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15

This is a temperamental difference more than a physical one, but I feel like this question deserves an answer with a lot less formalism than what Urs is using. The physical point that you should never lose sight of is that gauge symmetries are not symmetries at all: they don't map one state to another one, but instead identify a priori different states as ...

12

The mystery here should disappear once one realizes that the BRST complex -- being a dg-algebra -- is the formal dual to a space , namely to the "homotopically reduced" phase space. For ordinary algebras this is more familiar: the algebra of functions $\mathcal{O}(X)$ on some space $X$ is the "formal dual" to $X$, in that maps $f : X \to Y$ correspond to ...

11

Dear Student, as Moshe says, the reason why the Faddeev Popov ghosts "decouple" is that they're designed to decouple. More precisely, they - and all the formulae that depend on them - are designed so that the excitations of these ghosts, as well as unphysical excitations of more ordinary physical fields - such as the time-like and longitudinal components ...

7

The main answer to the question is that the full generator $$L_n = L^{(\mathrm{m})}_n+L^{(\mathrm{g})}_n$$ in Bosonic string theory is a sum of a matter part with normal ordering constant $a=1$, $$L^{(\mathrm{m})}_n=\frac{1}{2} \eta_{\mu\nu}\sum_m:\alpha_{n-m}^{\mu}\alpha_m^{\nu}:-\hbar a\delta_n^0,$$ and a ghost part $$L^{(\mathrm{g})}_m=\sum_n(m-n):... 6 Why do we gauge-fix the path integral in the first place? If we were doing lattice gauge theory, we didn't need to gauge-fix. But in the continuum case, (the Hessian of) the action for a generalized^1 gauge theory has zero-directions that lead to infinite factors when performing the path integral over gauge orbits. In a BRST formulation (such as, e.g., the ... 6 Here's part of my answer to the derivvation of the EM tensor for the ghost action. It does not match the expression you gave, but I may have made a mistake. CAn you check my work? We start with the action \begin{split} S_{gh} &= - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} \nabla_\mu c^\beta \\ \end{split} \... 6 I) First of all, note that although gauge theory and BRST formulation originally only referred to Yang-Mills theory (and hence QED), they nowadays apply to general theories with so-called local gauge symmetry, cf. e.g. this Phys.SE post. The Lagrangian and Hamiltonian BRST formalism are known as Batalin-Vilkovisky (BV) formalism and Batalin-Fradkin-... 5 Your mechanism of "one ghost falling in" is a different way of talking about the "evaporation of FP ghosts by a black hole". So do black holes evaporate ghosts? This is not a well-defined question because FP ghosts are unphysical, too. They're just a mathematical method to deal with gauge symmetries, in this case the diffeomorphism symmetry. In this sense, ... 5 This is an interesting question. And I know that Lubos has already written a nice and complete answer but I think I can add some of my own words here. I'll concentrate here on the non-abelian YM case, but the results here are quite general. The idea is that the BRST quantum action can be written as: S(qu) = S(cl) + \int\; d^4x\; s\Psi \end{... 5 As you already wrote, the (3/2)\partial^2 c term is needed for the current to be a one-form i.e. (1,0) tensor field; see also page 131 of Polchinski's String Theory, volume 1. This means that if you compute the OPE$$ T(z) j^{BRST}(0)\sim \dots, $$you want to get$$\dots \sim \frac{1}{z^2} j^{BRST}(0)+\frac{1}{z}\partial j^{BRST}(0), $$see e.g. ... 4 In Minkowski spacetime, the action has to be real. Btw, that's necessary for the classical limit to give principle of least action. Yes, such sums are ill-defined, so some might say that the theory is mathematically defined by analytically continuing (Wick rotating) to Euclidean time, where you have nice exponentially decaying weights. You'll get saddle ... 4 The solution to this problem comes from the sneaky fact (Kugo, 1978) that while the FP ghost field is hermitian c^\dagger (x) = c(x) while the anti-ghost field is anti-hermitian \bar c^\dagger (x)=-\bar c (x) . As a result, the plane wave expansion for the ghost/anti-ghost fields (Becchi, 2008), Scholarpedia are:$$ c^a(x)={1 \over(2 \pi)^{3/2}} \...

4

We only have one contribution from each gauge-equivalent matter field configuration: Let $P$ be the principal $G$-bundle associated to our gauge theory on the spacetime $\mathcal{M}$ (for simplicity, assume it is $\mathcal{M} \times G$. The matter fields are constructed as sections of an associated vector bundle $P \times_G V_\rho$, where $V_\rho$ is a ...

4

I) The gauge-fixed pure Maxwell action is $$\tag{1} S[A,c,\bar{c}]~=~\int \! d^4x~ {\cal L}$$ with Lagrangian density$^1$ $$\tag{2} {\cal L}~=~{\cal L}_0 -\frac{\chi^2}{2\xi}-d_{\mu}\bar{c}~d^{\mu}c, \qquad {\cal L}_0~:=~-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}, \qquad \chi~:=~d_{\mu} A^{\mu}, \qquad \xi~>~0,$$ consisting of (i) the Maxwell term, (ii) ...

3

On the torus there are two real moduli $\tau_1$ and $\tau_2$, and two conformal Killing vectors corresponding to translations. This means that you need two insertions of $b$ and two insertions of $c$ in order to saturate the zero mode path integral.

