1. As shown in Figure 3, the side length of square $A B C D$ is 1, and an equilateral triangle $\triangle A B E$ is constructed inside the square with $A B$ as one side. Point $P$ is the center of this equilateral triangle, and $P D$ intersects $A E$ at point $F$. Then $P F=$ $\qquad$
2.1. $\sqrt{\frac{11}{6}-\sqrt{3}}$.
Let $PE$ intersect $CD$ at point $M$, and intersect $AB$ at point $N$. Then $PM=1-PN=1-\frac{\sqrt{3}}{6}, PE=\frac{\sqrt{3}}{3}$, $PD=\sqrt{DM^{2}+PM^{2}}=\sqrt{\frac{4-\sqrt{3}}{3}}$, $PE \parallel AD$,
$$
\frac{PF}{FD}=\frac{PE}{AD}=\frac{\sqrt{3}}{3}, \frac{PF}{PD}=\frac{\sqrt{3}}{3+\sqrt{3}}=\frac{\sqrt{3}-1}{2} \text {. }
$$

Thus, $PF=\frac{\sqrt{3}-1}{2} PD=\sqrt{\frac{11}{6}-\sqrt{3}}$.