4
$\begingroup$

I'm learning BRST symmetry for Yang-Mills theory and I see that there are two ways of writing BRST differential. In some books (for example Ryder's and Ramond's textbooks) BRST differential acts as \begin{gather} \delta A_\mu^a =-D_\mu c^a, \\ \delta c^a= -\frac{1}{2} f^a_{bc} c^b c^c ,\\ \delta \bar{c}^a= f^a, \end{gather} where I skipped coupling constant, and $f^a$ is a gauge-fixing function, for example $f^a=\partial^\mu A_\mu^a$.

But in Srednicki's or Peskin and Schoeder's textbooks differential $\delta$ acts on $\bar{c}$ as $$ \delta \bar{c}^a= B^a, $$ where $B^a$ is the auxiliary field.

For me it seems that first approach is a simple elimination of the auxiliary field $B^a$ from the differential and from the action using condition $f^a=B^a$. Is that so? I just want to be sure I'm not missing something.

What form of BRST is "preferable" i.e. what are reasons to choose BRST transformation with or without auxiliary field?

$\endgroup$
4
$\begingroup$

On one hand, by including the Lautrup-Nakanishi field $B^a$, we have an off-shell BRST formulation, i.e. we can prove the nilpotency of the BRST transformation without using the (Euler-Lagrange) equations of motion.

On the other hand, for some applications, a simpler on-shell BRST formulation (where the Lautrup-Nakanishi field $B^a$ has been integrated out/eliminated) would suffice.

Remark: A BRST-invariant amplitude should be independent on whether we work in, e.g., Lorenz gauge, Coulomb gauge, axial gauge, etc, but this fact may in practice be difficult to trace if we work with on-shell BRST transformations, which depend explicitly on the gauge-fixing choice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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