How to Derive the Chern-Simons Contribution to the Effective Action?

Click For Summary
SUMMARY

The discussion focuses on deriving the Chern-Simons contribution to the effective action in quantum field theory (QFT) after coupling fermions to an external U(1) gauge field. The key expression for the coefficient of the Chern-Simons term in 2+1 dimensions is given by the integral involving the two-point Green's function, G. The effective action is expressed as Seff[A] = (C2/2π) εμνρ Aμν Aρ. The discussion highlights the necessity of understanding Feynman diagrams and the path-integral formalism to derive this result, with references to works by A. Zee and Dunne providing foundational insights.

PREREQUISITES
  • Quantum Field Theory (QFT) fundamentals
  • Feynman diagram techniques
  • Path-integral formalism
  • Understanding of Green's functions
NEXT STEPS
  • Study A. Zee's "Quantum Field Theory in a Nutshell," particularly chapters IV.3 and VI.1
  • Review Dunne's notes on effective action calculations, focusing on section 5
  • Examine Golterman's paper for a general approach to the Chern-Simons term
  • Explore Kaplan's homework problem and hints related to the Chern-Simons contribution
USEFUL FOR

This discussion is beneficial for students and researchers in quantum field theory, particularly those interested in effective actions, Chern-Simons theory, and the mathematical techniques involved in deriving such contributions.

Kirjava
Messages
26
Reaction score
1
Hi PF,

I'm still very much a novice when it comes to QFT, but there's a particular calculation I'd like to understand and which (I suspect) may be just within reach. In short, the result is that after coupling a system of fermions to an external U(1) gauge field, one obtains a Chern-Simons contribution to the low-energy effective action of the gauge field by integrating out the fermions. Supposedly the coefficient of this (presumably lowest order) term can be computed from a single Feynman diagram to give the expression (here for d = 2+1)

C_2 = \int \mathrm{d}^3 k ~ \epsilon_{\mu \nu \rho} G \partial_{\mu} G^{-1} G \partial_{\nu} G^{-1} G \partial_{\rho} G^{-1}
with G the two-point greens function. The Feynman diagram in question is depicted on the review article page I refer to below.

Sticking to d=2+1 the effective action of this gauge field then looks like

S_{eff}[A] = \frac{C_2}{2 \pi} \epsilon^{\mu \nu \rho} A_{\mu} \partial_{\nu} A_{\rho}

I'd basically like to know how this calculation is done. To be precise, starting with the expression
e^{iS_{eff}[A]} = \int D[c] D[c^{\dagger}] e^{iS(c,c^{\dagger}) + iA^{\mu}j_{\mu}}
and by assuming that A is slowly varying, how would one go about deriving the above?

I know a bit about Feynman diagrams and the path-integral formalism, but apparently not enough to understand how this is done. For those who might be intersted, the context for my question is given on page 36 of this review: http://arxiv.org/pdf/1008.2026v1.pdf, but I haven't found the references cited there particularly helpful. Given the way this result is talked about there and elsewhere in the literature I'm presuming that the techniques required are fairly standard, and that somebody with more experience around here might therefore be in a position to help.

Cheers,
a student
 
Last edited:
Physics news on Phys.org
I think the book by A. Zee, Quantum Field Theory in a Nutshell, Princeton, contains all you need.
In chapter IV.3 he describes how to calculate the effective action and in chapter VI.1 he shows how to calculate the term you are looking for.
 
Of all the references I find Dunne's approach on page 52 of http://arxiv.org/pdf/hep-th/9902115v1.pdf the most understandable. He shows exactly how the Chern-Simons term arises through an explicit computation and obtains its coefficient. Unfortunately these computations are carried out for Dirac propagator \frac{1}{(i{\not \partial} + m)} <br /> and it's not clear that the same reasoning can be easily extended to get the result for more general G posted above.

On the other hand Golterman on page 4 of http://arxiv.org/pdf/hep-lat/9209003v1.pdf seems to obtain the desired result in full generality with elegance and hardly any effort - although he doesn't explain in detail how the Chern-Simons term arises in the first place. Instead the idea seems to be
(1) Assume that a Chern-Simons term exists in the effective action, and calculate its coefficient by writing down a single Feynman diagram(?).
(2) Rewrite the contribution of this diagram by replacing everywhere the "photon vertex" with a partial derivative of the propagator (by invoking what looks like local charge conservation).

Unfortunately I don't understand either of these steps very well. For instance, I'm not sure what a "photon vertex" could mean given that our gauge field is fixed. I also lack the clarity to see that the Chern-Simons term coefficient should be given by the diagram he writes down. Maybe I was wrong about all this being in reach of my knowledge - perhaps I'll have to come back when I've got a sturdier grasp on the fundamentals (I suppose Zee would be a good place to start). But it's hard to let go!

So maybe someone will be interested in either pointing out the flaw in or carrying this meagre start on a derivation a little bit further:

We want to compute
S_{eff} = \mathrm{log} \, \mathrm{det} (G^{-1}+ {\not A}) = \mathrm{tr} \, \mathrm{log} (G^{-1}+ {\not A}) <br />
where G is a propagator which could carry the name of Dirac, or perhaps be something more appropriate to a lattice model (in this case "A-slash" would have to be suitably re-interpreted). Now, suspecting that it will lead to the CS form, expand the logarithm and extract the second order term:
S_{eff}^{(2)} = \frac{1}{2} \mathrm{tr} (G {\not A} G {\not A}) .
Now we (read: I) scratch our heads thinking about how the products of A in this term will turn into the desired product of A \partial A...
 
Last edited:

Similar threads

  • · Replies 0 ·
Replies
0
Views
883
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
724