I want to find the ghost terms (2.16) for the action in this paper. The gauge field action is given by

$$ \begin{align}S_{A} =& i \int d\tau \Big(\frac{1}{2}A_{1}(\partial_{\tau}^2 - r^2)A_{1} + \frac{1}{2}A_{2}(\partial_{\tau}^2 - r^2)A_{2}+ \frac{1}{2}A_{3}\partial_{\tau}^2 A_{3}+ 2 \epsilon^{ab3}\partial_{\tau}B^{i}_{3}A_{a} Y^{i}_{b} \cr &+ \sqrt{g}\epsilon^{abc} \partial_{\tau} Y^{i}_{a}A_{b}Y^{i}_{c} -\sqrt{g} \epsilon^{a3x} \epsilon^{bcx}B^{i}_{3}A_{a}A_{b}Y^{i}_{c}-\frac{g}{2}\epsilon^{abx}\epsilon^{cdx}A_{a} Y^{i}_{b} A_{c} Y^{i}_{d} \Big),\end{align}\tag{2.11}$$

where $A_{\mu}, \mu = 0,1, \dots,9$ is a $U(2)$ gauge field.

Here, we use the background gauge choice

$$ \overline{D}^{\mu} A_{\mu} = \partial^{\mu}A_{\mu} + [B^{\mu}, A_{\mu}]\tag{2.2}.$$ In the first step of the Faddeev-Popov trick,

$$ 1 = \int D\alpha \delta(G[A^\alpha])\det\Big(\frac{\delta G[A^{\alpha}]}{\delta \alpha}\Big).$$

Now the form of the functional derivative in the determinant depends on $\alpha$, which depends on the gauge transformation of $A_{\mu}$. I'm unsure about how the $A_{\mu}$ gauge transforms in terms of $\alpha$.

EDIT: Here is my work. I haven't managed to work out why there is a $\sqrt g$ on the $\epsilon^{abc}(\partial_{\tau} \overline{c}^{a}) c^{b} A^{c} $ term. How do I get the $\sqrt g$?

$$G^{a}t^{a} = \partial^{\mu} A_{\mu}^{a} t^{a} + [\ B^{\mu r} t^{r}, A_{\mu}^{s} t^{s} ]$$ The gauge condition is therefore given by: $$ G^{a} = \partial^{\nu} A_{\nu}^{a}+ B^{\nu r} A_{\nu}^{s} \epsilon^{rsa}t^{a}$$ From Srednicki's book, we have the expression for the ghost term in the Lagrangian is given by: $$ \mathcal{L}_{GH} = \overline{c}^{a} \frac{\partial G^{a}}{\partial A_{\mu}^{b}} D_{\mu}^{bc}c^{c} $$ Where the gauge covariant derivative is: $$ D_{\mu}^{bc} c^{c} = (\delta^{bc} \partial_{\mu} + \epsilon^{bsc} A_{\mu}^{s})c^{c} = \partial_{\mu} c^{b} + \epsilon^{bsc} A_{\mu}^{s} c^{c}$$ And $$ \frac{\partial G^{a}}{\partial A_{\mu}^{b}} = \delta^{ab} \partial^{\mu} + B^{\mu r} \epsilon^{rba}$$ Putting it all together, $$ \mathcal{L}_{GH} = \overline{c}^{a}(\delta^{ab} \partial^{\mu} + B^{\mu r} \epsilon^{rba})(\partial_{\mu} c^{b} + \epsilon^{bsc} A_{\mu}^{s} c^{c})$$

