11. Given that point $P$ is on the curve $y=\mathrm{e}^{x}$, and point $Q$ is on the curve $y=\ln x$. Then the minimum value of $|P Q|$ is $\qquad$ .
11. $\sqrt{2}$.

Notice that the curve $y=\mathrm{e}^{x}$ and $y=\ln x$ are symmetric about the line $y=x$.

Therefore, the minimum value of $|P Q|$ is twice the minimum distance from a point on the curve $y=\mathrm{e}^{x}$ to the line $y=x$.

Let $P\left(x, \mathrm{e}^{x}\right)$ be any point on the curve $y=\mathrm{e}^{x}$. Then the distance from $P$ to the line $y=x$ is
$$
\begin{array}{l}
d(x)=\frac{\left|\mathrm{e}^{x}-x\right|}{\sqrt{2}}=\frac{\mathrm{e}^{x}-x}{\sqrt{2}} \\
\Rightarrow d^{\prime}(x)=\frac{\mathrm{e}^{x}-1}{\sqrt{2}} .
\end{array}
$$

When $x>0$, $d^{\prime}(x)>0 \Rightarrow d(x)$ is monotonically increasing; when $x<0$, $d^{\prime}(x)<0 \Rightarrow d(x)$ is monotonically decreasing.
Thus, $d(x)_{\text {min }}=d(0)=\frac{\sqrt{2}}{2}$.
Hence, $|P Q|_{\min }=\sqrt{2}$.