In QFT it is assumed that every Heisenberg field has free field as weak limit of $x_0 \rightarrow \pm \infty $. $$ \phi(x) \rightarrow \sqrt{Z}\phi^{as}(x) (as = in,out) $$ where $Z$ is renormalization factor. To find the factor $Z$, you can check one particle pole of two point Green's function of $\phi$. $$ <0|T\phi(x)\phi(y)|0>|_{1particle}=Z\Delta_{F} $$ where $\Delta_F$ is free propagator.

  My question is how this method is applied to arbitrary polynomial of Heisenberg fields. For example, consider $\lambda \phi^4$ theory($V(\phi)=\frac{1}{2}\mu^2\phi^2+\frac{1}{4!}\lambda\phi^4$) and $\Phi(x)=(\phi(x))^3$.One can calculate Green's function of $\phi$ and $\Phi$ and see its one particle pole, say, $A\Delta_F$.Then is it true that $\Phi$ goes to $(A /\sqrt{Z})\phi^{as}$ in the weak limit of $x_0 \rightarrow \pm \infty$?

  In the textbook I'm reading , the above argument was used to show that action of conserved charge $Q$ on asymptotic fields is linear, where it was applied by setting $\Phi(x) = [Q,\phi(x)]$. The proof completes by taking a limit of both sides of $\Phi(x) = [Q,\phi(x)]$.

  I'm skeptical about whether this stands for arbitrary $\Phi$ for two reasons. One is that in the proof of the textbook it seems to me that the fact that $Q$ is conserved charge is not used. Secondary , in originally free theory it does not work because $\phi^2$ does not converge to 0.

  I'm new to QFT and I'm confused about this. I really appreciate it if you solve this question.

  • $\begingroup$ "action of conserved charge Q on asymptotic fields is linear," Could you be more precise on this point? I do not understand. $\endgroup$ – Valter Moretti Oct 26 '16 at 11:48
  • $\begingroup$ Sorry for ambiguity. It means [Q,\phi_i}=a_{ij}\phi_j where \phi_i is asymptotic field and a_{ij} is coefficient matrix which can contain derivative of finite order. $\endgroup$ – user126948 Oct 26 '16 at 12:47

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