11. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M, N$ are points on the diagonals $A C$ and $B F$ respectively. And $A M=F N$, if $A B=1$, then the range of values for $M N$ is $\qquad$
11. $\left[\frac{\sqrt{3}}{2}, 1\right]$

As shown in the figure, draw $M P \perp A B$ at $P$, connect $P N$. According to the problem, we can get $P N \perp A B$,
so $\angle M P N$ is the dihedral angle of the plane $D-A B$
$-E$, that is, $\angle M P N=120^{\circ}$,
Let $\Lambda M=F N=x$, then $M P=\frac{\sqrt{2}}{2} x, P \mathrm{~V}=\frac{\sqrt{2}-x}{\sqrt{2}}$, so
$$
\begin{aligned}
M N^{2} & =\frac{x^{2}}{2}+\left(\frac{\sqrt{2}}{\sqrt{2}}\right)^{2}-2 \frac{\sqrt{2}}{2} x \cdot \frac{\sqrt{2}-x}{\sqrt{2}} \cdot \cos 120^{\circ} \\
& =\frac{1}{2}\left(x^{2}-\sqrt{2} x+2\right) \\
& =\frac{1}{2}\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\frac{3}{4} \cdot(0 \leqslant x \leqslant \sqrt{2})
\end{aligned}
$$

Therefore, when $x=\frac{\sqrt{2}}{2}$, $M \mathrm{~N}_{\text {min }}=\frac{\sqrt{3}}{2}$.
When $x=0$ or $\sqrt{2}$, $M N_{\text {max }}=1$.