whose zero point energy is given by $$\frac{1}{2} {\rm{Tr}} \sqrt{ P\left(-\Delta + \alpha 
\phi(x)^2 \right)P} 
\label{eq:2.24}$$ and hence $$\frac{1}{2} {\rm{Tr}} \sqrt{ P\left(-\Delta + \alpha 
\phi(x)^2 \right)P} -\frac{1}{2} {\rm{Tr}} \sqrt{P (-\Delta) P}\ ,
\label{eq:2.25}$$ is an exact expression for the ground state energy. Using the operator monotonicity of the square root we get as an upper bound on (\[eq:2.23\]) $$\frac{1}{2} \int|\nabla \phi(x)|^2 {\rm d}x + \frac{1}{2} \sqrt{\alpha} 
\ {\rm{Tr}}\sqrt{ P \phi(x)^2 P} \ .
\label{eq:2.26}$$ As a trial function we use the positive function $$\phi(x) = {\rm{const.}} K^{-3/2} \int \left(1 - \frac{|k|}{K} \right)^3_+ e^{ik \cdot
x} {\rm d} x \ .
\label{eq:2.27}$$ Optimizing the resulting expression over $K$ yields the stated result. For details we refer the reader to [@LL].

It is natural to ask, how good this upper bound is. If we neglect the cross terms in $(p+A)^2$, i.e., we replace the kinetic energy by $p^2 +
\alpha A(x)^2$, then we have Theorem \[theo:2.3\], which we prove next.

[*Theorem \[theo:2.3\]:*]{} The upper bound was already given in Theorem \[theo:nonrel\] because $< p\cdot A > = 0$ in the state (\[eq:2.21\]). Loosely speaking equation (\[eq:2.12\]) expresses the Heisenberg uncertainty principle for the field operators. An uncertainty principle that is quite a 