11. (20 points) Let the hyperbola $C_{1}: x^{2}-y^{2}=2 m^{2}$, and the parabola $C_{2}$ has its vertex at $N(n, 0)$, with its focus being the left focus $F$ of the hyperbola $C_{1}$. A line $l$ with a slope of 1 passes through $F$ and intersects the parabola $C_{2}$ at points $P$ and $Q$. If there exist real numbers $m$ and $n$ such that $S_{\triangle P N Q}=4 \sqrt{2} m^{2}$, find $\frac{n}{m}$.
11. Clearly, $m \neq 0$.

It is easy to see that $F(-2|m|, 0)$.
Since the vertex of the parabola $C_{2}$ is $N(n, 0)$ and the focus is $F(-2|m|, 0)$, its equation is
$$
y^{2}=-4(2|m|+n)(x-n) .
$$

The equation of the line $l$ passing through point $F$ with a slope of 1 is $y=x+2|m|$.
From $\left\{\begin{array}{l}y^{2}=-4(2|m|+n)(x-n), \\ y=x+2|m|\end{array}\right.$, eliminating $y$ and rearranging, we get
$$
x^{2}+(12|m|+4 n) x+4 m^{2}-8|m| n-4 n^{2}=0 \text {. }
$$

Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=-(12|m|+4 n), \\
x_{1} x_{2}=4 m^{2}-8|m| n-4 n^{2} .
\end{array}
$$

Thus, $|P Q|=\sqrt{2\left(x_{1}-x_{2}\right)^{2}}$
$$
\begin{array}{l}
=\sqrt{2\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]} \\
=8|2| m|+n| .
\end{array}
$$

The distance from point $N(n, 0)$ to $x-y+2|m|=0$ is
$$
d=\frac{|n+2| m||}{\sqrt{2}} \text {. }
$$

Therefore, $S_{\triangle P N Q}=\frac{1}{2}|P Q| \cdot d=4 \sqrt{2} m^{2}$
$$
\begin{array}{l}
\Rightarrow(n+2|m|)^{2}=2 m^{2} \\
\Rightarrow \frac{n}{|m|}= \pm \sqrt{2}-2 \\
\Rightarrow \frac{n}{m}= \pm \sqrt{2}-2,2 \mp \sqrt{2} .
\end{array}
$$