When we want to compute correlation functions $\langle\Omega|\,T\hat{\phi}(x_1)\ldots|\Omega\rangle$ in an interacting quantum field theory, we relate it to the free-field objects $|0\rangle$ and $\hat\phi_I(x)$ using the interaction-picture time-evolution operator in the limit $T\rightarrow\infty$.

Eventually, we arrive at an expression like (see Peskin and Schroeder eqn. 4.30) $$\langle\Omega|\,T\hat{\phi}(x)\hat{\phi}(y)|\Omega\rangle=\lim_{T\rightarrow\infty}\mathcal{N}\langle 0|U(T,t_x)\phi_I(x)U(t_x,t_y)\phi_I(y)U(t_y,-T)|0\rangle.$$

The $U(T,t_x)$ and $U(t_y,-T)$ sitting at the end of the correlator look very much like the Møller wave operators of non-relativistic scattering theory

$$\Omega_\pm=\lim_{t\rightarrow\mp\infty}U(t)_\text{full}U_0(t),$$ that relate the in-asymptote and out-asymptote states to the actual state at $t=0$.

So my question is, are these two like the same thing, with the same properties? i.e. they are isometric, etc...


1 Answer 1


The non-relativistic expression for the wave operators

$$ Ω_±=\lim_{{t→∓∞}}U(t)_\text{full}U{_{0}}(t), $$

needs revision in field-theoretic situations since usually the free and interacting fields act in different Hilbert spaces. An early example is given in

"Asymptotic conditions and infrared divergences in quantum electrodynamics", P. P. Kulish and L. D. Faddeev. Theoretical and Mathematical Physics 1970, Volume 4, Issue 2, pp 745-757..

Thus I expect that there is no simple answer to this question.

  • $\begingroup$ Very helpful response! So, am I correct in saying that the the two operators are analogs, but the relativistic one requires revision? $\endgroup$
    – QuantumDot
    Commented Dec 9, 2012 at 0:38
  • $\begingroup$ From a mathematical point of view there are two things that must be proven for a given situation. The first is (obviously) the existence of the wave operators and the second is completeness, which is needed for the unitarity of the associated S-operator. The second is a rather non-trivial matter. In a field-theoretical situation it is already difficult to obtain a correct mathematical setting (divergencies, renormalisation) to start from. $\endgroup$
    – Urgje
    Commented Dec 9, 2012 at 10:42

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