I have a silly question in derivation of LSZ reduction formular, I can go directly with the derivation until I found a assumption that I can't convince myself.

In the book Quantum Field Theory and the Standard Model (Schwartz's book), it says (Section 6.1, discussing the LSZ reduction formula):

By construction, our $a_\vec p (t)$ and $a^\dagger_\vec p (t)$ operators are time independent at late and early times.

This assumption is equal to the claim that

$$\frac{\partial a_\vec p (t)}{\partial t}=0,\qquad \frac{\partial a^\dagger_\vec p (t)}{\partial t}=0$$

when $t\to \infty.$ This assumption is important since without this condition the derivation can't continue. I don't know what the physical meaning of this assumption is; is it not an assumption, but strictly pertaining to computational methods? Can somebody give me advice?

  • 1
    $\begingroup$ It is an assumption. This is equivalent to the statement that at late times the theory is non-interacting. In other words, all interactions last only for a finite amount of time. $\endgroup$
    – Prahar
    Oct 27, 2022 at 20:24

1 Answer 1


First, note that we are talking about Heisenberg picture of these operators.

The state, \begin{equation} |\psi_1\rangle=a_{\vec{p}_1}^\dagger(t_1)\ldots a_{\vec{p}_n}^\dagger(t_1)|\Omega\rangle \end{equation} may be described as this: at time $t_1$ there were $n$ particles with momenta $\{\vec{p}_k\}$.

If the creation operators depend on time then, \begin{equation} |\psi_2\rangle=a_{\vec{p}_1}^\dagger(t_2)\ldots a_{\vec{p}_n}^\dagger(t_2)|\Omega\rangle,\quad |\psi_1\rangle\neq |\psi_2\rangle \end{equation} This means that the state $|\psi_1\rangle$ at time $t_2$ describes a different situation than at $t_1$.

So something must have happened with those particles. Some interaction took place.

However with that condition at large times all the nontrivial dynamics stops. If you had a specific set of particles you will have the same set of particles later. Note that this is what you have in free QFT where the creation and annihilation operator in the Heisenberg picture do not depend on time.

I.e. this condition means that the at large times the interaction "turns off". The idea is that the particles flown so far away from each other that the interaction may be neglected.

The important thing to remember though is that those asymptotic creation and annihilation operators do not equal to such operators of the naive bare free QFT.


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