My Quantum Field Theory lecturer just said that $\hat{a}(p)|0\rangle=0$ because the vacuum state contains no particles. Now, according to Wikipedia,

"according to quantum mechanics, the vacuum state is not truly empty but instead contains fleeting electromagnetic waves and particles that pop into and out of existence".

So is this argument incorrect?

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    $\begingroup$ I would interpret it the other way round. As soon as you have defined an annihilation operator, you can DEFINE the vacuum by this definition. The definition of the vacuum then is that it vanishes if you apply the annihilation operator on it $\endgroup$
    – Noldig
    Oct 30 '15 at 12:24
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    $\begingroup$ Just curious: if there's a conflict between the rigorous treatment given by your lecturer and a hand-waving statement in Wikipedia, why is your first assumption that Wikipedia is correct to the letter and the rigorous treatment wrong? $\endgroup$ Oct 30 '15 at 12:24
  • $\begingroup$ Because the Wikipedia article cites three different sources for the quote and my lecturer's argument was not rigorous, as I've said in the question he said "because the vacuum state contains no particles". $\endgroup$ Oct 30 '15 at 12:29
  • $\begingroup$ Related: physics.stackexchange.com/q/34049/2451 , physics.stackexchange.com/q/145418/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 30 '15 at 12:39

In field theory, there are two vacua. The non-perturbative vacuum $|\Omega\rangle$ and the vacuum of the free theory $|0\rangle$. The wikipedia article makes reference to $|\Omega\rangle$ in terms of $|0\rangle$ and its excitations.

The true vacuum is annihilated by the (dressed) annihilation operators, and can be thought of perturbatively in terms $|0\rangle$ plus fluctuations. Thus $|\Omega\rangle$ is not "empty" if we define as empty $|0\rangle$. But one should keep in mind that this is an artifact of the perturbation theory.

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    $\begingroup$ You are correct that the Wikipedia article is about the non-perturbative vacuum, but it is, strictly speaking, not correct that $\lvert \Omega \rangle$ can be expressed as a sum of $\lvert 0 \rangle$ and other stuff, since the free and dressed/interacting Hilbert spaces are unitarily inequivalent by Haag's theorem, and so $\Omega$ and $0$ lie in different spaces - one can't be obtained from the other by adding something. $\endgroup$
    – ACuriousMind
    Oct 30 '15 at 14:31
  • $\begingroup$ @ACuriousMind: Yeah ok. I didn't try to be rigorous (but I've edited the answer), and I don't think anyone really cares if one can do that or not. The heuristic picture of their relationship is good enough. $\endgroup$
    – Adam
    Oct 30 '15 at 14:45

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