# Finding ghost terms for super Yang-Mills theory in background field gauge

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: $$\epsilon^{arx} \epsilon^{cbx}B^{ir}(B_{i}^{c} + \sqrt{g}Y_{i}^{c}) \overline{c}^{a} c^{b}$$

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

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.
• $$G^{a} t^{a} = D^{\mu}A_{\mu}^{a} t^{a}= \partial_{\mu} A_{\mu}^{a} t^{a}+ [B^{\mu c} t^{c}, A_{\mu d} t^{d}] = \partial_{\mu} A_{\mu a} t^{a}+B^{\mu c} A_{\mu d} f^{cda} t^{a}$$ Is this correct? – saad Jan 13 at 6:04
• ♦ 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 $\epsilon^{abc}(\partial_{\tau} \overline{c}^{a}) c^{b} A^{c}$. – saad Jan 17 at 11:08