# A Gauge breaking and Faddeev-Popov ghost particles

#### Elias1960

When the gauge symmetry is explicitly broken the theory is unrenormalizable, which means that (unlike in the renormalizable case) the manifold of perturbatively (and canonically) renormalized field theories associated with it is infinite-dimensional, due to the infinitely many counterterms. Nothing at all is known about the resulting nonperturbative situation.
This sounds strange, given that in the Wilsonian approach it is quite clear what to do with this sort of infinity. The non-renormalizable theory has to be understood as an effective field theory, and there is a critical length at which the effective field theory fails. One can now assume that all the terms have a similar order of magnitude at this critical distance. And then one can see what survives in the large distance limit. These are, first of all, renormalizable components. Then there are a few lowest order non-renormalizable ones.

Given this purely qualitative consideration, one can already make a qualitative guess. The renormalizable part of a massive gauge theory is the massless, gauge-invariant theory. The massive part gives a short distance force, thus, vanishes for large distances anyway (as suggested above).

We have to do this anyway, given that gravity is non-renormalizable, and this gives already a nice suggestion for the critical length. So, all that remains to be done would be to consider the lowest order non-renormalizable terms and what they give, say, for the massive gauge theories at Planck scale.

If this has been done somewhere, I would be very interested in the results. It not, I would be interested to understand why this has not been done.

(The same question appears for gauge theories with anomalies. Here I would guess that one can even predict qualitatively with more certainty that the anomalous part will be suppressed, thus, becomes very weak in comparison with the non-anomalous gauge fields. Here I would wonder why this method has not been used to extend the SM gauge group, given that one would not even have to invent a mechanism to suppress the additional gauge fields.)

#### A. Neumaier

The term I gave is the modification to the equations of motion, not the Lagrangian. The Lagrangian term giving it is essentially the same as your term there.

If you add such a term then the inner product cannot be positive semi-definite due to essentially a Reeh-Schlideder argument . See Strocchi "An Introduction to Non-perturbative Foundations of Quantum Field Theory" Proposition 3.3 p.155.
I have currently no access to this book, but there cannot be any rigorous results of this kind on a theory that is by definition not gauge invariant. As I remember Strocchi assumes from the start that there is gauge invariance, and then shows that it cannot be locally realized in a Hilbett space. But once the extra term is added, the theory is manifestly gauge noninvariant, hence Strocchi's assumption is not met.

#### vanhees71

Gold Member
One gets the "negative-norm states" not only in the path-integral formalism but of course in the equivalent operator formalism too. You use the BRST symmetry of the complete gauge-fixed Lagrangian including FP ghosts to define the constraint conditions for the physical states leading to a positive definite norm and a unitary S-matrix. For details, see

T. Kugo, I. Ojima, Manifestly Covariant Canonical
Formulation of the Yang-Mills Field Theories. I, Progress of
Theoretical Physics 60 (1978) 1869.
http://dx.doi.org/10.1143/PTP.60.1869

T. Kugo, O. Ojima, Manifestly Covariant Canonical
Formulation of Yang-Mills Field Theories. II: SU (2)
Higgs-Kibble Model with Spontaneous Symmetry Breaking,
Progress of theoretical physics 61 (1979) 294.
http://dx.doi.org/10.1143/PTP.61.294

T. Kugo, I. Ojima, Manifestly Covariant Canonical
Formulation of Yang-Mills Field Theories. III—Pure
Yang-Mills Theories without Spontaneous Symmetry
Breaking, Progress of Theoretical Physics 61 (1979) 644.

#### DarMM

Gold Member
I have currently no access to this book, but there cannot be any rigorous results of this kind on a theory that is by definition not gauge invariant. As I remember Strocchi assumes from the start that there is gauge invariance, and then shows that it cannot be locally realized in a Hilbett space. But once the extra term is added, the theory is manifestly gauge noninvariant, hence Strocchi's assumption is not met.
Perhaps I'm not really understanding what is meant here then. To me this all goes back to @vanhees71 post where vector fields have too many components to model massless particles directly as they are reps of $ISO(2)$. For that reason modelling massless particles with local fields means there will be gauge symmetry. Unless you mean to model massive particles?

#### A. Neumaier

This sounds strange, given that in the Wilsonian approach it is quite clear what to do with this sort of infinity. The non-renormalizable theory has to be understood as an effective field theory,
Yes, on the level of perturbation theory, one can say a lot. But I had meant nothing is known rigorously.
(DarMM is a mathematical physicist.)
One gets the "negative-norm states" not only in the path-integral formalism but of course in the equivalent operator formalism too.
But only if one adds ghosts (or other unphysical degrees of freedom) to preserve manifest gauge symmetry. Not for the family of modified theories where this is explicitly broken by nonrenormalizable terms.

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#### vanhees71

Gold Member
It's not clear to me, what you mean. Are you referring to something like the "unitary gauge"?

Also in the simpler abelian case, where you can use the Gupta-Bleuler formalism, you have "negative-norm states" first but then introduce the constraint conditions for physical states. In the original form of the Gupta-Bleuler formalism there are no FP ghosts introduced. This has been done for QED by Feynman first. Using the FP formalism for abelian gauge fields in the standard covariant gauges, the FP ghosts are non-interacting and thus play no role in the Feynman rules for physical S-matrix elements.

Of course all this is defined (though mathematically not rigorously) within perturbative QFT only.

#### A. Neumaier

It's not clear to me, what you mean. Are you referring to something like the "unitary gauge"?
No; it was a general remark. If there is no gauge symmetry then there is nothing to fix. Adding a gauge symmetry violating term of any sort, in particular the one the OP requested,
explicit symmetry breaking by adding a gauge breaking/fixing term (e.g. to enforce the Feynman gauge) to the Lagrangian.
changes the family of field theories considered to a bigger class, of which only a set of measure zero corresponds to the original gauge invariant family. Thus gauge considerations become irrelevant for the bigger class, since there the longitudinal component is physical (but with a singularity at vanishing couplng). Thus it doesn't make sense to introduce ghosts or unphysical Gupta-Bleuler degrees of freedom. Thus also no indefinite inner product.

#### DarMM

Gold Member
Thus it's describing a theory with massive bosons right?

#### A. Neumaier

Thus it's describing a theory with massive bosons right?
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.

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#### DarMM

Gold Member
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.

As you said we know very little about the nonperturbative behaviour of such theories, they're usually ignored due to being nonrenormalizable, but that might not be valid.

#### 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).

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#### DarMM

Gold Member
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.

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#### 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

Gold Member
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:

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

Gold Member
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.

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#### 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....

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#### 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.

"Gauge breaking and Faddeev-Popov ghost particles"

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