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I'm studying QFT and in a trouble about asymptotic assumption. It states every Heisenberg field converges to free field (asymptotic field) if one takes a limit of $x_0\rightarrow \pm \infty$. $$ \phi(x)\rightarrow \sqrt{Z}\phi^{\text{as}}(x) \\ \phi^{\text{as}}(x)=\int_{-\infty}^{+\infty}\frac{d^{3}p}{\sqrt{(2\pi)^{3}2p_0}}[a_{\text{as}}(\vec{p})e^{-ipx}+a_{\text{as}}^{\dagger}(\vec{p})e^{ipx}]\\ (\text{as}=\text{in},\text{out}) $$ Where $a_{\text{as}}(\vec{p})$is annihilation operator of asymptotic field.

I'm faced with a contradiction that this makes the theory trivial. Let $|n\rangle,|m\rangle$ eigenvectors of momentum operator$P_{\mu}$. Then $$ \langle n|\phi(x)|m\rangle=\langle n|e^{iPx}\phi(0)e^{-iPx}|m\rangle\\ =e^{i(p_{n}-p_{m})x}\langle n|\phi(0)|m\rangle $$

And its x-dependence is only phase. Because of asymptotic assumption and the fact that momentum eigenvectors form a complete set, $\phi(x)=\sqrt{Z}\phi^{\text{as}}(x)$ must holds.

What is wrong with this reasoning?

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Well, the point is that you are forcing the theory. The actual LSZ asymptotic condition says that (I omit the factor $Z$ for the sake of simplicity) $$\langle \Psi_1| \Phi(f,t)| \Psi_2\rangle \to \langle \Psi_1 |\Phi^{\pm}(f,t)| \Psi_2\rangle \quad \mbox{for $t \to \pm \infty$}\tag{0}$$ where $$\langle\Psi_1 |\Phi(f,t)| \Psi_2\rangle := \int_{x^0=t} \langle\Psi_1 |\Phi(x)| \Psi_2\rangle \: \overleftrightarrow{\partial}_{x^0} \: f(x) d^3x \tag{1}$$ and $f= f(x)$ is any solution of the free field equation and rapidly vanishes at spatial infinity. Similarly $$\langle\Psi_1 |\Phi^{\pm}(f,t)| \Psi_2\rangle := \int_{x^0=t} \langle\Psi_1|\Phi^\pm(x)| \Psi_2\rangle \: \overleftrightarrow{\partial}_{x^0} \: f(x) d^3x \tag{2}$$

Since $\Phi$ in (1) does not satisfy the free theory, the right hand side of (1) depends on $t$ and (0) may have sense.

Conversely $\Phi^{\pm}$ satisfies the free field equation and thus the right-hand side of (2) does not depend on $t$. The states $\Psi_i$ belong to a dense set in the Hilbert space generated by repeated application of respectively $a^\dagger_{in/out}(g)$ on the asymptotic vacuum states, where $g$ are smooth solutions of KG equation rapidly vanishing at spatial infinity.

You are quite far from the hypotheses written above.

ADDENDUM. There is another way, less rigorous, to state the LSZ condition into a fashion more familiar to physicists. First observe that, if $\Phi^\pm$ is the free field in the remote future/past, then $$i\int_{x^0=t} \Phi^\pm(x) \overleftrightarrow{\partial}_{x^0} \frac{e^{-ikx}}{\sqrt{(2\pi)^32k^0}} d^3x = a_\pm^\dagger(\vec{k})$$ where it is evident that the right-hand side is independent from $t$.

If we replace $\Phi^\pm$ for $\Phi$, the identity above fails because the interacting field $\Phi$ satisfies an equation different from Klein-Gordon's one. The condition LSZ just says that this is however true if (a) taking the limit for large $|t|$ and (b) referring to matrix elements (I am not sure on signs and coefficients and I omitted the factor $Z$) $$i\langle \Psi_1|\int_{x^0=t} \Phi^\pm(x) \overleftrightarrow{\partial}_{x^0} \frac{e^{-ikx}}{\sqrt{(2\pi)^32k^0}} d^3x |\Psi_2 \rangle \to \langle \Psi_1|a_\pm^\dagger(\vec{k})|\Psi_2 \rangle\quad \mbox{for $t \to \pm \infty$}\tag{4}$$ and $$-i\langle \Psi_1|\int_{x^0=t} \Phi^\pm(x) \overleftrightarrow{\partial}_{x^0} \frac{e^{ikx}}{\sqrt{(2\pi)^32k^0}} d^3x |\Psi_2 \rangle \to \langle \Psi_1|a_\pm(\vec{k})|\Psi_2 \rangle\quad \mbox{for $t \to \pm \infty$}\tag{5}$$ In all computations with LSZ reduction formulas only (4) and (5) are exploited. The popular naive formulation $$\Phi(x)\rightarrow \Phi^{\pm}(x)$$ is wrong from several viewpoints and if literally assumed easily leads to evidently false results as the one pointed out by the OP.

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  • $\begingroup$ I repeated my calculation with your correction and found it doesn't lead to free theory ,as <0|\Phi(f,t)|n> goes to 0 when |n> is multiparticle state. I'm yet to understand the meaning of LSZ asymptotic condition, it seems that considering only wave packet is important. And the condition is surely enough to deduce its consequences such as LSZ formula. I need $\endgroup$ – user126948 Nov 5 '16 at 8:04
  • $\begingroup$ to be more careful about usage of this condition. Thank you for your clear answer. $\endgroup$ – user126948 Nov 5 '16 at 8:07
  • $\begingroup$ Sorry, my formulas are affected by some typos. Let me some time to correct them $\endgroup$ – Valter Moretti Nov 5 '16 at 8:11
  • $\begingroup$ Well, barring coefficients, I think they are correct. I suggest you to have a look at Sect. II.3.2 of Haag's book "local quantum physics", where these things are discussed with a quite rigorous approach. $\endgroup$ – Valter Moretti Nov 5 '16 at 8:17
  • $\begingroup$ Also wikipedia gives a nice account en.wikipedia.org/wiki/LSZ_reduction_formula $\endgroup$ – Valter Moretti Nov 5 '16 at 8:20

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