1$, it has $N-1$ zeros. All of them are exactly at other particle positions. So it does not have any wasted zeros. In other words, if one additional particle, say $z_{N+1}$, is added to the system, and you necessarily need to create at least one more additional zero in the original wavefunction to ensure the Pauli exclusive principle between $z_1$ and $z_{N+1}$. However from the one particle quantum mechanics, we know that in general, adding one more node to the wavefunction requires finite additional kinetic energy. In other words, to change the density of particle by infinitesimal amount, $1/N$ here, you need to input finite amount of energy. This, by definition, simply tells you it is incompressible.
Incompressibility simply means the spectrum is gapped, as pointed out in these discussions Incompressible quantum liquid. And I think having the wavefunction itself does not tell us whether it is compressible or incompressible since the spectrum depends on the Hamiltonian. In the current case, the Hamiltonian contains only kinetic energy. Therefore the above arguments applies. If the Hamiltonian also contains interaction term, the arguments in the previous paragraph do not necessarily lead to a finite spectrum gap.  
A: Whether 'compressible' or 'incompressible' dependent on the energy gap separating the ground state and first excited state is finite or not in the thermal dynamical limit.
Laughlin wavefunction is the exact ground state of some 'pseudo-potential' Hamiltonian.
There are strong numerical evidence (but not analytical proof yet) that, Laughlin wavefunction's parent Hamiltonian is gapped.
