1
$\begingroup$

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.)

$\endgroup$
  • 2
    $\begingroup$ 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
1
$\begingroup$

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 $2/N - 2/N^2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.