Finding the Minimum of the Scalar Potential in F-Term SUSY Breaking

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SUMMARY

The minimum of the scalar potential in F-term SUSY breaking is defined by the equation V = ∑i=1N|Fi|2. To determine this minimum, one must apply the equations of motion for Fi, which introduces a dependency on the scalar fields, expressed as Fi = fij). The condition for a minimum is established by solving ∂V/∂φi = 0, ensuring that the second derivatives meet the necessary criteria.

PREREQUISITES
  • Understanding of F-term SUSY breaking
  • Familiarity with scalar potentials in supersymmetry
  • Knowledge of equations of motion in field theory
  • Basic calculus, specifically partial derivatives
NEXT STEPS
  • Study the derivation of scalar potentials in supersymmetry
  • Learn about the implications of the equations of motion in field theory
  • Explore the properties of second derivatives in optimization problems
  • Investigate other methods of SUSY breaking beyond F-term
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The discussion is beneficial for theoretical physicists, particularly those specializing in supersymmetry, as well as graduate students studying advanced field theory concepts.

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In F term SUSY breaking, is the minimum of V just:

[itex]V=\sum^{N}_{i}|F_i|^2[/itex]
 
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pleasehelpmeno said:
In F term SUSY breaking, is the minimum of V just:

[itex]V=\sum^{N}_{i}|F_i|^2[/itex]

That is the expression for the scalar potential. Before you can find a minimum, you need to use the equation of motion for [itex]F_i[/itex], which leads to a dependence on the scalar fields, [itex]F_i = f_i(\phi_j),[/itex] so that [itex]V = V(\phi_j)[/itex]. A minimum is the solution to [itex]\partial V/\partial\phi_i=0[/itex] with the correct properties on the 2nd derivatives.
 

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