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**Homework Statement**

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.