$$\mathcal{L}_{GH} = \overline{c}^{a} \Box c^{a} + \epsilon^{asc}\overline{c}^{a} \partial^{\mu} (A_{\mu}^{s} c^{c}) + B^{\mu r} \epsilon^{rca} \overline{c}^{a} \partial_{\mu} c^{c} + B^{\mu r} \epsilon^{bsc} \epsilon^{rba} A_{\mu}^{s} \overline{c}^{a} c^{c}$$ $\because$ this is a dimensionally reduced Yang-Mills theory, the space derivatives all disappear. Also, $B^{0} = 0$. $$ \mathcal{L}_{GH} = \overline{c}^{a} \partial^{t} \partial_{t} c^{a} + \epsilon^{asc}\overline{c}^{a} \partial^{t} (A^{s} c^{c}) + \cancelto{0}{B^{0 r} \epsilon^{rca} \overline{c}^{a} \partial_{t} c^{c}} + \cancelto{0}{B^{0 r} \epsilon^{bsc} \epsilon^{rba} A^{s}} + B^{i r} \epsilon^{bsc} \epsilon^{rba} A_{i}^{s} \overline{c}^{a} c^{c}$$ We make a Wick rotation $t \rightarrow -i\tau \implies \partial_{t} \rightarrow i \partial_{\tau}$. Also, $A^{c} \rightarrow -i A^{c}$. Then, upto a total derivative, in Euclidean space, $$ \mathcal{L}_{GH} = -\overline{c}^{a} \partial_{\tau}^{2} c^{a} + \epsilon^{abc}(\partial_{\tau} \overline{c}^{a}) c^{b} A^{c} + B^{i r} \epsilon^{cbx} \epsilon^{arx} A_{i}^{c} \overline{c}^{a} c^{b}$$ We can expand the last term above about the background field: $$ \epsilon^{arx} \epsilon^{cbx}B^{ir}(B_{i}^{c} + \sqrt{g}Y_{i}^{c}) \overline{c}^{a} c^{b}$$

$$ = (\delta^{ac} \delta^{rb} -\delta^{ab}\delta^{rc})B^{ir} (B_{i}^{c} + \sqrt{g} Y_{i}^{c}) \overline{c}^{a} c^{b}$$


1 Answer 1


Supersymmetry only plays a minor role here, so we will suppress it in this answer. The Faddeev-Popov gauge-fixing procedure and the background field method $$ A_{\mu}^a~=~\underbrace{\overline{A}_{\mu}^a}_{\text{background}} +\underbrace{{\cal A}_{\mu}^a}_{\text{quant. fluct.}} \tag{A}$$ are e.g. explained for Yang-Mills theory in Ref. 1, chapter 71 & 78, respectively. Using the notation & conventions of Ref.1, the resulting Lagrangian density (without matter fields) becomes $${\cal L}~=~-\frac{1}{4}F_{\mu\nu}^a F^{\mu\nu a} + \underbrace{\overline{c}^a\overline{D}^{\mu} D_{\mu}c^a}_{\text{Faddeev-Popov term}} - \underbrace{\frac{1}{2\xi} G^aG^a}_{\text{gauge-fixing term}},\tag{B}$$ with gauge-covariant derivative $$ D_{\mu}~=~\partial_{\mu}-igA_{\mu},\qquad \overline{D}_{\mu}~=~\partial_{\mu}-ig\overline{A}_{\mu},\tag{C}$$ and $R_{\xi}$ gauge-fixing function $$G^a~:=~\overline{D}^{\mu}{\cal A}_{\mu}^a.\tag{D}$$ We leave it to the reader to translate the Lagrangian density (B) into the notation of the Becker sisters.


  1. M. Srednicki, QFT, 2007; Chapter 71 & 78. A prepublication draft PDF file is available here.
  • $\begingroup$ $$G^{a} t^{a} = D^{\mu}A_{\mu}^{a} t^{a}= \partial_{\mu} A_{\mu}^{a} t^{a}+ [B^{\mu c} t^{c}, A_{\mu d} t^{d}] = \partial_{\mu} A_{\mu a} t^{a}+B^{\mu c} A_{\mu d} f^{cda} t^{a} $$ Is this correct? $\endgroup$
    – saad
    Jan 13, 2020 at 6:04
  • $\begingroup$ ♦ Thank you for the reference. I have one question: In the final answer for the ghost term (2.16), why are the only derivatives that hit the ghost and anti ghost fields w.r.t. Euclidean time (∂τ)? – $\endgroup$
    – saad
    Jan 15, 2020 at 8:26
  • $\begingroup$ Spatial derivatives disappear under the dimensional reduction. $\endgroup$
    – Qmechanic
    Jan 15, 2020 at 9:41
  • $\begingroup$ ♦ Could you please have a look at my work? I can break up the last term to get r-squared but I can't seem to figure out why there is a $\sqrt g$ in front of $\epsilon^{abc}(\partial_{\tau} \overline{c}^{a}) c^{b} A^{c}$. $\endgroup$
    – saad
    Jan 17, 2020 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.