Let $L$ be the left shift operator on $\ell^2(\mathbb{Z})$ with trace $\tau(T) := \langle T \delta\_0, \delta\_0 \rangle$. 


How can I show that the Brown measure of $L$ is the uniform measure on the unit circle?


The Brown measure is defined as follows: For each $z \in \mathbb{C}$ define $\nu\_{z}$ to be the spectral measure of the (self-adjoint) operator $(L - z)^\*(L-z)$. Then let $$f(z) = \frac{1}{2} \int\_0^\infty \log x \,d\nu\_z(x).$$ Then the Brown measure is defined as $$\mu\_L := \frac{1}{2\pi} \Delta f,$$ where the Laplacian is taken in the sense of distributions. Motivation for the definition can be found at [Terence Tao's notes on the circular law for random matrices](https://terrytao.wordpress.com/2010/03/14/254a-notes-8-the-circular-law/).


Edit: posted at math stackexchange <https://math.stackexchange.com/questions/2258115/brown-measure-of-left-shift-operator>



Brown measure of left shift operator
Since the two-sided shift is a normal operator this is not really a question about the Brown measure, but about the spectral measure, so it suffices to compute the $\*$-moments $\tau(L^nL^{\*m})$. For an introduction to Brown measure you might also see Chapter 11 of the book
<https://www.math.uni-sb.de/ag/speicher/publikationen/Mingo-Speicher.pdf>


