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) $\sqrt{2}$ Hint: Since the curves $y=\mathrm{e}^{x}$ and $y=\ln x$ are symmetric about the line $y=x$. 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 $d(x)=\frac{\left|\mathrm{e}^{x}-x\right|}{\sqrt{2}}=\frac{\mathrm{e}^{x}-x}{\sqrt{2}}, d^{\prime}(x)=\frac{\mathrm{e}^{x}-1}{\sqrt{2}}$.
When $x>0$, $d^{\prime}(x)>0$, $d(x)$ is monotonically increasing; when $x<0$, $d^{\prime}(x)<0$, $d(x)$ is monotonically decreasing.
Therefore, $d(x)_{\text {min }}=d(0)=\frac{\sqrt{2}}{2}$, i.e., $|P Q|_{\text {min }}=\sqrt{2}$.