11 Let $f(x)=x^{2}+a x$, and
$$
\{x \mid f(x)=0, x \in \mathbf{R}\}=\{x \mid f(f(x))=0, x \in \mathbf{R}\} \neq \varnothing,
$$

then the range of real numbers $a$ that satisfy the condition is . $\qquad$
11 Because the roots of $f(x)=0$ are $x=0$ or $x=-a$, $f(f(x))=0$ can be transformed into $f(x)=0$ or $f(x)+a=0$. According to the problem, $f(x)+a=0$ has no solution, or the roots of $f(x)+a=0$ are $x=0$ or $x=-a$. And $f(x)+a=0$, which is $x^{2}+a x+a=0$, by $\Delta=a^{2}-4 a<0$ or $a=0$, we know $0 \leqslant a<4$.