# Why is there a derivative with respect to psi of potential there in klein gordon equation?

In the Wikipedia article on the Klein Gordon equation there is a section titled Klein–Gordon equation in a potential that gives the equation for a field $$\psi$$ in potential $$V$$ as:

$$\Box\psi + \frac{dV}{d\psi} = 0$$

Why is there a derivative of the potential in this equation? What does it mean?

Because this is the way, the equation of motion is derived from the action principle. You have some action functional and look for the stationary points of this functional, such that for any infinitesimal in some sense function $$\delta \phi$$, satisfying certain conditions, like differentiability up to given order, and vanishing on the boundaries of the integration domain.
More precisely, action for a Klein-Gordon theory with some potential is: $$S = \int d^{D} x \left(\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - V(\phi) \right)$$ We look for the stationary points of the functional, such that for $$\phi + \delta \phi$$, variation vanishes in the first order: $$\delta S = \int d^{D} x \left(\partial_\mu \delta \phi \partial^\mu \phi - \frac{\partial V(\phi)}{\partial \phi} \delta \phi \right)$$ Here note, that the leading order term in $$V(\phi + \delta \phi) - V(\phi)$$ is proportional to derivative of $$V(\phi)$$ with respect to field $$\phi$$. Then after integration by parts, one has: $$\delta S = \int d^{D} x \left( \partial_\mu \partial^\mu \phi + \frac{\partial V(\phi)}{\partial \phi} \right) \delta \phi$$ And in order for this variation $$\delta S$$ to be zero for all admissible variation of $$\phi$$, the following equation has to hold: $$\partial_\mu \partial^\mu \phi + \frac{\partial V(\phi)}{\partial \phi} = 0$$ Which is the Klein-Gordon equation with potential
• @sayanpurkayastha Klein-Gordon equation is essentially relativistic. In order to get non-relativistic equation, one has to replace $\partial_0 \phi \partial^0 \phi$ by $\phi \partial_t \phi$ May 1, 2021 at 16:15