# Why is the order parameter in N=2 Seiberg-Witten theory $\langle \text{tr} \phi^2 \rangle$? (And discussion of gauge-variant order parameter in general)

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In this paper, Seiberg and Witten use the gauge invariant order parameter $\langle \text{tr} \phi^2 \rangle$ to parametrize the breaking of gauge symmetry (I'm using the standard abuse of terminology here, of course gauge symmetry cannot be broken, what's broken is the corresponding global symmetry).  But since this is gauge invariant, how can it tell us at all whether the gauge symmetry is broken? Just like in QCD, the flavor-chiral order parameter $\langle \bar{\psi} \psi \rangle$ is QCD-gauge invariant, but it would've been absurd to conclude  QCD gauge symmetry is broken just because $\langle \bar{\psi} \psi \rangle\neq 0$. (In fact it puzzles me why some people take gauge invariance of the order parameter as a virtue instead of a deficiency in this case.)

A late edit: much of the discussion (which is enlightening to me) has been centered around aboout "does $\langle \phi \rangle$ make sense at strong coupling?", but this is actually somewhat a diversion of what I wanted to say. The thing is I'm not even convinced $\langle \phi^2 \rangle\approx\langle\phi\rangle^2$ at weak coupling region. For example, a weakly coupled Ising model is in its disordered phase, if we use $\phi$ to denote lattice spin, then $\langle\phi\rangle=0$, while $\phi^2=\text{Id}$ which is nonzero in any possible phase.

Anther point raised by 40227 is Elithur's theorem, however the theorem only applies in a gauge-invariant quantization scheme, such as lattice gauge theory, while unfortunately Seiberg and Witten never explicitly specify what kind of quantization scheme they are having in mind.

edited Jul 3, 2016

If $A$ is a non gauge invariant field in a gauge theory, what is the meaning of $<A>$?

@40227, it doesn't have to have a direct meaning, although at the very least it tells us if the vacuum condensate is "charged". Nonzero vev of a gauge-variant field gives non-trivial phenomenology, like Higgs mechanism.

In fact a more physical example would be field-theoretic treatment of superconductivity. In this case you want to know if the vacuum is electrically charged, so you have to look at the vev of a charged field, which must be gauge variant.

Superconductivity is nonrelativistic, which makes a big difference in QFT.

@ArnoldNeumaier, still, there's no reason to rule out relativistic superconductivity, in which case we still have to know if the vacuum is charged.

Can you give me a reference to relativistic superconductivity, so that I can investigate the matter?

It seems to me that having a solid around already breaks the Poincare-symmetry down to a 3D lattice symmetry $\times$ time translations, which would completely change the situation compared to what has been discussed in algebraic QFT. Therefore, at present I have no intuition of what might happen for such a symmetry group.

@ArnoldNeumaier, I think color-superconductivity would count. Although the formalism treating it is often not manifestly covariant (which is not surprising since e.g. it's often dealt with in thermal QFT context),  it's still a QCD phenomenon after all.

It would be good if you ask a separate question about color-superconductivity, as discussing this here would change the nature of the thread.

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To understand what is going on one has to make a difference between what is a full/quantum/non-perturbative quantum field theory and what is a Lagrangian and/or semiclassical/perturbative description of a theory. In a full QFT, one has an algebra $\mathcal{A}$ of (physical) fields of operators. A (global) symmetry group $G$ is a group of automorphisms of this algebra. A choice of vacua is a choice of realization of $\mathcal{A}$ on an Hilbert space $H$, space of (physical) states (the states in $H$ are obtained from a particular vector in $H$, the vacuum, by action of elements of $\mathcal{A}$). We can have different choices of vacua corresponding to different (inequivalent) representations of $\mathcal{A}$ on an Hilbert space. For a given choice of vacua, all the symmetries of $\mathcal{A}$ are not necessarely realizable by unitary transformation of the Hilbert space: the realizable symmetries form a subgroup $H$ of $G$ and if $H$ is strictly smaller that $H$, one has spontaneously symmetry breaking from $G$ to $H$. The spectrum of the theory in the given vacua contains one massless scalar (Goldstone boson) for each continuous direction in $G/H$.

The notion of gauge symmetry depends on a specific Lagrangian description of the theory. In such description, one starts with a classical field theory with a gauge symmetry and one defines a QFT by quantization, let's say by the path integral approach. In such picture only gauge invariant classical fields define corresponding fields of operators in the quantum theory. Indeed to define correlation functions in the quantum theory one has to take the path integral over gauge equivalence classes of  fields and so only gauge invariants quantities can be included in the integrand. One could try to define correlation functions of gauge variants fields by fixing a gauge and it is indeed possible perturbatively but the results depend on the gauge choice and fixing a gauge is anyway in general impossible at the non-perturbative level (Gribov ambiguity). So, very concretely, in the Seiberg-Witten example, $\phi$, which is a gauge-variant field in the classical starting point of the Lagrangian description, does not define a well-defined field of operators in the full QFT and in particular it does not make sense to talk about an expectation value $<\phi>$.

