11. In the Cartesian coordinate system $x O y$, the equation of the hyperbola $C$ is $x^{2}-y^{2}=1$. Find all real numbers $a$ greater than 1 that satisfy the following condition: For any two perpendicular lines $l_{1}$ and $l_{2}$ passing through the point $(a, 0)$, if $l_{1}$ intersects the hyperbola $C$ at points $P, Q$, and $l_{2}$ intersects $C$ at points $R, S$, then $|P Q|=|R S|$ always holds.
When the slopes of lines $l_{1}, l_{2}$ exist and are not 0,
let the line $l_{1}: x=k y+a$, and solve the system $\left\{\begin{array}{l}x^{2}-y^{2}=1, \\ x=k y+a\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow\left(k^{2}-1\right) y^{2}+2 k a y+a^{2}-1=0 \\
\Rightarrow \Delta=4 k^{2} a^{2}-4\left(k^{2}-1\right)\left(a^{2}-1\right) \\
=4\left(k^{2}+a^{2}-1\right)>0 .
\end{array}
$$

Thus, $|P Q|=\sqrt{1+k^{2}} \cdot \frac{2 \sqrt{k^{2}+a^{2}-1}}{\left|k^{2}-1\right|}$. By substituting $-\frac{1}{k}$ for $k$, we similarly get
$$
|R S|=\sqrt{1+\left(-\frac{1}{k}\right)^{2}} \cdot \frac{2 \sqrt{\left(-\frac{1}{k}\right)^{2}+a^{2}-1}}{\left|\left(-\frac{1}{k}\right)^{2}-1\right|}=\sqrt{1+k^{2}} \cdot \frac{2 \sqrt{1+\left(a^{2}-1\right) k^{2}}}{\left|k^{2}-1\right|} \text {. }
$$

Then, $|P Q|=|R S| \Rightarrow k^{2}+a^{2}-1=1+\left(a^{2}-1\right) k^{2} \Rightarrow\left(a^{2}-2\right) k^{2}=a^{2}-2$ holds for any $k \neq 0$, thus $a^{2}-2=0 \Rightarrow a=\sqrt{2}$.

That is, the point $(a, 0)$ is the focus $F$ of the hyperbola. When the slope of line $l_{1}$ does not exist, the slope of line $l_{2}$ is 0, in which case it is easy to see that $|P Q|=|R S|=2$ also satisfies the condition.

In summary, the only real number $a=\sqrt{2}$ greater than 1 that meets the requirement.