# Goldstone boson couple to conserved current

The Goldstone boson in spontaneous symmetry breaking problem couples naturally to the associated conserved current of the broken symmetry. How can I see a rigorous (mathematical) derivation for that?

• You can search for a proof of the Quantum Nambu-Goldstone theorem. Try Hanzel & Martin - Quarks & Leptons. – Flint72 Apr 18 '14 at 22:39
• No, that's not it, but thank you. I figure out the answer: it's just the very definition of the Goldstone (base on the parameter of the generated symmetry which is broken), and the associated conserved current. – user109798 Apr 18 '14 at 23:53

From Goldstone theorem we know that $\langle0|J^\mu|\pi\rangle$ isn't zero, that's all.
Adding some extra details, from Lorentz symmetry you have $\langle0|J^\mu|\pi\rangle\sim p^\mu e^{ip x}$ which you can get for a pion coupled derivatively to the current $\mathcal{L}\sim J^\mu \partial_\mu \pi$.