(4) Given $a>b, ab=1$, then the minimum value of $\frac{a^{2}+b^{2}}{a-b}$ is ( ).
(A) $2 \sqrt{2}$
(B) $\sqrt{2}$
(C) 2
(D) 1
(4) Let $a-b=t$, then $t>0$,
$$
\frac{a^{2}+b^{2}}{a-b}=\frac{t^{2}+2}{t}=t+\frac{2}{t} \geqslant 2 \sqrt{2},
$$

Equality holds if and only if $t=\sqrt{2}$, i.e., $a=\frac{\sqrt{6}+\sqrt{2}}{2}, b=\frac{\sqrt{6}-\sqrt{2}}{2}$. Therefore, the answer is $\mathrm{A}$.