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
5,354 answers , 22,791 comments
1,470 users with positive rep
820 active unimported users
More ...

  Why is the term $m\delta_2\delta_m\bar{\psi}\psi$ ignored in the QED Lagrangian?

+ 4 like - 0 dislike
1830 views

Consider the QED Lagrangian

$$\mathcal{L}=\bar{\psi}_0(i\gamma^{\mu}\partial_{\mu}-e_0\gamma^{\mu}A_{0\mu}-m_0)\psi_0-\frac{1}{4}(\partial_{\mu}A_{0\nu}-\partial_{\nu}A_{0\mu})^2$$

where the 0 subscript denotes bare fields. The bare fields are related with the renormalized fields via

$$\psi_0=\sqrt{Z_2}\psi\qquad{}A_{0\mu}=\sqrt{Z_3}A_{\mu}$$

$$m_0=Z_mm\qquad{}e_0=Z_ee$$

with these redefinitions the Lagrangian takes the form

$$\mathcal{L}=Z_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-eZ_eZ_2\sqrt{Z_3}\bar{\psi}\gamma^{\mu}A_{\mu}\psi-mZ_2Z_m\bar{\psi}\psi-\frac{1}{4}Z_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2$$


it is customary to define

$$Z_1\equiv{}Z_eZ_2\sqrt{Z_3}$$

leaving

$$\mathcal{L}=Z_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-eZ_1\bar{\psi}\gamma^{\mu}A_{\mu}\psi-mZ_2Z_m\bar{\psi}\psi-\frac{1}{4}Z_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2.$$

Moreover, the $Z$ renormalization constants are defined to be

$$Z_1=1+\delta_1\qquad{}Z_2=1+\delta_2\qquad{}Z_3=1+\delta_3\qquad{}Z_m=1+\delta_m$$

it is often said that this leaves the QED Lagrangian in the following form (see page 2 of these notes http://isites.harvard.edu/fs/docs/icb.topic1146665.files/III-5-RenormalizedPerturbationTheory.pdf).

$$\mathcal{L}=\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi+\delta_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-e\bar{\psi}\gamma^{\mu}A_{\mu}\psi-e\delta_1\bar{\psi}\gamma^{\mu}A_{\mu}\psi-m\bar{\psi}\psi-m(\delta_m+\delta_2)\bar{\psi}\psi-\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2-\frac{1}{4}\delta_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2.$$

Nonetheless, the carefull reader will notice that one term coming from $mZ_2Z_m\bar{\psi}\psi$ is completely ignored, namely

$$m\delta_2\delta_m\bar{\psi}\psi.$$

Why?

asked Oct 28, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ revision history ]
retagged Oct 29, 2015 by Dilaton

2 Answers

+ 4 like - 0 dislike

Because formally we treat the counterterms as small, so anything higher than first order in counterterms is negligible. Alternatively, any contributions that the extra factor of \(\delta_2\)would make to the propagator could be absorbed by a redefinition of \(\delta_2\) and \(\cancel{\delta_m}\). This would add extra complication without changing the end-result.

Let me qualify the first statement in the answer rather than leaving it in a comment. The counterterms are part of a perturbative expansion in the renormalized coupling constant \(e_R\) and we take this to be small, therefore higher order terms in the coupling, like the one this question is asking about, are negligible. This is in contrast to the expansion in bare perturbation theory where the bare coupling, a non-observable and un-physical quantity, is formally infinite.

answered Oct 29, 2015 by holomorphic (70 points) [ revision history ]
edited Oct 29, 2015 by holomorphic

Only the second explanation is correct, as counterterms are large. Only $\delta_m$ must be redefined to get the correct Lagrangian. See my answer.

Maybe I should have been more clear. Yes, obviously counterterms are large, but they are part of a perturbative expansion in the coupling constant. More specifically in QED the 1-loop counterterms are second order in \(e_R\), and therefore the product in question, using the counterterms as commonly defined, is quartic in the charge.

Let me note that divergences are independent of the charge value. It is the structure of interaction who is bad.

+ 3 like - 0 dislike

To define the $Z$ renormalization constants as

$$Z_1=1+\delta_1,~~Z_2=1+\delta_2,~~Z_3=1+\delta_3,~~Z_m=1+\delta_m$$

is just sloppiness. A more careful definition as

$$Z_1=1+\delta_1,~~Z_2=1+\delta_2,~~Z_3=1+\delta_3,~~Z_m=1+\delta_m/Z_2$$

removes the extra term in the renormalized Lagrangian.

Note that the counterterms are not tiny but huge (infinite if the cutoff is removed), hence a product of counterterms cannot be neglected.

answered Oct 29, 2015 by Arnold Neumaier (15,787 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:
$\varnothing\hbar$ysicsOverflow
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
...