II. (50 points) Given that the center of the square $ABCD$ is at the origin of the coordinate system, and the four vertices of the square are on the graph of the function $f(x)=x^{3}+a x$. Find the range of the real number $a$.

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Let $A\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$. Then $B\left(-y_{0}, x_{0}\right) 、 C\left(-x_{0},-y_{0}\right) 、 D\left(y_{0},-x_{0}\right)$.
According to the problem, we have
$y_{0}=x_{0}^{3}+a x_{0}$,
$-x_{0}=y_{0}^{3}+a y_{0}$.
(1) $\times y_{0}-$ (2) $\times x_{0}$ gives
$x_{0}^{2}+y_{0}^{2}=x_{0} y_{0}\left(x_{0}^{2}-y_{0}^{2}\right)$.
(1) $\times x_{0}+$ (2) $\times y_{0}$ gives
$$
x_{0}^{4}+y_{0}^{4}+a\left(x_{0}^{2}+y_{0}^{2}\right)=0 \text {. }
$$

Let $x_{0}=r \cos \theta, y_{0}=r \sin \theta\left(r>0, \theta \in\left(0, \frac{\pi}{2}\right)\right)$.
Substituting into equation (3) gives
$$
\begin{array}{l}
r^{2}=r \cos \theta \cdot r \sin \theta \cdot r^{2}\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \\
=\frac{1}{4} r^{4} \sin 4 \theta .
\end{array}
$$

Therefore, $r^{2} \sin 4 \theta=4$.
Substituting into equation (4) gives $r^{2}\left(\cos ^{4} \theta+\sin ^{4} \theta\right)+a=0$.
Simplifying gives $r^{2} \cos 4 \theta=-4 a-3 r^{2}$.
$(5)^{2}+(6)^{2}$ and simplifying gives
$$
r^{4}+3 a r^{2}+2 a^{2}+2=0 \text {. }
$$

By $\Delta=9 a^{2}-4\left(2 a^{2}+2\right)=a^{2}-8 \geqslant 0$, then $a \geqslant 2 \sqrt{2}$ or $a \leqslant-2 \sqrt{2}$.

Also, from equation (4) we know $a<0$, hence $a \in(-\infty,-2 \sqrt{2}]$.