5
$\begingroup$

Take for example a spin $2$ field $h_{\mu\nu}$ and some gauge-invariant Lagrangian.

  1. Does the Faddeev-Popov trick work here? by work I mean: does it lead to a consistent and unitary theory? is the theory tractable using the standard techniques (e.g., the ghost determinant exponentiates, etc.)?

  2. What is the gauge-fixing functional that would lead to generalised $R_\xi$ gauges? how many gauge parameters can/should we introduce? Disregarding convergence issues, are $S$-matrix elements $\xi$-independent?

  3. To what extent is the standard BRST theory applicable here?

I am led to think Faddeev-Popov does not work, because the gauge algebra is open, so one must use Batalin-Vilkovisky. Is this correct?

Same questions about a Rarita-Schwinger field.

$\endgroup$
  • $\begingroup$ Consistency and unitarity depend too much on the interaction content of the perturbative QFT. Free linearized gravity is consistent and unitary, whereas if you add interactions – well, you know what happens. Add an $R^2$ interaction – you will recover renormalizability but lose unitarity. Faddeev-Popov formal manipulations don't require unitarity or renormalizability, they can be applied to any gauge-invariant path integral. $\endgroup$ – Prof. Legolasov May 24 '17 at 23:49
  • $\begingroup$ @SolenodonParadoxus fair point. Im actually not that ambitious here: I don't really care about renormalisability yet. I could settle for a gauge invariant $S$ matrix. $\endgroup$ – AccidentalFourierTransform May 28 '17 at 14:14
  • $\begingroup$ But isn't these two ultimately related? I've always seen proofs of unitarity and gauge-invariance of the renormalized $n$-loop S-matrix. Without renormalizability there isn't actually an S-matrix to begin with... But I see your point. In this case, provided that you have a gauge-invariant regularizer (like the one with higher-order covariant derivatives, for example) the Faddeev-Popov method works perfectly for the spin-2 field. You are aware that in this case you are fixing the remnant diffeomorphism invariance, right? $\endgroup$ – Prof. Legolasov May 28 '17 at 23:38
2
$\begingroup$

Let's consider perturbative quantum gravity as an example, with full metric $g_{\mu\nu}^f=g_{\mu\nu}+\kappa h_{\mu\nu}$. The Nakanishi-Lautrup auxiliary field and Faddeev-Popov ghost and antighost are vector fields. The BRST-quantised scalar Lagrangian density is$$R-2\Lambda+\frac{\xi}{2}B_\mu B^\mu-(\delta_\mu^\rho\delta_\nu^\sigma-kg_{\mu\nu}g^{\rho\sigma})(\nabla^\mu B^\nu \kappa h_{\rho\sigma}+i\nabla^\mu\bar{c}^\nu £_{c} g_{\rho\sigma}^f),$$where the covariant derivative is compatible with the unperturbed metric. You'll see the FP-ghost term contains a Lie derivative, which BRST-transforms the full metric. The most common gauge choice is $k=\frac{1}{2}$, for which the theory is anti-BRST invariant.

For more information, you may benefit from excerpts of my PhD thesis. In Secs 2.6.1-2.6.4, I explain the theory's BRST quantisation. (The formalism I have used above is not the more popular vielbein formalism, which is harder to compare by eye to BRST-quantised Yang-Mills theory; see 2.6.4.) In Appendix F (basically a rehash of Sec. 15.9 of Weinberg's The Quantum Theory of Fields, Volume 2: Modern Applications), I explain the motivation for the Batalin-Vilkovisky formalism, as well as why it wasn't ultimately needed for any of my thesis research. In short, you need BV when considering Hamiltonian constraints not originating in the Lie algebra (this is one way the case of perturbative gravity is unlike Yang-Mills), to repair BRST nilpotency due to an open algebra (which IIRC is not an issue here), or to address some quantum anomalies, especially in BRST or anti-BRST symmetries.

$\endgroup$
  • $\begingroup$ Thanks, this is nice!. I'll give your thesis a look, it seems very interesting. $\endgroup$ – AccidentalFourierTransform Jul 14 '17 at 21:24

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.