# Infinite tensor product identity

Consider the Hilbert space $$H = \mathbb{R}^2$$ spanned by $$\mathbf{e}_1$$ and $$\mathbf{e}_2$$. A number of authors have considered the infinite tensor product $$\mathcal{H} = \otimes_{i=1}^N H_i$$, $$N\to\infty$$ which has an uncountable basis isomorphic to the set of all binary sequences $$(x_i), x_i \in \{0,1\}$$, and in particular, have considered the element $$\omega = \otimes_{i=1}^N v_i$$, $$N\to\infty$$, where $$v_i$$ is a unit vector $$v_i = a\mathbf{e}_1 + b \mathbf{e}_2$$, $$a^2+b^2 = 1$$.

By simple binomial expansion, the vector $$\omega$$ is a sum of terms of the form $$a^{N-\sum_i x_i} b^{\sum_i x_i} \mathbf{e}_{x_1} \otimes \mathbf{e}_{x_2} \dots \otimes \mathbf{e}_{x_N} \cong a^{N-\sum_i x_i} b^{\sum_i x_i} (x_1, x_2, \dots, x_N)$$.

We can consider the set of all sequences $$(x_i)$$ for which $$\lim_{N\to\infty} \left|\frac{1}{N}\sum_{i=1}^N x_i - b^2\right| < \epsilon$$ for some small positive $$\epsilon$$, as well its complement, the set of all sequences for which $$\lim_{N\to\infty}\left|\sum_{i=1}^N \frac{1}{N} x_i - b^2\right| \ge \epsilon$$. Corresponding to these are the subspaces $$\mathcal{H}_\parallel$$ and $$\mathcal{H}_\perp$$

Then we can project $$\omega$$ onto the subspaces of $$\mathcal{H}_\parallel$$ and $$\mathcal{H}_\perp$$, i.e., write $$\omega = \omega_\parallel + \omega_\perp$$. It is possible to show, using Hoeffding's inequality (or Chernoff's inequality), that $$\lVert \omega_\perp \rVert^2 \le \lim_{N\to\infty} 2e^{-2\epsilon^2 N} = 0.$$ From here it is stated that $$\omega = \omega_\parallel$$. However, I'm not sure that last conclusion is justified, and it seems that there's a snake in the grass. Is the fact that the norm of $$\omega_\perp$$ vanishes enough to conclude that "nothing is lost" by setting $$\omega = \omega_\parallel$$?

For context, this problem arises in the quantum mechanics of systems with infinitely many degrees of freedom:

Update:

I think what's confusing me is that the coefficients in the "binomial expansion" of $$\omega$$ go to zero as $$N \to \infty$$ but the norm squared of $$\omega$$ is the sum of the squares of these coefficients, and in order for that sum to be finite it must be the sum of countably many non-zero terms (as opposed to a bunch of zeros).

• I propose that this should be migrated to physics.stackexchange.com – alphacapture Feb 10 at 15:17
• @alphacapture. I disagree. This is certainly more mathematics than physics. – md2perpe Feb 10 at 17:01
• @md2perpe The question is about whether the bound on the norm of the error implies "nothing is lost" in quantum mechanics, is it not? If so, then it is a question about quantum mechanics; If not, I don't understand what the question is asking. – alphacapture Feb 10 at 17:10
• @alphacapture. Maybe you are right. To me the question seemed to be about the limit of $\lVert \omega_\perp \rVert,$ but I see now that it is written "It is possible to show" not "Is it possible to show". – md2perpe Feb 10 at 17:21
• Anyway, $\lVert \omega_\perp \rVert$ tends to $0$ very fast (exponentially), so I would say that the answer to the question is affirmative. – md2perpe Feb 10 at 17:24