The problem statement, all variables and given/known data Consider the lagrangian [tex] L=\delta_\mu \phi \delta^\mu \phi^* - m^2 \phi \phi^* [/tex] Show that the transformation: [tex] \phi \rightarrow \phi + a \,\,\,\,\,\,\,\,\,\, \phi^* \rightarrow \phi^* + a^* [/tex] is symmetry when m=0. The attempt at a solution Substituting the transformation into the lagrangian gives: [tex] L=\delta_\mu ( \phi + a ) \delta^\mu ( \phi^* + a^* )- m^2 (\phi + a) ( \phi^* + a^*) [/tex] [tex] = (\delta_\mu \phi + \delta_\mu a) ( \delta^\mu \phi^* + \delta^\mu a^*) - m^2 ( \phi \phi^* + a^* \phi + a \phi^* + aa^*) [/tex] I know that this is a symmetry when the action is invariant, so the langrangian is invariant up to a total derivative. [itex] L \rightarrow L + \delta_\mu J^\mu [/itex] I can't see how I could rewrite the above expression so that [itex] L \rightarrow L + \delta_\mu J^\mu [/itex] . Obviously the second term vanishes with m=0, but expanding the first gives: [tex] L = \delta_\mu \phi \delta^\mu \phi^* + \delta_\mu a \delta^\mu a^* + \delta_\mu a \delta^\mu \phi^* + \delta_\mu \phi \delta^\mu a^* [/tex] meaning the total derivative would have to be [tex] \delta_\mu J^\mu = \delta_\mu a \delta^\mu a^* + \delta_\mu a \delta^\mu \phi^* + \delta_\mu \phi \delta^\mu a^* [/tex] which is a form I can't rewrite it in to (what would [itex] J^\mu [/itex] be?) The only solution I can see would be for a & a* to be independent of spacetime, so that [itex] \delta_\mu a = \delta^\mu a^* = 0 [/itex] , but other than answering the question I don't know if there is any justification in saying that. So, are a/a* independent of spacetime, or should I be able to rearrange the above to form a total derivative term? Or am I missing something else completely? Thanks.