By Jia Yiyang


2013-07-12 16:25:13 8 Comments

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 $\partial_\mu J^\mu=0$(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?

3 comments

@Jia Yiyang 2013-07-19 16:44:56

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 $(l-k)_\mu\Gamma^\mu(k,l)=iS^{-1}(k')-iS^{-1}(l)$, where $\Gamma^\mu(k,l)$ 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 $l\to k$ in GWI; "Ward-Takahashi Identity(WTI)" refers to $k_\mu {\mathcal M}^\mu(k) = 0$.

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 $\Gamma^\mu$ 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 $\langle \mathcal{T}\{J^\mu(x)\Psi_n(y)\bar{\Psi}_m(z)\}\rangle$. 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 $\Gamma^\mu(k,l)$ by

$$\int d^4xd^4yd^4ze^{-ipx}e^{-iky}e^{ilz}\langle \mathcal{T}\{J^\mu(x)\Psi_n(y)\bar{\Psi}_m(z)\}\rangle\equiv-iqS_{nn'}(k)\Gamma^\mu_{n'm'}(k,l)S_{m'm}(l)\delta^4(p+k-l)$$

where $S_{nm}$ is the Fourier transform of $\mathcal{T}\{\Psi_n(y)\bar{\Psi}_m(z)\}\rangle$(and omit a delta function), so it is the full electron propagator. Now we can see $\Gamma^\mu$ 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 $\Gamma^\mu_P$ denotes the proper vertex, and we can get $\Gamma^\mu_P$ if we can further strip away the photon vacuum polarization part. The rest basically follows from calculating $\frac{\partial}{\partial x^\mu}\langle \mathcal{T}\{J^\mu(x)\Psi_n(y)\bar{\Psi}_m(z)\}\rangle$, applying $\partial_\mu J^\mu=0$ 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 $\Gamma^\mu_P$. The idea is to isolate $\Gamma^\mu_P$ from $\Gamma^\mu$. As can be easily seen from the 2nd figure, we can first add back the bare photon propagator(let's denote it by $G_0^{\mu\nu}$), and then remove a full photon propagator $G^{\mu\nu}$, that is,

$$\Gamma^\mu_P(k,l)=G^{-1}(p)^\mu_{\ \ \nu}G_0^{\nu\rho}(p)\Gamma_\rho(k,l),$$ where $p=l-k$.

So to mimic the LHS of GWI, we have$$(l-k)_\mu\Gamma^\mu_P(k,l)=p_\mu G^{-1}(p)^\mu_{\ \ \nu}G_0^{\nu\rho}(p)\Gamma_\rho(k,l)\cdots\cdots(*).$$ Now here is where gauge invariance comes into play:

$$\text{Statement: Gauge invariance}\implies p_\mu G^{-1}(p)^\mu_{\ \ \nu}G_0^{\nu\rho}(p)=p^\rho.$$

If the statement is true, we immediately get from equation $(*)$ that $$p_\mu\Gamma^\mu_P(k,l)=p_\rho \Gamma^\rho(k,l),$$ and since GWI holds for $ \Gamma^\rho(k,l)$, from here we can conclude it also holds for $\Gamma^\mu_P(k,l)$.

Here's the sketch of the proof of the above statement: With a gauge parameter $\xi$, we can write the inverse of bare propagator as $$G^{-1}_0(p)_{\mu\nu}=(g_{\mu\nu}p^2-p_\mu p_\nu)-\frac{1}{\xi}p_\mu p_\nu.$$ 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, $$G^{-1}(p)_{\mu\nu}=(g_{\mu\nu}p^2-p_\mu p_\nu)F(p^2)-\frac{1}{\xi}p_\mu p_\nu.$$ 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 $\frac{1}{\xi}\to m^2$), in there the full propagator will also alter the longitudinal part so it becomes $$G^{-1}(p)_{\mu\nu}=(g_{\mu\nu}p^2-p_\mu p_\nu)F(p^2)-\frac{1}{\xi}H(p^2)p_\mu p_\nu,$$ then if you carry out the calculation in the statement, you'll get something like $ p_\mu G^{-1}(p)^\mu_{\ \ \nu}G_0^{\nu\rho}(p)=H(p^2)p^\rho$(or maybe $\frac{1}{H(p^2)}p^\rho$, cannot quite remember). Then for the proper vertex GWI is modified to $$p_\mu\Gamma^\mu_P(k,l)=H(p^2)[iS^{-1}(k')-iS^{-1}(l)],$$ which is not of too much good, e.g., one can't obtain the nice renormalization relation $Z_1=Z_2$ 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.

