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 carefully, e.g. following [Jackiw 1988](https://books.google.com/books?hl=en&lr=&id=0m7mJEruqiUC&oi=fnd&pg=PA107&dq=R.+Jackiw,+%22Schrodinger+Picture+for+Boson+and+Fermion+Quantum+Field+Theories.%22+&ots=JzjfA9I3Bf&sig=Y1aZzH0fqKSGScfGyI94SFykFvI#v=onepage&q=2.16%20%20Schrodinger%20picture%20&f=false) or [Lüscher 1985](https://www.sciencedirect.com/science/article/pii/055032138590210X?via%3Dihub). 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](https://en.wikipedia.org/wiki/Schrödinger_functional) to the bitter end...