In the classical theory, it makes sense to say that the field $\phi$ has a non-zero value at infinity, the usual description of the Higgs mechanism applies and this story extends to the perturbative level. To understand the relation with the full non-perturbative theory, it is useful to think in terms of path integrals. A Lagrangian for a gauge theory defines a full QFT by path integral over gauge equivalence classes of classical fields. In particular, one has the choice of boundary conditions at infinity for the classical fields we are integrating over and this choice is mapped to the choice of vacuum of the full quantum QFT. But this mapping can be quite non-trivial. In the Seiberg-Witten story, the boundary condition on the field $\phi$ is specified by a complex number $a$, well-defined up to a sign. Classically, the moduli space of classical vacua is parametrized by $a$. For $a \neq 0$, the gauge symmetry is spontaneously broken from $SU(2)$ to $U(1)$ and for $a=0$ the $SU(2)$ gauge symmetry is unbroken. For big $a$, the classical theory is weakly coupled at the symmetry breaking scale and so one expects that for every such $a$ the path integral with boundary conditions prescribed by $a$ defines a vacuum of the full quantum theory, with an infrared behaviour looking like the classical one: a U(1) gauge theory with massive W bosons. But for small $a$, the classical theory is strongly coupled and it is unlikely that the quantum theory looks like the classical one. In fact the path integral has infrared divergences making the correspondence between $a$ and quantum vacua doubtful. The conclusion is that $a$, what would be a candidate for $<\phi>$, is not a good well defined coordinate on the moduli space of vacua. It is not very surprising precisely because $\phi$ is not an allowed observable in the full theory.

Breaking of gauge symmetry is not breaking of a corresponding global symmetry simply because in general there is no global symmetry associated to a gauge symmetry. More precisely the conserved current associated to a global gauge transformation is in general gauge variant and so cannot define a well defined charge on the Hilbert space of (physical) states. (A well known exception to this statement is QED where the current associated to the global $U(1)$ is gauge invariant and there is a well defined eletric charge but to have a spontaneous symmetry breaking one needs a charged scalar and the current associated to global $U(1)$ is not gauge invariant because of the term $A^\mu A^\nu \phi \phi^\dagger$ in the Lagrangian). If there were really a breaking of a global symmetry then one should see a Goldstone boson.

The conclusion is that the notion of spontaneous symmetry breaking of a gauge symmetry only makes sense given a Lagrangian/ classical/ perturbative description of the theory. It is not surprising as gauge symmetry is simply a redundancy in a given description of the theory (digression: physical consequences of a gauge symmetry description exist at the level of asymptotic symmetries but they are much more sublter objects that a global symmetry acting on the Hilbert space).  So asking the question: is there a spontaneously symmetry broken of a gauge symmetry in a given vacuum of a full non-perturbative QFT does not really make sense. A question which makes sense is: are there some massless spin 1 particles ? If yes then there is a natural gauge theory description. If no then it's no.

So the meaningful questions that Seiberg and Witten are trying to answer are: what is  the space of vacua of the theory and what is the infrared physics in each of these vacua? They start by the classical story, with a moduli space parametrized by $a$, a $U(1)$ unbroken gauge symmetry at $a\neq 0$ and a $SU(2)$ unbroken gauge symmetry at $a =0$. They argue that this picture is qualitatively correct at the quantum level for large $a$. To study the general case, one needs a good coordinate on the space of vacua. Natural functions on the space of vacua are vev of fields of operators. $tr \phi^2$ is a well-defined field of operator of the theory because it comes from a gauge invariant function in the path integral definition of the QFT and so it makes sense to consider $<tr \phi^2>$. The fact that it is a good choice is not obvious a priori, it could be a constant function on the space of vacua for example. But it is a good choice because it is clearly a good choice in the region where the classical approximation is good, for large $a$, $<tr \phi^2> \sim a^2$. In other words, $<tr \phi^2>$ is the simplest way to extend to the full quantum theory the variable $a$ natural from the classical point of view. All the work is then to determine the quantum corrections to the classical picture, and in particular to compute exactly $<tr \phi^2>$ as a function of $a$ in the region of the space of vacua where $a$ is still a good coordinate.

answered Jan 1, 2016 by (5,120 points)

very nice! Thank you and happy new year!

Thanks for the long reply, but I think we have some fundamental disagreements. Let me ask the most important one first: how do you know there are spontaneously broken vacuua even if $\langle \text{tr} \phi^2\rangle\neq 0$? For example also look at the chiral order parameter mentioned in my main post. (Also for example in Ising model if you use $s^2$ instead of $s$ as the order parameter, and use the same flawed logic, you'll reach the absurd conclusion that rotational symmetry is broken at all temperatures.)

@ Jia Yiyang : the point of my answer is that the notion of "spontaneously broken vacua" does not make sense away from the perturbative regime. In the perturbative regime, the classical story is a correct approximation and in the classical strory $tr \phi^2 \neq 0$ implies $\phi \neq 0$, hence spontaneously broken gauge symmetry. In QCD, the chiral condensate is already a non-trivial dynamical non-perturbative effect and at this level I don't know what gauge symmetry breaking means. But certainly I agree with you that, in a classical story, a non-trivial expectation value of a gauge invariant field does not imply in general a spontaneous gauge symmetry breaking. Sometimes, as for $tr \phi^2$, non-trivial value of a gauge invariant field is related to a non-trivial value of a gauge variant field and so is a signal for gauge symmetry breaking. Sometimes it is not.

the point of my answer is that the notion of "spontaneously broken vacua" does not make sense away from the perturbative regime.

This is another point of disagreement. For spontaneous symmetry breaking to make sense, you only need vacuum degeneracy and that vacuum is not annhilated by some global symmetry charge (and a gauge-variant charge is by no means "ill-defined"), both of these conditions make sense non-perturbatively.

For Higgs mechanism to work, one doesn't need a genuine SSB of gauge symmetry (which is impossible anyway), all you need is "global symmetry breaking in a gauge theory". In the usual proof of Goldstone theorem, positivity of Hilbert space and locality (i.e. charge is an integertaion of local products of fields) assumptions are crucial; in a gauge theory, either you quantize your theory convariantly but breaks the positivity, or you keep positivity but breaks manifest locality (so you see a gauge-variant charge is actually a necessary feature of Higgs mechanism), so the usual proof of Goldstone theorem can be invalidated. There's a quite thorough discussion on the non-perturbative meaning of Higgs mechanism in the lecture notes by Strocchi.

