# Tag Info

11

OP is asking how can phenomenologists get away with neglecting the FP ghosts. The answer is that they can't. In a talk you can omit irrelevant stuff to keep things simple. In the full computation, you must include the FP ghosts (or use a ghost-free formalism, which in general is much more cumbersome; e.g., the axial or unitary gauge). For an explicit ...

10

I asked Mark Srednicki about this, and he told me that it's not really correct to say that negative-norm states break unitarity, because negative-norm states don't exist by the definition of the inner product. It's often a convenient calculational trick to formally expand your state space so that it's no longer a Hilbert space by adding in negative-norm ...

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

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 ...

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

It's no accident that you invoked ghosts to find a counter-example. Ghosts are non-unitary and the standard proof of the primary / descendant classification uses unitarity. In particular, given a local operator which is not a quasi-primary, the state $\mathcal{O}(0) \left | 0 \right >$ has some scaling dimension $\Delta$. We can then show using the ...

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

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

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 ...

7

Note that the inverse of $\Delta_{FP}$ is the Green function $G_{FP}(x,y)$ obeying: $$\Delta_{FP}|_x G_{FP}(x,y)=\delta(x-y)$$ so, the action that you wrote down is non-local, should be of the form: $$S_{FP}^{bosonic}=i\int dx \int dy\, \bar{\psi}^{a}(x)G_{FP}(x,y) \psi^{a}(y)$$ this action is non-local, since we cannot get away with just one ...

6

Perturbative QCD computations can be done in ghost-free gauges, such as the axial gauge. As far as I understand, this is customary for the computations revolving around the parton density functions in the proton, or in the pion. As well as for the final state equivalent, the parton fragmentation functions. Combined, this makes for a rather large section of ...

4

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. What is the ...

4

1) Note that $p^\alpha p^\beta \Pi_{\alpha\beta}$ is gauge invariant, but $\Pi_{\alpha\beta}$ is not. This implies that $a$ and $b$ are not separately gauge invariant, only their sum (or difference, in your convention) is. For example, Diakonov and Eides [Sov. Phys. JETP 54 (2), 232-240] write down a spectral representation corresponding to $a=0$, but many ...

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

"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

You got just the wrong coefficient in front of $\partial_z(bc)$, you expected twice as high coefficient due to $\lambda=2$, right? But that's because since the very beginning, your Noether derivation was completely insensitive to the value of $\lambda$, so you got a random value of $\lambda$ by this derivation. Your derivation is not sensitive to the right ...

4

As discussed in Kugo and Ojima 1979, "ghost is Hermitian, anti-ghost is anti-Hermitian" is just a convention, another being that both fields are Hermitian, which results in a factor of $i$ in the FP-ghost term so that the Lagrangian is still Hermitian. In their notation $c,\,\overline{c}$ are both Hermitian while $C:=c,\,\overline{C}:=i\overline{c}$ provide ...

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 $$\partial_\mu\phi\partial^\mu\phi=g^{\mu\nu}\partial_\... 4 Qmechanic is right, but his answer doesn't explain why we can't just consider the ghosts as physical and be done with it. There are two main reasons why ghosts can't be considered physical. They violate spin-statistics (ghosts are scalar fermions). The S-matrix operator as it stands isn't unitary. The problem can be traced back to the kinematics of gauge-... 4 As long as the ghosts are not external particles they can give non-zero contributions to amplitudes, including for boson self-energies. Recall that the Feynman diagrams are just a graphical representation of a mathematical perturbation series. 3 Bosonic path integrals :$$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]}$$are not mathematically well-defined, because of the presence of the imaginary unit in the exponential. ... 3 answer for questions 1,2, and 3a 1) Looking at 2.7.22 to 2.7.24 (and also 2.7.18abc), one define the ghost number N^g = \frac{-1}{2\pi i} \int_0^{2 \pi}dz :b(z)c(z): , and that all the operators c_n increase the ghost number by one. [N^g,c_m] = c_m, So the field c(z), made of operators c_m, increase the ghost number by one unit.So c^\... 3 I am adding my own computation to the mix since it took me a while to follow the Noether method of deriving eq. (13.17) in Blumenhagen, Lüst, Theisen. The setup is the same as in Polchinski. The fields b and c have conformal weight \lambda and 1-\lambda and their action is given by$$S = \frac{1}{2\pi} \int d^2z \, b \bar \partial c$$Recall ... 3 I did not furnish all the details because it would be too long, but I give some hints at the end of the answer. I have used the formulae :T^g: ~= ~:2(\partial c) b + c(\partial b): and :\frac{1}{2}cT^g: ~= ~:bc \partial c:, when there is an ambiguity in the calculus. We begin by :$$j_B = cT^m+:\frac{1}{2}:cT^g:+\frac{3}{2}\partial^2c=cT^m+:bc\partial ...

3

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

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 ...

3

Ghosts also arise in the context of the classic theory through Ostrogradsky theorem. You should take a look at this paper: Woodard, Richard. “Ostrogradskys Theorem on Hamiltonian Instability.” Scholarpedia, vol. 10, no. 8, 2015, p. 32243., doi:10.4249/scholarpedia.32243. It's available on arXiv. The general ideia is that, given a non-degenerate ...

3

Here is a slightly different perspective: OP's question seems to (possibly indirectly) inquire whether a quadratic ghost action term of the form $$\bar{c}_i ~M^i{}_j~ c^j,\tag{i}$$ where $M^i{}_j$ is a (possible infinite-dimensional) matrix, can be recast into an action term of the form $$\frac{1}{2} C^I ~A_{IJ}~ C^J~? \tag{ii}$$ Here $A_{IJ}$ is an ...

3

On one hand, the S-matrix does not depend on the gauge-fixing condition. On the other hand, there exist a unitary gauge, where Faddeev-Popov ghosts decouple from the theory. References: M.D. Schwartz, QFT and the Standard Model, 2014; Section 28.4. C. Itzykson & J.B. Zuber, QFT, 1985; Subsection 12-5-5.

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