Why is is the vacuum expectation value (VEV) of the axion field such that it cancels the theta angle of QCD?
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Answered by
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Hey mate ^_^
In other words, why does this VEV minimize the effective potential of the axion field,VEff.:=−(a(x)+θfξ)ξ32fπ2GμνaG~aμνVEff.:=−(a(x)+θfξ)ξ32fπ2GaμνG~μνa
and lead to the last equality for this particular VEV.
⟨∂VEff.∂a⟩=−ξ32π2f⟨GμνaG~aμν⟩|⟨a⟩=(−faθξ)=0
#Be Brainly❤️
In other words, why does this VEV minimize the effective potential of the axion field,VEff.:=−(a(x)+θfξ)ξ32fπ2GμνaG~aμνVEff.:=−(a(x)+θfξ)ξ32fπ2GaμνG~μνa
and lead to the last equality for this particular VEV.
⟨∂VEff.∂a⟩=−ξ32π2f⟨GμνaG~aμν⟩|⟨a⟩=(−faθξ)=0
#Be Brainly❤️
Answered by
2
Hello mate here is your answer.
In other words, why does this VEV minimize the effective potential of the axion field,VEff.:=−(a(x)+θfξ)ξ32fπ2GμνaG~aμ?
Answer-You can't just handwave away the VEV, because the U(1)U(1) symmetry is spontaneously broken in its choosing of a complex VEV's phase before you try to translate it away. There will still be a circle of potential-minimizing field amplitudes, which gives a nontrivial vacuum contribution. To think of it another way, the EOM is nonlinear so the usual decomposition into ladder operators can't be used to prove the field averages to zero in vacuo.
Hope it helps you.
In other words, why does this VEV minimize the effective potential of the axion field,VEff.:=−(a(x)+θfξ)ξ32fπ2GμνaG~aμ?
Answer-You can't just handwave away the VEV, because the U(1)U(1) symmetry is spontaneously broken in its choosing of a complex VEV's phase before you try to translate it away. There will still be a circle of potential-minimizing field amplitudes, which gives a nontrivial vacuum contribution. To think of it another way, the EOM is nonlinear so the usual decomposition into ladder operators can't be used to prove the field averages to zero in vacuo.
Hope it helps you.
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