@Diego Mazón 2013-07-19 20:02:56

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?

@Jia Yiyang 2013-07-20 03:27:57

@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.

@Andrew McAddams 2014-07-25 22:29:59

Excuse me for this question, but I will be grateful if you'll answer. Weinberg assume that Green function $$ \tag 1 \int d^{4}xd^{4}yd^{4}ze^{ipx + iky - ilz}\langle |\hat {T}\left( \hat {J}^{\mu}(x)\hat {\Psi}_{n}(y)\bar {\hat {\Psi}}_{m}(z)\right) |\rangle $$ contains sum of the diagramms with two electronic full propagators and photonic external line. But I don't see $\hat {A}_{\mu}$ under the time ordering operation ($\hat {J}^{\mu}$ doesn't contain it) so I don't understand why diagrams $(1)$ contain external photonic line. Can you explain this?

@Jia Yiyang 2014-07-26 00:26:55

@AndrewMcAddams: (1) indeed does not contain an external photonic line. Where did Weinberg say so?

@Andrew McAddams 2014-07-26 00:33:27

At page 446: "Eq. (10.4.19) gives the sum of all such graphs with an extra photon line attached". Here "eq. (10.4.19)" coincides with my $(1)$.

@Jia Yiyang 2014-07-26 14:02:35

@AndrewMcAddams: he means with a full photon propagator attached but on top of that a bare photon propagator must be stripped away, just note he immediately added "...but with the complete Dirac external line propagators and the bare photon external line propagator stripped away."

@Andrew McAddams 2014-07-26 17:53:04

@JiaYiyang : he wrote it for the vertex function $\Gamma^{\mu}$, not for $(1)$.

@Jia Yiyang 2014-07-26 18:01:13

@AndrewMcAddams: To me that really is just imprecise language he used. That's the only logical way I see to understand it and it fits well with rest part of the book.

@Andrew McAddams 2014-07-26 18:38:00

@JiaYiyang : thank you, I think that you're right.

@LYg 2014-09-28 14:07:37

@JiaYiyang, excellent! this is a really impressive answer. I'm been wondering about the how the WI expressed in current and that in proper vertex are related for a long time. Yeah, now indeed your answer showed that the proper vertex WI depend on gauge symmetry, not only on global symmetry!

@Diego Mazón 2013-07-16 22:39:36

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 $k^{\mu}\mathcal M_{\mu}=0$, where $\mathcal M_{\mu}$ is defined by $\mathcal M=\epsilon_{\mu}\,\mathcal M^{\mu}$, 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:

$$\mathcal L=-{1\over 4}\,F^2+{a^2\over 2}A^2+i\,\bar\Psi\displaystyle{\not}D\, \Psi - {m^2\over 2}\bar\Psi\Psi \\ =-{1\over 4}\,F^2+{a^2\over 2}A^2+i\,\bar\Psi\displaystyle{\not}\partial \, \Psi - {m^2\over 2}\bar\Psi\Psi-e\,A_{\mu}\,j^{\mu}$$

Note that this theory has a global phase invariance $\Psi\rightarrow e^{-i\theta}\,\Psi$, with a Noether current

$$j^{\mu}={\bar\Psi\, \gamma^{\mu}}\,\Psi$$

such that (classically) $\partial_{\mu}\,j^{\mu}=0$. 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:

$$\langle i\,|\,f \rangle\sim \epsilon _{\mu}\int d^4x\,e^{-ik\cdot x}\, \left(\eta^{\mu\nu}(\partial ^2-a^2)-\partial^{\mu}\partial^{\nu}\right)...\langle 0|\mathcal{T}A_{\nu}(x)...|0\rangle$$

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

$$\left(\eta^{\mu\nu}(\partial ^2-a^2)-\partial^{\mu}\partial^{\nu}\right)A_{\nu}=ej^{\mu}$$

Then, quantumly,

$$\left(\eta^{\mu\nu}(\partial ^2-a^2)-\partial^{\mu}\partial^{\nu}\right)\langle 0|\mathcal{T}A_{\nu}(x)...|0\rangle = e\,\langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle + \text{contact terms, which don't contribute to the S-matrix}$$

And therefore

$$\langle i\,|\,f \rangle\sim \epsilon _{\mu}\int d^4x\,e^{-ik\cdot x}\,...\langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle +\text{contact terms, which don't contribute}\sim \epsilon_{\mu}\mathcal{M}^{\mu}$$

If one replaces $\epsilon_{\mu}$ with $k_{\mu}$, one obtains

$$k_{\mu}\mathcal{M}^{\mu}\sim k _{\mu}\int d^4x\,e^{-ik\cdot x}\,...\langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle$$

