Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  Field redefinition of a free scalar field theory

+ 2 like - 0 dislike
1591 views

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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...