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
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Chiral effective lagrangian and higher order terms

+ 4 like - 0 dislike
768 views

At scales below $\Lambda_{QCD}$ global QCD symmetry $SU_{L}(3)\times SU_{R}(3)$ is spontaneously broken down to $SU_{f}(3)$, and pseudogoldstone degrees of freedom called mesons arise. It is possible to obtain explicit for of lagrangian for them: we substract goldstone degrees of freedom from quarks field, $q = U \tilde{q}$, where $U = e^{i\gamma_{5}t^{a}\epsilon_{a}}$ and $\epsilon_{a}$ parametrise goldstone degrees of freedom; then we replace $\bar{\tilde{q}}\tilde{q}, \bar{\tilde{q}}\gamma_{5}\tilde{q}$ by their VEVs $v, 0$. In the result chiral effective theory arises.

The question: I understand how (by using method described above) to derive term $\text{Tr}[\partial_{\mu}U^{\dagger}\partial^{\mu}U]$, but I don't understand how terms like
$$
\tag 1 \text{Tr}[\partial_{\mu}U^{\dagger}\partial^{\mu}U\partial_{\nu}U^{\dagger}\partial^{\nu}U]
$$ arise, since in QCD lagrangian there are only bilinear combinations of quark fields (if I understand correctly, term $(1)$ arise without taking into account gluon field), so product of only two $U$ matrix can arise.

I would be grateful for explanation how terms like $(1)$ arise.

asked Oct 9, 2015 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Oct 9, 2015 by NAME_XXX

1 Answer

+ 4 like - 0 dislike

The method described at the beginning of the question seems to be a classical one: assume that we have a non-trivial expectation value for some bilinear combination of quark, expand the classical Lagrangian of QCD around this expectation value to derive a Lagrangian for the Goldstone bosons. This approach clearly does not work in the full quantum mechanical QCD: in this setting, the spontaneous symmetry breaking is a complicated dynamical non-perturbative effect and similarly the effective theory describing the Goldstone bosons should emerge from the QCD Lagrangian via a complicated path integral calculation and in general it is not known how to do it. But the whole point of the chiral effective theory is that it is possible to say something non-trivial on the low energy behavior of the Goldstone bosons only by symmetry considerations, without having to know the complicated underlying microscopic dynamics it comes from. One simply writes the most general Lagrangian compatible with the symmetries of the problem. At low energy, the most relevant terms are those with the smaller number of derivatives. Up to two derivatives, there is only one such a term: $\text{Tr}[\partial_{\mu}U^{\dagger}\partial^{\mu}U]$. But then one has next to leading order with four derivatives and so on.

answered Oct 9, 2015 by 40227 (5,140 points) [ revision history ]

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$\hbar$y$\varnothing$icsOverflow
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
...