\label{eq:nonrel}$$ \[theo:nonrel\]

We do not know how to get upper and lower bounds that are of the same order in $\Lambda$, but we suspect that $\Lambda^{12/7}$ is the right exponent. This is supported by the following theorem in which the $p\cdot A$ term is omitted.

The ground state energy $E_0$ of the operator $$\frac{1}{2}  \left[ p^2 + \alpha A(x)^2 \right] + H_f
\label{eq:2.28}$$ satisfies the bounds $$C_1  \alpha^{2/7} \Lambda ^{12/7}
\leq E_0 \leq C_2  \alpha^{2/7} \Lambda ^{12/7}
\label{eq:2.29}$$ \[theo:2.3\]

While these results are not of direct physical relevance (since $E_0$ is not observable), the many-body problem is of importance since it reveals a dramtic difference between bosons and fermions.

The ground state energy of $N$ bosons, $E^{boson}_0(N)$, with Hamiltonian $$H(N)=\sum_{j=1}^N \frac{1}{ 2}(p_j +\sqrt \alpha A(x_j))^2  
+  H_f\
\label{manyham}$$ satisfies the bounds $$C_1 \sqrt{N}  
\sqrt{\alpha} \Lambda^{3/2} \leq
 E^{boson}_0(N) \leq
C_2 N^{5/7}\alpha ^{2/7} \Lambda^{12/7}
\label{eq:nonrelboson}$$ \[theo:nonrelboson\]

Thus, the energy $E^{boson}_0(N)$ is not extensive, i.e., it costs a huge energy to separate bosons. This has to be contrasted with the next theorem about fermions. The Hamiltonian is the same as before but it acts on the Hilbert space $${\mathcal F}\otimes
\wedge_{j=1}^N L^2(\mathbb{R}^3; \mathbb{C}^2) ~ ,$$ where the wedge product indicates the antisymmetric tensor 