What is the name of this Lagrangian and how can I find its equations of motion? I would appreciate if someone tell me how I should go about finding eom. for the following Lagrangian:$$L=-\frac{1}{2}\phi(\Box + m^2)\phi$$
 A: The Euler-Lagrangian (E-L) equation is
$$
\frac{\partial}{\partial x_\mu}\frac{\partial{\cal L}}{\partial(\partial^\mu\phi)}~-~\frac{\partial{\cal L}}{\partial\phi}~=~0.
$$
For the standard Klein-Gordon field the Lagrangian, ignoring complex valued nature of the field, is
$$
{\cal L}~=~\frac{1}{2}\partial_\mu\phi\partial^\mu\phi~+~\frac{1}{2}m^2\phi^2
$$
the equation of motion is the Klein-Gordon equation $\square\phi~-~m^2\phi~=~0$.
The Lagrangian proposed here is a bit different and this requires a modified E-L equation. The Lagrangian is dependent on the second derivatives of the field or ${\cal L}~=~{\cal L}(\phi,~\partial_\mu\phi,~\partial_\mu\partial_\nu\phi)$. I am not going to reproduce the derivation, but it is a second order generalization of the derivation of the E-L equation from the action. This E-L equation is then
$$
\frac{\partial^2}{\partial x_\mu\partial x_\nu}\frac{\partial^2{\cal L}}{\partial(\partial^\mu\phi)\partial(\partial^\nu\phi)}~+~\frac{\partial}{\partial x_\mu}\frac{\partial{\cal L}}{\partial(\partial^\mu\phi)}~-~\frac{\partial{\cal L}}{\partial\phi}~=~0.
$$.
The calculation is then comparatively simple, where the $\frac{\partial^2{\cal L}}{\partial(\partial^\mu\phi)\partial(\partial^\nu\phi)}$ will eliminate the $\square\phi$ and the second order derivative will then reproduce $\square\phi$. This then returns the Klein-Gordon equation. There is no first order differential of the field in the Lagrangian, so the $\frac{\partial}{\partial x_\mu}\frac{\partial{\cal L}}{\partial(\partial^\mu\phi)}$ returns nothing.
