11. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M$ and $N$ 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 Figure 5, draw $MP \perp AB$ at $P$, and connect $PN$. By the given conditions, we have $PN \perp AB$.

Therefore, $\angle MPN$ is the plane angle of the dihedral angle $D \cdot AB-E$, i.e., $\angle MPN = 120^{\circ}$.
Let $AM = FN = x$, then $MP = \frac{\sqrt{2}}{2} x, PN = \frac{\sqrt{2} - x}{\sqrt{2}}$.
$$
\begin{aligned}
\therefore MN^{2} & = \frac{x^{2}}{2} + \left(\frac{\sqrt{2} - x}{\sqrt{2}}\right)^{2} - 2 \times \frac{\sqrt{2}}{2} x \times \frac{\sqrt{2} - x}{\sqrt{2}} \times \cos 120^{\circ} \\
& = \frac{1}{2}\left(x^{2} - \sqrt{2} x + 2\right) = \frac{1}{2}\left(x - \frac{\sqrt{2}}{2}\right)^{2} + \frac{3}{4} \\
& (0 \leqslant x \leqslant \sqrt{2}).
\end{aligned}
$$

Thus, when $\boldsymbol{x} = \frac{\sqrt{2}}{2}$, $MN_{\min} = \frac{\sqrt{3}}{2}$;
when $x = 0$ or $\sqrt{2}$, $MN_{\max} = 1$.