Gauge breaking and Faddeev-Popov ghost particles

[A. Neumaier]

Okay, if we are taking the gauge breaking term as a mass term not as a term to weaken the Gauss law, which it must be if it has physical longitudinal polarized states, then I agree.

The point is that even when we start with a pure gauge fixing divergence squared term only, the presence of a further mass term is probably forced by the renormalization procedure.

 

[samalkhaiat]

I think gauge symmetry is pretty unavoidable when dealing with massless vector fields (and I guess also for massless fields of higher spin too). That's because the little group in the massless case is $\text{ISO}(2)$ or rather its covering group, i.e., the symmetry group of the Euclidean plane. If you do not want to have continuously many intrinsic ("polarization like") degrees of freedom, you must make the translations of this group act trivially, and this is afaik only possible using gauge symmetry. Then only the rotations around the three momentum are realized non-trivially giving rise to the two physical polarization degrees (it's only two for massless particles with spin $\geq 1/2$ and not $(2s+1)$ as for massive particles, where the little group is $\text{SO}(3)$ or rather its covering group $\text{SU}(2)$) of freedom with helicity as the corresponding observable.

You are absolutely right. If the vector field is massive, it is possible to separate the negative-norm covariantly because of the compact nature of the corresponding little group, $SO(3)$. But, for a massless vector field, it is impossible to avoid negative norms without violating manifest Lorentz covariance. This is because of the fact that the corresponding little group $E(2)$ is noncompact. In other words, the reason for the presence of negative norms is mathematical and has nothing to do with the field content of the theory. Indeed, given a massless vector field $A_{\mu}(x)$ together with the Poincare algebra, one can prove (without using any field equations or Lagrangians) the impossibility of covariantly quantizing $A_{\mu}$ in the positive-metric Hilbert space. So, ghosts and/or gauge-fixing fields have nothing whatsoever to do with the appearance of negative norm states. In short, indefinite-metric Hilbert (Krein) space is indispensable in the covariant formalism of QFT.

 

[A. Neumaier]

In short, indefinite-metric Hilbert (Krein) space is indispensable in the covariant formalism of QFT.

... of QFT with massless vector fields (only).

 

[DarMM]

If one wants a highly mathematical take on ghosts.

If we consider the Yang-Mills action $S$, the space of connections (fields) $\mathcal{V}$ and gauge group $\mathcal{G}$, the naive path integral is given by:
$$
\int_{\mathcal{V}}{e^{-S}}
$$
However this is wrong since we are integrating over field configurations that simply correspond to coordinate shifts along the fibers of the bundle. That's what causes gauge symmetry. If the $A_{\mu}$ fields are fiber bundle connections (which ultimately means they are associated with massless fields) then several fields are simply the same field in different fiber coordinates.
For this reason we should really be integrating over:
$$
\int_{\mathcal{V}/\mathcal{G}}{e^{-S}}
$$
The problem is that $\mathcal{V}/\mathcal{G}$ is quite a complicated quotient of two infinite dimensional spaces and integration theory upon it is undeveloped.

However the space of functions on $\mathcal{V}/\mathcal{G}$ should be equivalent in some sense to the space of $\mathcal{G}$ invariant functions on $\mathcal{V}$.
From homological theory this is equivalent to the space of $\mathfrak{g}$ invariant functions on $\mathcal{V}$, i.e. we can consider the Lie-Algebra instead of the Lie-Group.

The space of such functions is known as the Chevalley-Eilenberg complex. Denoted:
$$
\mathcal{C}^{*}\left(\mathfrak{g} , \mathcal{V}\right)
$$
Again from homology theory we find that:
$$
\mathcal{C}^{*}\left(\mathfrak{g} , \mathcal{V}\right) = \hat{Sym}^{*}\left(\mathfrak{g}\left[1\right] \oplus \mathcal{V}\right)
$$
That is the space of Lie algebra invariant functions over $\mathcal{V}$ is isomorphic to the space of symmetric functions over $\mathfrak{g}\left[1\right] \oplus \mathcal{V}$. The $\left[1\right]$ indicates you give the field a grade, i.e. make it fermionic. Thus we can consider the integration to be over the space of gauge fields and a fermionic Lie-algebra valued scalar. That's basically the origin of ghosts.

We would still have the problem that the quadratic part of the action $S$ is not invertible on this space. Thus although it is the correct space in a sense, we cannot perform perturbation theory.

Fortunately the theory of homology tells us that an integral on a vector space is equivalent in a certain sense to an integral on its shifted cotangent bundle. Thus we can perform an integral on:
$$
T^{*}\left[-1\right]\left(\mathfrak{g}\left[1\right] \oplus \mathcal{V}\right)
$$
instead. It turns out that this space is isomorphic to:
$$
\mathfrak{g}\left[1\right] \oplus \mathcal{V} \oplus \mathfrak{g}^{\lor}\left[-2\right] \oplus \mathcal{V}^{\lor}\left[-1\right]
$$
Which gives you the extra antighost and anti-gauge fields of the BRST formalism and a quadratic part of the action which is invertible.

 

[A. Neumaier]

If one wants a highly mathematical take on ghosts.

Could you please give a reference where this is discussed in the terms you just described? Can one descrbe the Poisson bracket on functions of the quotient space in these terms?

