(2 \pi)^{-3} \int_{|k|<
\Lambda} |k|^{-1} 
dk = \Lambda^2/4\pi^2$. We can now let $f(x) \rightarrow
V^{-\frac{1}{2}}$ and take $\varepsilon = (\pi/ \alpha)^{1/2}
\Lambda^{-1}$, whence $$E_0 \leq (\alpha/4 \pi)^{1/2} \Lambda ~.
\label{eq:4.5b}$$

Now we turn to the lower bound for $H$.

[**Step 1**]{}: Since $x \rightarrow \sqrt{x}$ is operator monotone, $$T > T_1 = | p_1 + \sqrt{\alpha} A_1 (x) | ~,
\label{eq:4.6}$$ where the subscript 1 denotes the 1 component (i.e., the $x$-component) of a vector. By replacing $T$ by $T_1$, we are now in a position to remove $A_1$ by a gauge transformation - but it has to be an [*operator-valued gauge transformation*]{}. The use of such a gauge transformation is is a novelty, as far as we are aware, in QED.

To effect the gauge transformation, set $$\varphi (x) = \frac{1}{\sqrt{2V}}~ \sum_{h,\lambda} ~
\frac{\varepsilon^{\lambda}(k)_1}{\sqrt{(k)}} ~ \left[ a_{\lambda} (k)
+ a_{\lambda}^{\ast} (-k) \right] \frac{e^{ik_1 x_1} -1}{ik_1} ~
e^{ik_{\perp}x_{\perp}}
\label{eq:4.7}$$ with $x_{\perp} = (x_2, x_3$). Then $[A_j (x), \varphi (x) ] = 0, ~~j= 1,2,3$ and $p_1 \exp~ [i \varphi (x)] = -A_1(x)$. The unitary $U = \exp~ [i
\varphi (x)]$ is a gauge transformation, but it is [*operator-valued*]{}. We have $$\begin{aligned}
U^{-1} | p_1 + A_1 (x) | U &=& 