11. (20 points) Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 2, point $P$ is on the incircle $\odot I$ of square $A B C D$, and point $Q$ is on the circumcircle $\odot O$ of square $A A_{1} D_{1} D$. Find the maximum and minimum values of $|P Q|$.
11. On the plane $x D y$, the equation of the incircle $\odot I$ of the square $A B C D$ is $(x-1)^{2}+(y-1)^{2}=1$;

On the plane $x D z$, the equation of the circumcircle $\odot O$ of the square $A A_{1} D_{1} D$ is $(x-1)^{2}+(z-1)^{2}=2$.
Let $P(1+\cos \alpha, 1+\sin \alpha, 0)$, $Q(1+\sqrt{2} \cos \beta, 0,1+\sqrt{2} \sin \beta)$.
Thus, $|P Q|^{2}$
$$
\begin{array}{l}
=(\cos \alpha-\sqrt{2} \cos \beta)^{2}+(1+\sin \alpha)^{2}+ \\
(1+\sqrt{2} \sin \beta)^{2} \\
= 5+2(\sin \alpha+\sqrt{2} \sin \beta-\sqrt{2} \cos \alpha \cdot \cos \beta) . \\
\text { Let } M=2(\sin \alpha+\sqrt{2} \sin \beta-\sqrt{2} \cos \alpha \cdot \cos \beta) .
\end{array}
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
M \leqslant 2 \sin \alpha+2 \sqrt{2\left(1+\cos ^{2} \alpha\right)} \\
\leqslant 2|\sin \alpha|+\sqrt{2\left(1+\cos ^{2} \alpha\right)}+\sqrt{2\left(1+\cos ^{2} \alpha\right)} \\
\leqslant \sqrt{(1+1+1)\left(4 \sin ^{2} \alpha+2\left(1+\cos ^{2} \alpha\right)+2\left(1+\cos ^{2} \alpha\right)\right)} \\
=2 \sqrt{6},
\end{array}
$$

When $\alpha=\arcsin \frac{\sqrt{6}}{3}, \beta=\frac{2 \pi}{3}$, $M$ reaches its maximum value $2 \sqrt{6}$;
$$
\begin{array}{l}
M \geqslant 2 \sin \alpha-2 \sqrt{2\left(1+\cos ^{2} \alpha\right)} \\
\geqslant-2|\sin \alpha|-\sqrt{2\left(1+\cos ^{2} \alpha\right)}-\sqrt{2\left(1+\cos ^{2} \alpha\right)} \\
\geqslant-\sqrt{(1+1+1)\left(4 \sin ^{2} \alpha+2\left(1+\cos ^{2} \alpha\right)+2\left(1+\cos ^{2} \alpha\right)\right)} \\
=-2 \sqrt{6},
\end{array}
$$

When $\alpha=\pi+\arcsin \frac{\sqrt{6}}{3}, \beta=\frac{5 \pi}{3}$, $M$ reaches its minimum value $-2 \sqrt{6}$.
Thus, $5-2 \sqrt{6} \leqslant|P Q|^{2} \leqslant 5+2 \sqrt{6}$
$$
\Rightarrow \sqrt{3}-\sqrt{2} \leqslant|P Q| \leqslant \sqrt{3}+\sqrt{2} \text {. }
$$

Therefore, the maximum value of $|P Q|$ is $\sqrt{3}+\sqrt{2}$, and the minimum value
is $\sqrt{3}-\sqrt{2}$.