# Convergence of measure of products of random unitaries

I'm trying to read Convergence conditions for random quantum circuits by Emerson, Livine, Llyod (https://doi.org/10.1103/PhysRevA.72.060302), arXiv version: (https://arxiv.org/abs/quant-ph/0503210) which as I understand is a sort of "Central Limiting Theorem" for products of random unitaries.

Stated in not so rigorous, i.e. physics language it is the statement that if you compose unitaries drawn independently from a universal gate set (could be continuous, or discrete) then the distribution of such long products (i.e., "deep circuits") tends towards the Haar distribution.

In more mathematical language the measure of a product of unitaries $$g = g_m \cdots g_2 g_1$$ where each unitary $$g_i$$ is drawn independently from a probability measure $$f$$ over the full unitary group $$G$$, which generates the full group, converges exponentially fast in $$m$$ to the constant function w.r.t. the Haar measure on the group.

Broadly speaking I understand the logic is that they analyze the distribution of the product of unitaries, which is given by multiple convolutions of the underlying distribution $$f$$. A Fourier transform then turns the analysis of convolutions into an analysis of products of the Fourier components of the underlying distribution. The Fourier components $$\hat{f}^s$$ of $$f$$ are organized in terms of unitary irreps $$s$$ of $$G$$, where the trivial representation $$s=0$$ has dimensionality $$d_0 = 1$$. The claim is then (I guess for suitably well-behaved $$f$$s) Fourier components for irreps $$s \neq 0$$ upon raising them to a high power are subleading to the $$\hat{f}^{s=0}$$ component, so only this component survives, which is Fourier coefficient of the constant function.

The crux seems to be showing that Fourier components $$\hat{f}^s$$ for $$s\neq0$$ has norm less than dimension $$d_s$$ so that $$(\hat{f}^s/d_s) < 1$$ and hence $$(\hat{f}^s/d_s)^m \to 0$$ as $$m\to \infty$$ (where $$m$$ is the number of convolutions/products) --- this is Eq. (9).

Indeed, it is stated that for all vectors $$x$$ in the $$s$$ representation, \begin{align} |\hat{f}^s x| = |d_s \int_G d\mu(g) f(g) \overline{D}^{(s)}(g) x|_2 \leq d_s \int_G d\mu(g) f(g) \| \overline{D}^{(s)}(g) \| |x|_2 = d_s |x|_2 \end{align} where $$d\mu(g)$$ is the Haar measure on $$G$$ and $$\overline{D}^{(s)}(g)$$ is (as I understand) is the representation of element $$g \in G$$ in the $$s$$-irrep (so that $$\| \overline{D}^{(s)}(g)\|=1$$). Note the 1st term within the modulus is the (matrix-valued) Fourier coefficient $$\hat{f}^s$$ acting on vector $$x$$, and the 2nd term within the modulus without $$x$$ is the definition of $$\hat{f}^s$$ -- it is just extracting the Fourier component in the $$s$$-irrep like in standard Fourier analysis $$\hat{f}(k) = \int dx f(x)e^{-ikx}$$.

It is also stated (top of page 3 of the arXiv version) that equality holds iff $$D(g)x = \xi(g) y$$ is true for all $$g$$ for which $$f(g) > 0$$ where $$\xi(g) \in \mathbb{C}$$. And that since any $$g \in G$$ can be generated by the support of $$f$$, $$D(g) x = \xi(g) y$$ must hold also for all $$g \in G$$. This implies that we have a 1d rep of $$G$$ embedded in the irrep $$s$$.

Here is where I do not understand:

1. Why does equality hold iff $$D(g)x = \xi(g) y$$ is true for all $$g$$ for which $$f(g) > 0$$? If I plug this into the integral, I get $$\hat{f}^s x = d_s \int_G d\mu(g) f(g) \overline{\xi}(g) \overline{y}$$. But I don't see why the RHS has norm $$d_s \|x\|_2$$. I also don't see the other direction.

2. The second statement that since any $$g \in G$$ can be "generated by the support of $$f$$" the relation must hold for all $$g \in G$$ is cryptic to me. Could someone explain?

3. Lastly, why does equality imply we have a 1d rep of $$G$$ embedded in the irrep $$s$$?

Thank you.

• For 1, it seems like it is the continuous version of the following fact: in a Euclidean space if $v_1,\ldots,v_n$ are vectors such that $v_1\neq 0$ and $|v_1+\cdots+v_n|=|v_1|+\cdots+|v_n|$ then all $v_2,\ldots,v_n$ are scalar multiples of $v_1$. Mar 6, 2022 at 19:24
• Do you have a response to the answer below? Mar 14, 2022 at 0:05

For a given probability distribution $$P$$ [on a compact group $$G$$] the limit of $$P^{n*}$$, $$n\to\infty$$, exists if and only if [the support] $$s(P)$$ [of $$P$$] is not contained in any coset of any closed, proper, normal subgroup of $$G$$. The limit of $$P^{n*}$$ is normalized Haar measure on $$G$$.