Consider the Lagrangian:

$$\mathcal{L}~=~-\frac{1}{2}\bar{\phi} \square \phi - \frac{1}{4}F_{\mu \nu}F^{\mu \nu} + \lambda \bar{\phi} \phi F_{\mu \nu}F^{\mu \nu}$$$\hspace{200px}$

The vertex Feynman rule I obtained from this Lagrangian is $$i\lambda \left[(p\cdot k)g^{\mu \nu} - p^{\mu}k^{\nu} \right]$$

where $p^{\mu}$ and $k^{\nu}$ are the photon momenta.

When I consider the 1-loop diagrams of 4-photon interaction with incoming momenta $p_1^{\mu}, p_2^{\nu}$ and outgoing momenta $p_3^{\rho}, p_4^{\sigma}$ , I obtained the corresponding Feynman amplitude:

$$ \begin{alignat}{4} & \Big(\frac{s}{2} & \Big)^2 (g^{\mu \nu}g^{\rho \sigma}) & \int \frac{\mathrm{d}^4k}{\left(2\pi\right)^4}\frac{1}{k^2\left(k+p_s\right)^2} \\ +~ & \Big(\frac{t}{2} & \Big)^2 (g^{\mu \rho}g^{\nu \sigma}) & \int \frac{\mathrm{d}^4k}{\left(2\pi\right)^4}\frac{1}{k^2\left(k+p_t\right)^2} \\ +~ & \Big(\frac{u}{2} & \Big)^2 (g^{\mu \sigma}g^{\rho \nu}) & \int \frac{\mathrm{d}^4k}{\left(2\pi\right)^4}\frac{1}{k^2\left(k+p_u\right)^2} \end{alignat} $$

where $p_s = p_1 + p_2$, $p_t = p_1 - p_3$, and $p_u = p_1 - p_4$.

My problem is, I cannot verify if my answer satisfies the Ward identity, which states that the amplitude should vanish when I replace one of the polarization vector with its momenta.

Is my amplitude wrong or is my Feynman rule wrong or if I'm right and just need to work harder to check?


1 Answer 1


First of all, your Lagrangian has a weird prefactor of the kinetic term, there is usually no $1/2$ when we are dealing with complex scalar fields. Then, I would actually put a $1/2$ in front of the interaction term to cancel the combinatorial prefactors. Then

$$\mathcal{L} = \partial_\mu \phi^\ast \partial^\mu \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu} $$

The Feynman rule for the vertex (all momenta incoming) is then given by

$$ 2 i \lambda (p^\mu k^\nu - g^{\mu \nu} p \cdot k) $$

Considering the process $\gamma \gamma \to \gamma \gamma$ at 1-loop we have 3 diagrams

enter image description here

Their sum (where the photon polarization vectors are truncated) is given by

$$\begin{align} i \mathcal{M}^{\mu \nu \rho \sigma} &= \lambda^2 (s g^{\mu \nu} - 2 p_2^\mu p_1^\nu) (s g^{\rho \sigma} - 2 p_4^\rho p_3^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_s)^2} \\ & + \lambda^2 (t g^{\mu \rho} + 2 p_3^\mu p_1^\rho) (t g^{\nu \sigma} + 2 p_4^\nu p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_t)^2} \\ & + \lambda^2 (u g^{\mu \sigma} + 2 p_4^\mu p_1^\sigma) (u g^{\nu \rho} + 2 p_3^\nu p_2^\rho) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_u)^2} \end{align} $$

Now you can easily recognize that contracting this with $p_1^\mu$, $p_2^\nu$, $p_3^\rho$ or $p_4^\sigma$ yields zero. For example

$$ \begin{align} i \mathcal{M}^{\mu \nu \rho \sigma} p_{1,\mu} &= \lambda^2 (s p_1^\nu - 2 (s/2) p_1^\nu) (s g^{\rho \sigma} - 2 p_4^\rho p_3^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_s)^2} \\ & + \lambda^2 (t p_1^\rho + 2 (-t/2) p_1^\rho) (t g^{\nu \sigma} + 2 p_4^\nu p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_t)^2} \\ & + \lambda^2 (u p_1^\sigma + 2 (-u/2) p_1^\sigma) (u g^{\nu \rho} + 2 p_3^\nu p_2^\rho)\int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_u)^2} = 0 \end{align} $$

Your result does not satisfy the Ward identity, because the amplitude is incorrect. For some reason you have neglected the second piece of the vertex $2 i \lambda (p^\mu k^\nu)$.

  • $\begingroup$ Replace the term with the second order derivative by a square of first order derivatives with a minus sign to implement charge conservation. $\endgroup$
    – my2cts
    Jun 13, 2018 at 8:21
  • $\begingroup$ Check the indices of your vertex rule, and on the last sentence. $\endgroup$
    – ohneVal
    Jun 13, 2018 at 8:57
  • $\begingroup$ @my2cts Perhaps I missed your point, but this Lagrangian is already invariant under C-symmetry, you do not need to "implement" it. Since $$\int d^4 x (- \phi^\ast \Box \phi) = \int d^4 x \partial_\mu \phi^\ast \partial^\mu \phi $$ it is perfectly fine to write the kinetic term this or the other way. $\endgroup$
    – vsht
    Jun 13, 2018 at 9:36
  • $\begingroup$ The lagrangian and its action integral are both manifestly c-variant. The equality is only valid for real $\phi$. $\endgroup$
    – my2cts
    Jun 13, 2018 at 13:28
  • $\begingroup$ Why? $ \partial^\mu (\phi^\ast \partial_\mu \phi) = (\partial^\mu \phi^\ast ) (\partial_\mu \phi) + \phi^\ast \Box \phi $ What does this have to do with $\phi$ being real or complex ? $\endgroup$
    – vsht
    Jun 13, 2018 at 15:03

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