Self-adjointness I know I have posted this question before some time ago. But no one could help so I decided to put my problem in another background.
The Schrödinger equation of a free scalar field is given by
$i\partial_{t}\Psi[\Phi,t]~=~\underset{A}{\underbrace{\frac{1}{2}\int d^{3}x\left(-\frac{\delta^{2}}{\delta\Phi^{2}(\vec{x})}+|\nabla\Phi|^{2}+m^{2}\Phi^{2}\right)}}\Psi[\Phi,t]$.
This is the Schrödinger representation of QFT.
Now I want to know, whether the operator $A$ on the r.h.s is essentially self-adjoint?
Any idea or advise?
My problem is how to handle the functional derivative here.
 A: The so-called $Q$-space may not be the best representation to discuss your question. I suggest you look at the (unitarily) equivalent representation of Fock spaces. Let assume your field is bosonic, and consider the Fock space
$$\Gamma_s(L^2(\mathbb{R}^d))=\bigoplus_{n=0}^\infty L^2_s(\mathbb{R}^{nd})$$
where $L^2_s$ is the space of symmetric functions (with respect to the exchange of variables), and the convention $L^2_s(\mathbb{R}^0):=\mathbb{C}$. In this space, your operator $A$ is commonly defined as the second quantization of $\omega_x=\sqrt{-\Delta_x + m^2}$, or in Fourier transform of $\omega(k)=\sqrt{k^2 + m^2}$. We define symbolically $A'=d\Gamma(\omega)$ (in either case). Let $f\in L^2_s(\mathbb{R}^nd)$, then
$$ (d\Gamma(\omega) f)(x_1,\dotsc,x_n)=\sum_{j=1}^n \omega_{x_j}\, f(x_1,\dotsc,x_n)\; .$$
Define $D(\omega^n)$ to be the domain of the self adjoint operator $\sum_{j=1}^n \omega_{x_j}$ on $L^2_s(\mathbb{R}^{nd})$. So the second quantization acts as the sum of the action of the one-particle operators on each variable. It is a self adjoint unbounded operator, with domain of essential self-adjointness
$$D=\{\phi=(\phi_0,\dotsc,\phi_n,\dotsc)\in \Gamma_s(L^2) \text{ with each } \phi_n\in D(\omega^n)\text{ such that }\phi_n=0 \text{ for all but a finite number of $n$} \} $$
$A'$ is unitarily equivalent to $A$, so also $A$ is a self-adjoint operator.
See the second volume of Reed and Simon, the section on free quantum fields.
