Charge operator and the Goldstone boson

Question

Can you explain a question on the Goldstone theorem about the charge operator: what does it mean that "the charge operator annihilates the vacuum" and even it creates new vacuum state, different from the starting one?

Answer 1

You need to understand the language of symmetry in QM and QFT, so maybe a decent QFT book is in order.

A typical symmetry transformation is a unitary operator $U=e^{i\theta e Q}$ transforming ("rotating") states in your Hilbert space and operators such as quantum fields acting on such. We normally consider symmetries leaving the hamiltonian, Lagrangian, and physical amplitudes of the theory invariant... isometries. $Q$ is the conserved charge one may find from the Lagrangian; e.g., consider the electric charge, which rotates the singly charged state $|\Psi\rangle\mapsto e^{ie\theta}|\Psi\rangle$.

In an "unbroken symmetry" realization, the vacuum is not charged, so it is a singlet, or a scalar, under action of the symmetry transformation, so then $U|0\rangle =|0\rangle$, which amounts to $Q|0\rangle=0$, which makes the rotation exponential operator tantamount to the identity.

If, however, $Q|0\rangle\neq 0$, the vacuum is not invariant (even if the Lagrangian and the hamiltonian are): under a rotation it goes to another state, $|\alpha\rangle\equiv |0\rangle +ie\theta Q|0\rangle +...$. This realization is called a "spontaneously broken symmetry realization", or "nonlinear", or "Nambu-Goldstone", etc... The new state $|\alpha\rangle$ is called a different vacuum, degenerate with $|0\rangle$, because it too is a ground state of the Hamiltonian, always presumed to be invariant under the symmetry $U$, $UHU^{-1}=H$, since $H|\alpha\rangle=H U|0\rangle=UH |0\rangle=0$.

The Fabri-Picasso and Goldstone theorems quantify these realizations relying on Noether's theorem, and study the quantum fluctuations sloshing from $|0\rangle$ to all the (different $\theta$) $|\alpha\rangle$s created by the action of Q on the original vacuum.

Very loosely speaking, addition of a massless, zero-momentum goldston takes the system from one of these vacua to another.

Answer 2

The goldstone boson is used to manifest the symmetry invariable for the vacuum. For example a -(μφ)^2/2+λφ^4/4 so there are two cases case 1: where the vacuum is at <φ>=0 so there is no degeneracy in the vacuum solution otherwise if <φ> does not equal to zero there are degeneracy in the vacuum states. It is like a geometrical locus where there is some sort of symmetry preserving the lagrangian at each vacuum state but the symmetry is spontaneously broken where the particle has only one vacuum stat. For example for SU(2) symmetry the geometrical locus for the vacuum is a circle for U (1) it is the sign of the vacuum +'ve or-vs. The gold stone bosons are responsible for transition of the particle from one vacuum to another without any energy cost so they are mass less