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:


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).

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    $\begingroup$ I propose that this should be migrated to physics.stackexchange.com $\endgroup$ – alphacapture Feb 10 at 15:17
  • $\begingroup$ @alphacapture. I disagree. This is certainly more mathematics than physics. $\endgroup$ – md2perpe Feb 10 at 17:01
  • $\begingroup$ @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. $\endgroup$ – alphacapture Feb 10 at 17:10
  • $\begingroup$ @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". $\endgroup$ – md2perpe Feb 10 at 17:21
  • $\begingroup$ Anyway, $\lVert \omega_\perp \rVert$ tends to $0$ very fast (exponentially), so I would say that the answer to the question is affirmative. $\endgroup$ – md2perpe Feb 10 at 17:24

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