# Hodge index theorem without ampleness condition

I read the following statement:

Let $\mathcal L$ be a line bundle with $(\mathcal L.\mathcal L)>0$ on surface. If $(\mathcal L'.\mathcal L)=0$, then $(\mathcal L'.\mathcal L')\leq0$.

This looks very like the Hodge index theorem, except that the condition $\mathcal L$ being ample is weaken by $(\mathcal L.\mathcal L)>0$. I followed the proof of Hodge index theorem and I think the condition of ampleness is essential (I followed the Hartshorne's book, whose proof is based on Lemma 5.1.7, which I cannot see still holds true if ampleness is replaced by $(\mathcal L.\mathcal L)>0$).

So, how can I prove this statement? Is there any trick to avoid the ampleness in the proof?

Fix an ample divisor $H$, then $(\mathcal{L}\cdot H)\ne 0$. We may assume that $(\mathcal{L}'\cdot H)\ne 0$ otherwise it is clear. After taking a multiple, we may assume that $(\mathcal{L}\cdot H)=(\mathcal{L}'\cdot H)$. Hence $((\mathcal{L}-\mathcal{L}')\cdot H)=0$. This implies that $(\mathcal{L}-\mathcal{L}')^2\leq 0$ by HIT. Then it is easy to see that $(\mathcal{L}')^2<0$.