Making use of $k_{\mu}\sim \partial_{\mu}\,,e^{-ik\cdot x}$, 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

$$k_{\mu}\mathcal{M}^{\mu}\sim \int d^4x\,e^{-ik\cdot x}\,...\, \partial_{\mu}\,\langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle$$

One can now use the Ward identity for the global $\Psi\rightarrow e^{-i\theta}\,\Psi$ symmetry (classically $\partial_{\mu}\,j^{\mu}=0$ over solutions of the matter, $\Psi$, equations of motion)

$$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle = \text{contact terms, which don't contribute to the S-matrix}$$

And hence

$$k^{\mu}\mathcal M_{\mu}=0$$

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

$$\mathcal L=-{1\over 4}\,F^2+|d\phi|^2+\mu^2\,|\phi|^2-\lambda\, (\phi^*\phi)^2$$

where $d=\partial - ig\, B$ and $F$ is the field strength (Faraday tensor) for $B$. This Lagrangian is invariant under the gauge transformation

$$B\rightarrow B + (1/g)\partial \alpha (x)$$ $$\phi\rightarrow e^{i\alpha(x)}\phi$$

Let's take a polar parametrization for the scalar field $\phi$: $\phi\equiv {1\over \sqrt{2}}\rho\,e^{i\chi}$, thus

$$\mathcal L=-{1\over 4}\,F^2+{1\over 2}\rho^2\,(\partial_{\mu}\chi-g\,B_{\mu})^2+{1\over 2}(\partial \rho)^2+{\mu^2\over 2}\,\rho ^2- {\lambda\over 4}\rho^4$$

We may now make the following field redefinition $A\equiv B - (1/g)\partial \chi$ and noting that $F_{\mu\nu}=\partial_{\mu}B_{\nu}-\partial_{\nu}B_{\mu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ is also the field strength for $A$

$$\mathcal L=-{1\over 4}\,F^2+{g^2\over 2}\rho^2\,A^2+{1\over 2}(\partial \rho)^2+{\mu^2\over 2}\,\rho ^2-{\lambda\over 4}\, \rho^4$$

If $\rho$ has a vacuum expectation value different from zero $\langle 0|\rho |0\rangle = v=\sqrt{\mu^2\over \lambda}$, it is then convenient to write $\rho (x)=v+\omega (x)$. Thus

$$\mathcal L=-{1\over 4}\,F^2+{a^2\over 2}\,A^2+g^2\,v\,\omega\,A^2+{g^2\over 2}\,\omega ^2\,A^2+{1\over 2}(\partial \omega)^2-{\mu^2\over 2}\,\omega ^2-\lambda\,v\omega^3-{\lambda\over 4}\, \omega^4+{v^4\,\lambda^2\over 4}$$

where $a\equiv g\times v$. If we now take the limit $g\rightarrow 0$, $v\rightarrow \infty$, keeping the product, $a$, constant, we get

$$\mathcal L=-{1\over 4}\,F^2+{a^2\over 2}\,A^2+{1\over 2}(\partial \omega)^2-{\mu^2\over 2}\,\omega ^2-\lambda\,v\omega^3-{\lambda\over 4}\, \omega^4+{v^4\,\lambda^2\over 4}$$

that is, all the interactions terms between $A$ and $\omega$ disappear so that $\omega$ 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.

$$\mathcal L=-{1\over 4}\,F^2+{a^2\over 2}\,A^2$$

Note that in a non-abelian gauge theory must be non-linear terms such as $\sim g A^2\,\partial A\;$, $\sim g^2 A^4$, which prevent us from taking the limit $g\rightarrow 0$.

@Jia Yiyang 2013-07-17 12:57:02

Thanks for the comprehensive reply. It seems you do agree with my statement, that Ward identity is a consequence of current conservation, or equivalently $U(1)$ global symmetry, but not really derived from the gauged(or local) symmetry. And how is Stuekelberg mechanism relevant to my question?

@Jia Yiyang 2013-07-17 13:02:02

Besides, Ward identity seems to tell a bit more than " photon's polarizations parallel to photon's propagation don't contribute to scattering amplitudes", because for scattering amplitudes, the external photon's momentum under consideration is on shell, while ward identity even holds for off shell $k^\mu$

@Diego Mazón 2013-07-17 17:58:05

@JiaYiyang I don't agree. Gauge invariance is a requirement. Note that in a theory with a complex scalar field with a quartic self-interaction, there is a global $U(1)$ symmetry, but there is not any Ward identity similar in spirit to the QED one (there is a Ward identity that says that number of particles minus the number of antiparticles is conserved). The same issue happens if one replaces $F_{\mu\nu}$ with $\partial_{\mu}A_{\nu}-b\partial_{\nu}A_{\mu}$, with $b\neq 1$, in QED. One still has $\partial j=0$, but not QED-like Ward identity.

