How many particles in $\phi_0(x)^2|0\rangle$? In Schwartz's "QFT and the standard model" on pg 22 he writes:

A two or zero particle state as in $\phi_0(x)^2\left|0\right>$.

I was wondering how this can be proved? I tried checking if  $\phi_0(x)^2\left|0\right>$ was an eigenstate of the number operator 
$$N=\int \frac{ d^3 p a_p{}^{\dagger } a_p }{(2 \pi )^3}.$$
But just got:
$$  N \phi_0 (0)^2|0\rangle=\int  \frac{d^3 kd^3 q}{(2 \pi )^3 \sqrt{\omega _k \omega _q}}2 \left(a_{\overset{\rightharpoonup }{k}}{}^{\dagger } a_{\overset{\rightharpoonup }{q}}{}^{\dagger }\right)|0\rangle, $$
which I can't see how this translates into Schwartz's statement.
 A: The field operator can be divided in two parts, one with positive frequency and other with negative frequency $$\phi(x) = \phi^+(x) + \phi^-(x)$$ $$\phi^+(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2\omega_p}} a_p e^{-ipx}\qquad \phi^-(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2\omega_p}} a_p^\dagger e^{ipx}$$ As you can see, the positive frequency part $\phi^+(x)$ is a linear combination of annihilation operators $a_p$ (so it kills one particle), and the negative frequency part $\phi^-(x)$ is a combination of creation operators $a_p^\dagger$, so it creates a particle. 
\begin{align}N\phi^+(x)|0\rangle = 0 &\qquad N \phi^-(x)|0\rangle = \phi^-|0\rangle \\ N\phi^-(x)^2|0\rangle = 2 \phi^-(x)^2|0\rangle &\qquad N\phi^+(x)\phi^-(x)|0\rangle=0\end{align}
So your state is 
\begin{align}\phi(x)^2|0\rangle =& [\phi^+(x) + \phi^-(x)][\phi^+(x) + \phi^-(x)]|0\rangle\\ =& \phi^+(x)^2|0\rangle + \phi^+(x)\phi^-(x)|0\rangle + \phi^-(x)\phi^+(x)|0\rangle + \phi^-(x)^2|0\rangle \\ =& \phi^-(x)^2|0\rangle + \phi^+(x)\phi^-(x)|0\rangle\end{align}
which is not an eigenstate of the number operator, because it is a superposition of one state with two particles and one state with zero particles. This is what "or" means in your Schwartz's quote

A two or zero particle state as in $\phi(x)^2|0\rangle$

