11. Let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a cube with edge length 1. Then the minimum distance between a point $P$ on the incircle of the top face $A B C D$ and a point $Q$ on the circle passing through the vertices $A, B, C_{1}, D_{1}$ is $\qquad$.
11. $\frac{\sqrt{3}-\sqrt{2}}{2}$.

Let point $O$ be the center of the cube. It is easy to see that
$$
O Q=\frac{\sqrt{3}}{2}, O P=\frac{\sqrt{2}}{2} \text {. }
$$

By the triangle inequality, we have $P Q \geqslant O Q-O P=\frac{\sqrt{3}-\sqrt{2}}{2}$, with equality if and only if points $O$, $P$, and $Q$ are collinear.

Clearly, when point $P$ is the midpoint of segment $A B$, and the ray $O P$ intersects the circumcircle of rectangle $A B C_{1} D_{1}$ at point $Q$, the condition is satisfied.