@Jia Yiyang : A gauge variant charge is in general ill-defined non-perturbatively: I don't know how to define its action on the vacuum for example.

A gauge variant charge is in general ill-defined non-perturbatively: I don't know how to define its action on the vacuum for example.

I fail to see the source of this confusion. By any standard, when you say a field theory is quantized (perturbatively or non-perturbatively), at the very least you have know to how each field operator (in your field algebra) acts on the Hilbert space, and a charge is just a combination of fields, so of course you know how it acts on Hilbert space, modulo perhaps renormalization/operator ordering issues.

I understand gauge-variance is often ugly, but it doesn't mean it's ill-defined. In fact in general the Hamiltonian of a gauge theory also depends on gauge choice, would you conclude the Hamiltonian is also ill-defined?

when you say a field theory is quantized (perturbatively or non-perturbatively), at the very least you have know to how each field operator (in your field algebra) acts on the Hilbert space, and a charge is just a combination of fields

That's precisely the difficulty. Charged operators don't act on a Hilbert space; they map between different Hilbert spaces (superselection sectors). In algebraic quantum field theory, this phenomenon goes under the name of the DHR-theorem. (DHR = Doplicher - Haag - Roberts)

In particular, the Wightman axioms (which give a Hilbert space for QFT) apply only to the uncharged part of the algebra of operators. The superselection sectors  are different representations of this uncharged algebra. The bounded version of the uncharged algebra is a $C^*$-algebra, strictly smaller than the field algebra (containing also the charged operators) which is only a $*$-algebra, but without C.

On the other hand, in perturbation theory, one cannot see any difference between superselection sectors - all representation spaces look perturbatively like the free Fock space. (This is an intrinsic reason why the perturbation series cannot converge.)

All this is part of the poorly understood infrared problems in QFT.

in general the Hamiltonian of a gauge theory also depends on gauge choice, would you conclude the Hamiltonian is also ill-defined?

This doesn't follow. One gets many Hamiltonians in different but isomorphic gauge-dependent representations, all unitarily equivalent. (Edit: Actually, after renormalization, they may well be nonisomorphic, representing unitarily inequivalent superselection sectors.)

@ArnoldNeumaier

That's precisely the difficulty. Charged operators don't act on a Hilbert space; they map between different Hilbert spaces (superselection sectors).

This can't be right, with SSB a charge operator creates a goldstone boson from vacuum, in the same sector. I think you meant to say the exponential of a charge operator.

This doesn't follow. One gets many Hamiltonians in different but isomorphic gauge-dependent representations, all unitarily equivalent.

My example is more directed to 40227, if I understand him/her correctly (so correct me if I'm wrong), he/she seemed to hold the opinion that gauge-dependence for a conserved charge is somehow a deadly sin, regardless of whether there's SSB or not.

@JiaYiyang: Note the difference between charge operator and charged operator. A charge operator may itself be uncharged, but a gauge-variant operator is always charged. Without using this terminology, my statement can be phrased in the case of interest as follows:

Gauge-variant operators map out of the Hilbert space of any particular superselection sector into another superselection sector of the theory. This implies that the expectation of a gauge-variant operator (independent of whether or not it represents a charge) can be taken in none of the superselection sectors, and hence is always ill-defined.

This is unlike the Hamiltonian in a particular gauge, which is gauge-dependent because the representation in which the Hamiltonian is written depends on the gauge used to write it down. But the Hamiltonian is of course the generator of the time translations in the Poincare group, and as such well-determined in any fixed representation.

There's a quite thorough discussion on the non-perturbative meaning of Higgs mechanism in the lecture notes by Strocchi.

But this discussion doesn't affect the statements made by 40227.

The Hilbert space in which there are massless gauge bosons is necessarily inequivalent to the Hilbert space in which, due to the Higgs mechanism  the gauge symmetry is broken and the vector bosons are massive. This is the case because the content of unitary irreducible representations of the Poincare group is different, and no unitary transform between the Hilbert spaces can change this content.

Perturbation theory simply glosses over such differences, and pays for this violation of mathematical consistency with the resulting infrared divergences.

Strocchi, in his nonperturbative arguments in Chapter 19, pays instead with using an indefinite (Gupta-Bleuler type) Krein space in place of the Hilbert space. The gauge-variant operators act on Krein space. But there are no states on Krein space, and hence no vacuum expectation values., since these require a positive definite inner product. (cf. p.198) For these one has to go to a Hilbert space inside the Krein space defining the vacuum sector of the theory, and the vacuum sector is preserved by the observable algebra $F_{obs}$ of gauge-invariant operators only.

@ArnoldNeumaier,

Gauge-variant operators map out of the Hilbert space of any particular superselection sector into another superselection sector of the theory.

I don't understand what this means, you can have a gauge theory (hence gauge-dependent operators) without SSB, so no superselection sectors.

For these one has to go to a Hilbert space inside the Krein space defining the vacuum sector of the theory, and the vacuum sector is preserved by the observable algebra $F_{obs}$ of gauge-invariant operators only.

If we are reading the same book, on page 197 Strocchi explicitly used the vev of the gauge-dependent order parameter, in stating Theorem 19.1

I'm really not sure if you and 40227 are coming from the same perspective. So let me ask you, do you think in Higgs mechanism there are degenerate inequivalent vacuua?

@JiaYiyang:  Superselection sectors are not specific to broken symmetry. QED has them, and even a massless scalar field in 2D has them. They are related to charges and/or topological issues in the boundary conditions of the fields.

Strocchi: I have the 2nd edition; Theorem 19.1 is indeed on p.197. The vacuum state considered there is not a physical state but a nonpositive state in Krein space. You can see this since in a Hilbert space there are no operators with the properties required of $\cal L$ in the two lines after (19.4). See also his comments in the context of footnotes 170/171 on p.182 (second page of Chapter 17).

