What is the net ionic equation for the reaction between Ba(OH)2 and H2SO4?
The reaction between $\ce{Ba(OH)2, H2SO4}$ is known as acid-base neutralisation, as $\ce{Ba(OH)2}$ is a relatively strong base and $\ce{H2SO4}$ the strong acid.
The molecular reaction is given as,
$$\ce{Ba(OH)2 + H2SO4 -&gt; BaSO4 (s) + 2H2O}$$

Before moving to net-ionic reaction, let's see the reaction of $\ce{BaCl2, Na2SO4}$ [1]:
$$\ce{BaCl2 + Na2SO4 -&gt; BaSO4 (s) + 2NaCl}$$
So, breaking up into ions, we see:
$$\ce{Ba^{2+} + 2Cl- + 2Na^{+} + SO4^{2-} -&gt; BaSO4 (s) + 2Na^{+} + 2Cl^{-}}$$
$\ce{Na^{+}, Cl^{-}}$ remains completely dissociated in the aqueous solution as they're salts of strong acid and strong base (i.e., they're strong electrolytes). Hence, $\ce{Na^{+}, Cl^{-}}$ are spectator ions as they don't participate in ionic reaction. Let's cancel them out, and we get the net ionic reaction:
$$\ce{Ba^{2+} + SO4^{2-} -&gt; BaSO4 (s)}$$

Now, let's see the reaction given in the question. In this, $\ce{BaSO4}$ precipitates out, so $\ce{Ba^{2+}, SO4^{2-}}$ can't be spectator ions. Also, H2O has a very low dissociation contant ($\ce{K_W = 10^{-14}}$, at 25°C), so it can't dissociate into ions. Hence, $\ce{H+, OH-}$ are too not the spectator ion. Therefore, this reaction has no spectator ion and hence ${H_2O}$ will be included in the ionic reaction, which is given as:
$$\ce{Ba^{2+} + 2OH- + 2H+ + SO4^{2-} -&gt; BaSO4 (s) + 2H2O(l)}$$

Another example:
[1]: Complete ionic and net ionic equations, Khan Academy
Note: As mentioned by andselisk in comments, $\ce{BaSO4 (s)}$ is the recommended way to write the precipitate, as per IUPAC. Old school symbol for precipitate was &quot;↓&quot;.