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26

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 ...


9

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-...


8

I) On one hand, the Faddeev-Popov (FP) formalism assumes that The gauge algebra is "irreducible", meaning that there are not higher levels of gauge-symmetries among the gauge generators. This is aka. gauge-for-gauge symmetry. The gauge algebra closes off-shell. If the gauge-fixing conditions do not depend on ghosts, then the FP action is ...


8

In the BRST quantisation of quantum field theories, cohomology does show up. The conserved charge associated to BRST quantisation $Q_B$ implies that physical states must be BRST-invariant via $Q_B |\psi \rangle=0$. It's easy to prove that $Q_B$ is nilpotent: $Q_B^2=0$ such that one can add a null state to any physical state without changing its BRST ...


7

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 different ...


7

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) ...


7

The ghosts are not so much inserted, as they naturally arise. The path integral of a gauge theory naively defined will integrate over all fields, including those related by a gauge symmetry, which are seen by the theory as being equivalent. The Faddeev-Popov procedure provides a means to split our integration over physically distinct configurations and ...


6

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 ...


6

Only 1 Grassmann-odd global parameter $\epsilon\in \mathbb{R}^{0|1}$ is needed in the BRST formulation $\delta=\epsilon {\bf s}$ even if the underlying gauge theory contains several gauge parameters. The formal proof of the existence of a BRST formulation for an arbitrary Hamiltonian & Lagrangian gauge theory with possibly reducible and open gauge ...


5

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/...


5

The expectation of the axial current divergence in a $\theta$ shifted $QCD$ vacuum is given by $\partial_{\mu} \langle J^{\mu5}_{\mathrm{inv}} \rangle_{\theta} = 2m_q \langle \bar{q}i\gamma^5q \rangle_{\theta} + \langle \Xi \rangle_\theta,$ where the first term on the right hand side is the explicit breaking term due to the quark masses and the second ...


5

Comments to OP's question (v1): In superfield formalism, there is a long tradition in the literature to consider constructions that interpret geometrically BRST (& anti-BRST) transformations as translations of Grassmann-odd $\theta$ and $\bar{\theta}$ coordinates in various physical systems, see e.g. Ref. 3 and references therein. The earliest articles ...


5

Any integrable function $f$ will in principle do. But the calculations may become more cumbersome. It should be obvious why we normally choose the function $f$ to be Gaussian, because it is exponentially decaying (after Wick rotation), and the math involved is simple and can be done analytically. Finally let us mention that via the BRST formulation, or ...


5

If you want to understand the quantization of gauge theories and the BRST procedure in particular in detail, the best reference is probably "Quantization of Gauge Systems" (QoGS) by Henneaux and Teitelboim. Requirements for a BRST symmetry to exist is that we have a gauge theory - or equivalently a constrained Hamiltonian theory - and that this ...


4

"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 ...


4

I think 't Hooft and Kugo are solving different problems. 't Hooft addresses the issue that the anomaly involves a topological term. As a result, in perturbation theory there is no theta dependence and the anomaly equation by itself does not solve the $U(1)$ problem. He shows that topological objects (semi-classically, instantons) generate theta dependence....


4

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\} = ...


4

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 ...


4

We do not start from the gauge fixed path integral in the BRST construction. What you describe (once one adds the missing Faddeev-Popov determinant) is the original Faddeev-Popov trick to get the ghosts, not the systematic BRST construction. The (Hamiltonian) BRST construction crucially first introduces the ghosts as parts of the extended phase space, and ...


4

There's one serious flaw in your entire strategy. Since $\overline{c},\,c$ are fermions, they are Grassmann-number-valued. Thus any complex numbers $w,\,z$ satisfy $(w\overline{c}+zc)^2=0$. You therefore can't rewrite the result in the manner you intended. More precisely, if $\phi=w\overline{c}+zc$ then $$\partial_\mu\phi\partial^\mu\phi=g^{\mu\nu}\partial_\...


4

Qmechanic is right, but his answer doesn't explain why we can't just consider the ghosts as physical and be done with it. There are two main reasons why ghosts can't be considered physical. They violate spin-statistics (ghosts are scalar fermions). The S-matrix operator as it stands isn't unitary. The problem can be traced back to the kinematics of gauge-...


4

The statement that one may add a $Q$ exact term to the action $S\to S+Q\chi$ without altering the correlators is not always strictly true. One has to be careful to make a choice of $\chi$ that does not change the asymptotic behaviour of the action $S$ at the boundary of field space. The choice of $\chi=-\psi$ is clearly a choice that drastically alters the ...


4

The BRST operator, much like some supersymmetry generators that may be nilpotent, is a Grassmannian operator – a fermionic one. So it cannot have any nonzero real eigenvalues. The discussion of its "meaning" is therefore very special and inseparable from the BRST procedure. The fun is that whenever we have a nilpotent operator $$ Q^2 = 0 ,$$ it is very easy ...


4

That's a very good question. At the pragmatic level: The field of complex numbers $\mathbb{C}$ has been replaced by the set of supernumbers. Hilbert spaces and operators are not just (conjugate) linear wrt. $\mathbb{C}$ but are $\mathbb{Z}_2$-graded (conjugate) linear wrt. supernumbers. The BRST parameter $\epsilon$ is an independent infinitesimal ...


4

Why the path integral needs gauge-fixing is e.g. discussed in this, this, this & this Phys.SE posts. The currently most general quantization scheme is Batalin-Vilkovisky (BV) quantization. Within the BV formalism the gauge-fixing fermion is more or less arbitrary as long as it is Grassmann-odd, a Lorentz scalar, has ghost number -1, and certain rank ...


4

A supersymmetry is a Grassmann-odd symmetry that takes Grassmann-even objects into Grassmann-odd objects, and vice-versa. Main examples of supersymmetry: Poincare supersymmetry (often abbreviated as SUSY). The Poincare superalgebra is a $\mathbb{Z}_2$ graded extension of the Poincare algebra. The number of supercharges $Q_A$ are labelled by an integer ${\...


3

As a beginner for working on relevant topics, I just write few words about your question. I hope it helps you up. Localization Principle has been great role in computing superconformal index also it gives the exact calculation in susy gauge theories. From some excellent works by Pestun, Kapustin, Willet and so on(about a decade ago?), many researcher now ...


3

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 ...


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 ...


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