Probably this difference in what are considered acceptable assumptions is the reason why  you come to a different interpretation then 40227 and I. We assume a physical vacuum state; you seem to be content with the formal but unphysical manipulations.

Who was talking about degenerate inequivalent vacua?

when you say a field theory is quantized (perturbatively or non-perturbatively), at the very least you have know to how each field operator (in your field algebra) acts on the Hilbert space, and a charge is just a combination of fields,

Under quantization, "fields" of the classical theory are mapped to field of operators in the quantum theory. "Fields" of a classical gauge theory are gauge invariant fields. Only for gauge invariants fields one expects to have a corresponding field of operators.

I have indicated in the second paragraph of my answer how, at least formally, I know how to define the expectation value of a gauge invariant field by a path integral and why I don't know how to do the same for a gauge variant field.

In fact in general the Hamiltonian of a gauge theory also depends on gauge choice, would you conclude the Hamiltonian is also ill-defined?

I don't understand this remark. The Hamiltonian of a gauge theory is gauge invariant: how  a measurable quantity like the energy could depend on an unphysical choice of gauge? (I agree that this statement is only true up to a subtlety related to asymptotic symmetries).

@ArnoldNeumaier,

Superselection sectors are not specific to broken symmetry. QED has them, and even a massless scalar field in 2D has them. They are related to charges and/or topological issues in the boundary conditions of the fields.......

Who was talking about degenerate inequivalent vacua?

40227 and I were, I believe. This is the reason I assumed you were talking about the same thing. But I'd still like to ask the same question, do you think there are degenerate vacuua?

My disagreement with 40227 derived from this statement of 40227:

the point of my answer is that the notion of "spontaneously broken vacua" does not make sense away from the perturbative regime.

And I simply fail to see how can SSB not make sense non-perturbatively, since all you need is vacuum degeneracy and the fact that symmetry charge doesn't annihilate vacuum. I understand you were trying to say the 2nd condition might be ill-defined, but this could be just a semantic difference, because in any case it still doesn't annihilate vacuum. It's just in your way of phrasing it, it doesn't annihilate because it can't act (in Coloumb gauge) or it can act but creates unphysical states (in local gauges).

Strocchi: I have the 2nd edition; Theorem 19.1 is indeed on p.197. The vacuum state considered there is not a physical state but a nonpositive state in Krein space. You can see this since in a Hilbert space there are no operators with the properties required of LL in the two lines after (19.4). See also his comments in the context of footnotes 170/171 on p.182 (second page of Chapter 17).

I believe in his writing, the vacuum is a physical one, it's just the excitation mode the field or charge operator creates is not physical, which is precisely the point of Higgs mechanism. In any case, vev of the charged order-parameter field is mathematically well-defined (to futher minimize the effect of our semantic difference, see theorem 19.3 of Strocchi, where a non-covariant quantization is used, only physical states survive, but Strocchi still freely uses vev of the charged field)，and 40227 is disagreeing with this (see his/her comments).

I'm really arguing with two different perspectives from you and 40227 here, so let's try to avoid further conflation.

@40227

"Fields" of a classical gauge theory are gauge invariant fields. Only for gauge invariants fields one expects to have a corresponding field of operators.

I don't see why this is necessarily case.

I have indicated in the second paragraph of my answer how, at least formally, I know how to define the expectation value of a gauge invariant field by a path integral and why I don't know how to do the same for a gauge variant field.

What's the difficulty? Just use

$$\langle \phi \rangle= \text{lim}_{J\to 0}\int \mathcal{D}U \mathcal{D}\phi e^{-S[U, \phi, J]}\phi,$$

where $J$ is a source, and $U$ is gauge link.

I don't understand this remark. The Hamiltonian of a gauge theory is gauge invariant: how  a measurable quantity like the energy could depend on an unphysical choice of gauge?

I thought you were saying any symmetry charge that is formally gauge choice dependent cannot exist, no? My point is the explicit forms of a gauge theory Hamiltonian is typically different in different gauge choices, but this is not a big problem since in different gauges the quantization procedures will also differ, and all the resulting theories should be unitarily equivalent. (Maybe we should also distinguish the terms "gauge-choice dependent" which is about gauge fixings and "gauge variant" which is about gauge transformations)

I'm really arguing with two different perspectives from you and 40227

Well, this is because infrared physics (and this includes symmetry breaking) is intrinsically nonperturbative. There is no uniform understanding of it in the literature since it is poorily understood. Thus how one understands things depends on one's background, and one gets different partial pictures from the different basic views (path integral, canonical quantization, functional Schroedinger approach, axiomatic quantum field theory). Sometimes these pictures are partially in conflict, but in general the more views one understands the better the total informal picture. My view is a mix. I had no difficulties understanding 40227; so I had assumed we have similar views. But now I see that he thinks more in terms of path integrals than I do. Nevertheless (and I find this reassuring for my point of view), what he says is consistent with the axiomatic approach based on physical states.

do you think there are degenerate vacuua?

Yes, but unlike what you had asked before, they are all equivalent through the action of the remaining symmetry group. That's why I was irritated.

vev of the charged order-parameter field is mathematically well-defined

Possibly, possibly not. We haven't yet a consistent axiomatic framework for gauge theory, not even in lower dimensions. Strocchi just makes assumptions that look plausible in a canonical (hence for gauge theories necessarily indefinite) quantization framework; but as Weinberg's treatise shows, canonical quantization has other difficulties in the gauge case.

In any case, it is clear now that you mean the vev defined with the indefinite inner product.

see theorem 19.3 of Strocchi, where a non-covariant quantization is used, only physical states survive, but Strocchi still freely uses vev of the charged field

Strocchi says explicitly on p.205 (end of 2nd par.) that the observable subalgebra (which acts on the physical Hilbert space, i.e., on each sector separately) contains the relevant order parameter.

