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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.

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    $\begingroup$ Schematically $\phi(x)\sim a_p + a_p^\dagger$ (with some $e^{ipx}$ factors, not important for counting). So $\phi^2\sim a^\dagger a^\dagger + a^\dagger a + aa^\dagger + aa$. Since $\phi^2$ acts on $|0\rangle$, we can throw away $aa$ and $a^\dagger a$ (i.e. normal ordering), and we are left with $a^\dagger a^\dagger + aa^\dagger$. The first term creates two particles, the second term does not change the particle number. $\endgroup$ – Meng Cheng Jun 7 '15 at 6:37
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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 $\phi(x)^2|0\rangle$

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  • $\begingroup$ Thanks, that's very helpful. Should the + and - be swapped around in that last line? $\endgroup$ – Virgo Jun 7 '15 at 18:06
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    $\begingroup$ No. Since $\phi^+(x)$ is a annihilation operator, $\phi^+(x)|0\rangle = 0$, and therefore $\phi^-(x)\phi^+(x)|0\rangle = 0$. In the other hand, $\phi^+(x)\phi^-(x)|0\rangle \neq 0$ $\endgroup$ – Bosoneando Jun 7 '15 at 18:49
  • $\begingroup$ One more question: shouldn't $$N{\phi^+}^2(x)|0\rangle = 2 {\phi^+}^2(x)|0\rangle$$ be $$N{\phi^-}^2(x)|0\rangle = 2 {\phi^-}^2(x)|0\rangle$$ ? $\endgroup$ – Virgo Jun 7 '15 at 23:20
  • $\begingroup$ @physicsphile Yes. Fixed! $\endgroup$ – Bosoneando Jun 8 '15 at 6:09

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