4. In rectangle $A B C D$, $A B=2, A D=1$, point $P$ on side $D C$ (including points $D, C$) and point $Q$ on the extension of $C B$ (including point $B$) satisfy $|\overrightarrow{D P}|=|\overrightarrow{B Q}|$, then the dot product $\overrightarrow{P A} \cdot \overrightarrow{P Q}$ of vector $\overrightarrow{P A}$ and vector $\overrightarrow{P Q}$ has a minimum value of $\qquad$.
Let $\overrightarrow{P A}=\overrightarrow{P D}+\overrightarrow{D A}=-t \overrightarrow{D C}+\overrightarrow{D A}$, 
then we have $\overrightarrow{P Q}=\overrightarrow{P C}+\overrightarrow{C Q}=(1-t) \overrightarrow{D C}+(1+2 t) \overrightarrow{D A}$. 
Thus, $\overrightarrow{P A} \cdot \overrightarrow{P Q}=t(t-1) \overrightarrow{D C}^{2}+(1+2 t) \overrightarrow{D A}^{2}=4 t(t-1)+1+2 t$
$=4 t^{2}-2 t+1=4\left(t-\frac{1}{4}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4}$, equality holds when $\overrightarrow{D P}=\frac{1}{4} \overrightarrow{D C}$.
Therefore, the minimum value of the dot product $\overrightarrow{P A} \cdot \overrightarrow{P Q}$ of vector $\overrightarrow{P A}$ and vector $\overrightarrow{P Q}$ is $\frac{3}{4}$.