What's the difficulty? Just use

$\langle \phi \rangle= \text{lim}_{J\to 0}\int \mathcal{D}U \mathcal{D}\phi e^{-S[U, \phi, J]}\phi$

where $J$ is a source, and $U$ is gauge link.

In this formula, what is the range of integration for the variables $U$ and $\phi$ ?

@40227,

I'd say over all their ranges. Or may I answer you this way: Over the same range that makes you think $\langle \text{tr} \phi^2 \rangle$ is well defiend?

@ArnoldNeumaier,

Yes, but unlike what you had asked before, they are all equivalent through the action of the remaining symmetry group. That's why I was irritated.

I said those vacuua are inequivalent because there are no implementable unitary maps linking them, i.e. the statement was precisely about the broken symmetries, not the remaining ones. In any case, we are on the same side about this particular point, but 40227 might not share the same opinion.

Strocchi says explicitly on p.205 (end of 2nd par.) that the observable subalgebra (which acts on the physical Hilbert space, i.e., on each sector separately) contains the relevant order parameter.

I think that's because on page 205 he's talking about the global U(1)-flavor symmetry, so a suitable order parameter can be gauge invariant (since the gauge group is a completely independent group), this is not surprising. But we are debating on the global symmetry that has a local counter part (gauge symmetry), so Theorem 19.3 is the relevant one here, in which he explicitly states the order parameter is in Coulomb algebra not the observable algebra.

Possibly, possibly not. We haven't yet a consistent axiomatic framework for gauge theory, not even in lower dimensions. Strocchi just makes assumptions that look plausible in a canonical (hence for gauge theories necessarily indefinite) quantization framework; but as Weinberg's treatise shows, canonical quantization has other difficulties in the gauge case.

Well, I would rather trust the results that are physically robust. It's profoundly weird if gauge dependence can eliminate the legitimacy of an order parameter. For example, if it were true, in QCD $\bar{\psi} \psi$ would be a perfectly ok order parameter (to study flavor-chiral SSB), but the moment we decide to also consider electroweak interaction, this order parameter becomes gauge-dependent, and suddenly it becomes unspeakable? Does this mean fundamentally it's flawed to study QCD chiral symmetry breaking without electroweak considerations? This is too absurd. If there's a genuine mathematical difficulty as you said (which I'm not convinced of yet), then probably mathematics has to change, not the physics.

@JiaYiyang:

But we are debating on the global symmetry that has a local counter part (gauge symmetry), so Theorem 19.3 is the relevant one here, in which he explicitly states the order parameter is in Coulomb algebra not the observable algebra.

I looked in more detail into the matter. From the algebraic point of view one has the Coulomb field algebra and the observable algebra. What I hadn't see before but became slowly apparent through our discussion is that the observable algebra is different depending on whether or not a gauge group is broken. It contains a Lie algebra $L$ of  charges. But which of these are realized as gauge charges depends on the representation.

In a field algebra in an unbroken representation, all charges in $L$ are gauge charges, the charged vector fields are represented in a massless representation, the observable algebra is the centralizer of $L$, and expectations make sense only for this small observable algebra.

In a field algebra in a broken representation, only the elements of a Lie subalgebra $L_0$ of $L$ are realized as gauge charges, the vector fields corresponding to this subalgebra are represented in a massless representation. The remaining ones are represented in a massive representation with longitudinal modes, and are therefore no longer gauge fields. Thus the observable algebra is the centralizer of the smaller $L_0$, and expectations make sense only for this bigger observable algebra.

This explains the validity of Theorem 19.3, since the assumption is made that the symmetry is broken.But in this case $L_0$ is trivial, so that the observable algebra coincides with the field algebra.

in QCD, $\bar\psi\psi$ would be a perfectly ok order parameter (to study flavor-chiral SSB), but the moment we decide to also consider electroweak interaction, this order parameter becomes gauge-dependent, and suddenly it becomes unspeakable?

With the new insight from the dependence of the observable algebra on the brokenness, this problem disappears, as the electroweak interaction is not a true gauge symmetry but a broken one.

It is unbroken at high temperatures, but at high temperature, there is no vacuum state in the sense of canonical QFT: The ground state at positive temperature is not Poincare invariant.

I'd say over all their ranges. Or may I answer you this way: Over the same range that makes you think ⟨trϕ2⟩ is well defiend?

If you take all the range then you probably obtain an infinite result because along the gauge orbits the action is constant and so there is no exponential suppression of the integrand.

To give a path integral definition of a gauge invariant quantity like $<tr \phi^2>$, one has to integrate over the quotient space by the gauge transformations, i.e. on the gauge equivalence classes of fields. This makes sense because a gauge invariant quantity naturally defines a function on this space whereas it is not the case for a gauge variant quantity.

Let me give an elementary finite dimensional analogue: take the real line $\mathbb{R}$ as analogue of the space of fields, the additive group $\mathbb{Z}$ as analogue of the group of gauge transformations, acting on $\mathbb{R}$ by translation. The quotient space $\mathbb{R}/\mathbb{Z}$ is a circle $S^1$ and is the analogue of the space of gauge equivalence classes of fields. The analogue of a gauge invariant quantity is a function on $\mathbb{R}$ invariant under integral translations, i.e.  a 1-periodic function. The analogue of a general gauge variant quantity is a general function on $\mathbb{R}$. It is clear what is the mean value of a 1-periodic function: it is the integral over any interval of length $1$. It is unclear what is the mean value of a general function: the integral over the real line will in general diverge.

@40227,

This makes sense because a gauge invariant quantity naturally defines a function on this space whereas it is not the case for a gauge variant quantity.

But the action is already not gauge invariant due to the source term. And also I don't see what goes wrong, let's say, if I calculate $\langle \phi \rangle$ using integration over the quotient manifold.

