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  Is it really proper to say Ward identity is a consequence of gauge invariance?

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Many (if not all) of the materials I've read claim Ward identity is a consequence of gauge invariance of the theory, while actually their derivations only make use of current conservation μJμ=0(which is only equivalent to a global phase symmetry). I'm aware of the fact that a gauge field has to be coupled to a conserved current to keep gauge invariance, but a non-gauge field can also be(though not must be) coupled to a conserved current and in that case Ward identity should still hold. So do you think it is at least misleading, if not wrong, to claim Ward identity is a consequence of gauge invariance?

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang
asked Jul 12, 2013 in Theoretical Physics by Jia Yiyang (2,640 points) [ no revision ]
Perhaps you could show an example where the current J is exactly conserved and couples to a field A via J, but the resulting theory does not have a gauge symmetry?

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user BebopButUnsteady
And where the Ward identity is obeyed as well.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user BebopButUnsteady
@BebopButUnsteady: how about a massive vector field Aμ coupled to some conserved current Jμ? The mass term m2AμAμ violates gauge symmetry, but Ward identity is still true because Jμ is conserved.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang

3 Answers

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This answer partially disagrees with Motl's. The crucial point is to consider the difference between the abelian and non-abelian case. I totally agree with Motl's answer in the non-abelian event — where these identities are usually denominated Slavnov-Taylor's rather than Ward's, so that I will refer to the abelian case.

First, a few words about terminology: Ward identities are the quantum counterpart to (first and second) Noether's theorem in classical physics. They apply to both global and gauge symmetries. However, the term is often reserved for the U(1) gauge symmetry in QED. In the case of gauge symmetries, Ward identities yield real identities, such as kμMμ=0, where Mμ is defined by M=ϵμMμ, in QED, that tell us that photon's polarizations parallel to photon's propagation don't contribute to scattering amplitudes. In the case of global symmetries, however, Ward identities reflect properties of the theory. For example, the S-matrix of a Lorentz invariant theory is also Lorentz invariant or the number of particles minus antiparticles in the initial state is the same as in the final state in a theory with a global (independent of the point in space-time) U(1) phase invariance.

Let's study the case of a massive vectorial field minimally coupled to a conserved current:

L=14F2+a22A2+iˉΨDΨm22ˉΨΨ=14F2+a22A2+iˉΨΨm22ˉΨΨeAμjμ

Note that this theory has a global phase invariance ΨeiθΨ, with a Noether current

jμ=ˉΨγμΨ

such that (classically) μjμ=0. Apart from this symmetry, it is well-known that the Lagrangian above is equivalent to a theory: i)that doesn't have an explicit mass term for the vectorial field. ii) that contains a scalar field (a Higgs-like field) with a different from zero vacuum expectation value, which spontaneously break a U(1) gauge symmetry (this symmetry is not the gauged U(1) global symmetry mentioned previously). The equivalence is in the limit where vacuum expectation value goes to infinity and the coupling between the vectorial field and the Higgs-like scalar goes to zero. Since one has to take this last limit, the charge cannot be quantized and therefore the U(1) gauge symmetry must be topologically equivalent to the addition of real numbers rather than the multiplication of complex numbers with unit modulus (a circumference). The difference between both groups is only topological (does this mean then that the difference is irrelevant in the following?). This mechanism is due to Stueckelberg and I will summarize it at the end of this answer.

In a process in which there is a massive vectorial particle in the initial or final state, the LSZ reductio formula gives:

i|fϵμd4xeikx(ημν(2a2)μν)...0|TAν(x)...|0

From the Lagrangian above, the following classical equations of motion may be obtained

(ημν(2a2)μν)Aν=ejμ

Then, quantumly,

(ημν(2a2)μν)0|TAν(x)...|0=e0|Tjμ(x)...|0+contact terms, which don't contribute to the S-matrix

And therefore

i|fϵμd4xeikx...0|Tjμ(x)...|0+contact terms, which don't contributeϵμMμ

If one replaces ϵμ with kμ, one obtains

kμMμkμd4xeikx...0|Tjμ(x)...|0

Making use of kμμ,eikx, integrating by parts, and getting ride of the surface term (the plane wave is an idealization, what one actually has is a wave packet that goes to zero in the spatial infinity), one gets

kμMμd4xeikx...μ0|Tjμ(x)...|0

One can now use the Ward identity for the global ΨeiθΨ symmetry (classically μjμ=0 over solutions of the matter, Ψ, equations of motion)

μ0|Tjμ(x)...|0=contact terms, which don't contribute to the S-matrix

And hence

kμMμ=0

same as in the massless case.

Note that in this derivation, it has been crucial that the explicit mass term for the vectorial field doesn't break the global U(1) symmetry. This is also related to the fact that the explicit mass term for the vectorial field can be obtained through a Higgs-like mechanism connected with a hidden (the Higgs-like field decouples from the rest of the theory) U(1) gauge symmetry.

A more careful calculation should include counterterms in the interacting theory, however I think that this is the same as in the massless case. We can think of the fields and parameters in this answer as bare fields and parameters.

