Consider a family of $N \times N$ matrices $M(t)$ labeled by time $t$. At all times, the off diagonal components of the matrix are independently Gaussianly distributed according to $\mathcal{N}(0,1/N)$. The diagonal components, on the other hand, are distributed according to $\mathcal{N}(1,1/N)$. Now we consider the commutator of two matrices at different times: $$ G(t,t') = [M(t),M(t')] $$ In general, $G(t,t')$ does not vanish. However, when we take the limit as $N$ goes to infinity, I would hope there is a sense in which $G(t,t')$ is small because the off diagonal components are suppressed by $1/\sqrt{N}$ (that is the standard deviation).

My question: How small is $G(t,t')$? Is there a general result that we can prove about the probability distribution of $G(t,t')$? How about higher commutators like $[[M(t),M(t')],M(t'']$? (The answer to the second question would probably allow me to answer the third question by induction.)

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    $\begingroup$ Bad idea to use $O$ for this, because it conflicts with Big O notation $\endgroup$ – Robert Israel Jul 21 '17 at 21:12
  • $\begingroup$ You told us about the distribution of elements of $M(t)$ at a particular time $t$, but what about the elements at different times? $\endgroup$ – Robert Israel Jul 21 '17 at 21:18
  • $\begingroup$ The distribution is independently Gaussian at each time with no correlation between different times. $\endgroup$ – Zhengyan Shi Jul 21 '17 at 21:19
  • $\begingroup$ Fixed the big O notation. Thanks! @RobertIsrael $\endgroup$ – Zhengyan Shi Jul 21 '17 at 21:21

Let $M(t) = I + A/\sqrt{N}$ and $M(t') = I + B/\sqrt{N}$, where the entries of $A$ and $B$ are independent standard normal random variables. Then $G(t,t') = [M(t), M(t')] = [A,B]/N$.

$$ [A, B]_{ij} = \sum_{k=1}^N A_{ik} B_{kj} - \sum_{k=1}^N B_{ik} A_{kj} $$

$\sum_{k=1}^N A_{ik} B_{kj}$ is the sum of $N$ iid random variables of mean $0$ and variance $1$, and therefore has mean $0$ and variance $N$. Similarly with $\sum_{k=1}^N B_{ik} A_{kj}$. These two sums are uncorrelated (although dependent) except in the case $i=j$, where the term $A_{ii} B_{ii}$ is common to both. Thus if $i \ne j$, $[A, B]_{ij}$ has variance $2N$, while $[A, B]_{ii}$ has variance $2N - 2$. Correspondingly, $G(t,t')_{ij}$ has variance $2/N$ and $G(t,t')_{ii}$ has variance $2/N - 2/N^2$.

  • $\begingroup$ By uncorrelated, do you just mean zero covariance? $\endgroup$ – Zhengyan Shi Jul 21 '17 at 22:20
  • $\begingroup$ Yes, that is what uncorrelated means. $\endgroup$ – Robert Israel Jul 21 '17 at 23:36
  • $\begingroup$ Very nice! It's good to see that $G(t,t')$ is suppressed. I thought the generalization to higher commutators would be easy, but then I realized that the components of $G(t,t')$ are not independently distributed, which means we cannot apply the same technique. Any idea how to generalize? $\endgroup$ – Zhengyan Shi Jul 22 '17 at 0:15

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