3

"Something is conserved for an action" simply means that the action carries a zero overall value of "something" (for an additive quantity). In this case, the action has $N_{gh}=0$. It follows that the equations of motion derived from the action imply $dN_{gh}/dt=0$. Most typically, they imply $\partial_\mu J^\mu_{gh}=0$ i.e. the local continuity law for a ...

3

answer for questions $1$,$2$, and $3a$ 1) Looking at $2.7.22$ to $2.7.24$ (and also $2.7.18abc$), one define the ghost number $N^g = \frac{-1}{2\pi i} \int_0^{2 \pi}dz :b(z)c(z):$ , and that all the operators $c_n$ increase the ghost number by one. $[N^g,c_m] = c_m$, So the field $c(z)$, made of operators $c_m$, increase the ghost number by one unit.So $c^\... 3 I did not furnish all the details because it would be too long, but I give some hints at the end of the answer. I have used the formulae$:T^g: ~= ~:2(\partial c) b + c(\partial b):$and$:\frac{1}{2}cT^g: ~= ~:bc \partial c:\$, when there is an ambiguity in the calculus. We begin by : $$j_B = cT^m+:\frac{1}{2}:cT^g:+\frac{3}{2}\partial^2c=cT^m+:bc\partial ... 3 The Higgs ghosts are not Faddeev-Popov ghosts. (For starters, the Faddeev-Popov ghosts in the standard model are Grassmann-odd, while the Higgs ghosts are Grassmann-even.) The Higgs ghosts are Goldstone bosons for the spontaneously broken part of the electroweak symmetry SU(2) \times U(1), which, popularly speaking, get eaten by the massive gauge bosons ... 3 The central charge counts the number of degrees of freedom only for matter fields living on a flat manifold (or supermanifold in the case of superstrings). An example where this counting argument fails for matter fields is the case of strings moving on a group manifold G whose central charge is given by the Gepner-Witten formula: c = \frac{k\mathrm{dim}(... 3 Most higher derivative theories —and in particular Lee-Wick's model— do not have ghost excitations but they are unstable (Hamiltonian unbounded from bellow). Yes, almost everyone says the opposite but all them are unfortunately wrong. They do not quantize the theory properly. Whenever a degree of freedom has negative energy at the classical level, it must ... 3 Caveat: The first part of this answer takes a very technical stance on the BRST procedure, and additionally works with a finite-dimensional phase space for convenience. It could appear quite far from the understanding of ghosts in the average application of BRST transformations or ghosts as a tool. The general conception of ghosts There are many ... 2 I thought at first this question was more general involving the nature of the Virasoro algebra. As a result the first two paragraphs are boiler plate discussions on that. The actual question is addressed in the third paragraph. The vector fields T^m~=~z^{m+1}\partial z satisfy the Witt algebra or Virasoro algebra without central extension for each index ... 2 Bosonic path integrals :$$Z = \int D\phi ~e^{-i \large \int ~ dx [\frac{1}{2}\phi (\square+m^2)\phi]}$$or Femionic path integrals (like Fadeev-Popov ghosts) :$$Z = \int D\eta D \tilde \eta ~e^{-i \large \int ~ dx [\tilde \eta^a \square \eta^a]}$$are not mathematically well-defined, because of the presence of the imaginary unit in the exponential. ... 2 I am afraid, I am just going to provide the standard lore here. I will do so nonetheless as not many have attempted answering this question yet. Allow me look at the questions from simplistic QM(not QFT) picture. I'll take the question in reverse order: When did we quantize? Well, we never have to, that is the beauty(?) of path integrals! You just pretend ... 2 In a nutshell, the Grassmann-odd Faddeev-Popov ghosts fields appear from the exponentiation of the Faddeev-Popov determinant, i.e., when we write the determinant as a Gaussian integral over Grassmann-odd variables. The Faddeev-Popov determinant can roughly be viewed as a Jacobian factor in the path integral that appears because the path integral variables ... 2 The behaviour of a vector field v^z at z=\infty can be described by first moving to coordinates that are well-defined in that coordinate patch. Defining w = \frac{1}{z}, we find$$ c^w = \frac{dw}{dz} c^z = - z^{-2} c^z $$Now, c^w must be well-defined at w=0. This implies that z^{-2} c^z must be well-defined as z \to \infty. Thus c^z cannot ... 2 I got it. \epsilon anticommutes with b_A. 2 It seems OP's main question concerns the systematics of gauge-fixing. We interpret/reformulate OP's questions as essentially the following. The original gauge-invariant action S_0 is unsuitable for quantization, so we add a non-gauge invariant gauge-fixing term to the action. Obviously we cannot add any non-gauge invariant term to the action. ... 2 Hints: Note that radial ordering {\cal R} and normal-ordered/non-singular terms are implicitly implied in OP's eq. (2), see eq. (A) below. From (i) the OPEs$$ {\cal R} [\beta(z)\gamma(w)]~=~\frac{1}{z-w} ~+~ :\beta(z)\gamma(w):,\tag{A}  {\cal R} [\beta(z)\beta(w)]~=~ :\beta(z)\beta(w):, \qquad{\cal R} [\gamma(z)\gamma(w)]~=~:\gamma(z)\gamma(w):, \...

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