In this other SE post: Is it really proper to say Ward identity is a consequence of gauge invariance? it is shown that the on-shell Ward identity is a consequence of global $U(1)$ symmetry for QED. This is very logical because the Ward identity is a consequence of the global symmetry and the contact terms are zero once you use the LSZ formula to extract the S-matrix matrix element. But it is also important to know how the the Ward identity behaves off-shell. For example when we consider the self-energy of a photon we know that

$$q_{\mu} \Pi^{\mu \nu} (q) = 0 \tag{1}$$

even if $q$ is off-shell. This is what protects the mass of the photon after renormalization.

Let us therefore write the Ward identity in the most general case:

$$\begin{equation} \label{eq1} \partial_{\mu} \langle T j^{\mu} (x) F_{1} (x_1)...F_{N} (x_n)\rangle = -i \sum_{j=1}^{n} \delta^4 (x-x_i) \langle TF_{1} (x_1)...\delta F_{j}(x_j)...F_{N} (x_n)\rangle,\tag{2} \end{equation}$$

where $F$ can be either a fermion field $\psi$ or a photon field $A^{\mu}$. Let us consider the case that for every fermion field there is an antifermion field. This is what happens in all diagrams since fermion fields inside Green's function are appear in couples inside currents. By using global $U(1)$ we get that the variations of photon fields are automatically zero, every fermion brings a $+i\theta$, and every antifermion brings a $-i\theta$, therefore the right hand side of Ward's identity is zero. This implies that even off-shell $q_{\mu} \Pi^{\mu \nu} (q)$. Therefore it is not gauge symmetry that protects the mass of the photon but global symmetry. If this is true it would imply that QED need not be a gauge theory to be renormalizable and we can add a mass term for the photon: $ \mathcal{L}_{\gamma mass}=\frac{1}{2} m_{\gamma}^2 A^{\mu}A_{\mu} $ that breaks gauge symmetry but preserves global symmetry. Also, because of the Ward identity the photon propagator after renormalization would be:

$$\begin{equation} D^{\mu \nu} = \frac{-i g^{\mu \nu}}{(q^2-m_{\gamma}^2)\left(1+ \Pi(q^2)\right)},\tag{3} \end{equation}$$

where $\Pi^{\mu \nu} (q) \equiv (g^{\mu \nu} q^2-q^{\mu} q^{\nu}) \Pi(q^2)$. This would imply that $U(1)$ symmetry preserves the mass of the photon even if $m_{\gamma} \neq 0$. All of this looks extremely suspicious to me and I am pretty sure I must have made a mistake somewhere in the derivation of the off-shell Ward identity but I can't tell where, and why assuming gauge symmetry would fix such mistake.

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    $\begingroup$ I don't understand how you conclude that the RHS is zero? A fermion does pull down a $+i\theta$ at $x_i$ and an anti-fermion pulls down $-i\theta$ at $x_j$ for $i \neq j$. How do these two cancel? $\endgroup$
    – Prahar
    Commented Jan 20, 2022 at 18:55
  • $\begingroup$ @Prahar precisely considering global transformations. $\theta$ is not a function of spacetime but a constant (this is also the hypothesis we use to derive Ward's identity). Therefore at $x_i$ and at $x_j$ we have the same $\theta$. $\endgroup$ Commented Jan 20, 2022 at 19:48
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    $\begingroup$ The $x_i$ dependence is coming from the Dirac delta function, NOT from the $\theta$. For instance, given a single fermion and anti-fermion, the LHS takes the form $\theta [ \delta^4(x-x_1) - \delta^4(x-x_2) ]< \psi(x_1) {\bar \psi}(x_2) >$ which is NOT zero. $\endgroup$
    – Prahar
    Commented Jan 20, 2022 at 19:50
  • $\begingroup$ @Prahar, could you please reply with this same statement to my question and add how assuming gauge invariance would make Ward's identity work off-shell so I can accept your answer? Thanks :) $\endgroup$ Commented Jan 20, 2022 at 19:53

1 Answer 1

  1. The Ward identities (2) for connected$^1$ correlators (with appropriate contact terms, derived via the Schwinger-Dysons equations for a global symmetry) hold off-shell, cf. Ref. 1.

  2. However, the transversality (1) of the photon 1-particle irreducible (1PI) vacuum polarization/self-energy is a consequence of local gauge symmetry (and appropriate class of gauge-fixing conditions), as explained in e.g. this and this Phys.SE posts.


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 9.6.


$^1$ The Ward identity (2) is originally for not-necessarily-connected correlators, but one can identify connected components, cf. the linked cluster theorem, to make it a statement about connected correlators.

  • $\begingroup$ Thank you for your kind reply, but while I see (for example in the first post) that they prove $k_{\mu} \mathcal{M}^{\mu} =0$ for amplitudes, I don't see where they prove that gauge invariance is needed to prove the transversality of $\Pi^{\mu \nu} (q)$. Furthermore my argument seems to work well for the photon self energy. By Ward identity and global $U(1)$ symmetry: $\partial_{\mu} \Pi^{\mu \nu} (x, 0) = \partial_{\mu} <T j^{\mu} (x) j^{\nu}(0)> = -i \delta(x) <T \delta j^{\nu}(0)> = 0$, therefore in $k$-space: $k_{\mu} \Pi^{\mu \nu}=0$. $\endgroup$ Commented Jan 21, 2022 at 11:16
  • $\begingroup$ One must distinguish between connected and 1PI correlators. $\endgroup$
    – Qmechanic
    Commented Jan 21, 2022 at 11:23
  • $\begingroup$ Could you please elaborate? Isn't $<T j^{\mu}(x) j^{\nu} (0) >$ the connected 1PI self-energy of the photon? I expect all disconnected parts of this matrix element to be zero. $\endgroup$ Commented Jan 21, 2022 at 11:31
  • $\begingroup$ $\uparrow$ It is not 1PI. $\endgroup$
    – Qmechanic
    Commented Jan 21, 2022 at 11:34
  • $\begingroup$ Sorry if I insist but which 1PR diagrams are generated by the expansion in Feynman diagrams? Because for example Peskin and Schroeder in formula (18.88) (page 617) write exactly $\Pi^{\mu \nu} = <T j^{\mu} j^{\nu} >$. $\endgroup$ Commented Jan 21, 2022 at 11:44

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