6. For squares $A E C D$ and $A B E F$ with side length $a$, the angle between the planes they lie in is $120^{\circ}$. Points $M$ and $N$ are on the diagonals $A C$ and $B F$ respectively, and $A M = F N$. The range of $M N$ is ( ).
(A) $\left(\frac{\sqrt{2}}{2} a, a\right]$
(B) $\left(\frac{\sqrt{3}}{2} a, a\right]$
(C) $\left[\frac{\sqrt{2}}{2} a, a\right]$
(D) $\left[\frac{\sqrt{3}}{2} a, a\right]$
6. (D).

As shown in Figure 1, draw $MP \perp AB$ at $P$, and connect $PN$, then $PN \perp AB$. Therefore, $\angle MPN$ is the plane angle of the dihedral angle $ABCD - A \sqrt{-} ABEF$, and $\angle MPN = 120^{\circ}$.

Let $AM = FN = x$, then
$$
\begin{aligned}
MP & = \frac{\sqrt{2}}{2} x, \quad PN = \frac{\sqrt{2} a - x}{\sqrt{2}}, \\
MN^{2} & = MP^{2} + PN^{2} - 2 MP \cdot PN \cos 120^{\circ} \\
& = \frac{1}{2} \left(x^{2} - \sqrt{2} a x + 2 a^{2}\right).
\end{aligned}
$$

According to the problem, $0 \leqslant x \leqslant \sqrt{2} a$, hence when $x = \frac{\sqrt{2}}{2} a$, $MN_{\text{min}} = \frac{\sqrt{3}}{2} a$.
Therefore, when $x = 0$ or $\sqrt{2} a$, $MN_{\operatorname{max}} = a$. Thus, the range of $MN$ is $\left[\frac{\sqrt{3}}{2} a, a\right]$.