It is unclear what is the mean value of a general function: the integral over the real line will in general diverge.

In general case, even the quotient manifold is non-compact. And we don't expect such integral to diverge just for non-compactness (for one lattice point) because the integrand is controlled by a Gaussian. But indeed gauge equivalent copies may give a divergence, but isn't the effect eliminated by $Z^{-1}[J]$ (which I forgot to write in my formula)?

But the action is already not gauge invariant due to the source term.

Indeed, I should have been more careful. As suggested in my answer, I am not thinking in term of source term: I specify a vacuum by specifying a boundary condition on the fields $\phi$ on which we do the path integral.

And also I don't see what goes wrong, let's say, if I calculate ⟨ϕ⟩⟨ϕ⟩ using integration over the quotient manifold.

Simply because $\phi$ is not a well defined function on the quotient manifold. In the toy example, if $x$ is a coordinate on $\mathbb{R}$, then $x$ is not a well-defined function on $\mathbb{R}/\mathbb{Z}$ because not 1-periodic.

In general case, even the quotient manifold is non-compact. And we don't expect such integral to diverge just for non-compactness (for one lattice point) because the integrand is controlled by a Gaussian. But indeed gauge equivalent copies may give a divergence, but isn't the effect eliminated by Z−1[J]Z−1[J] (which I forgot to write in my formula)?

I agree that non-compactness is not the main issue. The quotient manifold is non-compact but one expects convergence thanks to the exponential suppression of the configurations with big action. If we don't take the quotient then the problem is non-compactness along the gauge orbits: there is no exponential suppression in these directions precisely because the action is gauge invariant. I don't think that adding the normalization helps (it formally gives the right answer if we start by integrating a gauge invariant quantity but in general, I don't think it gives more that $\infty/\infty$: a formula like $<x^2>:=\frac{\int_\mathbb{R} x^2 dx}{\int_\mathbb{R} dx}$ does not sound very promising...)

Indeed, I should have been more careful. As suggested in my answer, I am not thinking in term of source term: I specify a vacuum by specifying a boundary condition on the fields ϕϕon which we do the path integral.

Putting how to define path integral aside for the moment, if you say you already know how to specify vacuua by boundary conditions, doesn't this mean you already know if there's SSB? For example, if in your list of vacuua some have nonvanishing boundary conditions, this already means such vacuua break the global gauge symmetry. How come you still think the notion of SSB doesn't make sense?

Simply because ϕϕ is not a well defined function on the quotient manifold. In the toy example, if xx is a coordinate on RR, then xx is not a well-defined function on R/ZR/Z because not 1-periodic.

Ok, I see you point now, thanks for the clarification.

I agree that non-compactness is not the main issue. The quotient manifold is non-compact but one expects convergence thanks to the exponential suppression of the configurations with big action. If we don't take the quotient then the problem is non-compactness along the gauge orbits: there is no exponential suppression in these directions precisely because the action is gauge invariant. I don't think that adding the normalization helps (it formally gives the right answer if we start by integrating a gauge invariant quantity but in general, I don't think it gives more that ∞/∞∞/∞: a formula like <x2>:=∫Rx2dx∫Rdx<x2>:=∫Rx2dx∫Rdx does not sound very promising...)

I see what you are saying now, but I think you are wrong here. The noncompact part of the field is taken care of by Gaussian suppression, but the gauge orbit only gives a finite Haar-volume contribution, so long as we are talking about compact gauge groups. So your $x^2$ example is flawed, a correct toy example would be like the following: consider a non-propagating action

$$S=\int_x \phi(x)^*\phi(x).$$

Clearly this model has a local U(1) symmetry, and it's a easy exercise to show that

$$\langle \phi(x_1)^2\rangle = \int \mathcal{D}\phi e^{-S} \phi(x_1)^2/Z$$

is completely well defined (note $\phi(x_1)^2$ is gauge-variant), in particular, on a finite lattice both the numerator and the denominator are finite, but $Z$ in the denominator helps a lot if you want to take infinite volume limit (and that was what I meant to say). Formally this carries over to the continuum theory, as far as the matter field integration is concerned. What's really distinguishing is the gauge field measure part: in the continuum theory the gauge orbit for gauge field is non-compact, so the integral diverges even for a single space-time point, and this is why we need gauge-fixing, ghost and all that. While in lattice gauge theory we integrate over gauge link variables, so even for the gauge field the gauge orbit becomes compact, and this is the standard argument of why lattice gauge theory doesn't necessarily need gauge-fixing.

I agree that non-compactness is not the main issue.

This can be seen by taking in the toy example a plane and cylinder in place of the real line and the circle. The cylinder is noncompact, but any continuous function fast decaying along the unbounded axis is integrable, while integrating the periodic extension of the same function to the plane is never integrable.

The gauge orbit only gives a finite Haar-volume contribution, so long as we are talking about compact gauge groups.

This would be the case if you just integrate over the gauge group, which is a finite-dimensional manifold. But you integrate over maps from space-time to the gauge group, which is an infinite-dimensional manifold. Thus compactness is lost.

Edit  @JiaYiyang:  If you first go to the lattice, you need to work with a finite 4D space-time, in order to have a compact integration domain. But this means that dynamics is lost. Moreover, since 4D lattice calculations are in periodic Euclidean time, the results correspond to a finite temperature, whereas so far, I thought we were discussing vacuum expectations. (At least Strocchi did.)

To preserve the vacuum (approximately), you can only discretize space and must work in real time; but then the gauge integrations are over maps from  (finite space) times (real times) to the gauge group, and this is again an infinite-dimensional, noncompact manifold. Thus the problems persist.

@ArnoldNeumaier,

But you integrate over maps from space-time to the gauge group, which is an infinite-dimensional manifold. Thus compactness is lost.

