Infinite Tensor Products

A countably infinite tensor product is not determined by repeatedly writing the finite tensor symbol. One useful Hilbert-space construction chooses a normalized reference vector in each factor and completes finite deviations from that reference. Different reference preparations can lead to different representations of the same local observable algebra.

Helpful background. Algebraic Tensor Products and Completion constructs finite tensor products. Normal States supplies monotone continuity. Fock Spaces in Rigorous Language distinguishes a particle-sector direct sum from an infinite product.

Reference vectors define compatible embeddings

Let $(\mathcal H_n)_{n\geq1}$ be separable Hilbert spaces and choose unit vectors $e_n\in\mathcal H_n$. Form the finite products $\mathcal K_N=\bigotimes_{n=1}^N\mathcal H_n$ and embed them isometrically by

$$\mathcal K_N\longrightarrow\mathcal K_{N+1}, \qquad x\longmapsto x\otimes e_{N+1}.$$

Identify vectors under these embeddings and complete the resulting increasing union. Denote the Hilbert space by $\bigotimes_n^{\,e}\mathcal H_n$. It is often called an incomplete infinite tensor product relative to the reference sequence.

An elementary vector has arbitrary entries in finitely many factors and reference entries thereafter. Inner products are finite products once both vectors are written at a common cutoff. The distinguished unit vector $\Omega_e$ has reference entries everywhere.

The space is separable because it is the closure of a countable union of separable finite products. The reference choice is part of this construction, not disposable notation.

Infinite product vectors require a tail test

Given another unit sequence $(f_n)$, consider the finite replacements

$$\Psi_N=f_1\otimes\cdots\otimes f_N\otimes e_{N+1}\otimes e_{N+2}\otimes\cdots.$$

These are defined elements of the reference space. For $M>N$,

$$\langle\Psi_N,\Psi_M\rangle =\prod_{n=N+1}^M\langle e_n,f_n\rangle,$$

so

$$\|\Psi_M-\Psi_N\|^2 =2-2\operatorname{Re} \prod_{n=N+1}^M\langle e_n,f_n\rangle.$$

The precise convergence requirement is therefore uniform convergence of these tail products to one. The sufficient condition $\sum_n|1-\langle e_n,f_n\rangle|<\infty$ guarantees it, but arbitrary phase choices must be handled with care.

If the phases of the $f_n$ are chosen so the overlaps are nonnegative real, the condition becomes

$$\sum_n\bigl(1-|\langle e_n,f_n\rangle|\bigr)<\infty.$$

This is necessary and sufficient for the phase-adjusted sequence to be Cauchy. Beyond any finitely many zero overlaps, the equivalence follows from convergence of products in $(0,1]$ and the comparison $-\log t\sim1-t$ near one. A finite number of zero overlaps is harmless: it represents finitely many exactly orthogonal replacements.

One should not instead demand a nonzero overlap with $\Omega_e$. A valid finite excitation can be orthogonal to the reference vacuum.

Small rotations accumulate over infinitely many sites

For qubits, take $e_n=|0\rangle$ and

$$f_n=\cos\theta_n|0\rangle+\sin\theta_n|1\rangle, \qquad 0\leq\theta_n\leq\pi/2.$$

When $\theta_n\to0$, the overlap criterion is equivalent to $\sum_n\theta_n^2<\infty$. Rotations of size $1/n$ therefore give a product-vector limit in the reference space, whereas rotations of size $1/\sqrt n$ do not.

A fixed nonzero angle at every site also fails. Every finite truncation is a legitimate normalized vector, yet the sequence of growing preparations has no norm limit in this representation. Finite-system implementability does not establish an infinite-system limit.

This example concerns tensor-product sectors. The inequivalent CCR representations page studies a related representation issue through a different, bosonic implementability theorem.

A valid product state can lie outside the reference sector

Again take the all-zero qubit reference. Replacing the first $N$ sites by $|1\rangle$ gives mutually orthogonal vectors as $N$ varies. They do not approach an all-one vector in this Hilbert space.

Nevertheless the all-one preparation defines consistent expectations on every finite set of sites. For a local product observable, assign the product of its one-site expectations in $|1\rangle$. These positive normalized functionals agree when identities are appended, and extend continuously to the norm closure of the finite-site observable algebras.

The resulting state is not represented by a density operator in the all-zero reference representation. To see this, let $R_N$ be the local projector requiring the first $N$ sites all to be one. On every finite-excitation basis vector of the reference Hilbert space, $R_N$ eventually vanishes. Uniform boundedness and density give $R_N\to0$ strongly.

A normal state in that representation must therefore satisfy $\operatorname{Tr}(\rho R_N)\to0$. The all-one product state instead assigns value one to every $R_N$. The contradiction proves that it is outside the normal state space of the chosen representation.

There is no failure of the infinite product preparation itself. The failure is representing every such preparation as a density operator in one selected reference Hilbert space.

Full products and local algebras answer different questions

Von Neumann’s full infinite tensor-product construction includes many reference sectors and can be nonseparable even when all factors are two-dimensional. The reference completion used here deliberately selects one sector. Neither should be confused with a Fock direct sum over finite particle numbers.

For infinite lattice systems, the norm-closed local observable algebra provides a reference-independent starting point. A state then selects a Hilbert-space representation through its expectation values. Whether two such representations are equivalent is additional information; an abstract isomorphism of their Hilbert spaces is not enough.

Exercises

1. For $\theta_n=n^{-\alpha}$, $\alpha>0$, determine when the phase-adjusted product vector exists in the all-zero reference space.

Solution

The criterion is convergence of $\sum_nn^{-2\alpha}$, which holds exactly for $\alpha>1/2$. At $\alpha=1/2$, the logarithmic divergence prevents tail products from approaching one uniformly.

2. Why does changing only the first factor to an orthogonal vector not obstruct existence?

Solution

The replacement sequence is constant after the first cutoff. All sufficiently late tail overlaps equal one. The resulting unit vector is orthogonal to $\Omega_e$, showing why convergence depends on tails rather than a nonzero full overlap with the reference.

References

• Bratteli, Ola, and Derek W. Robinson, Operator Algebras and Quantum Statistical Mechanics 1, second edition, Springer, 1987 — local observable algebras, states, and representations. Publisher.
• von Neumann, John, “On Infinite Direct Products,” Compositio Mathematica 6, 1–77 (1939) — full and incomplete infinite tensor-product constructions. Journal archive.