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When one talks about field strength renormalization, one defines the renormalized field $\psi^R(x)$ in the following way (I'm using the notation from Matthew Schwartz's book): $\psi^R(x) \equiv \frac{1}{\sqrt{Z_2}} \psi^0$, where $\psi^0$ is the bare free field (an operator acting on the usual Fock space) and $Z_2$ is some formally infinite number.

Now, even though it is the Green's function of the renormalized fields that should have finite physical values, should we be worried that such a definition takes us outside of Fock space? Are we changing the operators' amplitudes by making such a definition?

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closed as unclear what you're asking by ACuriousMind, Daniel Griscom, Gert, JamalS, user36790 Jan 20 '16 at 2:25

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    $\begingroup$ What do you mean by "taking us outside of Fock space"? An interacting theory, strictly speaking, doesn't have a Fock space as its space of states to begin with - we're always in the asymptotic free field case when talking about Fock spaces in QFT. $\endgroup$ – ACuriousMind Jan 18 '16 at 23:32
  • $\begingroup$ @ACuriousMind, I'm not talkng about interacting theory yet. $\psi^0$ is the free field. $\endgroup$ – some1 Jan 18 '16 at 23:54
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    $\begingroup$ We define $\psi^{R}$ to deal with interaction effects. For a free theory, $Z_{2}=1$, and $\psi^{R} = \psi^{0}$. $\endgroup$ – higgsss Jan 19 '16 at 0:40
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    $\begingroup$ Where did you find field-strength renormazation defined for a free field? Schwartz's book certainly introduces this concept for interacting fields. For free fields, nothing renormalizes the paramters in the theory. Even if you introduce $\psi^{R} = \psi^{0}/\sqrt{Z_2}$, it is a mere redefinitionof your field variable. $\endgroup$ – higgsss Jan 19 '16 at 14:10
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    $\begingroup$ @higgsss is correct - the field strength renormalization is $1$ if and only if the theory is free, so your question dissolves either way: Either you don't have to renormalize at all, or you don't have a Fock space. I don't know what you're asking. $\endgroup$ – ACuriousMind Jan 19 '16 at 15:01