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If a Lagrangian has some symmetry leaving the action invariant, we can introduce the corrsponding currents and charges. Now, the transformation can also be implemented as a unitary transformation of the fields in the form (let's consider a scalar field)
[tex] \phi \rightarrow \phi' = U \phi U^\dagger [/tex]
Now a couple of stupid questions.
What happens if we consider a correlation function of, say, [itex] \phi \phi^\dagger [/itex]? It seems like we won't have that [itex] \langle \phi' \phi^{\dagger '} \rangle = \langle \phi \phi^\dagger \rangle [/itex].
Also, why do we have to impose that the commutator of two transformations give a transformation? what would go wrong if it didn't?
To take a specific example, susy transformations leave a lagrangian invariant (modulo total derivatives) even before introducing auxiliary fields. And yet, we must introduce auxiliary fields to get the algebra to close. I am trying to understand why this is necessary.
Thanks in advance
[tex] \phi \rightarrow \phi' = U \phi U^\dagger [/tex]
Now a couple of stupid questions.
What happens if we consider a correlation function of, say, [itex] \phi \phi^\dagger [/itex]? It seems like we won't have that [itex] \langle \phi' \phi^{\dagger '} \rangle = \langle \phi \phi^\dagger \rangle [/itex].
Also, why do we have to impose that the commutator of two transformations give a transformation? what would go wrong if it didn't?
To take a specific example, susy transformations leave a lagrangian invariant (modulo total derivatives) even before introducing auxiliary fields. And yet, we must introduce auxiliary fields to get the algebra to close. I am trying to understand why this is necessary.
Thanks in advance