Q1: Results obtained on 2018-05-30

Given $\langle J\rangle$ and $\Delta J$, characterize resultant dynamics with eigenvalues $\lambda_n$ of covariance or correlation matrix $C$.

We use a covariance matrix to calculate the principal components analysis (PCA). To fit the log-eigenvalues of PCA (blue), we use least-squares to find the optimize line (red).

$\langle J\rangle$ = 1 $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\alpha$ -0.40 -0.39 -0.39

In the first case, we set the parameter $h$ equal to zero and the parameter $J$ is a normal distribution n(1.0,0.1) with different random numbers in the nearest-neighbor spin glass model.

In the second case, we set the parameter $h$ equal to zero and the parameter $J$ is a normal distribution n(1.0,0.2) with different random numbers in the nearest-neighbor spin glass model.

In the third case, we set the parameter $h$ equal to zero and the parameter $J$ is a normal distribution n(1.0,0.3) with different random numbers in the nearest-neighbor spin glass model.

We use a covariance matrix to calculate the principal components analysis (PCA). To fit the log-eigenvalues of PCA (blue), we use least-squares to find the optimize line (red).

$\langle J\rangle$ = 1 $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\alpha$ -0.32 -0.315 -0.314

In the first case, we set the parameter $h$ equal to zero and the parameter $J$ is a normal distribution n(1.0,0.1) with different random numbers in the all spin glass model.

In the second case, we set the parameter $h$ equal to zero and the parameter $J$ is a normal distribution n(1.0,0.2) with different random numbers in the all spin glass model.

In the third case, we set the parameter $h$ equal to zero and the parameter $J$ is a normal distribution n(1.0,0.3) with different random numbers in the all spin glass model.


We use a covariance matrix to calculate the principal components analysis (PCA).

$\langle J\rangle$ = n([0.8-1.2]) $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\lambda_n$ -0.4 -0.4 -0.4

In the first case, we set the parameter $h$ equal to zero and the different parameters $J$, $\Delta J$ = 0.1 with the same random number in the nearest neighbor spin glass model.

In the second case, we set the parameter $h$ equal to zero and the different parameters $J$, $\Delta J$ = 0.2 with the same random number in the nearest neighbor spin glass model.

In the third case, we set the parameter $h$ equal to zero and the different parameters $J$, $\Delta J$ = 0.3with the same random number in the nearest neighbor spin glass model.


In the cases, we set the parameter $h$ equal to zero and the parameter $J$ is a normal distribution with the the random number (seed(1)) in the all spin glass model.

$\langle J\rangle$ = n([0.8-1.2]) $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\lambda_n$ -0.325 -0.3165 -0.32

In the first case, we set the parameter $h$ equal to zero and the different parameters $J$, $\Delta J$ = 0.1 with the same random number in the all spin glass model.

In the second case, we set the parameter $h$ equal to zero and the different parameters $J$, $\Delta J$ = 0.2 with the same random number in the all spin glass model.

In the third case, we set the parameter $h$ equal to zero and the different parameters $J$, $\Delta J$ = 0.3 with the same random number in the all spin glass model.

Q2

2. How much are the eigenvalues influenced by the realized coupling configuration? (With an ensemble created with different seeds of random number generator, do the eigenvalues change significantly? What are their variances?)

A2:

we have taken the 100 eigenvalues of covariance matrix, see the part-A1: we took log-scale, the pca fitting lines are taken from the 2nd-31th eigenvalues (red line).

Results of the nearest-neighbor spin glass model: We use a covariance matrix to calculate the principal components analysis (PCA) in log-log scale.

$\langle J\rangle$ = 1 $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\alpha$ -0.40 -0.39 -0.39
variance 0.06 0.06 0.06

Results of all spin glass model: We use a covariance matrix to calculate the principal components analysis (PCA) in log-log scale.

$\langle J\rangle$ = 1 $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\alpha$ -0.32 -0.315 -0.314
variance 0.08 0.09 0.08

Q3

3. Fit the eigenvalues with $$\lambda_n\sim n^{-\alpha}.$$ Perform systematic calculations to find the relationship between $\alpha$ and $\langle J\rangle$, $\Delta J$.

A3:

we have 100 eigenvalues token from the covariance matrix, see A1: we took log-scale, the pca values fitting curves are taken from the 2nd-31th eigenvalues. To fit the log-eigenvalues of PCA (blue), we use least-squares to find the optimize line (red), see part:A1.

Resilts of the nearest-neighbor spin glass model: we use a covariance matrix to calculate the principal components analysis (PCA).

$\langle J\rangle$ = 1 $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\alpha$ -0.40 -0.39 -0.39

Results of all spin glass model: we use a covariance matrix to calculate the principal components analysis (PCA).

$\langle J\rangle$ = 1 $\Delta J$ = 0.1 $\Delta J$ = 0.2 $\Delta J$ = 0.3
$\alpha$ -0.32 -0.315 -0.314

Q4

4. Are there other ways to characterize the activity data beside $\alpha$?

We calculate the probability distribution of the covariances in the data of all spin model. The positive long tail distribution has some strongly positive correlations, their distribution of J, see part: A5.

Q5

5. Fixing $\langle J\rangle$ and $\Delta J$, does forms of the distribution $P(J)$ influence $\alpha$? For example, does having a long tail matters?

A5:

We compare the normal distribution J=1.0 with $\Delta J =0.1$ to the normal distribution J=1.0 with $\Delta J =0.1$ having a long tail. In this case, the long tail has larger $\alpha = -0.29$ then normal distribution $\alpha = -0.32$.