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I've just begun reading a book on QFT. In the introduction, the author says that particles are no longer localized at a point but are excitations of a (possibly quantized) field. He defines a field operator, and then plays with a specific example where the field operator is a linear combination of annihilation/creation operators that acts on the Fock Hilbert space.

I thought the Fock Hilbert space was only for point particles, but I guess you can use it as an abstract way to represent excitations. I'm curious if there are functions down the road that can represent these excitations, like the excited states of the hydrogen atom. But what is weird to me is that these excitations aren't some smearing of a point particle, but perturbations in a field itself? So can a field in QFT act on itself (is that what we are doing)? Are there situations where the surround field is weak enough and the field of the photon is strong enough that there is some feedback? Or am I missing the point entirely?

A. A. Grigorian
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  • What is the book you are using? This may help provide an answer. – gabe Jun 03 '19 at 22:01
  • Possibly related. Fields act on other fields, etc... and they all collectively act on the Fock vacuum. – Cosmas Zachos Jun 03 '19 at 22:05
  • Not sure I understand the question. QFT has observables represented by operators that act on a Hilbert space, just like in any quantum theory. One special feature of QFT is that the observables are constructed using field operators. It sounds like you might be confusing field operators (which are used to construct observables) with state-vectors (which encode what we know about the state of the system, like the fat that it's an excited state of an atom). Is this relevant to your question? – Chiral Anomaly Jun 04 '19 at 00:53
  • @ChiralAnomaly "QFT has observables represented by operators that act on a Hilbert space, just like in any quantum theory" is incorrect: QFT has operators-valued distributions (you must first smear it against a function to get an actual operator). – gented Jun 04 '19 at 11:48
  • @gented Good comment. This may be a language issue, and now that you mention it, I wonder if this language issue is really at the root of the OP's question. QFT, if it's well-defined at all, certainly includes obserables that are ordinary (even bounded) operators on the Hilbert space. However, sometimes we relax the word "observable" to inculde things that are localized at a point, like the un-smeared field operators are. Like you said, those are not ordinary operators, and they're also not observables in the sense of the general postulates of quantum theory. – Chiral Anomaly Jun 04 '19 at 12:55
  • The problem is, the Hilbert space that you obtain eventually depends very much on the choice of the functions you smear the operators with, so in this sense it is important to be precise if you are constructing, say, commuting or anti-commuting particle spaces. – gented Jun 04 '19 at 13:38

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