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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.

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1 Answer 1

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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.

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  • $\begingroup$ I see that makes sense now. Thanks. That is stated on page 22 but for some reason I interpreted it incorrectly. $\endgroup$
    – user480172
    Commented Dec 5, 2023 at 14:31
  • $\begingroup$ @user480172: You're welcome! If this answered your question, please accept this answer by clicking on the check-mark button to its left. (You might want to wait 24 hours to see if any other answers come in, but that's more of a "best practices" thing than a strict requirement.) $\endgroup$ Commented Dec 5, 2023 at 16:48

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