4. Let $P(0, a)$ be any point on the $y$-axis other than the origin, and let the line through point $P$ and parallel to the $x$-axis intersect the curve $y=\frac{1}{a} \ln x$ at point $Q$. The tangent line to the curve $y=\frac{1}{a} \ln x$ at point $Q$ intersects the $y$-axis at point $R$. Then the minimum value of the area of $\triangle P Q R$ is $\qquad$.
Given the figure, let $\frac{1}{a} \ln x=a \Rightarrow \ln x=a^{2} \Rightarrow x=\mathrm{e}^{a^{2}}$,
then $Q\left(\mathrm{e}^{a^{2}}, a\right)$. Also, $y^{\prime}=\frac{1}{a x} \Rightarrow k_{Q R}=\frac{1}{a \mathrm{e}^{a^{2}}}$,
thus $Q R: y-a=\frac{1}{a \mathrm{e}^{a^{2}}}\left(x-\mathrm{e}^{a^{2}}\right)$. Let $x=0$,
we get $R\left(0, a-\frac{1}{a}\right),|P R|=\frac{1}{|a|} \Rightarrow S_{\triangle P Q R}=\frac{\mathrm{e}^{a^{2}}}{2|a|}$.
Let $f(a)=\frac{\mathrm{e}^{a^{2}}}{2|a|}$, we only need to consider the case when $a>0$.
At this point, $f^{\prime}(a)=\frac{2 a \cdot \mathrm{e}^{a^{2}} \cdot a-\mathrm{e}^{a^{2}}}{2 a^{2}}=\frac{\mathrm{e}^{a^{2}}\left(2 a^{2}-1\right)}{2 a^{2}}$.
Therefore, when $a=\frac{\sqrt{2}}{2}$, the minimum area of $\triangle P Q R$ is $\frac{\mathrm{e}^{\frac{1}{2}}}{\sqrt{2}}$.