decomposition of Hilbert space into tensor product $L^2([0,\tfrac{1}{2}]) \otimes L^2([\tfrac{1}{2},1]) \simeq L^2([0,1])$

Question

The definition of entanglement entropy in Quantum Field Theory involves decompositing a Hilbert space into a tensor product $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B$.

As an example, is it possible to decompose the $L^2$ hilbert space $L^2([0,1], dx)$ into the tensor product of two Hilbert spaces just by splitting the interval?

Is there "mutual information"? To what extent is it possible to construct the value of an $L^2$ function on $[\tfrac{1}{2},1]$ using information from $[0,\tfrac{1}{2}]$?

Or is there no mutual information, and it is possible to have an $L^2$ function on $[0,1]$ gluing together such functions on the left and right halves?

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I am looking for "reasonable" Hilbert spaces to plug into $\mathcal{H}_A, \mathcal{H}_B$ in the above discussion on Entanglement entropy. Likely it is the bosonic fock space. I would then wonder how to take the "partial trace" over $A$.

An interesting issue was raised as to what a natural isomorphism might be for $L^2([0,t]) \otimes L^2([t,1]) \simeq L^2([0,1])$ with $t \in [0,1]$. In general I am pleased to learn that at the isomorphism $$L^2(A) \otimes L^2(B) \simeq L^2(A \cup B)$.

Answer 1

I think the suggested example is not a good fit for illustrating a tensor product decomposition, because $L^2$ functions on an interval are most naturally identified with states of a single particle in the interval (with some potential). The tensor product is then identified with the states of a pair of particles in two intervals, or a single particle in the product of intervals. A particle in the disjoint union of intervals is best described by the Hilbert direct sum.

If you chose your Hilbert spaces to be Fock spaces instead of function spaces, i.e., allowing multi-particle states, then the tensor product decomposition is more natural. More concretely, we take $L^2(I)$, and apply some exponential functor like "symmetric algebra". If we ignore subtleties with completion, the direct sum becomes a tensor product, because $Sym(V \oplus W) \cong Sym(V) \otimes Sym(W)$. The phrase "second quantization" sometimes appears in situations like this (although people have warned me that it is outdated and confusing).

At any rate, once this is done, you'll find that there is no mutual information, as long as the overlap has measure zero.

Answer 2

This has essentially been cleared up by Christian Remling but let me be more precise since there still seems to be some confusion. We are dealing with two categories---measure spaces (say, finite) and Hilbert spaces, and the natural functor from the former to the latter which assigns to a probability space the corresponding $L^2$-space. Both spaces have finite sums and products---disjoint unions and cartesian products in the first case, Hilbert space sums and tensor products in the second. The above functor preserves sums and products in the obvious sense. The case in question exploits the simple fact that the unit interval is naturally identifiable with the measure theoretical sum of the two half intervals. There are, of course, more sophisticated versions which apply to infinite constructions---sums and products.