Of course, this is the difficult infinite volume/continuum problem, but 40227 and I were debating on a much simpler problem.

If you first go to the lattice, you need to work with a finite 4D space-time, in order to have a compact integration domain. But this means that dynamics is lost. Moreover, since 4D lattice calculations are in periodic Euclidean time, the results correspond to a finite temperature, whereas so far, I thought we were discussing vacuum expectations. (At least Strocchi did.)

To preserve the vacuum (approximately), you can only discretize space and must work in real time; but then the gauge integrations are over maps from  (finite space) times (real times) to the gauge group, and this is again an infinite-dimensional, noncompact manifold. Thus the problems persist.

Again this is a limit problem, to get to 0 temperature we need to take infinite-volume limit in the time direction.

If one wants to talk about continuum formal set up from the beginning, the best we have is gauge-fixing, ghost, BRST and all that. Gauge symmetry is explicitly broken by gauge fixing, which would've rendered much of the discussion between 40227 and I less motivated.

Putting how to define path integral aside for the moment, if you say you already know how to specify vacuua by boundary conditions, doesn't this mean you already know if there's SSB? For example, if in your list of vacuua some have nonvanishing boundary conditions, this already means such vacuua break the global gauge symmetry. How come you still think the notion of SSB doesn't make sense?

It means that I already know what are the vacua and how to compute physical quantities in each of them. It is because we know the classical vacua and due to the large amount of supersymmetry ($N=2$), they cannot be lifted by a potential. I agree that in more complicated cases, the determination of the space of vacua is much more complicated. Functions on the space of vacua are given by expectation values of operators. To compute them, one introduces a source (as in your suggested definition of $<\phi>$), computes the path integral as a function of the source and does a Legendre transform to obtain an effective potential whose critical points are the different possible expectation values of the operator. Again, in order for this story to make sense, I am talking about gauge invariant operators. These expectation values of gauge invariant operators parametrize the possible vacua of the theory. In each of them, there is a physics to study. In some of them, there is a valid semiclassical description and it makes sense in this description to say if there is a SSB of gauge symmetry or not. For the others, I don't know what it means to talk about a SSB of gauge symmetry.

Of course, this is the difficult infinite volume/continuum problem, but 40227 and I were debating on a much simpler problem.

I think that we are talking about the same problem and on this point as on most of the above ones, I agree with Arnold Neumaier. The path integral I was referring to was really the infinite volume one (the BRST formalism is just one way to express the quotient measure on the quotient space and this does not require gauge fixing. One of the point of my answer was that there is no possible gauge fixing beyong perturbation theory due to Gribov ambiguity, i.e. the non-triviality of the space of fields seen as a bundle over the quotient space of gauge equivalence classes of fields, with fiber the group of gauge transformations). The whole question of existence of different vacua happens only in the infinite volume limit due to a non-trivial dynamics towards the infrared. I agree that on a lattice in a box, with everything discrete and finite, integrating over all gauge fields and then dividing by the volume of the group of gauge transformations makes sense for any, gauge invariant or not, fields (as $\int_{-N}^N x^2 dx /\int_{-N}^{N} dx$ makes sense for finite $N$). But the non-trivial issues happen in the infinite volume limit.

Again, in order for this story to make sense, I am talking about gauge invariant operators.

I think that we are talking about the same problem and on this point as on most of the above ones, I agree with Arnold Neumaier. The path integral I was referring to was really the infinite volume one

So far I still don't see a reason for the insistence on gauge invariance.  Lattice is pretty much the only useful non-perturbative framework so far, and I've argued on a finite lattice $\langle \phi \rangle$ is well defined, the rest of the question is whether the infinite-volume/continuum limit exists. So are you suggesting a gauge-variant expectation value necessarily fails to have a good limit?

I don't see why that should be a priori true and it certainly isn't the true for the simple U(1) gauged toy example I used. A complete calculation goes like this

$$\langle \phi(x_k) \rangle =\langle \phi_1 + i \phi_2 \rangle\\ =\lim_{J\to 0}\lim_{N\to \infty} Z^{-1}[J]\int \Big[ \prod^{N}_{n=1} \text{d}\phi_1(x_n) \text{d}\phi_2 (x_n) \Big] \phi(x_k)\exp[\sum_{1}^{N}(\phi^*(x_n)\phi(x_n)+J^*\phi+J\phi^*)]\\ =\lim_{J\to 0}e^{|J|^2} J =0.$$

So this is perfectly well-defined (continuum limit is also very trivial).

So are you suggesting a gauge-variant expectation value necessarily fails to have a good limit?

I am not saying that it necessarely fails but that it can fail.

The calculation that you have done, starting on a lattice with a field in a representation of the gauge group containing no trivial subrepresentation, always gives 0. See for example Itzykson,  Drouffe, Statistical field theory. Vol. 1, section 6.1.3 called "Order parameter and Elitzur's theorem".

Related comments by Seiberg can be found in this video talk : https://video.ias.edu/PiTP-Seiberg-2of3 , starting from 53:30 for comments on gauge symmetry and from 1:03:00 on gauge symmetry breaking.

Ok, I dug into Elithur theorem a bit and it indeed seems to be a genuine problem (the original Elithur theorem and Itzykson's exposition only applies to bounded fields like $\langle \cos \phi \rangle$, but it was generalized to $\langle \phi \rangle$ for abelian Higgs lattice model by this paper, using some very nontrivial estimate.)

So in light of EDDG(Elithur-De Angelis-de Falco-Guerra) theorem, the problem is not that $\langle \phi \rangle$ is ill-defined, in fact what's shown by EDDG is that it's almost always well-defined and equal to 0 (so instead we should really say $\langle \phi \rangle$ is too well-defined so that it becomes trivial......), and the latter property is a serious problem.

