6. Given the parabola $y^{2}=2 p x(p>0), P$ is a moving point on the negative half-axis of the $x$-axis, $P A 、 P B$ are tangent lines to the parabola, and $A 、 B$ are the points of tangency. Then the minimum value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $\qquad$ .
6. $-\frac{p^{2}}{4}$.

Let the equation of the tangent line be $x=t y+m$.
Substituting into the parabola equation yields
$$
y^{2}-2 p t y-2 p m=0 \text {. }
$$

From the condition that the line is tangent to the parabola, we get
$$
\begin{array}{l}
\Delta=4 p^{2} t^{2}+8 p m=0 \\
\Rightarrow m=-\frac{p t^{2}}{2}, y=p t, x=\frac{p t^{2}}{2} .
\end{array}
$$

Thus, $A\left(\frac{p t^{2}}{2}, p t\right), B\left(\frac{p t^{2}}{2},-p t\right), P\left(-\frac{p t^{2}}{2}, 0\right)$.
Therefore, $\overrightarrow{P A} \cdot \overrightarrow{P B}=\left(p t^{2}, p t\right) \cdot\left(p t^{2},-p t\right)$
$$
=p^{2} t^{4}-p^{2} t^{2}=p^{2}\left(\left(t^{2}-\frac{1}{2}\right)^{2}-\frac{1}{4}\right) \text {. }
$$

When $t= \pm \frac{\sqrt{2}}{2}$, $\overrightarrow{P A} \cdot \overrightarrow{P B}$ attains its minimum value $-\frac{p^{2}}{4}$.