# A Gauge breaking and Faddeev-Popov ghost particles

#### Michael Price

Summary
In QFT, if we add a gauge breaking term to the Lagrangian, do we still need to introduce ghost particles?
Summary: In QFT, if we add a gauge breaking term to the Lagrangian, do we still need to introduce Faddeev-Popov ghost particles?

Ghosts seems to be introduced to maintain gauge invariance. But 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?

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#### Demystifier

2018 Award
Ghosts can be thought of as non-physical degrees of freedom that appear in mathematical expressions. They are non-physical in the sense that different values of these degrees cannot be distinguished physically. For instance, different values of the angle $\varphi$ in a rotation-invariant theory cannot be distinguished physically, so the variable $\varphi$ can be thought of as a kind of "ghost". When the symmetry is explicitly broken, those degrees of freedom do not go away. Instead, they become physical. The ghosts "materialize", so to speak.

#### Michael Price

Ghosts can be thought of as non-physical degrees of freedom that appear in mathematical expressions. They are non-physical in the sense that different values of these degrees cannot be distinguished physically. For instance, different values of the angle $\varphi$ in a rotation-invariant theory cannot be distinguished physically, so the variable $\varphi$ can be thought of as a kind of "ghost". When the symmetry is explicitly broken, those degrees of freedom do not go away. Instead, they become physical. The ghosts "materialize", so to speak.
But why would the ghosts be introduced if there is no gauge symmetry, if we have destroyed it by added a gauge breaker to the Lagrangian? Haven't we just restricted the dynamics of the gauge bosons a bit more? I.e. removed a degree or degrees of freedom.

#### Dr.AbeNikIanEdL

I am not sure what you are asking.

First, with your reference to gauge theories it makes most sense you are talking about Faddeev-Popov ghosts, but then the wiki page you linked lists a bunch of other possible meanings of ghost, so which one are you asking about?

Second, assuming you are talking about Faddeev-Popov, do you mean why we need ghosts after adding a gauge fixing term, or are you talking about explicit breaking like adding a mass to W and Z bosons in SU(2) theory in a naive way?

#### Michael Price

I am not sure what you are asking.

Second, assuming you are talking about Faddeev-Popov, do you mean why we need ghosts after adding a gauge fixing term, or are you talking about explicit breaking like adding a mass to W and Z bosons in SU(2) theory in a naive way?
Yes, I mean Faddeev-Popov ghosts and explicit symmetry breaking by adding a gauge breaking/fixing term (e.g. to enforce the Feynman gauge) to the Lagrangian. Not about W/Z mass terms.

Sorry should have linked to the more specific article.

PS I've edited my original post to clarify things.

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#### DarMM

Gold Member
In Gauge Theories the fields $A_{\mu}$ overdescribe the physics. In the path integral approach the space of all such fields $\mathcal{V}$ is too large. This is essentially due to the fact that the dynamics of massless spin-1 bosons don't exactly correspond to that of local fields. Note: The dynamics are still local, but just not in direct correspondance with local fields. However since local fields are so mathematically neat to work with and allow you perform renormalization easily we continue to use them.

An entire family of fields $A^{G}_{\mu}$ which are transformations of some fixed $A_{\mu}$ by a guage transformation $G$ all describe the same physical situation. Such families are called gauge orbits.

So we should be integrating over $\mathcal{V}/\mathcal{G}$ and we do this by selecting out a surface on $\mathcal{V}$ by using a delta function or some similar trick. The delta function then selects out one field from each gauge orbit, defining a surface $\mathcal{S}$ which should give the same physics as $\mathcal{V}/\mathcal{G}$.

However it turns out that the gauge orbits are more "clumped" at some points on the surface than others, i.e. there can be areas of the surface with a higher density of orbits intersecting it there. The density of orbits is given by $\det\left(\partial^{\mu}D_{\mu}\right)$. To use perturbation theory one needs a way to evaluate this density perturbatively.

Luckily we know that functional determinants can be evaluated with fermionic path integrals. Thus we replace the determinant with a path integral over a new fermionic set of fields.

So in essence ghosts are ways of perturbatively evaluating the orbit density of your gauge fixed surface. There will be an orbit density to this surface regardless of whether you have a gauge breaking term or not. The gauge breaking term is present to prevent a contradiction between the Maxwell equations, non-triviality and local fields.

#### Dr.AbeNikIanEdL

Yes, I mean Faddeev-Popov ghosts and explicit symmetry breaking by adding a gauge breaking/fixing term (e.g. to enforce the Feynman gauge) to the Lagrangian. Not about W/Z mass terms.

Ok, in that case you still need ghosts. From a pragmatic point of view, there are contributions form unphysical states, and you need to subtract them out, and adding ghost particles is the way to do that. Without introducing the gauge fixing term, you would not even get to the stage where one can see this problem. It is probably better to view the introduction of ghosts as part of the gauge fixing procedure for non-abelean theories.

