# Is the 1PI self-energy of a massive photon transverse? EDIT: Upsetting consequences for the photon mass and renormalizability

Suppose we had the Lagrangian:

$$\mathcal{L} = -\frac{1}{4} F^{\mu \nu}F_{\mu \nu} + \overline{\psi} (i \gamma^{\mu}\partial_{\mu} -m)\psi -e\overline{\psi} \gamma_{\mu} \psi A^{\mu} +\frac{1}{2} m_{\gamma}^2 A_{\mu} A^{\mu},$$

which is just the Lagrangian of QED with a massive photon in the $$(+,-,-,-)$$ sign convention. The lagrangian does not have a local $$U(1)$$ symmetry but it does have global $$U(1)$$ symmetry, therefore the Ward identity for global symmetries holds.

Let us consider the correlator:

$$\mathcal{\Pi}^{R}_{\alpha \beta} (x) \equiv $$,

In Feynman diagrams this correlator is the very much reducible full self-energy of the photon.

By the Ward identity we have:

$$\partial^{\alpha}\mathcal{\Pi}^{R}_{\alpha \beta} (x) = -i\delta^4 (x) <\delta j(0)> = 0$$, the last passage is by the invariance under global $$U(1)$$ of the current. In Fourier space:

$$q^{\alpha}\mathcal{\Pi}^{R}_{\alpha \beta} (q) = 0 \; \; \; \; \; (1)$$

One can the write the fully dressed photon propagator as:

$$G^{\mu \nu} (q) = G_{0}^{\mu \nu} (q) + G_{0}^{\mu \alpha} (q) \Pi^{R}_{\alpha \beta} (q) G_{0}^{\beta \nu} (q) \; \; \; \; \; \; (2)$$

The explicit form of the bare photon propagator is:

$$G_0^{\mu \nu} (q) = \frac{-i \left( g^{\mu \nu} - \frac{q^{\mu} q^{\nu}}{m_{\gamma}^2} \right)}{q^2-m_{\gamma}^2+i \varepsilon}$$.

By contracting with $$q_{\mu}$$ both sides of equation $$(2)$$ we get:

$$q_{\mu} G^{\mu \nu} (q) = q_{\mu} G_0^{\mu \nu} (q) + \frac{-i \left( 1 - \frac{q^2}{m_{\gamma}^2} \right)}{q^2-m_{\gamma}^2+i \varepsilon} q^{\alpha} \mathcal{\Pi}^{R}_{\alpha \beta} (q) G_0^{\beta \nu} (q)$$,

and by $$(1)$$ the second term on the LHS is $$0$$, therefore:

$$q_{\mu} G^{\mu \nu} (q) = q_{\mu} G_0^{\mu \nu} (q) \; \; \; \; \; (3)$$.

But since by definition we have (calling $$\mathcal{\Pi}_{\alpha \beta} (q)$$ the 1PI photon self-energy):

$$G^{\mu \nu} (q) = G^{\mu \nu}_0 (q) + G^{\mu \alpha}_0 (q) \mathcal{\Pi}_{\alpha \beta} (q) G^{\beta \nu} (q)$$. By contracting both sides with $$q_{\mu}$$ we get:

$$q_{\mu} G^{\mu \nu} (q) = q_{\mu} G_0^{\mu \nu} (q) + q_{\mu} G^{\mu \alpha}_0 (q) \mathcal{\Pi}_{\alpha \beta} (q) G^{\beta \nu} (q)$$,

and by (3) we have:

$$q_{\mu} G^{\mu \alpha}_0 (q) \mathcal{\Pi}_{\alpha \beta} (q) G^{\beta \nu} (q) = 0$$, so that using the explicit expression of $$G_0^{\mu \alpha}$$ we get:

$$q^{\alpha} \mathcal{\Pi}_{\alpha \beta} (q) G^{\beta \nu} (q) = 0$$. Therefore we conclude:

$$q^{\alpha} \mathcal{\Pi}_{\alpha \beta} (q) = 0$$.

This looks so strange to me. Why should the photon self-energy be transverse even without gauge invariance? Is my math wrong somewhere?

