- #1

Kirjava

- 27

- 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)

[tex] 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} [/tex]

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

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

I'd basically like to know how this calculation is done. To be precise, starting with the expression

[tex] e^{iS_{eff}[A]} = \int D[c] D[c^{\dagger}] e^{iS(c,c^{\dagger}) + iA^{\mu}j_{\mu}} [/tex]

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

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)

[tex] 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} [/tex]

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

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

I'd basically like to know how this calculation is done. To be precise, starting with the expression

[tex] e^{iS_{eff}[A]} = \int D[c] D[c^{\dagger}] e^{iS(c,c^{\dagger}) + iA^{\mu}j_{\mu}} [/tex]

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: