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I am reading Schwartz's "Quantum field theory and the standard model". I have a question on how he derives the Feynman rules for an interacting scalar field from a Lagrangian formalism.

In particular, he needs to generalize the commutation relations of free fields to the case of interacting fields. For free fields it holds

$$ \begin{eqnarray} [\phi(\vec x), \phi(\vec y)]&=& 0\tag{7.3} \\ [\phi(\vec x), \pi(\vec y)]&=&i\delta^3(\vec x-\vec y) \tag{7.4}\end{eqnarray} $$ For interacting fields, he says (page 79, section 7.1):

We will also assume Eqs. (7.3) and (7.4) [the free fields commutation relations written above] are still satisfied. This is a natural assumption, since at any given time the Hilbert space for the interacting theory is the same as that of a free theory.

I do not understand the last sentence about the Hilbert space. Can you please help me understanding it?

In this question, it is shown that the creation and annihilation operators at any given time satisfy the same commutation relations as those in the free theory. But still I do not understand why the Hilbert space is the same. In an interacting theory, I expect states that are not present in the free theory, so the Hilbert space must be different.

Moreover, Schwartz statement is written as if it were an obvious statement. But without the previous question, it would have not been so obvious. So I am wondering if I am missing an easier, more intuitive, explanation of why the Hilbert space is the same.

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  • $\begingroup$ What states do you have in mind? They are all multiexcitation states deformed analytically w.r.t. the perturbative coupling constant. It's all an infinity of oscillators, rearranged continuously. $\endgroup$ Commented Mar 6, 2022 at 18:59

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