Stueckelberg mechanism

Consider the following Lagrangian

L=14F2+|dϕ|2+μ2|ϕ|2λ(ϕϕ)2

where d=igB and F is the field strength (Faraday tensor) for B. This Lagrangian is invariant under the gauge transformation

BB+(1/g)α(x)

ϕeiα(x)ϕ

Let's take a polar parametrization for the scalar field ϕ: ϕ12ρeiχ, thus

L=14F2+12ρ2(μχgBμ)2+12(ρ)2+μ22ρ2λ4ρ4

We may now make the following field redefinition AB(1/g)χ and noting that Fμν=μBννBμ=μAννAμ is also the field strength for A

L=14F2+g22ρ2A2+12(ρ)2+μ22ρ2λ4ρ4

If ρ has a vacuum expectation value different from zero 0|ρ|0=v=μ2λ, it is then convenient to write ρ(x)=v+ω(x). Thus

L=14F2+a22A2+g2vωA2+g22ω2A2+12(ω)2μ22ω2λvω3λ4ω4+v4λ24

where ag×v. If we now take the limit g0, v, keeping the product, a, constant, we get

L=14F2+a22A2+12(ω)2μ22ω2λvω3λ4ω4+v4λ24

that is, all the interactions terms between A and ω disappear so that ω becomes an auto-interacting field with infinite mass that is decoupled from the rest of the theory, and therefore it doesn't play any role. Thus, we recover the massive vectorial field with which we started.

L=14F2+a22A2

Note that in a non-abelian gauge theory must be non-linear terms such as gA2A, g2A4, which prevent us from taking the limit g0.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user drake
answered Jul 16, 2013 by drake (885 points) [ no revision ]
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By the way, I enjoy your questions. They are good. @JiaYiyang

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user drake
Ok, I'll take a look at the book later tonight. I'm glad you like the questions, I like to make people happy:) @drake

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang
L. Brown's book is so beautifully written that I can't stop reading it. Now I see gauge invariance is indeed crucial, but I slightly disagree with you about the role of gauge invariance. I'll write an answer later, it might take some time though. Thanks a lot! @drake

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang
@JiaYiyang You're welcome. I'm looking forward to your answer.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user drake
More on Stueckelberg formalism in Ron Maimon's Phys.SE answer here.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Qmechanic
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The sketch of the proof is quite simple. Since kM=0, ϵ and ϵ+ck, with c a constant, describe the same physical polarization. Then an in or out A field must be physically equivalent to A+ω. @JiaYiyang

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user drake
It seems we can apply the same proof to massive vector field, but massive vector field doesn't have such gauge symmetry(not sure if it's related to the hidden gauge symmetry you mentioned). @drake

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang
+ 2 like - 0 dislike

The Ward identity follows from the gauge symmetry and it's possible to see these things without mentioning any current whatsoever. The Ward identity says kμMμ(k)=0 which really says that the longitudinal polarization of the gauge boson, one with the pure-gauge polarization vector proportional to the momentum, ϵμkμ, "decouples" i.e. its interactions (scattering amplitudes) with any collection of physical particles vanish.

This vanishing implies a symmetry – now yes, kμMμ(k) may also be interpreted as a correlator including μJμ, a conserved current – and this symmetry is a gauge symmetry because the gauge field may only have a nonzero kμ i.e. dependence on the spacetime if we allow the symmetry parameter to depend on spacetime.

Fields with an extra m2AμAμ etc. are no longer coupled to a conserved current because the current is modified by an extra m2Aμ – because this appears as a factor multiplying Aμ in a term you just added – which also means that the Ward identity won't hold if you break the symmetry in this explicit way (the Ward identity will broken "more controllably" if you break the symmetry spontaneously, not explicitly, becausethe full Lagrangian still has the gauge symmetry i.e. the gauge field coupled to a conserved current).

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Luboš Motl
answered Jul 14, 2013 by Luboš Motl (10,278 points) [ no revision ]
I fully agree with your 1st paragraph, i.e. gauge symmetry implies Ward identity. But I don't understand your 2nd paragraph, i.e. Ward identity implies gauge symmetry, so I'd be grateful if you can elaborate more. I don't agree with your 3rd paragraph, if I understand you correctly, you mean m2AμAμ+AμJμ=Aμ(m2Aμ+Jμ). However, shouldn't m2AμAμ be interpreted as part of the free Lagrangian rather than the current?

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang
I have just written an answer that partially disagrees with yours. It would be nice if you could point out any mistake you see in my derivation or reasoning.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user drake
+ 2 like - 0 dislike

Let me try to answer my own question after spending quite some time reading L.Brown's "quantum field theory", but I'll not stick to his notations.

Let me clarify a bit on the terminology I'll use: "Generalized Ward Identity(GWI)" refers to (lk)μΓμ(k,l)=iS1(k)iS1(l), where Γμ(k,l) is an electron-electron-photon vertex function, S is an (full) electron-electron propagator.I'll come back to this in detail later; "Ward identity(WI)" refers to the special case when one lets lk in GWI; "Ward-Takahashi Identity(WTI)" refers to kμMμ(k)=0.

