# Anti-ghost translation invariance$.$

The Faddeev-Popov gauge-fixed Yang-Mills Lagrangian is invariant under $$\bar c\to\bar c+\chi$$ for any odd constant $\chi$. What is the physical interpretation of this invariance? What does this translation transformation correspond to, in practical terms (e.g., at the level of Feynman diagrams)?

Perhaps relevant: this invariance forbids terms quadratic in $\bar cc$, which are necessary for renormalisability if we pick e.g. a gauge-fixing condition of the form $\partial\cdot A+\alpha A^2\equiv 0$ rather than the standard $\partial\cdot A\equiv 0$ one.

• Do you have any reason to believe this has a physical interpretation? The (anti-)ghosts are artificial degrees of freedom to begin with, so I would not expect any symmetries/conserved quantities that act purely in the ghost part of the phase space to have any meaning at all. – ACuriousMind Aug 8 '17 at 12:29
• $c\to \mathrm e^{i\theta}c,\bar c\to\mathrm e^{-i\theta}\bar c$ corresponds to ghost number conservation, which has a rather clear physical meaning (at least in terms of Feynman diagrams). $\bar c\to\bar c+\chi$ eludes me though. – AccidentalFourierTransform Aug 8 '17 at 12:33
• I'd say that ghost number conservation is a necessary consequence of only quantities with zero ghost number being truly physical observables - meaning it is inherent in the BRST/ghost construction, since its non-conservation would make the condition of "zero ghost number" unstable. – ACuriousMind Aug 8 '17 at 12:36
• The corresponding Noether current is just the canonical momenta field for the antighost: $J_{\mu}^{\;a} = D_{\mu} c^{a}$. Current conservation is thus just the Euler-Lagrange equation for the ghost field: $\partial^{\mu} D_{\mu} c = 0$. I don't see any far-reaching physical significance. The same happens for the massless Klein-Gordon field, btw. – Prof. Legolasov Aug 8 '17 at 23:30
• It corresponds to an ambiguity in BRS transformation of anti-ghost field and corresponds to an auxiliary degree of freedom (non-dynamical) which arises after fixing one of the term in $\bar{\delta}c + {\delta} \bar{c}=0$ by hand, where \bar{\delta} corresponds to anti-BRS transformation. – ved Aug 9 '17 at 7:41

Like any continuous symmetry, it represents a conservation law. The symmetry is manifest in the Lagrangian density term $-i\partial_\mu\bar{c}D^\mu c$, where $\bar{c}$ is cyclic. But if we add a total derivative to get $i\bar{c}\partial_\mu D^\mu c$, we see $\partial_\mu D^\mu c=0$ so $D^\mu c$ is a Noether current. If you prefer to work with physical fields, it's best to work in the Landau gauge so $A^\mu$ is conserved, since the above ghost-sector result is just the BRST-transform of $\partial_\mu A^\mu =-\xi B$, which for $\xi = 0$ is a conservation law.
• nice +1, two comments: 1) I guess $B$ is the Lautrup-Nakanishi field, right? 2) what does the conservation law $\partial_\mu D^\mu c=0$ imply at the level of Feynman diagrams? [all Noether currents have an immediate consequence: $\partial_\mu T^{\mu\nu}=0$ implies that the total momentum is conserved (so that the correlation functions depend on the relative positions $x_i-x_j$); $\partial_\mu j^\mu=0$ implies that charge is conserved (so that the correlation functions vanish unless we have the same number of fields and its hermitian conjugates); etc] – AccidentalFourierTransform Aug 10 '17 at 13:39
• @AccidentalFourierTransform Yes, $B$ is the L-N field. – J.G. Aug 10 '17 at 14:14