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The standard proof that photons don't become massive in 4d QED is calculating the photon propagator and checking that it has a pole at momentum $q^2=0$.

The propagator is obtained summing loop diagrams of the type:

enter image description here

enter image description here

But this does not consider more complicated diagrams like:

enter image description here

Is there a way to include all possible diagrams? Can the proof be extended to include the rest of the EW sector and QCD?

Ward identity is often mentioned as a proof of massless photons, but it seems that it only indicates transverse photons.

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  • $\begingroup$ Based on the fact that the PDG book includes photon mass searches, I wold expect the answer is no. But that's just a guess. $\endgroup$ – probably_someone Jun 15 '17 at 10:19
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    $\begingroup$ The search for photon mass is for the eventuality of the QED model being wrong. $\endgroup$ – Slereah Jun 15 '17 at 10:21
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    $\begingroup$ "But this does not consider more complicated diagrams like:" Yes it does. The pole at $p^2=0$ corresponds to the one-particle irreducible function, which includes your "more complicated diagram" as well. In fact, and as the name indicates, the one-particle irreducible function includes all one-particle irreducible diagrams. In the general case (non-abelian gauge theories), you need Slavnov-Taylor instead of Ward, but the philosophy is the same $\endgroup$ – AccidentalFourierTransform Jun 15 '17 at 10:32
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    $\begingroup$ There is an equivalence between being massless and being a gauge field with unbroken gauge symmetry. So the Ward identity is proof that the photon remains massless - the photon acquiring mass is equivalent to the gauge symmetry being broken is equivalent to the Ward identity being violated. Is that an acceptable answer or do you really want to do diagrammatics here? $\endgroup$ – ACuriousMind Jun 15 '17 at 11:12
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    $\begingroup$ @ACuriousMind I've heard that statement, but how is that compatible with 2d being massive yet gauge invariance is preserved? $\endgroup$ – jinawee Jun 15 '17 at 11:14
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There is a prove in the lecture notes 12 of Relativistic Quantum Field Theory II from MIT OCW based on functional method. I will outline the prove here. The exact propagator of photon is $$\mathcal{G}(x)_{\mu\nu} = \langle \Omega | T A_{\mu}(x)A_{\nu}(0)| \Omega \rangle_C.$$ It can be represented by the following diagram enter image description here Let us define $i\Pi^{\mu\nu}$ to be the sum of all 1-particle-irreducible insertions into the photon propagator. So, we have $$\mathcal{G}(k) = G_{\rm F}(k) + G_{\rm F}(k)(i\Pi(k))G_{\rm F}(k) + \cdots = G_{\rm F}(k) \frac{1}{1-i\Pi(k)G_{\rm F}(k)}.$$ $G_{\rm F}(p)_{\mu\nu}$ is the free propagator of photon and so we have $$iG_{\rm F}(p)_{\mu\nu} = \frac{\eta_{\mu\nu}}{k^2-i\epsilon} - (1-\xi)\frac{k_{\mu}k_{\nu}}{(k^2-i\epsilon)^2} = \frac{1}{k^2-i\epsilon}(P^T_{\mu\nu} + \xi P^L_{\mu\nu}),$$ where $$P^T_{\mu\nu} \equiv \eta_{\mu\nu} - \frac{k_{\mu}k_{\nu}}{k^2}, \quad P^L_{\mu\nu} \equiv \frac{k_{\mu}k_{\nu}}{k^2}.$$ ($\xi = 1$ is the so-called Feynman gauge)

It is easy to derive that $$(iG_{\rm F})^{-1}_{\mu\nu} = k^2 (P^T_{\mu\nu} + \frac{1}{\xi} P^L_{\mu\nu}).$$ We may also decompose $i\Pi^{\mu\nu}$ as $$\Pi^{\mu\nu} = P_T^{\mu\nu}f_T(k^2) + P_L^{\mu\nu}f_L(k^2) = \eta^{\mu\nu}f_T + \frac{k^{\mu}k^{\nu}}{k^2}(f_L-f_T)$$ Therefore, $$(i\mathcal{G})^{-1}_{\mu\nu} = (k^2-f_T(k^2))P^T_{\mu\nu} + (\frac{k^2}{\xi}-f_L(k^2)) P^L_{\mu\nu},$$ $$\mathcal{G}(k)_{\mu\nu} = \frac{-i}{k^2-f_T(k^2)}P^T_{\mu\nu} + \frac{-i}{\frac{k^2}{\xi}-f_L(k^2)} P^L_{\mu\nu}.$$ If $f_{T,L}(k^2 = 0) \neq 0$, a mass will be generated for the photon. Because $\Pi(k)$ comes from 1PI diagrams, it should not be singular at $k^2 =0 $, and so $f_L - f_T = O(k^2)$, as $k \to 0$.


