Why can two annihilation operators or two creation operators always be neglected? In quantum field theory, my teacher often neglects terms like $a_p a_p$ and $(a_p)^\dagger (a_p)^{\dagger }$. I know that when we evaluate $\langle 0\rvert(a_p){}^{\dagger }(a_p){}^{\dagger }\rvert0\rangle$ and $\langle0\lvert a_p a_p\rvert0\rangle$, we will get $0$.
But what if we do not do this, what if we have $\langle 2\rvert a_p a_p\lvert2\rangle$, what will we get?
 A: The action of the creation and annihilation operators are given as
$$ a\lvert n\rangle = \sqrt{n}\,\lvert n-1\rangle,\quad n≥1\\
a^\dagger\lvert n \rangle = \sqrt{n+1}\,\lvert n+1\rangle,\quad n≥0$$
and the annihilation operator $a$ annihilates the vacuum: $a\lvert0\rangle=0.$ 
Now in your case $n=2$ so you get
$$ \langle 2\lvert aa\lvert 2\rangle=\sqrt{2}\,\langle 2\lvert a \lvert 1\rangle=\sqrt{2}\,\langle 2 \lvert 0\rangle=0,$$
assuming that the states are normalized as $\langle n\lvert m\rangle=\delta_{nm}$.
More generally, when dealing with quantum field theory where you have a many-particle state, the action of creation and annihilation operators are given as
$$a(\mathbf{p}_j)\rvert n_1,\dots,n_j,\dots\rangle=\sqrt{n_j}\,\lvert n_1,\dots n_j-1,\dots\rangle\\
a(\mathbf{p}_j)^\dagger\rvert n_1,\dots,n_j,\dots\rangle=\sqrt{n_j+1}\,\lvert n_1,\dots n_j+1,\dots\rangle$$
A: $$\left\langle n | a_p a_p | n \right\rangle \propto \left\langle n+1 | n-1 \right\rangle = 0, \, \left\langle n | a^\dagger_p a^\dagger_p | n \right\rangle \propto \left\langle n-1 | n+1 \right\rangle = 0.$$