#### A. Neumaier

Yes, I mean Faddeev-Popov ghosts and explicit symmetry breaking by adding a gauge breaking/fixing term (e.g. to enforce the Feynman gauge) to the Lagrangian.
With this term added you get an infinite family of inequivalent theories, parameterized by the new coupling constant $\lambda$ needed. The original theory is recovered only in the limit where $\lambda\to 0$. All other values give theories with explicitly broken gauge symmetry.

In Gauge Theories the fields $A_{\mu}$ overdescribe the physics.
Only if there is no explictly broken gauge invariance, i.e., when $\lambda=0$. If $\lambda\ne0$, the physics is changed, and all degrees of freedom of the fields $A_{\mu}$ are physical. Then no gauge can be fixed as there is no gauge symmetry in the classical Lagrangian with $\lambda\ne0$.
So in essence ghosts are ways of perturbatively evaluating the orbit density of your gauge fixed surface. There will be an orbit density to this surface regardless of whether you have a gauge breaking term or not.
But this orbit density is irrelevant except in the limit where the gauge breaking term vanishes. Of course, this is the situation where the original gauge invariant theory would be recovered.

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#### DarMM

Gold Member
Only if there is no explictly broken gauge invariance, i.e., when $\lambda=0$. If $\lambda\ne0$, the physics is changed, and all degrees of freedom of the fields $A_{\mu}$ are physical. Then no gauge can be fixed as there is no gauge symmetry in the classical Lagrangian with $\lambda\ne0$.

But this orbit density is irrelevant except in the limit where the gauge breaking term vanishes. Of course, this is the situation where the original gauge invariant theory would be recovered.
When the gauge symmetry is broken the state space is full of unphysical states with negative norm and forms a Krein space. So are these degrees of freedom really "physical" in any sense, the resultant space has no sensible physical interpretation, it's just an intermediate stage used to permit calculations in the physical theory with local operators.

If you don't implement the conditions that restore the physical theory then the path integral is just an abstract stochastic process with no relation to any sensible quantum field theory. So yes if you keep the gauge broken then the orbit density doesn't matter, but why is that fact even relevant?

#### vanhees71

Gold Member
Ghosts can be thought of as non-physical degrees of freedom that appear in mathematical expressions. They are non-physical in the sense that different values of these degrees cannot be distinguished physically. For instance, different values of the angle $\varphi$ in a rotation-invariant theory cannot be distinguished physically, so the variable $\varphi$ can be thought of as a kind of "ghost". When the symmetry is explicitly broken, those degrees of freedom do not go away. Instead, they become physical. The ghosts "materialize", so to speak.
A local gauge symmetry cannot be spontaneously broken (Elitzur's theorem). What happens instead is the "Anderson-Higgs-et-al mechanism", and you get massive gauge bosons by "eating up" the would-be-Goldstone modes. In non-unitary gauges the would-be-Goldstone bosons become ghosts as the Faddeev-Popov ghosts. These are good ghosts, cancelling unphysical degrees of freedom in gauge theories.

#### A. Neumaier

When the gauge symmetry is broken the state space is full of unphysical states with negative norm and forms a Krein space.
No. This is the case only assuming that you have already adjoined the ghost field. Without doing this, the situation is completely different:

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 rigorously known about the resulting nonperturbative situation.

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#### Michael Price

When the gauge symmetry is explicitly broken the theory is unrenormalizable,[...]
Is this the reason for Faddeev-Popov ghosts? I.e. To keep the theory renormalisable we mustn't explicitly break the gauge symmetry, and so, to maintain gauge symmetry, we are forced to go down the FP ghost route?

#### A. Neumaier

Is this the reason for Faddeev-Popov ghosts? I.e. To keep the theory renormalisable we mustn't explicitly break the gauge symmetry, and so, to maintain gauge symmetry, we are forced to go down the FP ghost route?
More or less, yes. Actually the gauge symmetry is lost anyway but instead the BRST supersymmetry appears.

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#### Elias1960

When the gauge symmetry is broken the state space is full of unphysical states with negative norm and forms a Krein space. So are these degrees of freedom really "physical" in any sense, the resultant space has no sensible physical interpretation, it's just an intermediate stage used to permit calculations in the physical theory with local operators.
Not necessarily. The unphysical states with negative norms have been introduced into the theory intentionally, by design, to make relativistic symmetry "manifest", by Gupta and Bleuer. They replaced another, earlier, quantization scheme proposed by Fermi and Dirac, which introduced different commutation rules for the temporal $A_0$ and the spatial $A_i$ components. The resulting observable predictions were nonetheless Lorentz-covariant, but this destroyed the covariance at the fundamental level.

Once one does not insist on gauge invariance being fundamental (else, the question would not make sense), but considers it as something approximate, emergent, one can as well consider relativistic symmetry as emergent. Instead, negative norms make no sense at all. In this case, it makes sense to use the older approach, given that it avoids any negative norms.

