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I've been reading Girvin's lecture notes on quantum hall effect and in a section on Haldane pseudo-potentials (paragraphs beneath equation 1.108) he says:

Because the relative angular momentum of a pair can change only in discrete (even integer) units, it turns out that this hard core model has an excitation gap. For example for $m = 3$, any excitation out of the Laughlin ground state necessarily weakens the nearly ideal correlations by forcing at least one pair of particles to have relative angular momentum 1 instead of 3 (or larger). This costs an excitation energy of order $v_1$.

The thing that confuses me is why there has to be a pair in the state with relative angular momentum 1? My explanation is that because of fixed $m$ if we have states in $m'>m$ then we would need at least another one in a state $m'' < m$ so on the average the total angular momentum would be $m$?

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  • $\begingroup$ I don't buy his argument, because he says the P's don't commute with each other (which is true) but also seems to assume that in the eigenstates all the relative angular momenta are well-defined, in other words all the P's are diagonalized. $\endgroup$
    – Ryan Thorngren
    Commented Sep 15, 2018 at 16:32
  • $\begingroup$ Actually I don't see why the P's don't commute with each other. There is a $U(1)^N$ abelian symmetry group $z_j \mapsto e^{i\theta_j}z_j$. Since the symmetries all commute, all the charges (hence the relative charges) may be defined. There is something missing from his discussion involving the confining potential. $\endgroup$
    – Ryan Thorngren
    Commented Sep 15, 2018 at 16:48

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I think that I've figured it out. So if you have a state with an $m'>m$ then automatically your mean radius of a state is larger because $r \sim \sqrt{2m}l_B$. But know because of the fixed sample surface and inability to jump to the next LL, by increasing the mean radius of one pair inevitably some of the electrons will be closer to at least one of particles in a pair with higher $m'$ and they will form a pair with an angular momentum $m'<m$, automatically that pair costs an energy. So by this we can see that Laughlin ground state is indeed gaped from the rest of the spectrum. Correct me if I am wrong.

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