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  Field redefinition of a free scalar field theory

+ 2 like - 0 dislike

In Srednicki's QFT book problem 10.5, he made a field redefinition to make the free scalar field Lagrangian transform to a theory that looks like an interacting field theory, then he used the new Lagrangian to compute the tree level contribution to the $\varphi\varphi\rightarrow\varphi\varphi$ scattering amplitude, of course it's zero because the new Lagrangian still discribes the same physics of no interaction. However, I am confused by the last sentence in the parentheses of the problem which says $``$At the loop level, We also have to take into account the transformation of the functional measure $\mathcal D\varphi$; see section 85.$"$ In my opinion, after the field redefinition, the new field $\varphi$ no longer satisfies the normalization conditions $\langle0|\varphi(x)|0\rangle=0$ and $\langle k|\varphi(x)|0\rangle=e^{-ikx}$ for the validity of the LSZ formula, so we should shift and rescale the field, then we get the Lagrangian

$\mathcal L=-\frac 12Z_\varphi\partial^\mu\varphi\partial_\mu\varphi-\frac 12Z_mm^2\varphi^2$

$-2Z_1\lambda\varphi\partial^\mu\varphi\partial_\mu\varphi-Z_2\lambda m^2\varphi^3-2Z_3\lambda^2\varphi^2\partial^\mu\varphi\partial_\mu\varphi-\frac 12Z_4\lambda^2m^2\varphi^4+Y\varphi$

This Lagrangian is equivalent to the free field Lagrangian, the scattering amplitude computed by this Lagrangian must be non-divergent, so we don't need to add more counterterms to cancel the divergence. If we want to compute the loop correction to the scattering amplitude, we can just draw the diagram and compute, why should we take into account the transformation of the functional measure $\mathcal D\varphi$? Did I misunderstand something?

Or, if we call the original free field $\varphi_0$, is he still computing the correlation function of $\varphi_0$ by the generating functional $Z_0(J)=\int\mathcal D\varphi_0\ e^{i\int d^4x(-\frac 12\partial^\mu\varphi_0\partial_\mu\varphi_0-\frac 12m^2\varphi^2_0+J\varphi_0)}$ and the field inside the LSZ formula is still $\varphi_0$?

If so, he just used the field redefinition $\varphi_0=\varphi+\lambda\varphi^2$ to change the integration variable, then there is no renormalization $Z$ factors inside the path integral, after making the substitution of the field, the path integral looks like

$Z_0(J)=\int\mathcal D\varphi\ \prod_x[1+2\lambda\varphi(x)]\times$

$e^{i\int d^4x[-\frac 12\partial^\mu\varphi\partial_\mu\varphi-\frac 12m^2\varphi^2-2\lambda\varphi\partial^\mu\varphi\partial_\mu\varphi-\lambda m^2\varphi^3-2\lambda^2\varphi^2\partial^\mu\varphi\partial_\mu\varphi-\frac 12\lambda^2m^2\varphi^4+J(\varphi+\lambda\varphi^2)]}$

The $\prod_x$ factor can be turned into $e^{\sum_x ln[1+2\lambda\varphi(x)]}\rightarrow e^{\int d^4x\ ln[1+2\lambda\varphi(x)]}$. 

But if we want to turn this generating functional into a functional derivative prefactor multiplied by the free generating functional $\int\mathcal D\varphi\ e^{i\int d^4x(-\frac 12\partial^\mu\varphi\partial_\mu\varphi-\frac 12m^2\varphi^2+J\varphi)}$, we should move the term $J\lambda\varphi^2$ into the functional derivative prefactor, then this term becomes a two-line vertex containing the source $2i\lambda\int d^4x\ J(x)$, so the source $J$ can appear on the internal line which is different from the ordinary scalar field theory of which the sources only appear on the external line. Does Srednicki mean this way? 

asked Jun 2, 2017 in Theoretical Physics by leili (10 points) [ no revision ]

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