We define the generating functional $E[J,\eta,\overline{\eta}]$ for connected diagrams in QED by $$Z[J,\eta,\overline{\eta}] = e^{-iE[J,\eta,\overline{\eta}]}$$ So, $$\mathcal{G}(x-y)_{\mu\nu} = i \frac{\delta^2 E[J,\eta,\overline{\eta}]}{\delta J^{\mu}(x) \delta J^{\nu}(y)}\bigg|_{J,\eta,\overline{\eta}=0}$$ For infinitesimal gauge transformations, we have $\delta A_{\mu} = \partial_{\mu} \lambda $, $\delta \Psi = ie_0\lambda\Psi$ and $\delta \overline{\Psi} = -ie_0 \lambda \overline{\Psi}$. Under a change of variables in the path integral, $Z[J,\eta,\overline{\eta}]$ will remain the same. Recall that $$Z[J,\eta,\overline{\eta}] = \int \mathcal{D}A \mathcal{D}\overline{\Psi} \mathcal{D}\Psi e^{i\int d^4x [\mathcal{L} + JA + \overline{\eta}\Psi + \overline{\Psi}\eta]} $$ where $$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \overline{\Psi} (i\gamma^{\mu}\partial_{\mu}-m_0) \Psi + e_0j^{\mu} A_{\mu} - \frac{1}{2\xi}(\partial_{\mu}A^{\mu})^2$$

The change of action is $$\delta S = -\frac{1}{\xi} \int d^4x \partial_{\mu} A^{\mu} \partial^2 \lambda + \int d^4x J^{\mu}\partial_{\mu}\lambda + ie_0\overline{\eta}\Psi\lambda - ie_0\overline{\Psi}\eta\lambda$$

Hence, we must have $$\int d^4x \lambda(x) \int \mathcal{D}A \mathcal{D}\overline{\Psi} \mathcal{D}\Psi e^{iS} \left[ -\frac{1}{\xi} \partial^2 \partial_{\mu} A^{\mu} - \partial_{\mu}J^{\mu} + ie_0(\overline{\eta}\Psi - \overline{\Psi}\eta)\right] = 0 $$ Since $$\langle A_{\mu}(x) \rangle_{J,\eta,\overline{\eta}} = - \frac{\delta E}{\delta J^{\mu}} \quad \langle \Psi(x) \rangle_{J,\eta,\overline{\eta}} = - \frac{\delta E}{\delta \overline{\eta}} \quad \langle \overline{\Psi}(x) \rangle_{J,\eta,\overline{\eta}} = \frac{\delta E}{\delta \eta}$$ The above equation can be written as $$\frac{1}{\xi} \partial^2 \partial^{\mu}\frac{\delta E}{\delta J^{\mu}} - \partial_{\mu}J^{\mu} - ie_0\left[ \overline{\eta}\frac{\delta E}{\delta \overline{\eta}} + \frac{\delta E}{\delta \eta} \eta \right]=0$$ By differentiation with $\delta J$ at $J,\eta,\overline{\eta} = 0$, we can get $$\frac{1}{\xi} \partial^2 \partial^{\mu} \frac{\delta^2 E[J,\eta,\overline{\eta}]}{\delta J^{\mu}(x) \delta J^{\nu}(y)}\bigg|_{J,\eta,\overline{\eta}=0} - \partial_{\nu} \delta(x-y) = 0$$ that is, $$\frac{i}{\xi}\partial^2 \partial^{\mu} \mathcal{G}(x-y)_{\mu\nu}+ \partial_{\nu} \delta(x-y) = 0 $$ or, written in momentum-space, $$-\frac{i}{\xi}k^2 k^{\mu} \mathcal{G}(k)_{\mu\nu}+ k_{\nu} = 0$$ So $$- \frac{k^2}{k^2-\xi f_L(k^2)} k_{\nu} + k_{\nu} = 0$$ Which means $f_L(k^2) =0$ and so, we have $f_T(k^2) \to O(k^2)$ as $k^2 \to 0$. The exact propagator of photon is $$\mathcal{G}(k)_{\mu\nu} = \frac{-i}{k^2(1-\pi(k^2))}P^T_{\mu\nu} + \frac{-i\xi}{k^2} P^L_{\mu\nu}$$ where $\pi(k^2) \equiv \frac{f_T(k^2)}{k^2}$. The exact propagator has a pole at $k^2=0$, so photon remain massless after quantum correction.

The discussion concerning on QCD corrections is beyond my knowledge and I am looking forward to a better answer.

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