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When working on the lattice it is easy to define a trivial product state. A state $|\psi\rangle$ is a trivial product state if it admits the following tensor decomposition, \begin{equation} |\psi\rangle=\bigotimes_{i}^N |\psi_i\rangle, \end{equation} where $|\psi_i\rangle$ is the state of a single a lattice site and $N$ is the number of lattice sites. A trivial product state is one with no entanglement since every lattice site is essentially doing its own thing.

This definition works perfectly when our space is a lattice with separate sites, but is there an analogous definition of trivial product state that works well in the continuum without ever using the lattice?

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Your lattice apparently has finite size (and we could assume cyclic boundaries for convenience). If you let $N\rightarrow\infty$ but keep finite size the limit is a continuum with cyclic boundaries.

To keep the energy finite in the limit process, each site on the lattice should decrease it's particle number expectation, e.g. we can choose: $$ |\psi_i\rangle = \frac1{\sqrt{1+\frac1N}}\ \big(1+{ \frac{a_i^\dagger}{\small\sqrt N}}\big)\ |0\rangle $$ so a superposition of (mainly) the vacuum and (a little bit of) a single particle state. The total continuum state is then: \begin{align} |\psi\rangle&=\lim_{N\rightarrow\infty}\ \frac1{(1+\frac1N)^{N/2}}\ \otimes_{i}^N \big(1+{ \frac{a_i^\dagger}{\small\sqrt N}}\big)\ |0\rangle \\ &= \frac1{\sqrt e}\ \lim_{N\rightarrow\infty}\ \otimes_{i}^N \big(1+{ \frac{a_i^\dagger}{\small\sqrt N}}\big)\ |0\rangle \end{align}

A mathematical technicality is that we do not have a notation for the continuum limit of the product (like the integral is the continuum limit of a summation) so it is then merely a problem of how to write down the limit in a closed form expression.

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  • $\begingroup$ Why you pick the states as superpositions? $\endgroup$
    – lcv
    Commented May 31 at 11:33
  • $\begingroup$ From your notation it seems that $a_i^\dagger$ creates a fermion or a boson. But taking tensor products is a bit of a contradiction then. A standard product would be the case. Unless of course the particles have no statistics $\endgroup$
    – lcv
    Commented May 31 at 11:38
  • $\begingroup$ Thank you for your answer, but I would like to know how one can construct the analogous of a trivial product state without ever using the lattice. I have edited the question to clarify this point. $\endgroup$
    – Truth and Beauty and Hatred
    Commented May 31 at 14:47
  • $\begingroup$ @lcv The question was for an "analogous definition" to the tensor decomposition given. The usual notation like $a^\dagger_i a^\dagger_2 \dots a^\dagger_N$ would of course suffice, but it is only hiding the fact that for bosons the total space is really a tensor product space and each $a^\dagger_i$ works in its own subspace. If fermions are present that is not true but then the original expression (for which we were to create an analogue) would have exactly that same problem. $\endgroup$
    – Jos Bergervoet
    Commented May 31 at 15:11
  • $\begingroup$ Well for bosons it is the symmetric tensor product and for fermion the antisymmetric one. That was my point. In any case no big deal, these symmetry considerations are just additional complications. $\endgroup$
    – lcv
    Commented May 31 at 16:29
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You can formally define such a state, but if the Hamiltonian has a term like $|\nabla \phi|^2$ in it, then that state will have an infinite energy. The reason is that a totally unentangled state will have no correlation between spatial regions, including infinitesimally separated ones. See this answer for more details.

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