not surprising because we havent yet fixed a gauge the culprit is the mixing

Not surprising because we havent yet fixed a gauge

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not surprising because we haven’t yet fixed a gauge the culprit is the mixing term between A μ and μ π efA μ μ π — simplest way to eliminate this mixing is to go to Landau gauge efA μ μ π = - efπ∂ μ A μ + total derivative total derivative in Landau gauge nice because it is renormalizable by simple power counting and the ghosts are the same as before and don’t depend on the vector boson masses 31
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the quadratic terms for A and π are - 1 4 ( μ A ν - ν A μ )( μ A ν - ν A μ )+ 1 2 ( μ π + efA μ ) ( μ π + efA μ )+ A μ s μ + πs μ ( μ A ν - ν A μ ) = ν π + efA ν + s ν μ ( μ π + efA μ ) = s ν ( ν π + efA ν + s ν ) = 0 μ s μ + s = 0 the equations can’t be satisfied for general sources which means that the quadratic terms are not invertible — not surprising because we haven’t yet fixed a gauge the culprit is the mixing term between A μ and μ π efA μ μ π — simplest way to eliminate this mixing is to go to Landau gauge efA μ μ π = - efπ∂ μ A μ + total derivative total derivative in Landau gauge nice because it is renormalizable by simple power counting and the ghosts are the same as before and don’t depend on the vector boson masses 32
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the quadratic terms for A and π are - 1 4 ( μ A ν - ν A μ )( μ A ν - ν A μ )+ 1 2 ( μ π + efA μ ) ( μ π + efA μ )+ A μ s μ + πs μ ( μ A ν - ν A μ ) = ν π + efA ν + s ν μ ( μ π + efA μ ) = s ν ( ν π + efA ν + s ν ) = 0 μ s μ + s = 0 the equations can’t be satisfied for general sources which means that the quadratic terms are not invertible — not surprising because we haven’t yet fixed a gauge the culprit is the mixing term between A μ and μ π efA μ μ π — simplest way to eliminate this mixing is to go to Landau gauge efA μ μ π = - efπ∂ μ A μ + total derivative total derivative in Landau gauge nice because it is renormalizable by simple power counting and the ghosts are the same as before and don’t depend on the vector boson masses 33
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+ + + · · · where is an insertion of the vector boson mass term - m 2 A μ A μ / 2 numerators of landau gauge propagators just projections operator - g μν + k μ k ν /k 2 k 2 + trace of Landau gauge propagator gives 3 3 2 Z ( m 2 ) ( k 2 + ) + 1 2 ( m 2 ) 2 ( k 2 + ) 2 + 1 3 ( m 2 ) 3 ( k 2 + ) 3 + · · · d 4 k (2 π ) 4 3 2 Z log ( k 2 + ) ( k 2 - m 2 + ) d 4 k (2 π ) 4 easier to calculate the second derivative with respect to m 2 2 ( m 2 ) 2 3 2 Z log ( k 2 + ) ( k 2 - m 2 + ) d 4 k (2 π ) 4 = 3 2 Z 1 ( k 2 - m 2 + ) 2 d 4 k (2 π ) 4 34
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+ + + · · · 2 ( m 2 ) 2 3 2 Z log ( k 2 + ) ( k 2 - m 2 + ) d 4 k (2 π ) 4 = 3 2 Z 1 ( k 2 - m 2 + ) 2 d 4 k (2 π ) 4 DR MS gives - i 3 32 π 2 log m 2 μ 2 thus 3 2 Z log ( k 2 + ) ( k 2 - m 2 + ) d 4 k (2 π ) 4 = - i 3 64 π 2 m 4 log m 2 μ 2 + 3 4 + A m 2 + B corresponding to a contribution to the potential 3 64 π 2 m 4 log m 2 μ 2 summed over all massive gauge bosons this is the Coleman-Weinberg potential back to Mathematica file su3.nb 35
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SU (3) coupled to an octet φ 3 × 3 matrix
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  • Spring '10
  • GEORGI
  • Quantum Field Theory, ... ..., wA, Quantum chromodynamics, Gauge theory, Dµ Dµ

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