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?

  • 1
    $\begingroup$ Please, provide more details about your question add the exact equation $\endgroup$ May 1, 2021 at 15:33
  • $\begingroup$ I saw in wikipedia that the klein gordon equation was in the form box(psi)+del(V)/del(psi)=0.why is there a derivative with respect to psi? $\endgroup$ May 1, 2021 at 15:40
  • $\begingroup$ Thnkx for the edit but whats the answer @john rennie $\endgroup$ May 1, 2021 at 16:01
  • $\begingroup$ The field isn’t in a potential. The field has a potential, representing its self-interaction. $\endgroup$
    – G. Smith
    May 1, 2021 at 17:20

1 Answer 1


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

  • $\begingroup$ Thank you so much for your answer but when i try to resuce this equation for non relativistic case which is schrodinger eqn, how do i kill the derivative because theres no derivative term in schrodinger eqn? $\endgroup$ May 1, 2021 at 16:09
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    $\begingroup$ @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 $\phi \partial_t \phi$ $\endgroup$ May 1, 2021 at 16:15
  • $\begingroup$ Thnakx again for your time $\endgroup$ May 1, 2021 at 16:23

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