Goldstone Boson - Goldstone's Argument

Goldstone's Argument

The principle behind Goldstone's argument is that the ground state is not unique. Normally, by current conservation, the charge operator for any symmetry current is time-independent,

{d\over dt} Q = {d\over dt} \int_x J^0(x) =0 ~.

Acting with the charge operator on the vacuum either annihilates the vacuum, if that is symmetric; else, if not, as is the case in spontaneous symmetry breaking, it produces a zero-frequency state out of it, through its shift transformation feature illustrated above. (Actually, here, the charge itself is ill-defined; but its better behaved commutators with fields, so, then, the transformation shifts, are still time-invariant, d〈δφg〉/dt = 0 .)

So, if the vacuum is not invariant under the symmetry, action of the charge operator produces a state which is different from the vacuum chosen, but which has zero frequency. This is a long-wavelength oscillation of a field which is nearly stationary: there are states with zero frequency, so that the theory cannot have a mass gap.

This argument is clarified by taking the limit carefully. If an approximate charge operator is applied to the vacuum,

{d\over dt} Q_A = {d\over dt} \int_x e^{-x^2\over 2A^2} J^0(x) = -\int_x e^{-x^2\over 2A^2} \nabla \cdot J = \int_x \nabla(e^{-x^2\over 2A^2}) \cdot J ~ ,

a state with approximately vanishing time derivative is produced,

\| {d\over dt} Q_A |0\rangle \| \approx {1\over A} \| Q_A|0\rangle \|.

Assuming a nonvanishing mass gap m0, the frequency of any state like the above, which is orthogonal to the vacuum, is at least m0,

\| {d\over dt} |\theta\rangle \| = \| H |\theta\rangle \| \ge m_0 \| \; |\theta\rangle \| ~.

Letting A become large leads to a contradiction. Consequently m0 = 0.

This argument fails, however, when the symmetry is gauged, because then the symmetry generator is only performing a gauge transformation. A gauge transformed state is the same exact state, so that acting with a symmetry generator does not get one out of the vacuum.

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