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Why does a spurion analysis work independently of the UV physics?

+ 1 like - 0 dislike
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In short, my question is why does a spurion analysis work to produce the correct symmetry breaking terms regardless of the high energy physics?

The context that this question arose is from an Effective Field Theory course (for more context, see here, Eq. 5.50). Consider the QCD Lagrangian, \begin{equation} {\cal L} _{ QCD} = \bar{\psi} \left( i \gamma^\mu D_\mu - m \right) \psi \end{equation} The kinetic part is invariant under a chiral transformation: \begin{equation} \psi \rightarrow \left( \begin{array}{cc} L & 0 \\ 0 & R \end{array} \right) \psi \end{equation} however, the mass term is not. Now the claim I don't understand is as follows. Suppose the mass transformed as, \begin{equation} m \rightarrow L m R ^\dagger \end{equation} In that case the mass term would be invariant under such a transformation. To write down the correct chiral symmetry breaking terms in our Lagrangian we find the terms invariant given this transformation for $ m $ and then make $ m $ a constant again.

The way I understand this physically is that the breaking arises from a high energy spurion field, $ X $, which gets a VEV, $ m $. When we write down all possible chiral symmetry preserving terms using the transforming $m$, we are writing down all the terms that the spurion couples to. The VEV is then inserted and is equal to $ m $.

But this procedure assumes that the spurion obeys the chiral symmetry, $ SU(2) _L \times SU(2) _R $, and transforms as, $ X \rightarrow L X R ^\dagger $. How do we know this assumption is true? In fact it seems to fail for the case of QCD since the ``spurion field'' is really the Higgs field, which is a singlet under $ SU(2) _R $.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user JeffDror
asked May 2, 2014 in Theoretical Physics by JeffDror (650 points) [ no revision ]
retagged May 4, 2014
This post is also posted at physicsoverflow.org/15650/…, but I didn't get a response so I decided to repost here. I hope this isn't breaking any copy write laws...

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user JeffDror

1 Answer

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The spurion has nothing to do with UV physics and actual physical fields getting vevs. The field $X$ should be considered jut as a fictitious entity or a a tool to write down the correct effective lagrangian. The point is that the structure of the IR effective theory is largely independent of the particular UV completion that it originated from (as long as all the symmetries are respected and all light degrees of freedom kept in the theory). You can even mock up you own semplified UV theory to study the structure of the terms that are admissible in the IR lagrangian. A spurion exploits exactly this freedom.

Imagine instead you know the UV theory and want to assign quantum numbers to the spurion in order to recovery the symmetry and be able to eventually write down the explicit breaking terms in the IR theory. Say that the term breaking the symmetry in the lagrangian is $g \mathcal{O}$ where $\mathcal{O}$ is a non-singlet operator. Thus you will just need to promote the coupling $g$ to a spurion carrying a representation of the symmetry group such that the product $g\otimes\mathcal{O}$ contains a singlet. Only a finite number of irreducible representations below a certain dimension will do the job. Those will provide all the possible assignments of quantum numbers for the spurion you are after.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user TwoBs
answered May 2, 2014 by TwoBs (265 points) [ no revision ]
Thanks for your answer. So you are saying that any theory will give the same low energy physics as long as it has the same low energy field content. That is very surprising! In particular, for the pion case the mass term that we get assuming the spurion obeys a chiral symmetry is unique?

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user JeffDror
No, it is not unique. There are other quantum numbers that you can assign to the spurious. The bidoublet is just the simplest.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user TwoBs
Now that I read your comment I understand better your question. So you do have the UV theory at hand, and are actually asking how one should assign the quantum numbers to the spurion in order to recovery the symmetry and be able to eventually write down the explicit breaking terms in the IR theory. If this is the actual question the answer is very simple: g will need to carry a representation of the symmetry group such that the product $g\otimes\mathcal{O}$ contains a singlet, where $\mathcal{O}$ is the non-singlet operator in front of g in the lagrangian.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user TwoBs
I have just included this last comment of mine in the answer.

This post imported from StackExchange Physics at 2014-05-04 11:16 (UCT), posted by SE-user TwoBs

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