Consider the simplest eigenfunctionals of the (free) field Hamiltonian
$$ H= \frac{1}{2} \int d^3 x ~( \Pi (x)^{2}+(\nabla\Phi)^{2} +m^2\Phi ^{2}).$$
Assuming this Hamiltonian and the functional quantization
$$ \Pi \sim \frac{-i\delta}{\delta \Phi} $$
what would be the eigen-energies?
Well, they are all infinite, since free scalar field theory is simply a compactly repackaged array of quantum harmonic oscillators. You are simply supposed to follow the systematic translation rules for the Schrödinger representation from QM to QFT carefully, e.g. following Jackiw 1988 or Lüscher 1985.
As for one oscillator, the energy eigen-functionals are Gaussian in
form. I'll give you a jump-start: the ground state eigen-functional $\langle\Phi|0\rangle$ is just
\begin{equation}
\Psi[ \Phi] =\exp\left( -\frac{1}{2}\int d^3 x\,\Phi\left(
x\right) \sqrt{m^{2}-\nabla_{x}^{2}}~ \Phi\left( x\right) \right).
\end{equation}
Boundary conditions are assumed such that the $\sqrt{m^{2}-\nabla_{x}^{2}}$
kernel in the exponent is naively self-adjoint.
"Integrating by parts" one of the $\sqrt{m^{2}-\nabla_{z}^{2}}$ kernels,
functional derivation $\delta\Phi\left( x\right) /\delta\Phi\left( z\right)=\delta^3\left( z-x\right) $ leads to
\begin{equation}
\frac{\delta}{\delta\Phi\left( z\right) }\Psi [\Phi]
=-\left( \sqrt{m^{2}-\nabla_{z}^{2}}\,\Phi\left( z\right) \right)
\Psi[ \Phi] ,
\end{equation}
so that
$$
\frac{\delta^{2}\qquad }{\delta\Phi\left( w \right) \delta\Phi
\left( z\right) }\Psi[ \Phi]
=\left( \sqrt{m^{2}-\nabla_{w}^{2}}\,
\Phi\left( w\right) \right) \left( \sqrt{m^{2}-\nabla_{z}^{2}}\,
\Phi\left( z\right) \right) \,\Psi[ \Phi]\\ - \sqrt
{m^{2}-\nabla_{z}^{2}}\,\delta^3\left( w-z\right) \,\Psi[ \Phi]~.
$$
Note that the divergent zero-point energy density, \begin{equation} E_0= { 1\over 2 } \lim_{w\rightarrow z} \sqrt {m^{2}-\nabla_{z}^{2}}~ \delta^3\left( w-z\right), \end{equation} may be handled rigorously using $\zeta$-function regularization. It is but the sum of all zero-point energies of the infinity of oscillators. The price of operator-valued distributions.
Leaving this zero-point energy present leads to the standard energy eigenvalue equation, again through integration by parts, \begin{equation} \frac{1}{2} \int d^3 z\,\left( - \frac{\delta^{2}} {\delta\Phi\left( z\right) ^{2} }+\Phi\left( z\right) \left( m^{2}-\nabla_{z}^{2}\right) \Phi\left( z\right)-2E_0 \right) ~\Psi[ \Phi]=0 ~, \end{equation} that is the lowest eigenvalue of H is $\int d^3 x ~E_0$.
You might proceed to form functional ladder operators, etc... and pursue the finer aspects of Schrödinger functional theory to the bitter end...