@Diego Mazón 2013-07-17 18:04:39

I think that the case of QED plus a mass term for the photon is very special, because there is a hidden gauge symmetry. In my exposition, this hidden gauge symmetry is given by the transformation $B\rightarrow B+(1/g)\alpha$, $\chi\rightarrow \chi + \alpha$. I think this answers your last question too.

@Diego Mazón 2013-07-17 18:22:04

@JiaYiyang Concerning your second comment, $k\cdot \mathcal M=0$ is a consequence of gauge invariance. Besides this, I think that one can also prove that $k\cdot \mathcal M=0$ implies gauge invariance under some assumptions, such as Poincare invariance, unitarity, locality, and existence of a vacuum state.

@Jia Yiyang 2013-07-18 01:35:12

the $\phi^4$ theory is not a good example since nothing is coupled to the conserved current $i(\phi^*\partial_\mu\phi-\partial_\mu\phi^*\phi)$, so not really what I want to investigate. As for the theory of $\partial_{\mu}A_{\nu}-b\partial_{\nu}A_{\mu}$, why doesn't it have QED-like Ward identity?

@Jia Yiyang 2013-07-18 01:38:33

" I think that one can also prove that k⋅M=0 implies gauge invariance under some assumptions, such as Poincare invariance, unitarity, locality, and existence of a vacuum state." This seems to be a pretty convincing argument, if you can provide a proof.

@Diego Mazón 2013-07-18 02:09:24

The sketch of the proof is quite simple. Since $k\cdot \mathcal M=0$, $\epsilon$ and $\epsilon + c\, k$, with $c$ a constant, describe the same physical polarization. Then an in or out $A$ field must be physically equivalent to $A+\partial \omega$. @JiaYiyang

@Jia Yiyang 2013-07-18 02:25:44

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

@Diego Mazón 2013-07-18 02:51:51

Yes, if I'm right, the fact that for a non-self-interacting vectorial field an explicit mass term is equivalent to a one generated by a Higgs mechanism (at least for some topologies) is the the responsible of the Ward identity. In that case you must use the $B$ field in my notation (i.e., before fixing the unitary gauge). I think you'd enjoy reading the Ward identity section of L. Brown "Quantum field theory". Then, you will understand why gauge invariance is crucial. @JiaYiyang

@Diego Mazón 2013-07-18 02:57:33

By the way, I enjoy your questions. They are good. @JiaYiyang

@Jia Yiyang 2013-07-18 03:10:57

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

@Jia Yiyang 2013-07-19 06:27:47

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

@Diego Mazón 2013-07-19 06:31:19

@JiaYiyang You're welcome. I'm looking forward to your answer.

@Qmechanic 2013-09-28 19:42:09

More on Stueckelberg formalism in Ron Maimon's Phys.SE answer here.

@Name YYY 2016-12-29 10:35:09

Sorry, but does the presence of Ward identities in massive QED leads to the preserving of the renormalizability?

@Luboš Motl 2013-07-14 05:57:23

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 $k_\mu {\mathcal M}^\mu(k) = 0$ which really says that the longitudinal polarization of the gauge boson, one with the pure-gauge polarization vector proportional to the momentum, $\epsilon_\mu\sim k_\mu$, "decouples" i.e. its interactions (scattering amplitudes) with any collection of physical particles vanish.

This vanishing implies a symmetry – now yes, $k_\mu {\mathcal M}^\mu(k)$ may also be interpreted as a correlator including $\partial_\mu J^\mu$, a conserved current – and this symmetry is a gauge symmetry because the gauge field may only have a nonzero $k^\mu$ i.e. dependence on the spacetime if we allow the symmetry parameter to depend on spacetime.

Fields with an extra $m^2 A^\mu A_\mu$ etc. are no longer coupled to a conserved current because the current is modified by an extra $m^2 A_\mu$ – because this appears as a factor multiplying $A_\mu$ 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).

@Jia Yiyang 2013-07-14 08:55:02

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 $m^2 A^{\mu} A_\mu+A^\mu J_\mu=A_\mu(m^2 A^\mu+J^\mu)$. However, shouldn't $m^2 A^\mu A_\mu$ be interpreted as part of the free Lagrangian rather than the current?

@Diego Mazón 2013-07-16 22:41:40

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.