3
$\begingroup$

After going through an application of the von Neumann entropy(from quantum information theory) to certain problems in computational neuroscience [2], it occurred to me that this entropy might have applications to classical mechanics as it appears to be generally useful for dimensionality reduction.

Motivation:

Let's suppose we have an ergodic dynamical system with $N$ observables $x_i(t) \in \mathbb{R}$ which are sampled using a sequence of $n$ measurements so we have a dataset $X \in \mathbb{R}^{N \times n}$. Given $X$, we may compute the statistics $X_i = x_i - \langle x_i \rangle$ and $\sigma_i^2 = \langle X_i^2 \rangle$ which allows us to define the Pearson correlation matrix with entries:

\begin{equation} R_{i,j} = \frac{X_i \cdot X_j}{\sigma_i \cdot \sigma_j} \tag{1} \end{equation}

Given $R \in \mathbb{R}^{N \times N}$, we may define the density matrix $\rho = \frac{R}{N}$ which is positive semi-definite, Hermitian and has unit-trace. Thus, we may calculate the entropy of $R$ using the von Neumann entropy:

\begin{equation} S(\rho) = -\text{tr}(\rho \cdot \ln \rho) = - \sum_{i=1}^N \lambda_i \cdot \ln \lambda_i \tag{2} \end{equation}

where $\lambda_i$ are the eigenvalues of $\rho$.

If $N$ is large, in order to compress the dataset $X$ so that we keep the dimensions that contain $95\%$ of the statistical information it is sufficient to find the discrete subset $S \subset [1,N]$ that maximises:

\begin{equation} - \sum_{i \in S} \lambda_i \cdot \ln \lambda_i \tag{3} \end{equation}

subject to the constraint $\frac{- \sum_{i\in S} \lambda_i \cdot \ln \lambda_i}{- \sum_{i=1}^N \lambda_i \cdot \ln \lambda_i} \leq \frac{95}{100}$ which may be done using sorting algorithms such as Quick Sort. Moreover, the cardinality $\lvert S \rvert$ provides us with an approximate upper-bound on the intrinsic dimension of an ergodic dynamical system.

Question:

Has the von Neumann entropy ever been used in classical mechanics?

Note:

It is worth noting that the idea of viewing the Pearson correlation matrix as a density matrix is quite natural and not new [4].

References:

  1. von Neumann, John (1932). Mathematische Grundlagen der Quantenmechanik (Mathematical foundations of quantum mechanics) Princeton University Press., . ISBN 978-0-691-02893-4.

  2. H. Felippe et al. The von Neumann entropy for the Pearson correlation matrix: A test of the entropic brain hypothesis. Arxiv. 2021.

  3. E.T. Jaynes. Information theory and statistical mechanics. The Physical Review. 1957.

  4. linello (https://physics.stackexchange.com/users/10941/linello), Can a correlation matrix be regarded as a quantum density matrix?, URL (version: 2016-09-28): https://physics.stackexchange.com/q/282904

$\endgroup$

1 Answer 1

2
$\begingroup$

You are really asking just about the applications of the notion of entropy to an analysis of correlation matrices, so that this question is not necessarily related with classical mechanics. The entropy here is the plain Boltzmann - Planck - Shannon entropy rather than von Neumann's one, as the mere appearance of a positive definite symmetric matrix does not necessarily warrant referring to the whole quantum mechanics machinery. The keyword is "principal component analysis" (PCA). A quick search on "entropy + principal component analysis" returns quite a few articles (overwhelmingly paramathematical though). For example, Component retention in principal component analysis with application to cDNA microarray data where the entropy is used to provide an estimate of the number of interpretable components in a principal component analysis, or Entropy principal component analysis and its application to nonlinear chemical process fault diagnosis where "entropy principal component analysis" appears right in the title.

$\endgroup$
2
  • $\begingroup$ I initially thought that, but in order to correctly apply the Von Neumann entropy to dynamical systems we first need a general and robust method for the exact and global linearisation of nonlinear dynamical systems. This is provided by methods for approximating the Koopman Operator. Would you not agree? $\endgroup$ Oct 8, 2021 at 14:06
  • $\begingroup$ @Aidan Rocke - The Koopman operator is a completely different issue. In my opinion your answer should appear as a new separate question. $\endgroup$
    – R W
    Oct 8, 2021 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.