 

[DarMM]

Kevin Costello, Renormalization and effective field theory, Mathematical Surveys and Monographs, Volume 170, American Mathematical Society (2011)

Chapter 5 in particular and the references there in. Which I believe can be read in isolation. Most of the book concerns a "rigorous" attempt at perturbative Wilsonian effective field theory.

I think it's good to see the gauge antifields in physicists language first:
https://arxiv.org/abs/hep-th/0506098
The Poisson bracket can indeed be described. It's encapsulated in the Chevalley-Eilenberg differential and its equivalent on the final space:
$$
\mathfrak{g}\left[1\right] \oplus \mathcal{V} \oplus \mathfrak{g}^{\lor}\left[-2\right] \oplus \mathcal{V}^{\lor}\left[-1\right]
$$

 

[vanhees71]

I guess so. The added term will generate infinitely many other counterterms, most likely also including a mass term. In the resulting infinite-dimensional manifold of theories there is probably a distinguished finite-dimensional manifold of theories corresponding in a nonperturbative sense to the original family, but for lack of nonperturbative theory no one knows how to select it. Thus one needs to truncate the family and live with an effective theory.

Why are we discussing about gauge theory when the theory in question is none? The gauge-fixing terms introdcuced to gauge-symmetric Lagrangians via the FP path-integral quantization procedure do not break the gauge invariance but fix a gauge and this procedure thus necessarily involves the introduction of the FP ghost fields.

The abelian case is special, because there the FP ghosts decouple and can thus be lumped into the indefinite and irrelevant diverging factor of the path integral. That's the a-posteriori explanation within the FP formalism and its "translation" to the covariant operator formalism, why the Gupta-Bleuler quantization, starting from a non-gauge-invariant Lagrangian works out to describe QED as a gauge theory.

 

[A. Neumaier]

Why are we discussing about gauge theory when the theory in question is none?

Because I answered the question of the poster of #1, clarified in #5:

suppose we have eliminated the gauge invariance, from the start, by explicitly introducing a gauge breaking term to the Lagrangian? Do we still need to introduce ghosts?

 

[samalkhaiat]

... of QFT with massless fields (only).

1) If you are in the business of nitpicking, then you should have said “... of QFT with massless vector fields ”, because the QFT of massless scalar does not require indefinite-metric.
2) By QFT, I meant gauge-invariant quantum field theories such as QED and QCD (which come naturally with massless vector fields) or any field theory that admits a quantum BRST complex[*]

[*] A quantum BRST complex is a triple $\left( \mathcal{V} , Q , N \right)$ consisting of a vector space $\mathcal{V}$ (with non-degenerate norm) graded by the eigen-spaces of an anti-self-adjoint operator $N$ (the ghost number operator) with integer eigenvalues $\mathcal{V} = \bigoplus_{c \in \mathbb{Z}} \mathcal{V}_{c}$; and a non-trivial self-adjoint operator $Q$ (the BRST charge operator) which has a degree 1 with respect to the grading and satisfies $Q^{2} = 0$.
It is clear that for the BRST charge $Q$ to be non-trivial, the norm of $\mathcal{V}$ must be indefinite. Otherwise $\forall \varphi \in \mathcal{V}$, $$\lVert Q \varphi \rVert^{2} = \langle Q \varphi , Q \varphi \rangle = \langle \varphi , Q^{2} \varphi \rangle = 0 ,$$ and hence $Q$ would vanish identically. Thus $\mathcal{V}$ must contain zero-norm vectors. But because the norm is non-degenerate, $\mathcal{V}$ must also have negative norm vectors. Thus, indefinite-metric spaces are inherent to BRST quantization.

 

[A. Neumaier]

1) If you are in the business of nitpicking, then you should have said “... of QFT with massless vector fields ”, because the QFT of massless scalar does not require indefinite-metric.

Corrected.

By QFT, I meant gauge-invariant quantum field theories such as QED and QCD (which come naturally with massless vector fields)

You didn't say this before; it is a nonstandard usage of the term. If QFT were generally understood in your sense, Volume 1 of Weiberg's treatise on QFT would have missed almost totally its declared subject matter...

 

[samalkhaiat]

You didn't say this before;

I didn’t have to, because the whole of #37 was about the covariant quantization of the massless vector field $A_{\mu}(x)$. I wouldn’t use the term “covariant formalism” if my QFT was that of massless scalars or spinors which I did not even mention in #37.

 

[Elias1960]

... But, for a massless vector field, it is impossible to avoid negative norms without violating manifest Lorentz covariance. ... In other words, the reason for the presence of negative norms is mathematical and has nothing to do with the field content of the theory. Indeed, given a massless vector field $A_{\mu}(x)$ together with the Poincare algebra, one can prove (without using any field equations or Lagrangians) the impossibility of covariantly quantizing $A_{\mu}$ in the positive-metric Hilbert space. ... In short, indefinite-metric Hilbert (Krein) space is indispensable in the covariant formalism of QFT.

Sorry, but the decision to require manifest Lorentz covariance is nothing forced on us by mathematics, and also not forced on us by something physical - for physics, all one needs is covariance of the observable predictions, which is much less than manifest Lorentz covariance. It is a purely metaphysical decision to prefer such a theory.