# Tag Info

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 ...

11

Ghostly Lie algebra cohomology Let $\mathfrak{g}$ be our Lie algebra and $V_\rho$ a representation space with representation map $\rho : \mathfrak{g} \to \mathrm{End}(V_\rho)$. $V_\rho$ is, by the action through the representation, naturally a $\mathfrak{g}$-module (people missing the ring structure in $\mathfrak{g}$ - just embed it into the universal ...

8

You seem to be talking about the "old covariant quantization" in which $L_n$ for positive $n$ and $(L_0-a)$ annihilate physical ket states $|\psi\rangle$, right? It's analogous to the Gupta-Bleuler quantization http://en.wikipedia.org/wiki/Gupta-Bleuler_quantization which was a standard procedure used already in electromagnetism. The idea is that the ...

7

J.W. van Holten's "Aspects of BRST Quantization" arXiv:hep-th/0201124 might be what you're looking for...

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. ...

5

A modern treatment of this subject can be found in Segreev's book on the Kahler geometry of loop spaces also available online. This line of research started with the seminal work of Bowick and Rajeev: The holomorphic geometry of the closed bosonic string theory and $Diff S^1/S^1$ (Spires) (and independently Kirillov and Yuriev (Please see the reference in ...

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 ...

5

I) Let us first clarify the left and right derivatives. Left derivatives is explained between eq. (15.8.9) and (15.8.10) in Ref. 1. A left derivative means a derivative that acts from the left. E.g. if $F= \chi G$, where $G$ does not depend on $\chi$, then $\frac{\delta_LF}{\delta\chi}=G$. Similarly, a right derivative acts from the right. E.g. if $F= ... 3 The reason for the discrepancy is that the BRST operator is not self adjoint$\Omega^{\dagger} \ne \Omega$, This can easily be seen from its action on the ghosts: $$\delta c^{a} = \epsilon \{\Omega, c^a\} = i \epsilon f^{a}_{bc} c^b c^c$$ $$\delta \bar{c}^{a} = \epsilon \{\Omega, \bar{c}^a\} = - i \epsilon b^a$$ $$\delta b^{a} = \epsilon \{\Omega, b^a\} = ... 3 I) The gauge-fixed pure Maxwell action is$$\tag{1} S[A,c,\bar{c}]~=~\int \! d^4x~ {\cal L} $$with Lagrangian density^1$$\tag{2} {\cal L}~=~-\frac{1}{4}F_{\mu \nu}F^{\mu \nu} -\frac{\chi^2}{2\xi}-d_{\mu}\bar{c}~d^{\mu}c, \qquad \chi~:=~d_{\mu} A^{\mu}, \qquad \xi~>~0,$$consisting of (i) the Maxwell term, (ii) the gauge-fixing term, and (iii) ... 3 On the torus there are two real moduli \tau_1 and \tau_2, and two conformal Killing vectors corresponding to translations. This means that you need two insertions of b and two insertions of c in order to saturate the zero mode path integral. 3 answer for questions 1,2, and 3a 1) Looking at 2.7.22 to 2.7.24 (and also 2.7.18abc), one define the ghost number N^g = \frac{-1}{2\pi i} \int_0^{2 \pi}dz :b(z)c(z): , and that all the operators c_n increase the ghost number by one. [N^g,c_m] = c_m, So the field c(z), made of operators c_m, increase the ghost number by one unit.So ... 3 I did not furnish all the details because it would be too long, but I give some hints at the end of the answer. I have used the formulae :T^g: ~= ~:2(\partial c) b + c(\partial b): and :\frac{1}{2}cT^g: ~= ~:bc \partial c:, when there is an ambiguity in the calculus. We begin by :$$j_B = cT^m+:\frac{1}{2}:cT^g:+\frac{3}{2}\partial^2c=cT^m+:bc\partial ... 3 I) Since total divergence terms do not contribute to Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post, one could just integrate the Faddeev-Popov$\bar{c}c$term by part so that there are no more than first derivatives present and the standard form of the EL equations applies. II) Alternatively, in the presence of higher derivatives, the EL ... 3 Since$R[A]$is gauge invariant, the variation of$R[A]$is zero when$A^a_{\mu}$undergoes the infinitesimal gauge transformation$A^a_{\mu}\rightarrow A^a_{\mu} + \epsilon (D_{\mu}\alpha)^a$where$\alpha^a$is any Lie algebra valued field and$\epsilon$is an infinitesimal parameter. The variation of$R[A]$under this gauge transformation is $$0=\delta R ... 3 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 ... 3 The LSZ formula definitely applies to theories with massless gauge bosons, like QED and QCD. The S-matrix is given by the LSZ formula, which relates the former to correlation functions, which are in turn given by a path integral. The LSZ formula assumes asymptotic in- and out-states for particles of interest. In the path integral formalism, gauges can be ... 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 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 = ... 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 Kugo and Ojima's work was one of the major breakthroughs in understanding the role of BRST in the quantization of gauge theories. Historically BRST was discovered in the path integral formalism. The understanding of this theory as a cohomology theory started from the Kugo and Ojima's work. Now, the action is BRST invariant with and without the Gaussian ... 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]}$$are not mathematically well-defined, because of the presence of the imaginary unit in the exponential. ... 2 Hints: The left-hand side$$-\frac{1}{2}g^2f^{abc}f^{cde}\left(A_{\mu}^{b}c^{d}c^{e}+{\rm cycl}(b,d,e)\right)$$of eq. (16.47) can be relabelled as$$-\frac{1}{2}g^2\left( f^{abc}f^{cde}+{\rm cycl}(b,d,e)\right)A_{\mu}^{b}c^{d}c^{e}.$$P&S assume that the structure constants f^{abc} are totally antisymmetric, cf. text below eq. (15.79). 2 In a nutshell, the Grassmann-odd Faddeev-Popov ghosts fields appear from the exponentiation of the Faddeev-Popov determinant, i.e., when we write the determinant as a Gaussian integral over Grassmann-odd variables. The Faddeev-Popov determinant can roughly be viewed as a Jacobian factor in the path integral that appears because the path integral variables ... 2 You should think in terms of the norm of a state, and what happens here is that you have negative-norm states which you don't want, otherwise you can't construct a Hilbert space. Copying the formula in Wikipedia,$$ \langle\vec{k}_a;\epsilon_\mu|\vec{k}_b;\epsilon_\nu\rangle=(-\eta_{\mu\nu}){1\over 2|\vec{k}_a|}\delta(\vec{k}_a-\vec{k}_b)$you see that ... 2 You are always allowed to introduce a new integration variable as long as its not its already being summed over. This might be more clear in discrete form: \begin{align} \int d x \, f (x) & \rightarrow \Delta x\sum _i \,f ( x _i ) \\ & = \big( N \Delta y\sum _j g ( y _j ) \big) \Delta x \sum _i f ( x _i ) \\ \end{align} where N \Delta ...

2

You want to compute the integral $Z = \int d [A] e^{iS}$ and since it has a gauge symmetry, there are multiple values of $A$ that generating the same Action $S$, since $S(A_g)=S(A_{g'})$ for the two different choices of gauge $g,g'$. Now you have a gauge-fixing condition like $\partial^\mu A_\mu = 0$. The integral which contains enough physical ...

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