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Let $H$ be a separable Hilbert space and let $\{e_i\}$ be an orthonormal basis. My first question is:

Is there a probability measure on $B(H)$ such that for $T$ chosen uniformly randomly the matrix entries $\langle Te_i, e_j \rangle$ are i.i.d. Gaussians?

In case I didn't formulate that quite correctly, I'm trying to imitate in infinite dimensions the familiar idea that that one can choose a "random matrix" by viewing its entries as i.i.d. Gaussians.

In case a probability measure is too optimistic, is there some alternative notion of "random bounded linear map" on Hilbert space that one can use? Also, if it is too much to hope that all matrix entries are i.i.d. Gaussian, it might be enough for this to be true "approximately" in some sense, though I won't try to formulate what that would mean until I get a better sense of what can go wrong.


Edit: As pointed out in the comments, my requirements that $T$ be bounded and that the matrix elements are i.i.d. are inconsistent. So I would like to eliminate the requirement that that the distributions of the matrix elements are identical, though I still want them to be independent. Most importantly, I am hoping for an actual measure on $B(H)$ (or a proof that no such measure exists).

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    $\begingroup$ How could your $T$ possibly be bounded (assuming $H$ is infinite-dimensional of course)? $\endgroup$ – Martin Hairer May 14 '14 at 12:07
  • $\begingroup$ I don't quite get what do you mean by "chosen unifromly randomly". But if one multiplies a matrix with iid standard Gaussian entries by a non-random vector $h$ from $H$, the result is a well defined vector of iid centered Gaussians with variance $\|h\|^2$. Not sure if this helps you. $\endgroup$ – zhoraster May 14 '14 at 14:35
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    $\begingroup$ looks like you want to define an $N\times N$ GUE random matrix directly in the $N=\infty$ situation. You may look at free probability theory which does something of this kind although it is not what you had in mind, i.e., a probability measure on $B(H)$. $\endgroup$ – Abdelmalek Abdesselam May 14 '14 at 17:37
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    $\begingroup$ I guess you want some kind of non-degeneracy condition as well. Otherwise let $A$ be the operator that maps $e_1$ to $e_1$ and $e_2, e_3, \dots$ to 0, and let $T = \xi A$ where $\xi$ is real-valued Gaussian. $\endgroup$ – Nate Eldredge May 15 '14 at 3:36
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    $\begingroup$ Actually, without the "identically distributed" requirement, shouldn't it be pretty easy to let the entries of the "matrix" be independent Gaussians with variances chosen to guarantee that $T$ is bounded almost surely? $\endgroup$ – Nate Eldredge May 15 '14 at 3:44
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Here is a sort of arbitrary example that I just picked because it is easy to construct.

Let $\{\xi_{ij}\}$ be iid $N(0,1)$ random variables on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$, and let $\{a_{ij}\}$ be your favorite doubly-indexed sequence of real numbers satisfying $\sum_{i,j} |a_{ij}|^2 < \infty$. For instance, try $a_{ij} = 2^{-i-j}$. Then let the $ij$ entry of your random operator $T$ be $a_{ij} \xi_{ij}$. $T$ is then almost surely Hilbert-Schmidt and in particular bounded.

More formally, let $HS$ be the Hilbert space of Hilbert-Schmidt operators on $H$, and $\|\cdot\|_{HS}$ the Hilbert-Schmidt norm $\|A\|_{HS}^2 = \sum_{i,j} |\langle A e_i, e_j \rangle|^2$. Let $T_{ij}$ be the rank-one operator with $T_{ij} e_i = e_j$ and $T_{ij} e_k = 0$ for $k \ne i$. Note that $\|T_{ij}\|_{HS} = 1$. Consider the series $T = \sum_{i,j} a_{ij} \xi_{ij} T_{ij}$. We have $$ \sum_{i,j} \mathbb{E} \|a_{ij} \xi_{ij} T_{ij}\|^2_{HS} = \sum_{i,j} |a_{ij}|^2 \|T_{ij}\|_{HS}^2 \mathbb{E} \xi_{ij}^2 = \sum_{i,j} |a_{ij}|^2 < \infty$$ and therefore the series converges in the vector-valued Hilbert space $L^2(\Omega; HS)$, so $T$ makes sense as an $HS$-valued random variable. In particular, $\langle T e_i, e_j \rangle = a_{ij} \xi_{ij}$ which are independent Gaussians (and if you choose all $a_{ij}$ nonzero, they are all nondegenerate). Now the law of $T$ is a probability measure of the kind you desire.

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