11. The linear function $y=-\frac{\sqrt{3}}{3} x+1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. A square $A B C D$ is constructed with segment $A B$ as one side in the first quadrant (as shown in Figure 5). In the second quadrant, there is a point $P\left(a, \frac{1}{2}\right)$, which satisfies $S_{\triangle A B P}=S_{\text {square } A B C D}$. Then $a=$ $\qquad$
11. $\frac{\sqrt{3}}{2}-8$.

Let $x=0, y=0$, we get $A(\sqrt{3}, 0) 、 B(0,1)$. Then $O A=\sqrt{3}, O B=1, A B=2, S_{\text {square } A B C D}=4$.
Therefore, $S_{\triangle A B P}=4$.
As shown in Figure 12, connect

$P O$. Then
$$
\begin{array}{l} 
S_{\triangle A O P} \\
=\frac{1}{2} \times \frac{1}{2} \sqrt{3} \\
=\frac{\sqrt{3}}{4} ; \\
S_{\triangle B O P}=\frac{1}{2} \times(-a) \times 1=-\frac{a}{2} ; \\
S_{\triangle A O B}=\frac{1}{2} \times \sqrt{3} \times 1=\frac{\sqrt{3}}{2} ; \\
S_{\triangle A B P}=S_{\triangle B O P}+S_{\triangle A O B}-S_{\triangle A O P} .
\end{array}
$$

Thus, $4=-\frac{a}{2}+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{4}$.
Solving for $a$ gives $a=\frac{\sqrt{3}}{2}-8$.