Broken Symmetries and the Masses of Gauge Bosons
Physical Review Letters, 1964 · paper · Physics
Reports that when spontaneous symmetry breaking occurs in theories whose conserved currents couple to gauge fields, the spin-one gauge quanta acquire mass. Their longitudinal modes become the Goldstone bosons when the coupling vanishes, analogous to plasmons in superconductors.
Checked: PDF from web.archive.org, fetched 5 Oct 2026. Page numbers are those of that PDF.
1 finding
Reference or label Page 2 reads "The two Y = 0, I = 0 gauge fields remain massless", but an octet has only one Y=0, I=0 member. The two massless fields are the Y=0, I3=0 members, consistent with the earlier wording "the two Y = 0, I3 = 0 members of the octet" and with the residual symmetry "generated by Y and I3".
- p. 2 Main text, first column, SU(3) octet discussion
The two Y =0, I=O gauge fields remain massless
- p. 1 Main text, second column
may be chosen to be the two Y=0, I3=0 members of the octet
Judge's note: Crop C002-a confirms the page shows "I = 0" with no subscript 3. Crop C002-b shows "I3 = 0" in the earlier passage on page 1. An SU(3) octet contains exactly one Y=0, I=0 state. The preceding sentence makes the I3=±1 members of the Y=0, I=1 triplet massive, which leaves the singlet and the triplet's I3=0 member massless. The label should therefore read I3=0. This is a wrong quantum-number label.
Second check: another model agreed
Report a problem with this finding
1 minor (typos and the like; most checked by the judge only)
Typo Eq. (2c) is printed as "∂_ν F^{μν} = eφ_0{∂^μ(Δφ_1) − eφ_0 A_μ}": the free index μ is upper on the left-hand side and on ∂^μ, but lower on A_μ.
- p. 1 Eq. (2c)
∂_ν F^{μν} = eφ_0{∂^μ(Δφ_1) − eφ_0 A_μ}
Judge's note: On page 1 (column 2), Eq. (2c) clearly shows a subscript μ on A, while F^{μν} and ∂^μ carry a superscript μ. In Eq. (2a) the same bracket is contracted with ∂^μ, so the lower index is correct there. In (2c), however, μ is a free index and its placement is inconsistent. The physics is unaffected; this is notational only.
Second check: none (judge only)