Schroedinger equation for wave functional (QFT) As far as I'm aware you can solve for the wave functional $\Psi[\phi]$ of a field using the Schrodinger equation $$i\hbar\frac{\partial \Psi}{\partial t}=H\Psi.$$

*

*Should $H$ here be the Hamiltonian, or the Hamiltonian density?

*If it's the Hamiltonian, is there a version of this equation for the Hamiltonian density?

*Would assuming a certain form for the functional like $\Psi\sim e^{-\omega \phi^2}$ be of any use in simplifying the equation?

I'm pretty new to QFT so excuse me if the question is poorly formulated.
 A: Recall, from Hatfield's textbook (QFT of point particles and strings) & Jackiw's review that the functional equation you are emulating is just that, a functional equation the extension of an infinite sum of canonical pairs $[q_i,p_j]=i\hbar \delta_{ij}$ to
$$ [\phi(x),\pi(y)]\propto \delta(x-y). $$
So, just as the Hamiltonian in QM deals with all degrees of freedom, just so in QFT,
$$
H\psi[\phi]=  \int\!\! d^3 x  \left (-\frac{\delta^2}{\delta\phi(x)^2}  +\phi(x)O\phi(x)+... \right )\psi[\phi],
$$
where the ellipses (...) suggests cubic and higher terms in the potential, rarely used. $O$ is a normally nonlocal operator, i.e., $$
  O \phi(x) = \int \! d^3y ~O(x-y) \phi(y),
$$
such as $O=m^2-\nabla^2$, etc. I've left the time dependence implicit throughout.
It is then evident that the ground state of the quadratic Hamiltonian is
$$
\propto e^{-\tfrac{1}{2} \int d^3z ~\phi(z) \sqrt{O} \phi(z) }, 
$$
but you must attend to the δ-functions.

*

*When confused, try to consider uncoupled oscillators, i.e. $O$ a constant.


*PS if you insist on using functions instead of functionals, you can always convert the latter to the former by sticking in gonzo gratuitous delta functions,
$$
H'[\phi]=\int d^3 x ~~{\cal H}(x)  \delta (x-y)=  {\cal H}(y),
$$
but why??

A more explicit definition for $\sqrt O$ is clearest in 1d space. For
$$
\sqrt{O}\phi(x)\equiv \int\! dy~ K(x-y)\phi(y). ~~~\leadsto \\
O\phi[x]= \sqrt{O}\sqrt{O}\phi(x) =\int\! dydz~ K(x-y) K(y-z)\phi(z) ~~~\implies \\
O(x-z)= \int\! dy ~~K(x-y) K(y-z),
$$
the equation defining the kernel of the square root.
