In Weinberg, vol. 1, Section 9.2, Weinberg defines the in and out vacua as states with no particles (9.2.4): $$a_{\rm in}|{\rm VAC,in}\rangle=0$$ $$a_{\rm out}|{\rm VAC,out}\rangle=0$$ He does this by assuming that the Heisenberg field operators $\Phi(\vec{x},t)$ are asymptotic to the interaction-picture, free-field operators, because they can be expanded as plane waves with free-field dispersion relations (9.2.5): $$\Phi \to \int_p [a_p \exp(-i \omega_p t + i p\cdot x) + h.c. ]$$ This means that the asymptotic vacua are eigenstates of the free hamiltonian, $$H_0=\sum_p \omega_p a^\dagger_p a_p\,,$$ correct? But I thought that the path-integral corresponded to expectation values of time-ordered products in the full, interacting vacuum, i.e. in eigenstates of $H$. Moreover, if they are eigenstates of the asymptotic $H_0$ hamiltonian, how do we know that the in and out vacua are the same state? What happened to $$ \langle T\{ \mathcal{O}(t)\mathcal{O}(t')\}\rangle = \frac{\delta^2}{\delta J^2}({\rm path\, integral}[J])_{J=0}\quad?$$
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1$\begingroup$ The key point is that interaction can not transform vacuum, evolution of vacuum state under perturbation gives nothing more than phase factor $\endgroup$– Artem AlexandrovMar 9, 2020 at 13:48
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$\begingroup$ @ArtemAlexandrov That seems to contradict the arguments above. $\endgroup$– Eric David KramerMar 9, 2020 at 19:20
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$\begingroup$ I don’t have a copy of Weinberg's textbook on me, but I’m pretty sure he mentions that the in and out vacuum is the same thing somewhere. $\endgroup$– Prof. LegolasovMar 10, 2020 at 4:13
2 Answers
The 'in' and 'out' states, as Weinberg defined in Chapter 3, are eigenstates of the full Hamiltonian $H$, not the free Hamiltonian $H_0$.
$a_{\text{in}}$ and $a_{\text{out}}$ are the annihilation operators for the 'in' and 'out' states, respectively. They are not free-field operators. Any proof of the LSZ formalism would involve constructing such operators, if you want more detail or an alternative view.
Eqn. (9.2.5) simply claims that the limit of the Heisenberg field at $t\rightarrow \mp \infty$ can be expanded in terms of creation and annihilation operators at that time. You are free to do that at any given time, just that the operator won't be time-independent like with free fields.
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$\begingroup$ But why should the Heisenberg operator have a well-defined limit as $t\to \pm \infty$? $\endgroup$ Mar 12, 2020 at 11:05
As answered by JF132, the 'in' and 'out' states are eigenstates of the full Hamiltonian, and the $a_{\rm in}$ and $a_{\rm out}$ are not free-field operators.
The limit that Weinberg is using is known as the 'weak topology' limit. That means that the matrix element of the operator between normalizable states will have this limit. Intuitively, this is because a normalizable state always asymptotes to well-separated wave packets that are effectively 'free' from each other (cf. Haag: Local Quantum Physics, or https://en.wikipedia.org/wiki/LSZ_reduction_formula)
The above proof of course assumes that there are no massless particles. If there are, then the states will not really be asymptotically free and the operator cannot be so decomposed. This somehow shows up as infrared divergences when the usual formulae are used.
