# Clarification Needed for The Klein-Gordon Field Acting on the Vacuum State (Peskin and Schroeder)

In Peskin and Schroesder's Introduction to Quantum Field Theory, section 2.3, the Klein Gordon Field has the expression

$$\phi(x,t) := \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2\omega_{p}}} [a_{p} e^{ip \cdot x} + a_{p}^{\dagger} e^{-ip \cdot x}]\tag{2.25}$$

with $$a_{p}$$ and $$a^{\dagger}_{p}$$ the ladder operators for the quantized harmonic oscillator corresponding to momentum $$p$$.

A couple pages later (pg. 24), they say it follows that

$$\phi(x,t)\lvert 0 \rangle = \int \frac{d^{3}p}{(2\pi)^{3}} e^{-ip \cdot x}\lvert p \rangle$$

where

$$\lvert p \rangle = a^{\dagger}_{p} \lvert 0 \rangle$$

I don't quite understand how this follows. It seems like somehow the first term becomes $$\lvert 0 \rangle$$, but I'm not sure how. I understand that $$a_{p} \lvert 0 \rangle = \lvert 0 \rangle$$, but what role does $$e^{ip \cdot x}$$ play? I understand this is very similar to

The operation of scalar field $\phi (\vec x)$ on vacuum state

But this same step in not explained in that post.

I think your confusion is because you have confused the vacuum state $$\lvert 0 \rangle$$ with the zero vector. The two are not the same thing; the first one is a non-zero vector in the Fock space with non-zero norm ($$\langle 0 | 0 \rangle = 1$$), while the second one has zero norm. And it is also the case that $$a_{p} \lvert 0 \rangle = 0 \neq \lvert 0 \rangle.$$ So when $$\phi(x)$$ is applied to $$|0\rangle$$, all of the annihilation operators send the vacuum state state to zero, and the resulting terms just vanish from the final expression.