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#### vanhees71

Gold Member
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.

#### DarMM

Gold Member
When the gauge symmetry is explicitly broken the theory is unrenormalizable
Exactly what way are we breaking gauge symmetry here? Adding a mass?

#### vanhees71

Gold Member
The FP formalism fixes the gauge but doesn't break gauge symmetry. Choosing the background-field gauge you can even get a gauge-invariant effective action. Usually if you start with a gauge theory and break the gauge symmetry explicitly the theory becomes unphysical. Adding naively a mass term in non-Abelian gauge theories makes the theory unphysical, and thus you have to use the Anderon-Higgs-et-al mechanism to describe massive non-Abelian gauge bosons, and usually you have at least a physical Higgs boson left. (In)famously the predicted Higgs boson in the SM has been discovered. In the Abelian case you can introduce a mass term without breaking local gauge invariance (Stueckelberg formalism).

#### DarMM

Gold Member
I know all of that, but I don't really understand in what sense we are "adjoining ghosts". My understanding is that if you break gauge symmetry by adding the usual gauge breaking term to the Lagrangian:
$$\lambda\partial_{\mu}\partial^{\nu}A_{\nu}$$
then the state space immediately has negative norm states.

#### Demystifier

2018 Award
But why would the ghosts be introduced if there is no gauge symmetry, if we have destroyed it by added a gauge breaker to the Lagrangian? Haven't we just restricted the dynamics of the gauge bosons a bit more? I.e. removed a degree or degrees of freedom.
When you add a symmetry breaking term (don't confuse it with a gauge fixing term, that's different), you don't remove degrees of freedom. Just the opposite, you add new degrees.

#### vanhees71

Gold Member
I know all of that, but I don't really understand in what sense we are "adjoining ghosts". My understanding is that if you break gauge symmetry by adding the usual gauge breaking term to the Lagrangian:
$$\lambda\partial_{\mu}\partial^{\nu}A_{\nu}$$
then the state space immediately has negative norm states.
I've no clue what "adjoining ghosts" means, but we introduce them in FP procedure to get a perturbative handle on the functional determinant introduced when integrating over the gauge group. The unphysical gauge-field degrees of freedom and the FP ghost conspire such as to cancel the unphysical degrees of freedom in physical (i.e., gauge-invariant) quantities, leading finally (order by order perturbation theory) to a univary S-matrix. No negative-norm states are physical of course. Formally it's a bit tricky a proof making use of the BRST symmetry and the associated Slavnov-Taylor identities. Everything becomes much simpler in the background-field gauge, where you deal with gauge-invariant effective actions and you can do with Ward-Takahashi identities as in the Abelian (QED) case.

#### DarMM

Gold Member
As I said I know all that, that's what I'm saying. I don't understand what @A. Neumaier is saying.

#### vanhees71

Gold Member
I know that you know that. I answered for the rest of PF readers too ;-). I'm reminded of Einstein's dictum about mathematicians in 1908 after he read Minkowski's famous article, but as is well known this was a premature statement since Einstein later realized how important math really is for his work, and he regretted not to have attended Minkowski's lectures in Zürich. Usually you have to work hard to translate the mathematicians' jargon into physicists' jargon. Sometimes it's impossible though...

#### DarMM

Gold Member
I know that you know that. I answered for the rest of PF readers too ;-)
Ah of course, very sensible, my apologies.

#### A. Neumaier

Exactly what way are we breaking gauge symmetry here? Adding a mass?
It doesn't matter for my argument. Intended in the original question (made precise in post #5 was apparently a term proportional to $(\nabla\cdot A)^2$ or another gauge-fixing term.
My understanding is that if you break gauge symmetry by adding the usual gauge breaking term to the Lagrangian:
$$\lambda\partial_{\mu}\partial^{\nu}A_{\nu}$$
then the state space immediately has negative norm states.
This term is not a scalar, hence is not eligible. If you add $\lambda(\nabla\cdot A)^2$ with $\lambda\ne0$, the Lagragian is no longer renormalizable, and renormalized perturbation theory needs new constants at each order. Apart from that, everything is canonical and hence in a Hilbert space setting.
Negative norm states are absent; they never appear in canonical quantization.

Negative norm states enter only through the coupling with a ghost field in a path integral approach, There the integrand only goes over a quotient space and integrating formally over gauged-fixed representatives (ignoring Gribov copies) needs to be compensated by a functional determinant, which is modeled by an unphysical ghost field. Only at this stage, definiteness can no longer be maintained.

#### DarMM

Gold Member
This term is not a scalar, hence is not eligible. If you add $\lambda(\nabla\cdot A)^2$ with $\lambda\ne0$, the Lagragian is no longer renormalizable, and renormalized perturbation theory needs new constants at each order.
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.

"Gauge breaking and Faddeev-Popov ghost particles"

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