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Integrating over a gauge field in the field integral formalism

+ 5 like - 0 dislike
409 views

I'm currently trying to study a chapter in Altland & Simons, "Condensed Matter Field Theory" (2nd edition) and I'm stuck at the end of section 9.5.2, page 579.

Given the euclidean Chern-Simons action for a gauge field $a_µ$ that is coupled to a current $j_µ$

$$ S[a_µ,j_µ] = ∫d^3x (j_µ a_µ + \frac{iθ}4ε_{µνλ}a_µ ∂_ν a_λ) $$

the task is to integrate out the gauge field and obtain the effective action for the current.

Since this is a gauge field, we have to take care about the superfluous gauge degree of freedom. Altland & Simons note that one way to do it would be to introduce a gauge fixing term $α (∂_µ a_µ)^2$ and let $α\to ∞$ at the end.

However, this does not seem to work. In momentum space, the Chern-Simons action plus gauge fixing terms is proportional to

$$∫ d^3q\ a_µ(-q) \left( \begin{array}{ccc} \alpha q_0^2 & -i q_2 & i q_1 \\ i q_2 & \alpha q_1^2 & -i q_0 \\ -i q_1 & i q_0 & \alpha q_2^2 \end{array} \right)_{µν} a_ν(q) .$$

To get the effective action for the current, I just have to invert this matrix, which we call $A_{µν}$, and send $α\to ∞$. But this can't be. For instance, one entry of the inverse matrix reads

$$ A^{-1}_{01} = \frac{-q_0 q_1-i q_2^3 \alpha }{q_1^2 q_2^2 q_0^2 \alpha ^3- α(q_0^4+q_1^4+q_2^4) } $$

and this vanishes in the limit $α\to∞$. Same for the other entries. This is bad.

My question, hence

How to properly perform the functional integral over a gauge field $a_µ$ with a gauge fixing contribution $α(∂_µ a_µ)^2$ where $α\to ∞$?

I am aware that there are other methods, for instance to integrate only over the transverse degrees of freedom, as Altland & Simons note. I don't mind learning about them as well, but I would like to understand the one presented here in particular. Not to mention that I may have made a simple mistake in the calculation above.

This post imported from StackExchange Physics at 2014-04-05 17:31 (UCT), posted by SE-user Greg Graviton
asked May 12, 2012 in Theoretical Physics by Greg Graviton (675 points) [ no revision ]
Just in case people don't notice (I only noticed because of @Moshe 's answer): the mistake made above is that $\alpha(\partial_\mu a_\nu)^2 = \alpha(\partial_\mu a_\mu) (\partial_\nu a_\nu)$, so each matrix $A_{\mu \nu}$ entry must contain the term $\alpha q_\mu q_\nu$. Right now this only appears in the diagonal components.

This post imported from StackExchange Physics at 2014-04-05 17:31 (UCT), posted by SE-user Olaf
Oops, indeed. This term does and should correspond to the projection $L_{µν} a_ν = \frac{q_µq_ν}{q^2} a_ν$ onto the longitudinal degrees of freedom.

This post imported from StackExchange Physics at 2014-04-05 17:31 (UCT), posted by SE-user Greg Graviton

1 Answer

+ 6 like - 0 dislike

In the present case I think that it is more convenient to perform the propagator computation covariantly (and not in components).

The inverse propagator (in the momentum space) can be read from the Abelian Chern Simons action including the gauge fixing term as:

$ G^{-1}_{\mu\nu}(k) = \alpha q_{\mu} q_{\nu} + i \frac{\theta}{4} \epsilon_{\mu\nu\rho}q^{\rho}$

For the propagator we use the Ansatz:

$G_{\sigma\tau}(k) = \beta q_{\sigma} q_{\tau} + i \gamma \epsilon_{\sigma\tau\eta}q^{\eta}$

The parameters $\beta$ and $\gamma$ must be calculated fro the condition:

$ G^{-1}_{\mu\nu}(k) \delta^{\nu\sigma} G_{\sigma\tau}(k)= \delta_{\mu\tau} $

Please observe that the propagator cannot contain a term proportional to $\delta_{\nu\sigma}$, because this term would result a term proportional to $\epsilon_{\mu\sigma\tau}q^{\tau}$ after contraction with the inverse propagator which cannot be canceled with any other term.

We obtain:

$(\alpha \beta q^2 -\frac{\theta}{4} \gamma) q_{\mu} q_{\tau} + \frac{\theta}{4} \gamma q^2 \delta^{\mu\tau} = \delta^{\mu\tau}$

(Where, The following identity was used: $\delta^{\nu \sigma}\epsilon_{\mu\nu\rho} \epsilon_{\sigma\tau\eta} = \delta_{\mu \eta}\delta_{\rho \tau}- \delta_{\mu \tau}\delta_{\rho \eta}$)

Thus:

$ \gamma = \frac{4}{\theta q^2}$

$ \beta = \frac{\theta \gamma }{4 \alpha q^2} = \frac{1}{\alpha q^4}$

Thus $ \beta $ vanishes in the limit $ \alpha \to \infty$ and we are left with the effective action:

$ \frac{1}{\theta}\int d^3q J^{\sigma}(-q) \frac{\epsilon_{\sigma\tau\eta}q^{\eta}}{q^2} J^{\tau}(-q)$

The factor 4 cancels with a similar factor coming from the square completion.

This post imported from StackExchange Physics at 2014-04-05 17:31 (UCT), posted by SE-user David Bar Moshe
answered May 13, 2012 by David Bar Moshe (3,875 points) [ no revision ]
Thank you! I now also understand why the method of letting $α\to ∞$ works for any gauge invariant action. The reason is simply that a gauge invariant action projects onto the the transversal degrees of freedom while the expression $\frac{q_νq_µ}{q^2}$ projects on the longitudinal degree of freedom.

This post imported from StackExchange Physics at 2014-04-05 17:31 (UCT), posted by SE-user Greg Graviton

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