The immediate question is, if the continuum theory is in any sense a limit of a lattice theory, how can  a non-zero VEV ever emerge during this "taking continuum limit" process?

As a consequence, in case of Seiberg-Witten's $\langle \text{tr}\phi^2\rangle$, you still can't make the claim $\langle \text{tr}\phi^2 \rangle \sim a^2$ for large $a$, since EDDG tells us $\langle \phi\rangle$ is always 0 quantum mechanically (in other words, classical limit is never achieved smoothly). Indeed one can just retreat and say "let's just treat $\langle \text{tr}\phi^2 \rangle$ as some moduli parameter of our theory, nothing else." Then the problem becomes why $\langle \text{tr}\phi^2 \rangle$ has anything to do with Higgs phenomenology at all? After all Higgs phenomenology is based on a nonzero $\langle \phi \rangle$, and EDDG tells us there's absolutely no relation between  $\langle \text{tr}\phi^2 \rangle$  and $\langle \phi \rangle$ (at least in a gauge-invariant lattice formalism).

if the continuum theory is in any sense a limit of a lattice theory, how can  a non-zero VEV ever emerge during this "taking continuum limit" process?

probably by inventing suitable projectors $P$ to subspaces of the lattice Hilbert spaces where the limiting states are well-defined and give the expectation in an appropriate superselection sectors, $\langle f\rangle=\lim \langle PfP\rangle$.

@ArnoldNeumaier, but at what stage of the limit taking does one introduce such projection? It's a bit hard to even imagine since "introducing $P$ or not" is a binary choice while taking limit is a continuous process.

After scratching my head really hard I think I kinda understand what the deal is with EDDG theorem. In a gauge invariant formalism, one can locally change the field without energy cost, then in thermal equilibrium , even if we start with a configuration with all fields on all latiices aligned to a fixed direction (on gauge orbit),  lattice field on each single lattice will quickly equilibrate into a uniform superposition of all configurations on the gauge orbit, so any gauge invariant operator will have 0 VEV. This is different from a global symmetry situation, where if we start with all fields aligned, the phase will be "locked" in that direction because there will be an extensive energy cost if we want to "rotate" the fields locally from one point to the other.

However, there's one thing about continuum gauge theory that EDDG doesn't seem to capture, even putting aside gauge-fixing issue. That is, in continuum gauge theory a gauge transformation is required to preserve the boundary conditions at infinity, i.e. $\lim_{x\to \infty} G(x)=I$, and because of this one cannot say a global symmetry transformation is just a special case of gauged ones.  And clearly with this it's definitely meaningful to ask if there's a global symmetry breaking in a given gauge theory, because global symmetry breaking gives different boundary conditions at infinity which are stable under gauge transformations.

So if one imagine somehow we have a infinite-volume lattice gauge theory, to capture the same requirement one must somehow suppress gauge fluctuations at large distance, and note that this suppression doesn't have to be a gauge-fixing, although gauge-fixing can achieve this requirement (for the moment I don't know what else can it be, and I've been thinking about it, quite painfully...).

@40227, maybe I should ask a question that's more relevant to the title. How critical is the usage of $\langle \text{tr} \phi^2\rangle$ in Seiberg-Witten paper? Had they used $\langle \phi \rangle$, does it go terribly wrong?

at what stage of the limit taking does one introduce such projection?

maybe I should ask a question that's more relevant to the title. How critical is the usage of ⟨trϕ2⟩⟨trϕ2⟩ in Seiberg-Witten paper? Had they used ⟨ϕ⟩⟨ϕ⟩, does it go terribly wrong?

Again the point is that we don't know what $<\phi>$ means. In the Seiberg-Witten context, one could naively think that there is a way to define $<\phi>$. Indeed, classically, $<\phi>$ makes sense, is determined up to gauge transformations by its eigenvalues $(a,-a)$, and according to the classical Higgs mechanism, $|a|$ is (maybe up to some numerical constants) the mass of the $W$ bosons. This suggests a definition of $|a|$ in the full quantum theory: define it as the mass of the $W$ bosons. This works in the semiclassical region of the moduli space of vacua where there exists $W$ bosons but this fails in the strong coupling region: it is unclear a priori what is the spectrum of the theory and if there is in the spectrum something one could call $W$ bosons (and in fact it is part of the conclusion of the Seiberg-Witten analysis that the $W$ bosons existing at weak coupling disappear at strong coupling). One can try to do better: classically $a$ is the electric component of the central charge. The central charge is part of the supersymmetry algebra and so makes sense in the full quantum theory (we are looking at supersymmetric vacua). This suggests a definition of $a$ in the full quantum theory: define it as the electric part of the central charge. But this fails because the "electric part" is not well-defined because of the electromagnetic duality in 4d abelian gauge theory. More precisely, it is well-defined in the weak coupling region and one can "analytically continue" it in the strong coupling region but due to singularities in the moduli space of vacua, one can come back in the weak coupling region with a non-trivial monodromy. So $a$ is a multivalued function on the moduli space of vacua and so is not enough to describe the moduli space. Conversely, $u=<Tr \phi^2>$ is a good coordinate on the moduli space and the subject of the Seiberg-Witten paper is to determine the relation between $u$ and $a$: for a given vacuum $u$, what is the central charge, what is the spectrum...

@JiaYiyang The gauge invariant operator of the broken $SU(2)$ SYM is tr$\phi^2$.

@conformal_gk, yes, that's what the paper uses, but my disagreement is that such order parameter is not a good one, namely, it's insufficient to use nonzero tr$\phi^2$ to invoke Higgs mechanism.

@JiaYiyang you 're right the vev is non-zero. Thus it is sufficient.

@conformal_gk. no it's not, look the first example I used in my original post: chiral condensate in QCD, it's gauge invariant, it's nonzero, still no Higgs mechanism.

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