Field Theory: Prove transformations are a symmetry

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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.
 

Answers and Replies

  • #2
nrqed
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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.
Yes, "a" is a constant (independent of x). In the absence of gravity, that's all we can do. If we include gravity (by working withe formalism of GR and introducing the metric), then one may make space-time transformations local.
 
  • #3
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Great, thanks.
Is there a way to justify that a is a constant from the notation? (I assume not as there is no notation here to signify the dependence of [itex] \phi [/itex] on X, but the question does not explicitly say a has no dependence)
 

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