EDIT: Something must be wrong. I tried developing a bit more the math to see where it would lead and I got some upsetting results. If one has $$q_{\mu} \mathcal{M}^{\mu...}=0$$, where $$\mathcal{M}^{\mu}$$ is any amplitude with on-shell fermions and one off-shell photon with 4-momentum $$q$$, then one can disregard the term proportional to $$q^{\mu}q^{\nu}$$ in the bare propagator. This makes the theory seemingly renormalizable. Also, for the same reason, one can write the full photon propagator as:

$$G^{\mu \nu} = \frac{-ig^{\mu \nu}}{(q^2 - m_{\gamma}^2) \left[1-\Pi(q^2)\right]}$$, which would mean that the photon mass is protected from renormalization, which is even weirder.

• Related post by OP: physics.stackexchange.com/q/689809/2451 Related: physics.stackexchange.com/q/70882/2451 Feb 25 at 1:19
• I can only offer you my solidarity, i have been working exactly on this kind of questions about massive QED to understand the role of gauge invariance in deriving certain properties (such as Z1=Z2). But I found the same results for massive and massless QED. I must have made some errors, or maybe I should understand better the differnce between primary and secondary constraint of QED. Feb 28 at 10:04
• Maybe @Qmechanic can help both of us if he has a few minutes. I kinda rely on him a lot on this website :) Feb 28 at 10:40
• Feb 28 at 14:41

Here is one line of reasoning:

1. We can incorporate the Proca/massive photon field $$A_{\mu}$$ into a gauge theory via the Stuckelberg mechanism$$^1$$ $${\cal L}_S~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} -\frac{1}{2}(m_{\gamma} A^{\mu} +\partial^{\mu}\phi) (m_{\gamma} A_{\mu}+\partial_{\mu}\phi).$$ The fermionic matter is governed by the Dirac Lagrangian density \begin{align} {\cal L}_D~=~&\bar{\psi}(i\gamma^{\mu}D_{\mu}-m_e)\psi,\cr D_{\mu} ~=~&\partial_{\mu}-ieA_{\mu} .\end{align}

2. It is possible to choose a $$R_{\xi}$$-gauge-fixing condition $$G~=~\partial^{\mu}A_{\mu} + \xi m_{\gamma}\phi$$ so that the Stückelberg scalar $$\phi$$ becomes free (but massive). In detail, the gauge-fixed Lagrangian density is $${\cal L}_{GF}~=~-\frac{G^2}{2\xi},$$ while the Faddeev-Popov (FP) term reads $${\cal L}_{FP}~=~\bar{c}\left(\partial^2 -\xi m_{\gamma}^2 \right)c.$$ The total gauge-fixed Lagrangian density becomes$$^2$$ \begin{align}{\cal L}~=~&{\cal L}_S+{\cal L}_D+{\cal L}_{FP}+{\cal L}_{GF}\cr ~\sim~&\frac{1}{2}A_{\mu}\left(\delta^{\mu}_{\nu} (\partial^2-m_{\gamma}^2)+ (\frac{1}{\xi}-1)\partial^{\mu} \partial_{\nu} \right)A^{\nu}\cr ~+~&\frac{1}{2}\phi(\partial^2-\xi m_{\gamma}^2)\phi +{\cal L}_D+{\cal L}_{FP} . \end{align} Note in particular that the fermionic matter $$\psi$$ only couples to $$A_{\mu}$$, not to $$\phi.$$ In fact the $$\bar{\psi}A_{\mu}\psi$$ cubic vertex is the only interaction in the entire model! The Faddeev-Popov (FP) ghosts $$c$$ and $$\bar{c}$$ also decouple.

3. Then we can in principle construct the corresponding Ward identities for 1-particle irreducible (1PI) correlator functions, and show renormalizability. Explicit calculations show that the simplest such Ward identity confirms the transversality of the massive photon vacuum polarization tensor/self-energy, cf. OP's title question.

$$^1$$ This answer uses the $$(-,+,+,+)$$ sign convention.
$$^2$$ The $$\sim$$ symbol means equality up to total derivative terms.