Yang-Mills 3 boson Lagrangian term in Peskin and Schroeder

[Liany]

Hi all,

I'm not certain if this is the correct section of the forum for this thread but I'm trying to understand ghosts and BRST symmetry and my starting point is chapter 16 of Peskin and Schroeder where I've found a nagging issue. My issue is regarding the derivation of equation (16.6) on page 507. The starting point is the Yang-Mills Lagrangian (16.1)

\begin{equation}
\mathcal{L} = -\frac{1}{4}F^{a}_{\mu\nu}F^{a\mu\nu}+\overline{\psi}\left(i\displaystyle{\not}{D}-m\right)\psi
\end{equation}

where we have the field strength defined as (16.2)
\begin{equation}
F_{\mu\nu}^a = \partial_{\mu}A^{a}_{\nu}-\partial_{\nu}A^a_{\mu}+gf^{abc}A^b_{\mu}A^c_{\nu}
\end{equation}

and the covariant derivative is (16.3)
\begin{equation}
D_{\mu} = \partial_{\mu} - i g A^{a}_{\mu}t^{a}
\end{equation}

where I've already dropped the r subscript on the representation matrix.

From this the book goes on to equation (16.6) for the Lagrangian, my issue is with the 3 boson vertex term which is given as
\begin{equation}
-gf^{abc}\left(\partial_{\kappa}A_{\lambda}^a\right)A^{\kappa b}A^{\lambda c}.
\end{equation}

My issue is that the 3 boson vertex term I get is
\begin{equation}
- \frac{gf^{abc}}{2}\left(\partial_{\kappa}A^{a}_{\lambda}-\partial_{\lambda}A^a_{\kappa}\right)A^{\kappa b}A^{\lambda c}.
\end{equation}

So what this boils down to is that I can't immediately see why
\begin{equation}
\left(\partial_{\kappa}A^{a}_{\lambda}-\partial_{\lambda}A^a_{\kappa}\right) = 2\left(\partial_{\kappa}A_{\lambda}^a\right).
\end{equation}

I feel I'm missing something obvious or a previously stated result/assumption but I've scoured the internet for more verbose versions of these steps and read through relevant earlier sections of the book to no avail (not to say I haven't overlooked something). Of the other texts I've checked, Srednicki's draft QFT text does the exact same as Peskin and Schroeder where as Muta's "Foundations of Quantum Chromodynamics" and Kaku's "Quantum Field Theory (A Modern Introduction)" leave it in the form I find.

Thanks for any help.

 

[Orodruin]

Is there perhaps some convenient property of the structure constants $f^{abc}$ that you can use?

 

[i0j1k2l3]

your eq.6 is wrong, eq.5 comes from following simple trick:

$$ f^{abc} \partial_\mu A^a_\nu A^{b\nu}A^{c\mu} = f^{abc} \partial_\nu A^a_\mu A^{b\mu}A^{c\nu} = f^{acb} \partial_\nu A^a_\mu A^{c\mu}A^{b\nu} = -f^{abc} \partial_\nu A^a_\mu A^{c\mu}A^{b\nu} $$

 

[haushofer]

Is there perhaps some convenient property of the structure constants $f^{abc}$ that you can use?

It's been a while for me (a long, long while :P ), but isn't this just because f^{abc} is antisymmetric in b and c? So,

\begin{equation}
f^{abc} \left(\partial_{\kappa}A^{a}_{\lambda}-\partial_{\lambda}A^a_{\kappa}\right)A^{\kappa b}A^{\lambda c} = \\
f^{a[bc]}\left(\partial_{\kappa}A^{a}_{\lambda}-\partial_{\lambda}A^a_{\kappa}\right)A^{\kappa b}A^{\lambda c} = \\
f^{abc} \left(\partial_{\kappa}A^{a}_{\lambda}-\partial_{\lambda}A^a_{\kappa}\right)A^{[\kappa |b|}A^{\lambda] c} = \\
2f^{abc} \left(\partial_{[\kappa}A^{a}_{\lambda]} \right)A^{\kappa b}A^{\lambda c} = \\
2f^{abc} \left(\partial_{\kappa}A^{a}_{\lambda} \right)A^{\kappa b}A^{\lambda c}
\end{equation}

You can do this because both indices b and c (and kappa and lambda) are on the same field A. (the | b| denotes that you obviously don't antisymmetrize over b in the third line)

 

[Orodruin]

It's been a while for me (a long, long while :P ), but isn't this just because fabcfabcf^{abc} is antisymmetric in b and c?

Yes, which is the "convenient" property I wanted the OP to find and do the computation for himself, but #3 let the cat out of the box.

 

[haushofer]

Yes, which is the "convenient" property I wanted the OP to find and do the computation for himself, but #3 let the cat out of the box.

Ok. I read your post like you weren't sure.

 

[Orodruin]

Ok. I read your post like you weren't sure.

I teach this stuff and leading questions have been a modus operandi at PF as long as I can remember ...

 

[haushofer]

Ok, then it's my bad.