8. Given that $A$ and $B$ are points on $C_{1}: x^{2}-y+1=0$ and $C_{2}: y^{2}-x+1=0$ respectively. Then the minimum value of $|A B|$ is $\qquad$
8. $\frac{3 \sqrt{2}}{4}$.

Since the parabolas $C_{1}$ and $C_{2}$ are symmetric with respect to the line $y=x$, points $A$ and $B$ are also symmetric with respect to the line $y=x$. Otherwise, the symmetric points $A'$ and $B'$ of $A$ and $B$ with respect to $y=x$ would lie on the other parabola, and $|AB| = |A'B'|$.
Let $AB$ and $A'B'$ intersect at point $M$. Then
$$
\begin{array}{l}
|AB| + |A'B'| \\
= |MA| + |MB| + |MA'| + |MB'| \\
> |AA'| + |BB'| .
\end{array}
$$

Therefore, either $|AA'|$ or $|BB'|$ must be less than $|AB|$, which is a contradiction.

Thus, we only need to find twice the minimum distance from point $A$ to the line $y=x$. Then $A$ is the point on $y=x^2+1$ where the tangent line is parallel to $y=x$, solving which gives $A\left(\frac{1}{2}, \frac{5}{4}\right)$.
Therefore, the minimum value of $|AB|$ is $\frac{3 \sqrt{2}}{4}$.