related quantities to Subsection \[subsec:definitions\], but we state here the result that we aim to prove.

\[main2\] Let $(M,\tau)$ be a tracial $W^\ast$-probability space. Furthermore, let $X_1,\dots,X_n \in M$ be self-adjoint elements and assume that $\delta^\ast(X_1,\dots,X_n) = n$ holds.

Then, for any non-zero non-commutative polynomial $P$, there exists no non-zero element $w\in{\operatorname{vN}}(X_1,\dots,X_n)$ such that $$P(X_1,\dots,X_n) w = 0.$$

Recall that to each element $X=X^\ast\in M$ corresponds a unique probability measure $\mu_X$ on the real line ${\mathbb{R}}$, which has the same moments as $X$, i.e. it satisfies $$\tau(X^k) = \int_{\mathbb{R}}t^k\, d\mu_X(t) \qquad\text{for $k=0,1,2,\dots$}.$$ It is an immediate consequence of Theorem \[main2\] that the distribution $\mu_{P(X_1,\dots,X_n)}$ of $P(X_1,\dots,X_n)$ for any non-constant self-adjoint polynomial $P$ cannot have atoms, if $\delta^\ast(X_1,\dots,X_n)=n$. The precise statement reads as follows.

\[cor-main2\] Let $(M,\tau)$ be a tracial $W^\ast$-probability space and let $X_1,\dots,X_n \in M$ be self-adjoint with $\delta^\ast(X_1,\dots,X_n)=n$.

Then, for any non-constant self-adjoint non-commutative polynomial $P$, the distribution $\mu_{P(X_1,\dots,X_n)}$ of $P(X_1,\dots,X_n)$ does not have atoms.

Indeed, any atom $\alpha$ of the distribution $\mu_{P(X_1,\dots,X_n)}$, i.e. any $\alpha\in{\mathbb{R}}$ satisfying $\mu_{P(X_1,\dots,X_n)}(\{\alpha\})\neq 0$, leads by the spectral theorem for bounded self-adjoint operators on Hilbert spaces to a non-zero projection $u$ satisfying $(P(X_1,\dots,X_n) - \alpha 1)u = 0$. Thus, applying Theorem \[main2\] yields 