yields $$H_f \geq \frac{9 \pi^2}{4\Lambda ^3} \Pi(x)^2 - \frac{9}{8} \Lambda
\label{eq:2.19}$$ Choosing $a= (3 \pi)/(\sqrt{2} \Lambda^{3/2})$ yields the [*lower bound*]{} $$H \geq \frac{1}{3 \pi}\sqrt{\frac{\alpha}{2}} \Lambda ^{3/2} - 
\frac{9}{8}\Lambda
\label{eq:2.20}$$

The idea of using a commutator, as in (\[eq:2.15\]), (\[eq:2.16\]) to estimate the ground state energy, goes back to the study of the polaron [@LY].

For the upper bound we take a simple trial function of the form $$\phi(x) \otimes \Psi
\label{eq:2.21}$$ where $\Psi \in {\mathcal F}$ is normalized and $\phi(x)$ is a [*real*]{} function normalized in $L^2({\mathbb R}^3)$. An upper bound to the energy is thus given by $$\frac{1}{2} \int|\nabla \phi(x)|^2 {\rm d} x +\frac{ \alpha}{2}
\int \phi(x)^2 \left(\Psi, A(x)^2 \Psi\right) {\rm d} x + (\Psi,H_f
\Psi) ~ .
\label{eq:2.22}$$ It is not very difficult to see that the last two terms can be concatenated into the following expression. $$\frac{1}{2}\int \left(\Psi, \left[ \Pi(x)^2 +\alpha  A(x) (-\Delta
+\phi(x)^2) A(x)\right] \Psi\right) {\rm d}x ~ -\frac{1}{2} {\rm{Tr}}
\sqrt{P-\Delta P} \ .
\label{eq:2.23}$$ Here, $P$ is the projection onto the divergence free vector fields with ultraviolet cutoff $\Lambda$. This can be deduced by writing the field energy in terms of $\Pi(x)$ and $A(x)$. The first term in (\[eq:2.23\]) is a sum of harmonic oscillators 