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 vectors $\overrightarrow{P A}$ and $\overrightarrow{P Q}$ has a minimum value of $\qquad$
4. $\frac{3}{4}$.

Let's assume points $A(0,0), B(2,0), D(0,1)$.
Let point $P(t, 1)(0 \leqslant t \leqslant 2)$.
Then, by $|\overrightarrow{D P}|=|\overrightarrow{B Q}|$, we get $Q(2,-t)$.
$$
\begin{array}{l}
\text { Hence } \overrightarrow{P A}=(-t,-1), \overrightarrow{P Q}=(2-t,-t-1) \\
\Rightarrow \overrightarrow{P A} \cdot \overrightarrow{P Q}=(-t)(2-t)+(-1)(-t-1) \\
\quad=t^{2}-t+1=\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4} .
\end{array}
$$

When $t=\frac{1}{2}$, $(\overrightarrow{P A} \cdot \overrightarrow{P Q})_{\text {min }}=\frac{3}{4}$.