Feynman rule of $\bar{\phi} \phi F_{\mu \nu}F^{\mu \nu}$ and its corresponding 4-photon scattering amplitude - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-21T17:28:11Z https://physics.stackexchange.com/feeds/question/406455 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/406455 3 Feynman rule of $\bar{\phi} \phi F_{\mu \nu}F^{\mu \nu}$ and its corresponding 4-photon scattering amplitude LY3000 https://physics.stackexchange.com/users/106228 2018-05-17T14:03:56Z 2018-06-14T05:10:24Z <p>Consider the Lagrangian:</p> <p>$$\mathcal{L}~=~-\frac{1}{2}\bar{\phi} \square \phi - \frac{1}{4}F_{\mu \nu}F^{\mu \nu} + \lambda \bar{\phi} \phi F_{\mu \nu}F^{\mu \nu}$$$\hspace{200px}$<a href="https://i.stack.imgur.com/evAeI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/evAeI.png" width="200"/></a></p> <p>The vertex Feynman rule I obtained from this Lagrangian is $$i\lambda \left[(p\cdot k)g^{\mu \nu} - p^{\mu}k^{\nu} \right]$$</p> <p>where $p^{\mu}$ and $k^{\nu}$ are the photon momenta.</p> <p>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:</p> <p>\begin{alignat}{4} &amp; \Big(\frac{s}{2} &amp; \Big)^2 (g^{\mu \nu}g^{\rho \sigma}) &amp; \int \frac{\mathrm{d}^4k}{\left(2\pi\right)^4}\frac{1}{k^2\left(k+p_s\right)^2} \\ +~ &amp; \Big(\frac{t}{2} &amp; \Big)^2 (g^{\mu \rho}g^{\nu \sigma}) &amp; \int \frac{\mathrm{d}^4k}{\left(2\pi\right)^4}\frac{1}{k^2\left(k+p_t\right)^2} \\ +~ &amp; \Big(\frac{u}{2} &amp; \Big)^2 (g^{\mu \sigma}g^{\rho \nu}) &amp; \int \frac{\mathrm{d}^4k}{\left(2\pi\right)^4}\frac{1}{k^2\left(k+p_u\right)^2} \end{alignat}</p> <p>where $p_s = p_1 + p_2$, $p_t = p_1 - p_3$, and $p_u = p_1 - p_4$.</p> <p>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.</p> <p>Is my amplitude wrong or is my Feynman rule wrong or if I'm right and just need to work harder to check?</p> https://physics.stackexchange.com/questions/406455/feynman-rule-of-bar-phi-phi-f-mu-nuf-mu-nu-and-its-corresponding-4/411553#411553 1 Answer by vsht for Feynman rule of $\bar{\phi} \phi F_{\mu \nu}F^{\mu \nu}$ and its corresponding 4-photon scattering amplitude vsht https://physics.stackexchange.com/users/92787 2018-06-13T08:08:13Z 2018-06-14T05:10:24Z <p>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</p> <p>$$\mathcal{L} = \partial_\mu \phi^\ast \partial^\mu \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \lambda \phi^\ast \phi F^{\mu \nu} F_{\mu \nu}$$</p> <p>The Feynman rule for the vertex (all momenta incoming) is then given by</p> <p>$$2 i \lambda (p^\mu k^\nu - g^{\mu \nu} p \cdot k)$$</p> <p>Considering the process $\gamma \gamma \to \gamma \gamma$ at 1-loop we have 3 diagrams</p> <p><a href="https://i.stack.imgur.com/qPPvJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qPPvJ.png" alt="enter image description here"></a></p> <p>Their sum (where the photon polarization vectors are truncated) is given by</p> <p>\begin{align} i \mathcal{M}^{\mu \nu \rho \sigma} &amp;= \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} \\ &amp; + \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} \\ &amp; + \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}</p> <p>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 </p> <p>\begin{align} i \mathcal{M}^{\mu \nu \rho \sigma} p_{1,\mu} &amp;= \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} \\ &amp; + \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} \\ &amp; + \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}</p> <p>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 $2 i \lambda (p^\mu k^\nu)$.</p>