4 added 19 characters in body edited Jun 14 '18 at 5:10 vsht 16144 bronze badges 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$$$\mathcal{L} = \partial_\mu \phi^\ast \partial^\mu \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ 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 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 $$2 i \lambda (p^\mu k^\nu)$$. 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ 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 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 $$2 i \lambda (p^\mu k^\nu)$$. 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 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 $$2 i \lambda (p^\mu k^\nu)$$. 3 another typo edited Jun 13 '18 at 9:28 vsht 16144 bronze badges 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ The Feynman rule for the vertex (all momenta incoming) is then given by $$2 i \lambda (p^\mu k^\mu - g^{\mu \nu} p \cdot k)$$$$2 i \lambda (p^\mu k^\nu - g^{\mu \nu} p \cdot k)$$ Considering the process $$\gamma \gamma \to \gamma \gamma$$ at 1-loop we have 3 diagrams 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 $$2 i \lambda (p^\mu k^\mu)$$$$2 i \lambda (p^\mu k^\nu)$$. 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ The Feynman rule for the vertex (all momenta incoming) is then given by $$2 i \lambda (p^\mu k^\mu - g^{\mu \nu} p \cdot k)$$ Considering the process $$\gamma \gamma \to \gamma \gamma$$ at 1-loop we have 3 diagrams 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 $$2 i \lambda (p^\mu k^\mu)$$. 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ 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 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 $$2 i \lambda (p^\mu k^\nu)$$. 2 typos edited Jun 13 '18 at 8:14 vsht 16144 bronze badges 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ The Feynman rule for the vertex (all momenta incoming) is then given by $$2 i \lambda (p^\mu k^\mu - g^{\mu \nu} p \cdot k)$$ Considering the process $$\gamma \gamma \to \gamma \gamma$$ at 1-loop we have 3 diagrams 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 \nu} - 2 p_3^\mu p_1^\nu) (s g^{\rho \sigma} - 2 p_4^\rho p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_t)^2} \\ & + \lambda^2 (u g^{\mu \nu} - 2 p_4^\mu p_1^\nu) (s g^{\rho \sigma} - 2 p_3^\rho p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_u)^2} \end{align}\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^\nu - 2 (t/2) p_1^\nu) (s g^{\rho \sigma} - 2 p_4^\rho p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_t)^2} \\ & + \lambda^2 (u p_1^\nu - 2 (u/2) p_1^\nu) (s g^{\rho \sigma} - 2 p_3^\rho p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_u)^2} = 0 \end{align}\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 $$2 i \lambda (p^\mu k^\mu)$$. 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ The Feynman rule for the vertex (all momenta incoming) is then given by $$2 i \lambda (p^\mu k^\mu - g^{\mu \nu} p \cdot k)$$ Considering the process $$\gamma \gamma \to \gamma \gamma$$ at 1-loop we have 3 diagrams 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 \nu} - 2 p_3^\mu p_1^\nu) (s g^{\rho \sigma} - 2 p_4^\rho p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_t)^2} \\ & + \lambda^2 (u g^{\mu \nu} - 2 p_4^\mu p_1^\nu) (s g^{\rho \sigma} - 2 p_3^\rho p_2^\sigma) \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^\nu - 2 (t/2) p_1^\nu) (s g^{\rho \sigma} - 2 p_4^\rho p_2^\sigma) \int \frac{d^d k}{(2 \pi)^d} \frac{1}{k^2 (k+ p_t)^2} \\ & + \lambda^2 (u p_1^\nu - 2 (u/2) p_1^\nu) (s g^{\rho \sigma} - 2 p_3^\rho p_2^\sigma) \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 $$2 i \lambda (p^\mu k^\mu)$$. 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 $$\mathcal{L} = - \phi^\ast \Box \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$ The Feynman rule for the vertex (all momenta incoming) is then given by $$2 i \lambda (p^\mu k^\mu - g^{\mu \nu} p \cdot k)$$ Considering the process $$\gamma \gamma \to \gamma \gamma$$ at 1-loop we have 3 diagrams 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 $$2 i \lambda (p^\mu k^\mu)$$. 1 answered Jun 13 '18 at 8:08 vsht 16144 bronze badges