I should confess when I asked this question and when I put the words "...claim Ward identity is a consequence of gauge invariance of the theory.", I didn't know which of the three identities they were referring to, but now at least I can say GWI is really a consequence of gauge invariance, not global phase symmetry. In short, if Γμ in GWI is taken as an improper vertex(i.e. 1-particle reducible vertex), then GWI holds for theories which respect current conservation(or global phase symmetry). However, for theory with a gauge symmetry, we get a stronger GWI, that is, GWI holds not only for an improper vertex, but also the proper one(i.e. 1-particle irreducible vertex).

GWI of Improper Vertex

First let's see how to get the GWI for current conservation, and here I'll basically copy from Weinberg Vol I chap 10. Consider the vacuum time-ordered product T{Jμ(x)Ψn(y)ˉΨm(z)}. Diagrammatically, this is the sum all the diagrams with 1 external photon propagator and 2 external electron propagators, but with a bare external photon propagator stripped away. Now Weinberg defines Γμ(k,l) by

d4xd4yd4zeipxeikyeilzT{Jμ(x)Ψn(y)ˉΨm(z)}iqSnn(k)Γμnm(k,l)Smm(l)δ4(p+kl)

where Snm is the Fourier transform of T{Ψn(y)ˉΨm(z)}(and omit a delta function), so it is the full electron propagator. Now we can see Γμ is the vertex function after 2 full electron propagators and 1 bare photon propagator get stripped away, thus it is 1-particle reducible along the photon line(i.e. still contains the photon vacuum-polarization correction),hence improper. Diagrams involved are: enter image description here

where a dashed line means that the line has been stripped, and ΓμP denotes the proper vertex, and we can get ΓμP if we can further strip away the photon vacuum polarization part. The rest basically follows from calculating xμT{Jμ(x)Ψn(y)ˉΨm(z)}, applying μJμ=0 and then a Fourier transform.

GWI of Proper Vertex

Now I shall claim for theory with local gauge invariance, GWI holds also for proper vertex ΓμP. The idea is to isolate ΓμP from Γμ. As can be easily seen from the 2nd figure, we can first add back the bare photon propagator(let's denote it by Gμν0), and then remove a full photon propagator Gμν, that is,

ΓμP(k,l)=G1(p)μ  νGνρ0(p)Γρ(k,l),

where p=lk.

So to mimic the LHS of GWI, we have(lk)μΓμP(k,l)=pμG1(p)μ  νGνρ0(p)Γρ(k,l)().

Now here is where gauge invariance comes into play:

Statement: Gauge invariancepμG1(p)μ  νGνρ0(p)=pρ.

If the statement is true, we immediately get from equation () that pμΓμP(k,l)=pρΓρ(k,l),

and since GWI holds for Γρ(k,l), from here we can conclude it also holds for ΓμP(k,l).

Here's the sketch of the proof of the above statement: With a gauge parameter ξ, we can write the inverse of bare propagator as G10(p)μν=(gμνp2pμpν)1ξpμpν.

Because of the gauge invariance, the full propagator only differs from the bare one in the transverse part, and the longitudinal part remains the same, that is, G1(p)μν=(gμνp2pμpν)F(p2)1ξpμpν.
This theorem itself involves another not-so-short proof, and it can be found in Brown's book, anyway the key point is that to prove it one needs local symmetry, global symmetry is not enough. Plug in the general forms of the two propagators one can easily prove the statement.

This is in contrast with a theory without gauge invariance(e.g. you can get the propagtor of massive vector field by doing the replacement 1ξm2), in there the full propagator will also alter the longitudinal part so it becomes G1(p)μν=(gμνp2pμpν)F(p2)1ξH(p2)pμpν,

then if you carry out the calculation in the statement, you'll get something like pμG1(p)μ  νGνρ0(p)=H(p2)pρ(or maybe 1H(p2)pρ, cannot quite remember). Then for the proper vertex GWI is modified to pμΓμP(k,l)=H(p2)[iS1(k)iS1(l)],
which is not of too much good, e.g., one can't obtain the nice renormalization relation Z1=Z2 for proper vertex. Also, this means in gauge theory we can consider the (proper) vertex renormalization separately from vacuum polarization, while say in massive vector theory vacuum polarization has to be taken into account.

PS: Brown also gives a second proof of proper vertex WI by using effective action technique, which is in a way "shorter". However it needs much more preliminary knowledge about effective action, and also won't be so handy to contrast between the roles of gauge invariance and current conservation in GWI, so I didn't adopt the method here.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang
answered Jul 19, 2013 by Jia Yiyang (2,640 points) [ no revision ]
@drake: feel free to comment.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang
Nice answer! I think that if the mass term is produced by a Higgs-like mechanism, then the identity also holds, as NAKANISHI's paper (linked in Motl's answer) points out. What do you think?

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user drake
@drake: I can't fully understand the article since I haven't learnt spontaneous symmetry breaking and the article is not fully self contained. But I guess yes, since NAKANISHI was talking about the proper vertex.

This post imported from StackExchange Physics at 2014-03-31 22:25 (UCT), posted by SE-user Jia Yiyang

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