crucial role (even in in the absence of Coulomb attractions). We show that $E_0(N) \geq C N
E_0$ for fermions, but we are unable to show that the universal constant $C = 1$. Even if $C < 1$, the situation could still be saved by mass renormalization, which drives the bare mass to zero as $\Lambda$ increases, thereby pushing particles apart.

BOSONS
------

[*Theorem \[theo:nonrelboson\];*]{} This theorem concerns the ground state energy of $N$ charged bosons. the Hamiltonian is given by \[manyham\] acting on the Hilbert space of symmetric functions tensored with the photon Fock space $\mathcal F$. It states, basically, that $C_1 \sqrt{N} \sqrt{\alpha} \Lambda^{3/2} \leq 
E_0^{boson}(N) \leq  C_2 N^{5/7}  \alpha^{2/7} \Lambda^{12/7}$.

The proof follows essentially that of the one particle case. The interesting fact is that it implies [*binding*]{} of [*charged bosons*]{} (in the absence of the Coulomb repulsion). The binding energy is defined by $$\Delta E (N) = E_0 (N) - NE_0(1)$$ and satisfies the bounds $$\begin{aligned}
\Delta E(N) &\geq& C_1 \sqrt{N} \sqrt{\alpha}~ \Lambda^{3/2} -
C_2 N  \alpha^{2/7} \Lambda^{12/7}  \nonumber \\
\Delta E(N) &\leq& C_2 N^{2/7} 
\alpha^{2/7} \Lambda^{12/7} -  C_1 N \sqrt{\alpha}~ \Lambda^{3/2}
\label{eq:3.6}\end{aligned}$$ which can be made negative for appropriately chosen $N$ and $\Lambda$. There will 