How should I understand the Hilbert space of field theory on non-commutative spaces?

I have a naive question about quantum field theory on non-commutative spaces. I apologize if my question is somewhat vague. Also, I have a weak mathematical background, and I apologize if my question is mathematically unsound.

To start with, let us consider a real scalar field theory defined on a spatial lattice (the time direction is continuous). Each spatial lattice point $x$ is assigned a pair of conjugate operators $\phi(x)$ and $\pi(x)$, and a Hilbert space $\mathcal{H}_x$ on which $\phi(x)$ and $\pi(x)$ act. The Hilbert space for the entire theory can be thought of as the tensor product $\otimes_x\mathcal{H}_x$.

I have been reading the pedagogical review by Richard Szabo about quantum field theory on non-commutative space. It seems to me that the way to define a field theory on non-commutative space is simply replacing every product by a star product in the Euclidean action. Similar to a conventional field theory, we integrate over all possible configurations in the path integral. In some sense, a quantum field theory on a non-commutative space is simply a non-local field theory on a conventional space.

My question is, is it possible to find a Hamiltonian for quantum field theory on non-commutative space? I expect it is not going to be easy as the time direction is also non-local. If this is indeed possible, how should I think of the Hilbert space of such a theory?

Thank you,

Isidore

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If $x$ belongs to an infinite set as it seems from your question, there is nothing like $\otimes_x {\cal H}_x$. –  Valter Moretti Feb 25 '14 at 10:07
Hi @Moretti, I am content with a finite lattice of size $L\times L\times L$ with periodic boundary condition imposed. –  Isidore Seville Feb 25 '14 at 10:30
OK it works, otherwise there are two possibilities (1) use a Fock space, (2) define some notion of infinite tensor product relying upon some preferred choice of vectors, giving rise to non equivalent infinite tensor products. –  Valter Moretti Feb 25 '14 at 10:32
Hi @Moretti, how about the non-commutative ones? I expect it is perhaps impossible to think in terms of finite size lattices. –  Isidore Seville Feb 25 '14 at 10:35
Sorry, I do not know, it is too far from my research activity I think you are more informed than me on these topics! –  Valter Moretti Feb 25 '14 at 10:39