d^3x \left(\hbar {\delta \over \delta \psi(\vec{x})}c(-i{\bf
{\buildrel \rightarrow \over \alpha}  \cdot {\buildrel \rightarrow
\over \nabla} } + \beta {mc\over\hbar}) \psi(\vec{x})\right)
\Phi[\psi]= E\Phi[\psi]. \label{fdiraceq}$$

It is an easy matter to show that Eqs. (\[fpone\]) and (\[fptwo\]) at $t=0$ can be rewritten in terms of respective operators in the Schrödinger representation as follows $$\widehat{\psi}(\vec{x}) = \sum_{r=1}^4 \int {d^3p \over (2
\pi\hbar)^{3/2}} \sqrt{mc^2 \over E_{\vec{p}}}\widehat{\bf
b}(\vec{p},r)w(\vec{p},r) \exp \big(i\vec{p} \cdot \vec{x} / \hbar
\big),$$ $$\widehat{\psi}^{\dag}(\vec{x})= {\delta \over \delta
\psi(\vec{x})} = \sum_{r=1}^4 \int {d^3p \over (2 \pi\hbar)^{3/2}}
\ \sqrt{mc^2 \over E_{\vec{p}}} {\delta \over \delta {\bf
b}(\vec{p},r) } w^{\dag}(\vec{p},r) \exp\big(-i \vec{p} \cdot
\vec{x} /\hbar \big).$$ Inserting these formulae into (\[fdiraceq\]) one gets $$\sum_{r=1}^4 \int d^3p \varepsilon_r E_{\vec{p}} {\delta \over
\delta {\bf b}(\vec{p},r)} {\bf b}(\vec{p},r)\Phi[{\bf b}] = E
\Phi[{\bf b}], \label{functdeq}$$ where ${\bf b}(\vec{p},r)$ is the eigenvalue of $\widehat{\bf
b}(\vec{p},r)$.

The stable vacuum $|0 \rangle$ for the Dirac free field is the ground state in which all the negative energy states are filled. To be precise consider the case of having only one fermion state and remind that ${\bf b}$ and $\hbar {\delta \over \delta {\bf
b}}$ are the annihilation and creation operators of a single fermion. In addition, let $\Omega_0({\bf b})$ and $\Omega_1({\bf
b})$ be the 