save half of the kinetic energy and write $$H = \frac{1}{4} \sum^{N}_{j=1} (p_j + \sqrt{\alpha} A (x_j))^2 +
H' ~ .$$ We apply the commutator estimate (\[eq:2.14\]) to the pair $$i [p_j + \sqrt{\alpha} A (x_j), \frac{1}{\sqrt{N_j (\underbar{X}, R) +
1}} \Pi (x_j)]$$ and obtain the bound (with the caveat mentioned above), for all $\alpha > 0$, $$H' \geq a \sqrt{\alpha} \Lambda^3 \sum^N_{j=1}
\frac{1}{\sqrt{N_j(\underbar{X}, R) + 1}} - a^2 \sum^N_{j=1} \frac{1}{N_j
(\underbar{X}, R) + 1} {\rm{F}} (x_j)^2 + H_f
\label{eq:3.8}  $$ The next two steps are somewhat nontrivial and we refer the reader to [@LL]. First one notes that the modes $F(x_i)$ and $F(x_j)$ are essentially orthogonal (i.e., they commute) if $|x_i- x_j| > \Lambda^{-1}~ .$ Ignoring the technical details of how this is implemented, the key observation is that the last two terms in (\[eq:3.8\]) can be estimated from below by $ - N\Lambda$ provided $a = \Lambda^{-3/2}$.

The next ingredient is a new Lieb-Thirring type estimate involving the function $N_j(\underbar{X},R)$. It is here and only here that the Pauli exclusion principle is invoked.

On the space $\wedge^{N}_{j=1} L^2 {\mathbb R}^3; {\mathbb C}^q)$ of antisymmetric functions $$\sum_{j=1}^{N} (p_j + \sqrt{\alpha} A (x_j))^2 \geq \frac{C}{q^{2/3}}
\frac{1}{R^2} \sum^N_{j=1} N_j (\underbar{X}, R)^{2/3}$$ 