When considering the polarization vectors of a massive spin-1 field, like an $A_\mu$ with Lagrangian density $$ \tag{A} \mathscr{L} = - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}M^2 A_\mu A^\mu,$$ we expand the solutions of $A_\mu$ as $$ \tag{B} A^\mu(x) = \sum_\lambda \int d \tilde{k} [ \varepsilon^{\mu}(\lambda,k) a(\lambda,k) e^{-ikx} + \varepsilon^{\mu}(\lambda,k)^* a^\dagger(\lambda,k) e^{ikx} ], $$ where the three polarization vectors $\{ \varepsilon(\lambda,k) \}_\lambda$ satisfy the completeness relations $$ \tag{C} \sum_\lambda \varepsilon^\mu(\lambda,k) \varepsilon^{\nu*}(\lambda,k) = - \eta^{\mu\nu} + \frac{k^\mu k^\nu}{M^2}. $$

The questions:

  1. Where does the second term $k^\mu k^\nu/M^2$ come from?
  2. Why is it needed?
  3. Using (C) and going to the massless $M \to 0$ limit we get a singularity, which would make the relation ill defined. So how should the massless limit be handled?
  4. In what other circumstances (massive/massless spin-X fields) is a similar correction to the completeness relations needed?

The above formulas can be found for example in Srednicki, (85.16), with a somewhat different notation (he uses opposite Minkowski space metric convention than me).

  • $\begingroup$ the RHS is the most general symmetric 2-tensor you can write using the metric and k. it turns out that -1 and $1/M^2$ are the correct coefficients. $\endgroup$
    – Phoenix87
    Jan 9, 2015 at 14:17

1 Answer 1

  1. Observe that the "completeness" relation is just the projector onto the space spanned by the polarisation vectors. A priori, we know that $P_\epsilon^{\mu\nu} := \sum_\lambda \epsilon^\mu_\lambda(k)\epsilon^{*\nu}_\lambda(k)$ therefore projects onto the subspace orthogonal to the momentum, since $\zeta^\mu k_\mu = 0$ for any polarization vector $\zeta^\mu$. The most general symmetric 2-tensor depending on only the metric and the momentum is $P^{\mu\nu}_\epsilon = a\eta^{\mu\nu} + b k^\mu k^\nu$ with $a,b \in \mathbb{C}$ initially arbitrary. Applying the projector to the momentum yields $P^{\mu\nu}_\epsilon k_\mu = a k_\nu + b k^2 k_\nu \overset{!}{=} 0$, and using $k^2 = m^2$, the relation you quote is obtained.

  2. Because it is the condition that the vector is a polarization, and not any vector.

  3. The massless limit of the Proca action is indeed ill-defined. One can use the Stückelberg Lagrangian with an auxiliary scalar field $\phi$, which has a smooth massless limit as well as explicit gauge invariance for massive $\mathrm{U}(1)$ fields. (This is why massive photons are not a problem in QED)

$$ L = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} + \frac{1}{2}(\partial^\mu \phi + m A^\mu)(\partial_\mu \phi + m A_\mu)$$

  1. You need such a modification every time you have unphysical degrees of freedom, as we have with polarizations $\sim k$ in this case.
  • $\begingroup$ So, it is right to say that this condition (the (C) I mentioned) is not to be proven but rather to be imposed to ensure that polarization vectors used in the expansion of $A_\mu$ are such that the EOM $$ \left[ \square \eta^{\mu\nu} - (1-1/\xi) \partial^\nu \partial^\mu \right] A_\mu(x) $$ hold? Or in other words, they project onto the space of 4-vectors $\varepsilon^\mu$ satisfying $$ \left[ -k^2 \eta^{\mu\nu} + (1-1/\xi) k^\mu k^\nu \right] \varepsilon_\mu = 0 $$ $\endgroup$
    – glS
    Jan 10, 2015 at 16:35
  • $\begingroup$ @glance: They project onto the space of 4-vectors orthogonal to $k$. The restriction is due to the fact that the Proca equation implies the constraint $\partial_\mu A^\mu = 0$, which must be implemented on the Hilbert space just like the gauge condition in Gupta-Bleuler quantization for gauge bosons. In that sense it is indeed imposed to ensure that the $\epsilon$ span only the space of the physical states. $\endgroup$
    – ACuriousMind
    Jan 10, 2015 at 16:51
  • $\begingroup$ Yes, sorry, I mixed massive and massless cases. In the massless case its the same thing though, right? We still impose a Lorentz condition and thus require $k^\mu \varepsilon_\mu = 0$, while the additional gauge freedom comes from the identification of all polarization vectors differing by a multiple of the momentum 4-vector: $\varepsilon_\mu(k) \approx \varepsilon_\mu(k) + \alpha k_\mu$. Is this correct? $\endgroup$
    – glS
    Jan 10, 2015 at 17:57
  • 2
    $\begingroup$ @glance: Yes, that is correct. Dividing out the spurious states corresponding to multiples of the 4-momentum for massless vector bosons follows from the residual gauge symmetry/Ward identities, and we do not have these for massive fields. $\endgroup$
    – ACuriousMind
    Jan 10, 2015 at 18:07

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