# Calculation of the Abelian Induced Chern-Simons Term

In Gerald Dunne's paper "Aspects of Chern-Simons Theory" (http://arxiv.org/abs/hep-th/9902115) I'm a little confused as to how equation (225) on page 53 is obtained. Equation (225):

$$\tag{1} S^{\text{quad}}_{\text{eff}} [A, m] = \\ \dfrac{N_{f}}{2} \int \dfrac{d^{3}p}{(2\pi)^{3}} \left[ A^{\mu}(-p) \int \dfrac{d^{3}k}{(2\pi)^{3}} \text{tr} \left[ \gamma^{\mu} \dfrac{ \gamma^{\alpha}p_{\alpha} + \gamma^{\beta}k_{\beta} - m}{(p + k)^{2} +m^{2}} \gamma^{\nu}\dfrac{\gamma^{\gamma}k_{\gamma} - m}{k^{2} + m} \right] A^{\nu}(p) \right]$$

(Please forgive my lack of Feynman slash notation, I don't know how to write it on this site). It's explained that this term comes from the term:

$$\tag{2} \dfrac{N_{f}}{2} tr \left( \dfrac{1}{i\gamma^{\alpha}\partial_{\alpha} + m} \gamma^{\mu}A_{\mu} \dfrac{1}{i\gamma^{\beta}\partial_{\beta} + m} \gamma^{\nu}A_{\nu} \right)$$

found in equation (224). I think I understand how to get (1) from (2) but there are steps I'm unsure of:

Firstly I believe that (2) isn't complete. When I calculated the effective action from the original QED action I got something more like:

$$\tag{3} \dfrac{N_{f}}{2} tr \left( \dfrac{1}{i\gamma^{\alpha}\partial_{\alpha} + m} ie\gamma^{\mu}A_{\mu} \dfrac{1}{i\gamma^{\beta}\partial_{\beta} + m} ie\gamma^{\nu}A_{\nu} \right)$$

Using Peskin and Schroeder page 305, we see that equation (3) above corresponds to equation (9.80) (up to some minus signs). Page 305 of Peskin and Schroeder explains that this corresponds to the Fenyman diagram:




In the above diagram the crosses indicate that the gauge field corresponds to a background field. See this Stack Exchange post for details.

Now to obtain equation (1) it should be a simple case of writing the Feynamn rules for the above diagram in momentum space. This is where I get a bit unsure. If the Feynman diagram had been the following instead:




corresponding to a dynamical gauge field, then the Feynman rules would have given (see Peskin and Schroeder, page 245) something of the form:

$$\tag{4} (-ie)^{2}(-1) \int \dfrac{d^{3}k}{(2\pi)^{3}} tr \left[ \gamma^{\mu} \dfrac{i}{\gamma^{\alpha}k_{\alpha} - m} \gamma^{\nu} \dfrac{i}{\gamma^{\beta}k_{\beta} + \gamma^{\gamma}p_{\gamma} - m} \right]$$

This can be rewritten (See Peskin and Schroeder, page 63, equation (3.120):

$$\tag{5} (-ie)^{2}(-1) \int \dfrac{d^{3}k}{(2\pi)^{3}} tr \left[ \gamma^{\mu} \dfrac{i(\gamma^{\alpha} p_{\alpha} + m)}{k^{2} - m} \gamma^{\nu} \dfrac{i(\gamma^{\beta} p_{\beta} + \gamma^{\gamma}k_{\gamma} +m)}{(p+k)^{2} - m} \right]$$

This looks somewhat like the term between the $A^{\mu}(-p)$ and the $A^{\nu}(p)$ in equation (1) above. Differences might simply be due to differences in conventions (ie differences in metric signature between Dunne's paper and Peskin and Schroeder's book). In fact Gerald Dunne refers to that term in (1) as the 'kernel' and confirms that it corresponds to the second Feynman diagram drawn above (see page 53, equation (226), of http://arxiv.org/abs/hep-th/9902115).

Note that for the second diagram we only integrate over the internal momentum, k.

Now if we were to say that the gauge field is a background field (non-dynamical) this source explains that we would expect the extra $A^{\mu}$ and $A^{\nu}$ terms that appear in (1). Such non-dynamical gauge fields appear in the Feynman rule for the vertices.

Question 1: Why do we write $A^{\mu}(-p)$ and $A^{\nu}(p)$ in equation (1). Surely the Feynman diagram would suggest both incoming and outgoing photons have the same sign of momentum.

Question 2: Why, with the non-dynamical gauge field, would we need to then integrate over $p$ as well. We get such an integral over p in (1), and (1) corresponds to the the first Feynman diagram. We don't get such an integral over p for the second diagram (the one without background gauge fields).

I'm still very shaky on how to derive (1), and, as is probably quite obvious, I only roughly know how it is obtained. Any feedback would be greatly appreciated.

I only read the 5th chapter up to the point that confuses you in your cited paper, so I don't know what is written in the previous ones.

Q1: Both momenta point inward. This is no problem, since there is e.g. no flow of charge that would suggest a direction. You can even assign different momenta for both particles: $A^\mu(q)$ and $A^\nu(p)$. However, Energy-momentum conservation $\sim \delta^{(4)}(q+p)$ turns your $q$ into $-p$.

Q2: Since you have a background field, the $A^\mu$ and $A^\nu$ lines are internal ones and do not correspond to real particles (they are called virtual, however, i don't really like this terminology...). Since the background can create such a virtual particle at any momentum $p$, we have to integrate over all possibilities.

(It's not a very detailed answer, but maybe it helps anyway ;) )

• Ah I see, so is it correct that the first diagram I drew isn't complete, it's just the internal part of another (arbitrary) larger diagram? Would this be true also for the diagram that gives equation (9.80) on page 305 of Peskin and Schroeder? Also I don't completely understand how both momenta can point inwards without violating momentum conservation. I would have thought the momentum 'going in' has to equal the momentum 'going out'. That is to say, I would have though both gauge fields would need the same momentum (p for both). – Siraj R Khan Jan 7 '15 at 13:20
• @Siraj R Khan. First of all, momentum going in is just a choice of direction. The fact that the ingoing momentum is $-p$ tells you, it is the same as an outgoing momentum $p$. The diagram you drew is correct, the blobs on either ends indicate that there is a background 'source'. Internal line only means, that it is between a blob and a vertex. (I don't have the P&S book here, so I will have to comment on that later) – Clever Jan 7 '15 at 13:24
• I see what you're saying, but would you not have to stick to the same choice of direction for both the incoming and outgoing momenta (such that both would be p or both would be -p). It seems odd (cheating?) to change the direction arbitrarily from one part of the diagram to another. – Siraj R Khan Jan 7 '15 at 13:28
• It is not cheating. It's just like saying that i don't want to use subtraction, but instead use addition of negative numbers: $a-b=a+(-b)$ ;) You see, e.g. in a $\phi^4$ treelevel diagram, I can always choose all momenta as going into the vertex. However, e-m conservation $\delta(p_1+p_2+p_3+p_4)$ tells me that at least one momentum has to be negative, i.e. outgoing. – Clever Jan 7 '15 at 13:33
• Yes, exactly! :) – Clever Jan 7 '15 at 14:02